Python data types

| | | | | | | | | | | |

The following is the standard or built-in Python data type:

numeric

In Python, a numeric data type represents data that has a numeric value. The numeric value can be an integer, floating number, or even a complex number. These values ‚Äã‚Äãare defined in Python as int , float , and complex class.

  • Intergers — this value is represented by the int class. It contains positive or negative integers (no fractional or decimal). There is no limit to the length of the interger in Python.
  • Float — this value is represented by the float class. This is a real floating point number. Indicated by a decimal point. Optionally, the character e or E followed by a positive or negative integer can be added to indicate scientific notation.
  • Complex numbers — A complex number is represented by a complex class. Specified as (real) + (imaginary) j . For example — 2 + 3j

Note. The type () function is used to define the data type.

# Python program for
# show numeric value

a = 5

print ( "Type of a:" , type (a))

b = 5.0

print ( " Type of b: " , type (b))

c = 2 + 4j

print ( "Type of c:" , type (c ))

Output:

 Type of a: & lt; class ’int’ & gt; Type of b: & lt; class ’float’ & gt; Type of c: & lt; class ’complex’ & gt; 

Sequence type

In Python, the sequence — it is an ordered collection of similar or different data types. Sequences allow you to store multiple values ‚Äã‚Äãin an organized and efficient manner. There are several types of sequences in Python:

1) String

In Python, strings — they are arrays of bytes representing Unicode characters. The string — it is a set of one or more characters, enclosed in single, double, or triple quotes. There is no character data type in python, character is a string of length one. It is represented by the str class.

Creating a string

Strings in Python can be created using single or double quotes or even triple quotes.

# Python program for
# Create a line


# Create line
# with single quotes

String1 = ’Welcome to the Geeks World’

print ( " String with the use of Single Quotes: " )

print (String1)


# Create a line
# with double quotes

String1 = "I’ma Geek"

print ( "String with the use of Double Quotes:" )

print (String1)

print ( type (String1))


# Create a line
# with triple quotes

String1 = "I am a geek and I live in a geek world

print ( "String with the use of Triple Quotes:" )

print (String1)

print ( type (String1))


# Create a triple line
# Quotes allow multiple lines

String1 = & # 39; & # 39; & # 39; Geeks

For

Life ""

print ( "Creating a multiline String:" )

print (String1)

Exit :

 String with the use of Single Quotes: Welcome to the Geeks World String with the use of Double Quotes: I ’ma Geek & lt; class’ str’ & gt; String with the use of Triple Quotes: I’ma Geek and I live in a world of "Geeks" & lt; class ’str’ & gt; Creating a multiline String: Geeks For Life 

Accessing string elements

In Python, individual characters in a string can be accessed with using the indexing method. Indexing allows negative address references to access characters at the end of a line, for example, -1 refers to the last character, -2 refers to the second last character, and so on.
Raises IndexError when accessing an index out of range. Only integers are allowed, since index, floating point, or other types will cause TypeError .

# Python program for access
# line characters

String1 = "GeeksForGeeks"

print ( "Initial String:" )

print (String1)


# Print the first o character

print ( "First character of String is: " )

print (String1 [ 0 ])


# Print the last character

print ( " Last character of String is: " )

print (String1 [ - 1 ])

Output:

 Initial String: GeeksForGeeks First character of String is: G Last character of String is: s 

Remove / Update from Strings

In Python, updating or removing characters from a string is not allowed. This will throw an error because assigning an element or removing an element from a string is not supported. This is because strings are immutable, so the elements of a string cannot be changed once assigned. Only newlines can be reassigned to the same name.

# Python program to update / remove
# line character

String1 = "Hello, I’ma Geek"

print ( "Initial String:" )

print (String1)


# Character update
# lines

String1 [ 2 ] = ’p’

print ( "Updating character at 2nd Index: " )

print (String1)


# Deleting a character
# lines

del String1 [ 2 ]

print ( "Deleting character at 2nd Index:" )

print (String1)

Exit :

 Traceback (most recent call last): File "/home/360bb1830c83a918fc78aa8979195653.py‚", line 10, in String1 [2] = ’p ’TypeError:’ str ’object does not s upport item assignment Traceback (most recent call last): File "/home/499e96a61e19944e7e45b7a6e1276742.py‚", line 10, in del String1 [2] TypeError: ’str’ object doesn’t support item deletion 

Escape Sequencing in Python

When printing single and double quoted strings, it raises a SyntaxError because the string already contains single and double quotes and therefore does not can be printed using any of them. Hence, either triple quotes are used to print such a string, or Escape sequences are used to print such strings.
Escape sequences start with a backslash and can be interpreted in different ways. If single quotes are used to represent a string, then all single quotes present in the string must be escaped, and the same is done for double quotes.

