Printing words vertically in Python

| | | | |

The given string s. Return all words vertically in the same order as they appear in s.
Words are returned as a string list, padded with spaces when necessary. (Walking on the steps is not allowed).
Each word will only be placed in one column and there will be only one word per column.

Printing words vertically in Python

Example 1:

Input: s = "HOW ARE YOU"
Output: ["HAY","ORO","WEU"]
Explanation: Each word is printed vertically. 
 "HAY"
 "ORO"
 "WEU"

Example 2:

Input: s = "TO BE OR NOT TO BE"
Output: ["TBONTB","OEROOE","   T"]
Explanation: Trailing spaces is not allowed. 
"TBONTB"
"OEROOE"
"   T"

Example 3:

Input: s = "CONTEST IS COMING"
Output: ["CIC","OSO","N M","T I","E N","S G","T"]

Constraints:

  • 1 <= s.length <= 200
  • s contains only upper case English letters.
  • It’s guaranteed that there is only one space between 2 words.

Method #1 : [Horizontal to Vertical] using loop + ""

In this, we add newline character to after each character so that each element gets rendered at next line.

# Python3 code to demonstrate working of
# Interconvert Horizontal and Vertical String
# using [Horizontal to Vertical] using loop + "
"

# initializing string
test_str = ’geeksforgeeks’

# printing original String
print("The original string is : " + str(test_str))

# using loop to add "
" after each character
res = ’’
for ele in test_str:
	res += ele + "
"
	
# printing result
print("The converted string : " + str(res))

Output:

Output
The original string is : geeksforgeeks
The converted string : g
e
e
k
s
f
o
r
g
e
e
k
s

Method #2 : [Vertical to Horizontal] using replace() + ""

In this, we perform the task of conversion by removing "
" by replacement by empty string.

# Python3 code to demonstrate working of 
# Interconvert Horizontal and Vertical String
# using [Vertical to Horizontal] using replace() + "
"
  
# initializing string
test_str = ’g
e
e
k
s
f
o
r
g
e
e
k
s
’
  
# printing original String
print("The original string is : " + str(test_str))
  
# using replace() + "
" to solve this problem
res = test_str.replace("
", "")
      
# printing result 
print("The converted string : " + str(res)) 

Output:

The original string is : g
e
e
k
s
f
o
r
g
e
e
k
s

The converted string : geeksforgeeks
Printing words vertically in Python

Vertical Print String - Python3.2

StackOverflow question

I’m writing a script that will take as user inputed string, and print it vertically, like so:

input = "John walked to the store"

output = J w t t s
         o a o h t
         h l   e o
         n k     r
           e     e
           d

I’ve written most of the code, which is as follows:

import sys

def verticalPrint(astring):
    wordList = astring.split(" ")
    wordAmount = len(wordList)

    maxLen = 0
    for i in range (wordAmount):
        length = len(wordList[i])
        if length >= maxLen:
            maxLen = length

    ### makes all words the same length to avoid range errors ###
    for i in range (wordAmount):
        if len(wordList[i]) < maxLen:
            wordList[i] = wordList[i] + (" ")*(maxLen-len(wordList[i]))

    for i in range (wordAmount):
        for j in range (maxLen):
            print(wordList[i][j])

def main():
    astring = input("Enter a string:" + ’
’)

    verticalPrint(astring)

main()

I’m having trouble figure out how to get the output correct. I know its a problem with the for loop. It’s output is:

input = "John walked"

output = J
         o
         h
         n

         w
         a
         l
         k
         e
         d

Any advice? (Also, I want to have the print command used only once.)

