Saving image from url in PHP

| | | |
There are two different approaches to loading an image from a URL, which are listed below:
  • Using basic file processing.
  • Using the cURL HTTP library.
Both of these approaches have advantages and disadvantages.Using basic file processing:This is the basic and easiest way to accomplish a task. As with any file, start by creating an empty file and open it in "write" mode. After that, extract the content from the original URL and paste it into this file. And it’s as simple as it sounds.From the script, you can figure out what it does.
  • Declaring two variables named $urland $imgrepresenting the source URL and target file, respectively.
  • Use the file_put_contents() functionto write a string to a file that takes two arguments. One is the filename (or path) and the other - the contents of this file.
  • Use the file_get_contents()function to read the file into a string.
Example: $url = https://media.engineerforengineer.org/wp-content/ uploads / engineerforengineer-6-1.png ; $img = ’ logo.png’ ;
// Function for writing the image to a file file_put_contents ( $img , file_get_contents ( $url )); echo " File downloaded! "
? & Gt;
File downloaded! 
Note.Saves the image to the server named logo.png.Now the only problem with this method is that it requires the allow_url_fopen configuration, which is set to 1 by default. But sometimes project requirements do not allow this option. This could be due to some preventive security measures or just a design principle. In such cases, there is another way to save the image.Using the HTTP library, cURL:Strictly speaking, cURL - it is not just an HTTP library. It also has several other data transfer protocols. Since our image is on an HTTP server, we will limit ourselves to this small section of this library.cURL allows us to make HTTP requests in PHP. Start by initializing an instance of it and setting up some of the required parameters for the request, including the URL itself. Then run this query, which returns the contents of the file. After that, the rest of the procedure is the same. Once we get the data, put it in a file and save it.Match :
  • In this script, we have defined the file_get_contents_curl to replicate the file_get_contentsbehavior from the previously mentioned technique.
  • Within this function, we have initialized a cURL instance with usingthe curl_init to use it to fetch data.
  • After that, some parameters need to be set with curl_setopt,for this particular example to work. This function takes three arguments
    • A cURL instance
    • The corresponding option to be set
    • AND the value to which the option is set
    This example sets the following parameters:
    • CURLOPT_HEADER, which should guarantee whether you need to receive headers or not;
    • CURLOPT_RETURNTRANSFER, which passes data as the return value of the curl_exec,rather than outputthem directly.
    • There is another CURLOPT_URL option that sets the URL for the request.
  • We then extractdata from curl_execand return it from the parent function.
  • This data is then written to a file on your computer using file_put_contents .
Example : function file_get_contents_curl ( $url ) { $ch = curl_init(); curl_setopt ( $ch , CURLOPT_HEADER, 0); curl_setopt ( $ch , CURLOPT_RETURNTRANSFER, 1); curl_setopt ( $ch , CURLOPT_URL, $url ); $data = curl_exec ( $ch ); curl_close ( $ch ); return $data ; } $data = file_get_contents_curl ( https://media.engineerforengineer.org/wp-content /uploads/engineerforengineer-6-1.png ); $fp = ’ logo-1.png’ ; file_put_contents ( $fp , $data ); echo "File downloaded!"
?>
Output:
File downloaded! 
This method provides some flexibility when retrieving content from the Internet. As mentioned earlier, it is not limited to HTTP only, but can be used in many other cases as well. This allows you to customize the transmission the way you want. For example, file_get_contentsuses a simple GET request to retrieve data, but with cURL it can also use GET, POST, PUT, and other methods.

Saving image from url in PHP: StackOverflow Questions

Python"s equivalent of && (logical-and) in an if-statement

Question by delete

Here"s my code:

def front_back(a, b):
  # +++your code here+++
  if len(a) % 2 == 0 && len(b) % 2 == 0:
    return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] 
  else:
    #todo! Not yet done. :P
  return

I"m getting an error in the IF conditional.
What am I doing wrong?

Answer #1:

You would want and instead of &&.