# Program Python for
# Escape Sequencing
Line #


# Starting line

String1 = & # 39; & # 39; I’m & # 39; Geek & # 39; & # 39; & # 39; & # 39;

print ( " Initial String with use of Triple Quotes: " )

print (String1)


# Escaping one quote

String1 = ’I’ma" Geek "’

print ( "Escaping Single Quote: " )

print (String1)


# Avoiding double quotes

String1 = "I’ma" Geek ""

print ( " Escaping Double Quotes: " )

print (String1)


# Print paths with
# use escape sequences

String1 = "C: Python Geeks "

print ( "Escaping Backslashes:" )

print (String1)

Exit:

 Initial String with use of Triple Quotes: I’ma " Geek "Escaping Single Quote: I’ma" Geek "Escaping Double Quotes: I’ma" Geek "Escaping Backslashes: C: PythonGeeks 

Note. To to learn more about strings,

# Python program for demonstration
# Create a list


# Create a list

List = []

print ( "Intial blank List:" )

print ( List )


# Create a list with
# use a string

List = [ ’GeeksForGeeks’ ]

print ( "List with the use of String: " )

print ( List )


# Create a list with
# use multiple values ‚Äã‚Äã

List = [ "Geeks" , "For" , "Geeks" ]

print ( "List containi ng multiple values: " )

print ( List [ 0 ])

print ( List [ 2 ])


# Create a multidimensional list
# (Attaching a list to the list)

List = [[ ’Geeks’ , ’ For’ ], [ ’Geeks’ ]]

print ( "Multi-Dimensional List:" )

print ( List )

Exit :

 Intial blank List: [] List with the use of String: [’GeeksForGeeks’] List containing multiple values: Geeks Geeks Multi-Dimensional List : [[’Geeks’,’ For’], [’Geeks’]] 

Adding items to the list

Items can be added to the list using the append () built-in function. Only one element at a time can be added to the list using the append () method. To add an element to the desired position, use the insert () method. Besides the append () and insert () methods, there is another method for adding elements, extend () , this method is used to add multiple elements to the end of the list.

# Python program to demonstrate
# Add items to the list


# Create a list

List = []

print ( "Initial blank List:" )

print ( List )


# Adding elements
# in the list

List . append ( 1 )

List . append ( 2 )

List . append ( 4 )

print ( "List after Addition of Three elements:" )

print ( List )


# Add an element to
# specific position
# (using the insert method)

List . insert ( 3 , 12 )

List . insert ( 0 , ’Geeks’ )

print ( "List after performing Insert Operation:" )

print ( List )


# Adding multiple elements
# to the end of the list
# (using the Extend method)

L ist . extend ([ 8 , ’Geeks’ , ’ Always’ ])

print ( "List after performing Extend Operation: " )

print ( List )

Exit :

 Initial blank List: [] List after Addition of Three elements: [1, 2, 4] List after Insert Operation: [’Geeks’, 1, 2, 4, 12] List after performing Extend Operation: [’ Geeks’, 1, 2, 4, 12, 8, ’Geeks’,’ Always’] 

Accessing items from a list

To access items from a list, refer to the index number. Use the index operator [] to access an item in the list. The index must be an integer. The nested list is accessed using nested indexing. In Python, reverse sequence indices represent positions from the end of an array. Instead of calculating the offset, as in List [len (List) -3] , simply write List [-3] . Negative indexing means start at the end, -1 refers to the last item, -2 refers to the second last item, etc.

# Python program for demonstration
# access an item from the list


# Create a list with
# use multiple values ‚Äã‚Äã

List = [ "Geeks" , "For" , "Geeks" ]


# access to element from
# list by number py index

print ( "Accessing element from the list " )

print ( List [ 0 ])

print ( List [ 2 ])


# access the element using
# negative indexing

print ( "Accessing element using negative indexing " )


# print last list item

print ( List [ - 1 ])


# print the third last item in the list

print ( List [ - 3 ])

Exit:

 Accessing element from the list Geeks Geeks Accessing element using negative indexing Geeks Geeks 

Removing items from the list

Items can be removed from the list with using the built-in remove () function but an error occurs if the element does not exist in the set. The Pop () function can also be used to remove and return an item from a set, but by default it only removes the last item in the set, to remove an item from a specific position in the list, the item index is passed as an argument to the pop () method.

Note: The Remove method in the List will only remove the first occurrence of the item you are looking for.

# Python demo
# Remove items in the list


# Create a list

List = [ 1 , 2 , 3 , 4 , 5 , 6 ,

7 , 8 , 9 , 10 , 11 , 12 ]

print ( "Intial List:" )

print ( List )


# Removing items from the list
# using the Remove () method

List . remove ( 5 )

List .remove ( 6 )

print ( "List after Removal of two elements:" )

print Python data types __del__: Questions

How can I make a time delay in Python?

5 answers

I would like to know how to put a time delay in a Python script.

2973

Answer #1

import time
time.sleep(5)   # Delays for 5 seconds. You can also use a float value.

Here is another example where something is run approximately once a minute:

import time
while True:
    print("This prints once a minute.")
    time.sleep(60) # Delay for 1 minute (60 seconds).

2973

Answer #2

You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds

How to delete a file or folder in Python?

5 answers

How do I delete a file or folder in Python?

2639

Answer #1


Path objects from the Python 3.4+ pathlib module also expose these instance methods:

Python data types __dict__: Questions

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

5 answers

Carl Meyer 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.)

5839

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

5839

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}

5839

Answer #3

An alternative:

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

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