Answer:

Use itertools.zip_longest:

>>> from itertools import zip_longest
>>> text = "John walked to the store"
for x in zip_longest(*text.split(), fillvalue=’ ’):
    print (’ ’.join(x))
...     
J w t t s
o a o h t
h l   e o
n k     r
  e     e
  d      

Suppose we have string s. We have to find all the words vertically in the same order in which they appear in s. Here the words are returned as a list of strings, if necessary, we have to fill in spaces. (Trailing spaces are not allowed.) Each word will be placed in only one column, and there will be only one word in one column. Thus, if the input string is "LIKE YOU?", Then the output will be ["HAY", "ORO", "WEU"]

To solve this problem, we will follow the following steps:

  • s := make a list of strings split by the spaces, make one empty array x, set row = 0
  • for each word I in s, set row := max of row and length of i
  • col := length of s
  • make one array and filled with empty string, and its size is the row
  • for I in range 0 to col ‚Äì 1
  • j := 0
  • while j < length of s[i]
    • while i < length of ans[j] >= 1, do ans[j] := ans[j] concatenate " ‚Äù
    • ans[j] := ans[j] concatenate s[i, j]
    • increase j by 1
  • return ans
class Solution(object):
   def printVertically(self, s):
      s = s.split(" ")
      x = []
      row = 0
      for i in s:
         row = max(row, len(i))
      col = len(s)
      ans = ["" for i in range(row)]
      j = 0
      for i in range(col):
         j = 0
         while j < len(s[i]):
            #print(j, i)
            while i - len(ans[j]) >= 1:
               ans[j] += " "
            ans[j] += s[i][j]
            j += 1
      return ans
ob = Solution()
print(ob.printVertically("HOW ARE YOU"))
print(ob.printVertically("TO BE OR NOT TO BE"))

Input:

"HOW ARE YOU"
"TO BE OR NOT TO BE"

Output:

["HAY","ORO","WEU"]
["TBONTB","OEROOE"," T"]
Printing words vertically in Python

Printing words vertically in Python: StackOverflow Questions

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

Question by Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)

Answer #1:

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
    
  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

Answer #2:

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
{"a": 1, "c": 11, "b": 10}

Answer #3:

An alternative:

z = x.copy()
z.update(y)

Answer #4:

Another, more concise, option:

z = dict(x, **y)

Note: this has become a popular answer, but it is important to point out that if y has any non-string keys, the fact that this works at all is an abuse of a CPython implementation detail, and it does not work in Python 3, or in PyPy, IronPython, or Jython. Also, Guido is not a fan. So I can"t recommend this technique for forward-compatible or cross-implementation portable code, which really means it should be avoided entirely.

Answer #5:

This probably won"t be a popular answer, but you almost certainly do not want to do this. If you want a copy that"s a merge, then use copy (or deepcopy, depending on what you want) and then update. The two lines of code are much more readable - more Pythonic - than the single line creation with .items() + .items(). Explicit is better than implicit.

In addition, when you use .items() (pre Python 3.0), you"re creating a new list that contains the items from the dict. If your dictionaries are large, then that is quite a lot of overhead (two large lists that will be thrown away as soon as the merged dict is created). update() can work more efficiently, because it can run through the second dict item-by-item.

In terms of time:

>>> timeit.Timer("dict(x, **y)", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.52571702003479
>>> timeit.Timer("temp = x.copy()
temp.update(y)", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.694622993469238
>>> timeit.Timer("dict(x.items() + y.items())", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
41.484580039978027

IMO the tiny slowdown between the first two is worth it for the readability. In addition, keyword arguments for dictionary creation was only added in Python 2.3, whereas copy() and update() will work in older versions.

Printing words vertically in Python: StackOverflow Questions

Finding the index of an item in a list

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

Answer #1:

>>> ["foo", "bar", "baz"].index("bar")
1

Reference: Data Structures > More on Lists

Caveats follow

Note that while this is perhaps the cleanest way to answer the question as asked, index is a rather weak component of the list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about list.index follow. It is probably worth initially taking a look at the documentation for it:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give index a hint. For instance, in this snippet, l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:

>>> import timeit
>>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000)
9.356267921015387
>>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000)
0.0004404920036904514
 

Only returns the index of the first match to its argument

A call to index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.

>>> [1, 1].index(1)
0
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> next(g)
0
>>> next(g)
2

Most places where I once would have used index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for index, take a look at these excellent Python features.