Answer #2:

Python uses and and or conditionals.

i.e.

if foo == "abc" and bar == "bac" or zoo == "123":
  # do something

Answer #3:

I"m getting an error in the IF conditional. What am I doing wrong?

There reason that you get a SyntaxError is that there is no && operator in Python. Likewise || and ! are not valid Python operators.

Some of the operators you may know from other languages have a different name in Python. The logical operators && and || are actually called and and or. Likewise the logical negation operator ! is called not.

So you could just write:

if len(a) % 2 == 0 and len(b) % 2 == 0:

or even:

if not (len(a) % 2 or len(b) % 2):

Some additional information (that might come in handy):

I summarized the operator "equivalents" in this table:

+------------------------------+---------------------+
|  Operator (other languages)  |  Operator (Python)  |
+==============================+=====================+
|              &&              |         and         |
+------------------------------+---------------------+
|              ||              |         or          |
+------------------------------+---------------------+
|              !               |         not         |
+------------------------------+---------------------+

See also Python documentation: 6.11. Boolean operations.

Besides the logical operators Python also has bitwise/binary operators:

+--------------------+--------------------+
|  Logical operator  |  Bitwise operator  |
+====================+====================+
|        and         |         &          |
+--------------------+--------------------+
|         or         |         |          |
+--------------------+--------------------+

There is no bitwise negation in Python (just the bitwise inverse operator ~ - but that is not equivalent to not).

See also 6.6. Unary arithmetic and bitwise/binary operations and 6.7. Binary arithmetic operations.

The logical operators (like in many other languages) have the advantage that these are short-circuited. That means if the first operand already defines the result, then the second operator isn"t evaluated at all.

To show this I use a function that simply takes a value, prints it and returns it again. This is handy to see what is actually evaluated because of the print statements:

>>> def print_and_return(value):
...     print(value)
...     return value

>>> res = print_and_return(False) and print_and_return(True)
False

As you can see only one print statement is executed, so Python really didn"t even look at the right operand.

This is not the case for the binary operators. Those always evaluate both operands:

>>> res = print_and_return(False) & print_and_return(True);
False
True

But if the first operand isn"t enough then, of course, the second operator is evaluated:

>>> res = print_and_return(True) and print_and_return(False);
True
False

To summarize this here is another Table:

+-----------------+-------------------------+
|   Expression    |  Right side evaluated?  |
+=================+=========================+
| `True` and ...  |           Yes           |
+-----------------+-------------------------+
| `False` and ... |           No            |
+-----------------+-------------------------+
|  `True` or ...  |           No            |
+-----------------+-------------------------+
| `False` or ...  |           Yes           |
+-----------------+-------------------------+

The True and False represent what bool(left-hand-side) returns, they don"t have to be True or False, they just need to return True or False when bool is called on them (1).

So in Pseudo-Code(!) the and and or functions work like these:

def and(expr1, expr2):
    left = evaluate(expr1)
    if bool(left):
        return evaluate(expr2)
    else:
        return left

def or(expr1, expr2):
    left = evaluate(expr1)
    if bool(left):
        return left
    else:
        return evaluate(expr2)

Note that this is pseudo-code not Python code. In Python you cannot create functions called and or or because these are keywords. Also you should never use "evaluate" or if bool(...).

Customizing the behavior of your own classes

This implicit bool call can be used to customize how your classes behave with and, or and not.

To show how this can be customized I use this class which again prints something to track what is happening:

class Test(object):
    def __init__(self, value):
        self.value = value

    def __bool__(self):
        print("__bool__ called on {!r}".format(self))
        return bool(self.value)

    __nonzero__ = __bool__  # Python 2 compatibility

    def __repr__(self):
        return "{self.__class__.__name__}({self.value})".format(self=self)

So let"s see what happens with that class in combination with these operators:

>>> if Test(True) and Test(False):
...     pass
__bool__ called on Test(True)
__bool__ called on Test(False)

>>> if Test(False) or Test(False):
...     pass
__bool__ called on Test(False)
__bool__ called on Test(False)

>>> if not Test(True):
...     pass
__bool__ called on Test(True)

If you don"t have a __bool__ method then Python also checks if the object has a __len__ method and if it returns a value greater than zero. That might be useful to know in case you create a sequence container.

