Check if one list is a subset of another in Python

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A list is a subset of another list if all elements of the former are also elements of the latter. For example, [1, 2] is a subset of [1, 2, 3], but [1,4] is not. USE set.issubset() TO CHECK IF A LIST IS A SUBSET OF ANOTHER LIST Use set(list) to convert the lists to sets. Call set1.issubset(set2) to return a Boolean indicating whether set1 is a subset of set2.
print(list1)
OUTPUT
[1, 2]
print(list2)
OUTPUT
[1, 2, 3]
set1 = set(list1)
Convert lists to sets

set2 = set(list2)

is_subset = set1.issubset(set2)
Check if set1 is found in set2

print(is_subset)
OUTPUT
True

Check if one list is a subset of another in Python

Python check if list is subset of another

Sometimes we encounter the problem of checking if one list is just an extension of the list i.e just a superset of one list. This kind of problems are quite popular in competitive programming. Having shorthands for it helps the cause. Lets discuss various ways to achieve this particular task.

Check if one list is a subset of another Using all()

all() is used to check all the elements of a container in just one line. Checks for all the elements of one list for existence in other list.
# Python3 code to demonstrate
# to check if list is subset of other
# using all()

# initializing list
test_list = [9, 4, 5, 8, 10]
sub_list = [10, 5, 4]

# printing original lists
print ("Original list : " + str(test_list))
print ("Original sub list : " + str(sub_list))

# using all() to
# check subset of list
flag = 0
if(all(x in test_list for x in sub_list)):
	flag = 1
	
# printing result
if (flag) :
	print ("Yes, list is subset of other.")
else :
	print ("No, list is not subset of other.")

Output:

Original list : [9, 4, 5, 8, 10] Original sub list : [10, 5, 4] Yes, list is subset of other.

Check if one list is a subset of another Using set.issubset()

The most commonly used and recommended way to search for a sublist. This function is tailor-made to accomplish the particular task of checking whether one list is a subset of another.
# Python3 code to demonstrate
# to check if list is subset of other
# using issubset()
 
# initializing list
test_list = [9, 4, 5, 8, 10]
sub_list = [10, 5]
 
# printing original lists
print ("Original list : " + str(test_list))
print ("Original sub list : " + str(sub_list))
 
# using issubset() to
# check subset of list
flag = 0
if(set(sub_list).issubset(set(test_list))):
    flag = 1
     
# printing result
if (flag) :
    print ("Yes, list is subset of other.")
else :
    print ("No, list is not subset of other.")

Output:

Original list : [9, 4, 5, 8, 10]
Original sub list : [10, 5]
Yes, list is subset of other.

How to check if one list is a subset of another in Python

Check if one list is a subset of another Using iteration and counter

Use the number of items in both lists to verify that the second list is a subset of the first list.
# Python3 code to demonstrate
# to check if list is subset of other
 
#Importing
from collections import Counter
 
def checkInFirst(a, b):
     #getting count
    count_a = Counter(a)
    count_b = Counter(b)
 
    #checking if element exists in second list
    for key in count_b:
        if key not in  count_a:
            return False
        if count_b[key] > count_b[key]:
            return False
    return True
 
# initializing list
a = [1, 2,4,5]
b = [1, 2,3]
 
#Calling function
res = checkInFirst(a, b)
 
#Printing list
print ("Original list : " + str(a))
print ("Original sub list : " + str(b))
 
if res==True :
    print ("Yes, list is subset of other.")
else :
    print ("No, list is not subset of other.")
     
#Added by Paras Jain(everythingispossible)

Output:

Original list : [1, 2, 4, 5]
Original sub list : [1, 2, 3]
No, list is not subset of other.

Check if one list is a subset of another Using set.intersection()

Another method that deals with sets is to see if the intersection of both lists is the sublist that we are checking. This confirms that one list is a subset of the other.
# Python3 code to demonstrate
# to check if list is subset of other
# using intersection()
 
# initializing list
test_list = [9, 4, 5, 8, 10]
sub_list = [10, 5]
 
# printing original lists
print ("Original list : " + str(test_list))
print ("Original sub list : " + str(sub_list))
 
# using intersection() to
# check subset of list
flag = 0
if((set(sub_list) & set(test_list))== set(sub_list)):
    flag = 1
     
# printing result
if (flag) :
    print ("Yes, list is subset of other.")
else :
    print ("No, list is not subset of other.")

Output:

Original list : [9, 4, 5, 8, 10]
Original sub list : [10, 5]
Yes, list is subset of other.

Determine if one list is a subset of another in Python

How to check if one list is a subset of another Python?

Python Set issubset() method returns True if all elements of a set are present in another set (passed as an argument). If not, it returns False. Set A is said to be the subset of set B if all elements of A are in B.

How can I verify if one list is a subset of another?

StackOverflow question

I need to verify if a list is a subset of another - a boolean return is all I seek. Is testing equality on the smaller list after an intersection the fastest way to do this? Performance is of utmost importance given the number of datasets that need to be compared. Adding further facts based on discussions: Will either of the lists be the same for many tests? It does as one of them is a static lookup table. Does it need to be a list? It does not - the static lookup table can be anything that performs best. The dynamic one is a dict from which we extract the keys to perform a static lookup on. What would be the optimal solution given the scenario?

Answer:

Use set.issubset Example:
a = {1,2}
b = {1,2,3}
a.issubset(b) # True
a = {1,2,4}
b = {1,2,3}
a.issubset(b) # False

The performant function Python provides for this is set.issubset. It does have a few restrictions that make it unclear if it’s the answer to your question, however. A list may contain items multiple times and has a specific order. A set does not. Additionally, sets only work on hashable objects. Are you asking about subset or subsequence (which means you’ll want a string search algorithm)? Will either of the lists be the same for many tests? What are the datatypes contained in the list? And for that matter, does it need to be a list? Your other post intersect a dict and list made the types clearer and did get a recommendation to use dictionary key views for their set-like functionality. In that case it was known to work because dictionary keys behave like a set (so much so that before we had sets in Python we used dictionaries). One wonders how the issue got less specific in three hours.

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Check if one list is a subset of another in Python __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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