Throws if element not present in list

A call to index results in a ValueError if the item"s not present.

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If the item might not be present in the list, you should either

  1. Check for it first with item in my_list (clean, readable approach), or
  2. Wrap the index call in a try/except block which catches ValueError (probably faster, at least when the list to search is long, and the item is usually present.)

Answer #2:

One thing that is really helpful in learning Python is to use the interactive help function:

>>> help(["foo", "bar", "baz"])
Help on list object:

class list(object)
 ...

 |
 |  index(...)
 |      L.index(value, [start, [stop]]) -> integer -- return first index of value
 |

which will often lead you to the method you are looking for.

Answer #3:

The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use enumerate():

for i, j in enumerate(["foo", "bar", "baz"]):
    if j == "bar":
        print(i)

The index() function only returns the first occurrence, while enumerate() returns all occurrences.

As a list comprehension:

[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]

Here"s also another small solution with itertools.count() (which is pretty much the same approach as enumerate):

from itertools import izip as zip, count # izip for maximum efficiency
[i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]

This is more efficient for larger lists than using enumerate():

$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 174 usec per loop
$ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]"
10000 loops, best of 3: 196 usec per loop

Answer #4:

To get all indexes:

indexes = [i for i,x in enumerate(xs) if x == "foo"]

Answer #5:

index() returns the first index of value!

| index(...)
| L.index(value, [start, [stop]]) -> integer -- return first index of value

def all_indices(value, qlist):
    indices = []
    idx = -1
    while True:
        try:
            idx = qlist.index(value, idx+1)
            indices.append(idx)
        except ValueError:
            break
    return indices

all_indices("foo", ["foo";"bar";"baz";"foo"])

Printing words vertically in Python: StackOverflow Questions

Why is it string.join(list) instead of list.join(string)?

Question by Evan Fosmark

This has always confused me. It seems like this would be nicer:

my_list = ["Hello", "world"]
print(my_list.join("-"))
# Produce: "Hello-world"

Than this:

my_list = ["Hello", "world"]
print("-".join(my_list))
# Produce: "Hello-world"

Is there a specific reason it is like this?

Answer #1:

It"s because any iterable can be joined (e.g, list, tuple, dict, set), but its contents and the "joiner" must be strings.

For example:

"_".join(["welcome", "to", "stack", "overflow"])
"_".join(("welcome", "to", "stack", "overflow"))
"welcome_to_stack_overflow"

Using something other than strings will raise the following error:

TypeError: sequence item 0: expected str instance, int found

Answer #2:

This was discussed in the String methods... finally thread in the Python-Dev achive, and was accepted by Guido. This thread began in Jun 1999, and str.join was included in Python 1.6 which was released in Sep 2000 (and supported Unicode). Python 2.0 (supported str methods including join) was released in Oct 2000.

  • There were four options proposed in this thread:
    • str.join(seq)
    • seq.join(str)
    • seq.reduce(str)
    • join as a built-in function
  • Guido wanted to support not only lists and tuples, but all sequences/iterables.
  • seq.reduce(str) is difficult for newcomers.
  • seq.join(str) introduces unexpected dependency from sequences to str/unicode.
  • join() as a built-in function would support only specific data types. So using a built-in namespace is not good. If join() supports many datatypes, creating an optimized implementation would be difficult, if implemented using the __add__ method then it would ve O(n¬≤).
  • The separator string (sep) should not be omitted. Explicit is better than implicit.

Here are some additional thoughts (my own, and my friend"s):

  • Unicode support was coming, but it was not final. At that time UTF-8 was the most likely about to replace UCS2/4. To calculate total buffer length of UTF-8 strings it needs to know character coding rule.
  • At that time, Python had already decided on a common sequence interface rule where a user could create a sequence-like (iterable) class. But Python didn"t support extending built-in types until 2.2. At that time it was difficult to provide basic iterable class (which is mentioned in another comment).

Guido"s decision is recorded in a historical mail, deciding on str.join(seq):

Funny, but it does seem right! Barry, go for it...
Guido van Rossum

Answer #3:

Because the join() method is in the string class, instead of the list class?