See also 4.1. Truth Value Testing.

NumPy arrays and subclasses

Probably a bit beyond the scope of the original question but in case you"re dealing with NumPy arrays or subclasses (like Pandas Series or DataFrames) then the implicit bool call will raise the dreaded ValueError:

>>> import numpy as np
>>> arr = np.array([1,2,3])
>>> bool(arr)
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()
>>> arr and arr
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

>>> import pandas as pd
>>> s = pd.Series([1,2,3])
>>> bool(s)
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
>>> s and s
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

In these cases you can use the logical and function from NumPy which performs an element-wise and (or or):

>>> np.logical_and(np.array([False,False,True,True]), np.array([True, False, True, False]))
array([False, False,  True, False])
>>> np.logical_or(np.array([False,False,True,True]), np.array([True, False, True, False]))
array([ True, False,  True,  True])

If you"re dealing just with boolean arrays you could also use the binary operators with NumPy, these do perform element-wise (but also binary) comparisons:

>>> np.array([False,False,True,True]) & np.array([True, False, True, False])
array([False, False,  True, False])
>>> np.array([False,False,True,True]) | np.array([True, False, True, False])
array([ True, False,  True,  True])

(1)

That the bool call on the operands has to return True or False isn"t completely correct. It"s just the first operand that needs to return a boolean in it"s __bool__ method:

class Test(object):
    def __init__(self, value):
        self.value = value

    def __bool__(self):
        return self.value

    __nonzero__ = __bool__  # Python 2 compatibility

    def __repr__(self):
        return "{self.__class__.__name__}({self.value})".format(self=self)

>>> x = Test(10) and Test(10)
TypeError: __bool__ should return bool, returned int
>>> x1 = Test(True) and Test(10)
>>> x2 = Test(False) and Test(10)

That"s because and actually returns the first operand if the first operand evaluates to False and if it evaluates to True then it returns the second operand:

>>> x1
Test(10)
>>> x2
Test(False)

Similarly for or but just the other way around:

>>> Test(True) or Test(10)
Test(True)
>>> Test(False) or Test(10)
Test(10)

However if you use them in an if statement the if will also implicitly call bool on the result. So these finer points may not be relevant for you.

How do you get the logical xor of two variables in Python?

Question by Zach Hirsch

How do you get the logical xor of two variables in Python?

For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):

str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
    print "ok"
else:
    print "bad"

The ^ operator seems to be bitwise, and not defined on all objects:

>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: "str" and "str"

Answer #1:

If you"re already normalizing the inputs to booleans, then != is xor.

bool(a) != bool(b)

Answer #2:

You can always use the definition of xor to compute it from other logical operations:

(a and not b) or (not a and b)

But this is a little too verbose for me, and isn"t particularly clear at first glance. Another way to do it is:

bool(a) ^ bool(b)

The xor operator on two booleans is logical xor (unlike on ints, where it"s bitwise). Which makes sense, since bool is just a subclass of int, but is implemented to only have the values 0 and 1. And logical xor is equivalent to bitwise xor when the domain is restricted to 0 and 1.

So the logical_xor function would be implemented like:

def logical_xor(str1, str2):
    return bool(str1) ^ bool(str2)

Credit to Nick Coghlan on the Python-3000 mailing list.

Saving image from url in PHP: StackOverflow Questions

How can I open multiple files using "with open" in Python?

I want to change a couple of files at one time, iff I can write to all of them. I"m wondering if I somehow can combine the multiple open calls with the with statement:

try:
  with open("a", "w") as a and open("b", "w") as b:
    do_something()
except IOError as e:
  print "Operation failed: %s" % e.strerror

If that"s not possible, what would an elegant solution to this problem look like?