I agree it looks funny.

See http://www.faqs.org/docs/diveintopython/odbchelper_join.html:

Historical note. When I first learned Python, I expected join to be a method of a list, which would take the delimiter as an argument. Lots of people feel the same way, and there’s a story behind the join method. Prior to Python 1.6, strings didn’t have all these useful methods. There was a separate string module which contained all the string functions; each function took a string as its first argument. The functions were deemed important enough to put onto the strings themselves, which made sense for functions like lower, upper, and split. But many hard-core Python programmers objected to the new join method, arguing that it should be a method of the list instead, or that it shouldn’t move at all but simply stay a part of the old string module (which still has lots of useful stuff in it). I use the new join method exclusively, but you will see code written either way, and if it really bothers you, you can use the old string.join function instead.

--- Mark Pilgrim, Dive into Python

join list of lists in python

Question by Kozyarchuk

Is the a short syntax for joining a list of lists into a single list( or iterator) in python?

For example I have a list as follows and I want to iterate over a,b and c.

x = [["a";"b"], ["c"]]

The best I can come up with is as follows.

result = []
[ result.extend(el) for el in x] 

for el in result:
  print el

Answer #1:

import itertools
a = [["a","b"], ["c"]]
print(list(itertools.chain.from_iterable(a)))

Answer #2:

x = [["a";"b"], ["c"]]

result = sum(x, [])

Printing words vertically in Python: StackOverflow Questions

How do I merge two dictionaries in a single expression (taking union of dictionaries)?

Question by Carl Meyer

I have two Python dictionaries, and I want to write a single expression that returns these two dictionaries, merged (i.e. taking the union). The update() method would be what I need, if it returned its result instead of modifying a dictionary in-place.

>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

How can I get that final merged dictionary in z, not x?

(To be extra-clear, the last-one-wins conflict-handling of dict.update() is what I"m looking for as well.)

Answer #1:

How can I merge two Python dictionaries in a single expression?

For dictionaries x and y, z becomes a shallowly-merged dictionary with values from y replacing those from x.

  • In Python 3.9.0 or greater (released 17 October 2020): PEP-584, discussed here, was implemented and provides the simplest method:

    z = x | y          # NOTE: 3.9+ ONLY
    
  • In Python 3.5 or greater:

    z = {**x, **y}
    
  • In Python 2, (or 3.4 or lower) write a function:

    def merge_two_dicts(x, y):
        z = x.copy()   # start with keys and values of x
        z.update(y)    # modifies z with keys and values of y
        return z
    

    and now:

    z = merge_two_dicts(x, y)
    

Explanation

Say you have two dictionaries and you want to merge them into a new dictionary without altering the original dictionaries:

x = {"a": 1, "b": 2}
y = {"b": 3, "c": 4}

The desired result is to get a new dictionary (z) with the values merged, and the second dictionary"s values overwriting those from the first.

>>> z
{"a": 1, "b": 3, "c": 4}

A new syntax for this, proposed in PEP 448 and available as of Python 3.5, is

z = {**x, **y}

And it is indeed a single expression.

Note that we can merge in with literal notation as well:

z = {**x, "foo": 1, "bar": 2, **y}

and now:

>>> z
{"a": 1, "b": 3, "foo": 1, "bar": 2, "c": 4}

It is now showing as implemented in the release schedule for 3.5, PEP 478, and it has now made its way into the What"s New in Python 3.5 document.

However, since many organizations are still on Python 2, you may wish to do this in a backward-compatible way. The classically Pythonic way, available in Python 2 and Python 3.0-3.4, is to do this as a two-step process:

z = x.copy()
z.update(y) # which returns None since it mutates z

In both approaches, y will come second and its values will replace x"s values, thus b will point to 3 in our final result.

Not yet on Python 3.5, but want a single expression

If you are not yet on Python 3.5 or need to write backward-compatible code, and you want this in a single expression, the most performant while the correct approach is to put it in a function:

def merge_two_dicts(x, y):
    """Given two dictionaries, merge them into a new dict as a shallow copy."""
    z = x.copy()
    z.update(y)
    return z

and then you have a single expression:

z = merge_two_dicts(x, y)

You can also make a function to merge an arbitrary number of dictionaries, from zero to a very large number:

def merge_dicts(*dict_args):
    """
    Given any number of dictionaries, shallow copy and merge into a new dict,
    precedence goes to key-value pairs in latter dictionaries.
    """
    result = {}
    for dictionary in dict_args:
        result.update(dictionary)
    return result

This function will work in Python 2 and 3 for all dictionaries. e.g. given dictionaries a to g:

z = merge_dicts(a, b, c, d, e, f, g) 

and key-value pairs in g will take precedence over dictionaries a to f, and so on.

Critiques of Other Answers

Don"t use what you see in the formerly accepted answer:

z = dict(x.items() + y.items())

In Python 2, you create two lists in memory for each dict, create a third list in memory with length equal to the length of the first two put together, and then discard all three lists to create the dict. In Python 3, this will fail because you"re adding two dict_items objects together, not two lists -

>>> c = dict(a.items() + b.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for +: "dict_items" and "dict_items"

and you would have to explicitly create them as lists, e.g. z = dict(list(x.items()) + list(y.items())). This is a waste of resources and computation power.

Similarly, taking the union of items() in Python 3 (viewitems() in Python 2.7) will also fail when values are unhashable objects (like lists, for example). Even if your values are hashable, since sets are semantically unordered, the behavior is undefined in regards to precedence. So don"t do this:

>>> c = dict(a.items() | b.items())

This example demonstrates what happens when values are unhashable:

>>> x = {"a": []}
>>> y = {"b": []}
>>> dict(x.items() | y.items())
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: "list"

Here"s an example where y should have precedence, but instead the value from x is retained due to the arbitrary order of sets:

>>> x = {"a": 2}
>>> y = {"a": 1}
>>> dict(x.items() | y.items())
{"a": 2}

Another hack you should not use:

z = dict(x, **y)

This uses the dict constructor and is very fast and memory-efficient (even slightly more so than our two-step process) but unless you know precisely what is happening here (that is, the second dict is being passed as keyword arguments to the dict constructor), it"s difficult to read, it"s not the intended usage, and so it is not Pythonic.

Here"s an example of the usage being remediated in django.

Dictionaries are intended to take hashable keys (e.g. frozensets or tuples), but this method fails in Python 3 when keys are not strings.

>>> c = dict(a, **b)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings

From the mailing list, Guido van Rossum, the creator of the language, wrote:

I am fine with declaring dict({}, **{1:3}) illegal, since after all it is abuse of the ** mechanism.

and

Apparently dict(x, **y) is going around as "cool hack" for "call x.update(y) and return x". Personally, I find it more despicable than cool.

It is my understanding (as well as the understanding of the creator of the language) that the intended usage for dict(**y) is for creating dictionaries for readability purposes, e.g.:

dict(a=1, b=10, c=11)

instead of

{"a": 1, "b": 10, "c": 11}

Response to comments

Despite what Guido says, dict(x, **y) is in line with the dict specification, which btw. works for both Python 2 and 3. The fact that this only works for string keys is a direct consequence of how keyword parameters work and not a short-coming of dict. Nor is using the ** operator in this place an abuse of the mechanism, in fact, ** was designed precisely to pass dictionaries as keywords.

Again, it doesn"t work for 3 when keys are not strings. The implicit calling contract is that namespaces take ordinary dictionaries, while users must only pass keyword arguments that are strings. All other callables enforced it. dict broke this consistency in Python 2:

>>> foo(**{("a", "b"): None})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foo() keywords must be strings
>>> dict(**{("a", "b"): None})
{("a", "b"): None}

This inconsistency was bad given other implementations of Python (PyPy, Jython, IronPython). Thus it was fixed in Python 3, as this usage could be a breaking change.