Answer #1:

As of Python 2.7 (or 3.1 respectively) you can write

with open("a", "w") as a, open("b", "w") as b:
    do_something()

In earlier versions of Python, you can sometimes use contextlib.nested() to nest context managers. This won"t work as expected for opening multiples files, though -- see the linked documentation for details.


In the rare case that you want to open a variable number of files all at the same time, you can use contextlib.ExitStack, starting from Python version 3.3:

with ExitStack() as stack:
    files = [stack.enter_context(open(fname)) for fname in filenames]
    # Do something with "files"

Most of the time you have a variable set of files, you likely want to open them one after the other, though.

Answer #2:

For opening many files at once or for long file paths, it may be useful to break things up over multiple lines. From the Python Style Guide as suggested by @Sven Marnach in comments to another answer:

with open("/path/to/InFile.ext", "r") as file_1, 
     open("/path/to/OutFile.ext", "w") as file_2:
    file_2.write(file_1.read())

open() in Python does not create a file if it doesn"t exist

What is the best way to open a file as read/write if it exists, or if it does not, then create it and open it as read/write? From what I read, file = open("myfile.dat", "rw") should do this, right?

It is not working for me (Python 2.6.2) and I"m wondering if it is a version problem, or not supposed to work like that or what.

The bottom line is, I just need a solution for the problem. I am curious about the other stuff, but all I need is a nice way to do the opening part.

The enclosing directory was writeable by user and group, not other (I"m on a Linux system... so permissions 775 in other words), and the exact error was:

IOError: no such file or directory.

Answer #1:

You should use open with the w+ mode:

file = open("myfile.dat", "w+")

Answer #2:

The advantage of the following approach is that the file is properly closed at the block"s end, even if an exception is raised on the way. It"s equivalent to try-finally, but much shorter.

with open("file.dat";"a+") as f:
    f.write(...)
    ...

a+ Opens a file for both appending and reading. The file pointer is at the end of the file if the file exists. The file opens in the append mode. If the file does not exist, it creates a new file for reading and writing. -Python file modes

seek() method sets the file"s current position.

f.seek(pos [, (0|1|2)])
pos .. position of the r/w pointer
[] .. optionally
() .. one of ->
  0 .. absolute position
  1 .. relative position to current
  2 .. relative position from end

Only "rwab+" characters are allowed; there must be exactly one of "rwa" - see Stack Overflow question Python file modes detail.

Difference between modes a, a+, w, w+, and r+ in built-in open function?

In the python built-in open function, what is the exact difference between the modes w, a, w+, a+, and r+?

In particular, the documentation implies that all of these will allow writing to the file, and says that it opens the files for "appending", "writing", and "updating" specifically, but does not define what these terms mean.

Answer #1:

The opening modes are exactly the same as those for the C standard library function fopen().

The BSD fopen manpage defines them as follows:

 The argument mode points to a string beginning with one of the following
 sequences (Additional characters may follow these sequences.):

 ``r""   Open text file for reading.  The stream is positioned at the
         beginning of the file.

 ``r+""  Open for reading and writing.  The stream is positioned at the
         beginning of the file.

 ``w""   Truncate file to zero length or create text file for writing.
         The stream is positioned at the beginning of the file.

 ``w+""  Open for reading and writing.  The file is created if it does not
         exist, otherwise it is truncated.  The stream is positioned at
         the beginning of the file.

 ``a""   Open for writing.  The file is created if it does not exist.  The
         stream is positioned at the end of the file.  Subsequent writes
         to the file will always end up at the then current end of file,
         irrespective of any intervening fseek(3) or similar.

 ``a+""  Open for reading and writing.  The file is created if it does not
         exist.  The stream is positioned at the end of the file.  Subse-
         quent writes to the file will always end up at the then current
         end of file, irrespective of any intervening fseek(3) or similar.

Saving image from url in PHP: 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.

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