I submit to you that it is malicious incompetence to intentionally write code that only works in one version of a language or that only works given certain arbitrary constraints.

More comments:

dict(x.items() + y.items()) is still the most readable solution for Python 2. Readability counts.

My response: merge_two_dicts(x, y) actually seems much clearer to me, if we"re actually concerned about readability. And it is not forward compatible, as Python 2 is increasingly deprecated.

{**x, **y} does not seem to handle nested dictionaries. the contents of nested keys are simply overwritten, not merged [...] I ended up being burnt by these answers that do not merge recursively and I was surprised no one mentioned it. In my interpretation of the word "merging" these answers describe "updating one dict with another", and not merging.

Yes. I must refer you back to the question, which is asking for a shallow merge of two dictionaries, with the first"s values being overwritten by the second"s - in a single expression.

Assuming two dictionaries of dictionaries, one might recursively merge them in a single function, but you should be careful not to modify the dictionaries from either source, and the surest way to avoid that is to make a copy when assigning values. As keys must be hashable and are usually therefore immutable, it is pointless to copy them:

from copy import deepcopy

def dict_of_dicts_merge(x, y):
    z = {}
    overlapping_keys = x.keys() & y.keys()
    for key in overlapping_keys:
        z[key] = dict_of_dicts_merge(x[key], y[key])
    for key in x.keys() - overlapping_keys:
        z[key] = deepcopy(x[key])
    for key in y.keys() - overlapping_keys:
        z[key] = deepcopy(y[key])
    return z

Usage:

>>> x = {"a":{1:{}}, "b": {2:{}}}
>>> y = {"b":{10:{}}, "c": {11:{}}}
>>> dict_of_dicts_merge(x, y)
{"b": {2: {}, 10: {}}, "a": {1: {}}, "c": {11: {}}}

Coming up with contingencies for other value types is far beyond the scope of this question, so I will point you at my answer to the canonical question on a "Dictionaries of dictionaries merge".

Less Performant But Correct Ad-hocs

These approaches are less performant, but they will provide correct behavior. They will be much less performant than copy and update or the new unpacking because they iterate through each key-value pair at a higher level of abstraction, but they do respect the order of precedence (latter dictionaries have precedence)

You can also chain the dictionaries manually inside a dict comprehension:

{k: v for d in dicts for k, v in d.items()} # iteritems in Python 2.7

or in Python 2.6 (and perhaps as early as 2.4 when generator expressions were introduced):

dict((k, v) for d in dicts for k, v in d.items()) # iteritems in Python 2

itertools.chain will chain the iterators over the key-value pairs in the correct order:

from itertools import chain
z = dict(chain(x.items(), y.items())) # iteritems in Python 2

Performance Analysis

I"m only going to do the performance analysis of the usages known to behave correctly. (Self-contained so you can copy and paste yourself.)

from timeit import repeat
from itertools import chain

x = dict.fromkeys("abcdefg")
y = dict.fromkeys("efghijk")

def merge_two_dicts(x, y):
    z = x.copy()
    z.update(y)
    return z

min(repeat(lambda: {**x, **y}))
min(repeat(lambda: merge_two_dicts(x, y)))
min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
min(repeat(lambda: dict(chain(x.items(), y.items()))))
min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))

In Python 3.8.1, NixOS:

>>> min(repeat(lambda: {**x, **y}))
1.0804965235292912
>>> min(repeat(lambda: merge_two_dicts(x, y)))
1.636518670246005
>>> min(repeat(lambda: {k: v for d in (x, y) for k, v in d.items()}))
3.1779992282390594
>>> min(repeat(lambda: dict(chain(x.items(), y.items()))))
2.740647904574871
>>> min(repeat(lambda: dict(item for d in (x, y) for item in d.items())))
4.266070580109954
$ uname -a
Linux nixos 4.19.113 #1-NixOS SMP Wed Mar 25 07:06:15 UTC 2020 x86_64 GNU/Linux

Resources on Dictionaries

Answer #2:

In your case, what you can do is:

z = dict(list(x.items()) + list(y.items()))

This will, as you want it, put the final dict in z, and make the value for key b be properly overridden by the second (y) dict"s value:

>>> x = {"a":1, "b": 2}
>>> y = {"b":10, "c": 11}
>>> z = dict(list(x.items()) + list(y.items()))
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python 2, you can even remove the list() calls. To create z:

>>> z = dict(x.items() + y.items())
>>> z
{"a": 1, "c": 11, "b": 10}

If you use Python version 3.9.0a4 or greater, then you can directly use:

x = {"a":1, "b": 2}
y = {"b":10, "c": 11}
z = x | y
print(z)
{"a": 1, "c": 11, "b": 10}

Answer #3:

An alternative:

z = x.copy()
z.update(y)

Answer #4:

Another, more concise, option:

z = dict(x, **y)

Note: this has become a popular answer, but it is important to point out that if y has any non-string keys, the fact that this works at all is an abuse of a CPython implementation detail, and it does not work in Python 3, or in PyPy, IronPython, or Jython. Also, Guido is not a fan. So I can"t recommend this technique for forward-compatible or cross-implementation portable code, which really means it should be avoided entirely.

Answer #5:

This probably won"t be a popular answer, but you almost certainly do not want to do this. If you want a copy that"s a merge, then use copy (or deepcopy, depending on what you want) and then update. The two lines of code are much more readable - more Pythonic - than the single line creation with .items() + .items(). Explicit is better than implicit.

In addition, when you use .items() (pre Python 3.0), you"re creating a new list that contains the items from the dict. If your dictionaries are large, then that is quite a lot of overhead (two large lists that will be thrown away as soon as the merged dict is created). update() can work more efficiently, because it can run through the second dict item-by-item.

In terms of time:

>>> timeit.Timer("dict(x, **y)", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.52571702003479
>>> timeit.Timer("temp = x.copy()
temp.update(y)", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
15.694622993469238
>>> timeit.Timer("dict(x.items() + y.items())", "x = dict(zip(range(1000), range(1000)))
y=dict(zip(range(1000,2000), range(1000,2000)))").timeit(100000)
41.484580039978027

IMO the tiny slowdown between the first two is worth it for the readability. In addition, keyword arguments for dictionary creation was only added in Python 2.3, whereas copy() and update() will work in older versions.

Printing words vertically in Python: StackOverflow Questions

How do you split a list into evenly sized chunks?

Question by jespern

I have a list of arbitrary length, and I need to split it up into equal size chunks and operate on it. There are some obvious ways to do this, like keeping a counter and two lists, and when the second list fills up, add it to the first list and empty the second list for the next round of data, but this is potentially extremely expensive.

I was wondering if anyone had a good solution to this for lists of any length, e.g. using generators.

I was looking for something useful in itertools but I couldn"t find anything obviously useful. Might"ve missed it, though.

Related question: What is the most “pythonic” way to iterate over a list in chunks?

Answer #1:

Here"s a generator that yields the chunks you want:

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in range(0, len(lst), n):
        yield lst[i:i + n]

import pprint
pprint.pprint(list(chunks(range(10, 75), 10)))
[[10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
 [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
 [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
 [40, 41, 42, 43, 44, 45, 46, 47, 48, 49],
 [50, 51, 52, 53, 54, 55, 56, 57, 58, 59],
 [60, 61, 62, 63, 64, 65, 66, 67, 68, 69],
 [70, 71, 72, 73, 74]]

If you"re using Python 2, you should use xrange() instead of range():

def chunks(lst, n):
    """Yield successive n-sized chunks from lst."""
    for i in xrange(0, len(lst), n):
        yield lst[i:i + n]

Also you can simply use list comprehension instead of writing a function, though it"s a good idea to encapsulate operations like this in named functions so that your code is easier to understand. Python 3:

[lst[i:i + n] for i in range(0, len(lst), n)]

Python 2 version:

[lst[i:i + n] for i in xrange(0, len(lst), n)]

Answer #2:

If you want something super simple:

def chunks(l, n):
    n = max(1, n)
    return (l[i:i+n] for i in range(0, len(l), n))

Use xrange() instead of range() in the case of Python 2.x

Answer #3:

Directly from the (old) Python documentation (recipes for itertools):

from itertools import izip, chain, repeat

def grouper(n, iterable, padvalue=None):
    "grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

The current version, as suggested by J.F.Sebastian:

#from itertools import izip_longest as zip_longest # for Python 2.x
from itertools import zip_longest # for Python 3.x
#from six.moves import zip_longest # for both (uses the six compat library)

def grouper(n, iterable, padvalue=None):
    "grouper(3, "abcdefg", "x") --> ("a","b","c"), ("d","e","f"), ("g","x","x")"
    return zip_longest(*[iter(iterable)]*n, fillvalue=padvalue)

I guess Guido"s time machine works—worked—will work—will have worked—was working again.

These solutions work because [iter(iterable)]*n (or the equivalent in the earlier version) creates one iterator, repeated n times in the list. izip_longest then effectively performs a round-robin of "each" iterator; because this is the same iterator, it is advanced by each such call, resulting in each such zip-roundrobin generating one tuple of n items.

Answer #4:

I know this is kind of old but nobody yet mentioned numpy.array_split:

import numpy as np

lst = range(50)
np.array_split(lst, 5)
# [array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9]),
#  array([10, 11, 12, 13, 14, 15, 16, 17, 18, 19]),
#  array([20, 21, 22, 23, 24, 25, 26, 27, 28, 29]),
#  array([30, 31, 32, 33, 34, 35, 36, 37, 38, 39]),
#  array([40, 41, 42, 43, 44, 45, 46, 47, 48, 49])]

Answer #5:

I"m surprised nobody has thought of using iter"s two-argument form:

from itertools import islice

def chunk(it, size):
    it = iter(it)
    return iter(lambda: tuple(islice(it, size)), ())

Demo:

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]

This works with any iterable and produces output lazily. It returns tuples rather than iterators, but I think it has a certain elegance nonetheless. It also doesn"t pad; if you want padding, a simple variation on the above will suffice:

from itertools import islice, chain, repeat

def chunk_pad(it, size, padval=None):
    it = chain(iter(it), repeat(padval))
    return iter(lambda: tuple(islice(it, size)), (padval,) * size)

Demo:

>>> list(chunk_pad(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk_pad(range(14), 3, "a"))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, "a")]

Like the izip_longest-based solutions, the above always pads. As far as I know, there"s no one- or two-line itertools recipe for a function that optionally pads. By combining the above two approaches, this one comes pretty close:

_no_padding = object()

def chunk(it, size, padval=_no_padding):
    if padval == _no_padding:
        it = iter(it)
        sentinel = ()
    else:
        it = chain(iter(it), repeat(padval))
        sentinel = (padval,) * size
    return iter(lambda: tuple(islice(it, size)), sentinel)

Demo:

>>> list(chunk(range(14), 3))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13)]
>>> list(chunk(range(14), 3, None))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, None)]
>>> list(chunk(range(14), 3, "a"))
[(0, 1, 2), (3, 4, 5), (6, 7, 8), (9, 10, 11), (12, 13, "a")]

I believe this is the shortest chunker proposed that offers optional padding.

As Tomasz Gandor observed, the two padding chunkers will stop unexpectedly if they encounter a long sequence of pad values. Here"s a final variation that works around that problem in a reasonable way:

_no_padding = object()
def chunk(it, size, padval=_no_padding):
    it = iter(it)
    chunker = iter(lambda: tuple(islice(it, size)), ())
    if padval == _no_padding:
        yield from chunker
    else:
        for ch in chunker:
            yield ch if len(ch) == size else ch + (padval,) * (size - len(ch))

Demo:

>>> list(chunk([1, 2, (), (), 5], 2))
[(1, 2), ((), ()), (5,)]
>>> list(chunk([1, 2, None, None, 5], 2, None))
[(1, 2), (None, None), (5, None)]

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