How to check if all of the following items are in a list?

Question

I found, that there is related question, about how to find if at least one item exists in a list:
How to check if one of the following items is in a list?

But what is the best and pythonic way to find whether all items exists in a list?

Searching through the docs I found this solution:

>>> l = ['a', 'b', 'c']
>>> set(['a', 'b']) <= set(l)
True
>>> set(['a', 'x']) <= set(l)
False

Other solution would be this:

>>> l = ['a', 'b', 'c']
>>> all(x in l for x in ['a', 'b'])
True
>>> all(x in l for x in ['a', 'x'])
False

But here you must do more typing.

Is there any other solutions?

Answer 1

Operators like <= in Python are generally not overriden to mean something significantly different than "less than or equal to". It's unusual for the standard library does this--it smells like legacy API to me.

Use the equivalent and more clearly-named method, set.issubset. Note that you don't need to convert the argument to a set; it'll do that for you if needed.

set(['a', 'b']).issubset(['a', 'b', 'c'])

Answer 2

I would probably use set in the following manner :

set(l).issuperset(set(['a','b'])) 

or the other way round :

set(['a','b']).issubset(set(l)) 

I find it a bit more readable, but it may be over-kill. Sets are particularly useful to compute union/intersection/differences between collections, but it may not be the best option in this situation ...

Answer 3

I like these two because they seem the most logical, the latter being shorter and probably fastest (shown here using set literal syntax which has been backported to Python 2.7):

all(x in {'a', 'b', 'c'} for x in ['a', 'b'])
#   or
{'a', 'b'}.issubset({'a', 'b', 'c'})

Answer 4

What if your lists contain duplicates like this:

v1 = ['s', 'h', 'e', 'e', 'p']
v2 = ['s', 's', 'h']

Sets do not contain duplicates. So, the following line returns True.

set(v2).issubset(v1)

To count for duplicates, you can use the code:

v1 = sorted(v1)
v2 = sorted(v2)


def is_subseq(v2, v1):
    """Check whether v2 is a subsequence of v1."""
    it = iter(v1)
    return all(c in it for c in v2) 

So, the following line returns False.

is_subseq(v2, v1)

Answer 5

Not OP's case, but - for anyone who wants to assert intersection in dicts and ended up here due to poor googling (e.g. me) - you need to work with dict.items:

>>> a = {'key': 'value'}
>>> b = {'key': 'value', 'extra_key': 'extra_value'}
>>> all(item in a.items() for item in b.items())
True
>>> all(item in b.items() for item in a.items())
False

That's because dict.items returns tuples of key/value pairs, and much like any object in Python, they're interchangeably comparable

Answer 6

Another solution would be:

l = ['a', 'b', 'c']
potential_subset1 = ['a', 'b']
potential_subset2 = ['a', 'x']
print(False not in [i in l for i in potential_subset1]) # True
print(False not in [i in l for i in potential_subset2]) # False

What makes my solution great is that you can write one-liners by putting the lists inline.

Answer 7

An example of how to do this using a lambda expression would be:

issublist = lambda x, y: 0 in [_ in x for _ in y]

Answer 8

Short syntax

I discovered a very readable syntax while experimenting on the Python interpreter.

>>> my_list = [1, 2, 3, 4, 5]
>>> (6 or 7) in my_list
False
>>> (2 or 6) in my_list
True
>>> (2 and 6) in my_list
False
>>> (2 and 5) in my_list
True

List of items to search for

If you have a long list of objects to search for, held in a sub_list variable:

>>> my_list = [1, 2, 3, 4, 5]
>>> sub_list = ['x', 'y']

If any (at least one) item is contained in the superset (or statement):

>>> next((True for item in sub_list if next((True for x in my_list if x == item), False)), False)
False

>>> sub_list[0] = 3
>>> next((True for item in sub_list if next((True for x in my_list if x == item), False)), False)
True

If all items are contained in superset (and statement), then sub_list is a full subset. Also featuring a bit of De Morgan's Law:

>>> next((False for item in sub_list if item not in my_list), True)
False

>>> sub_list[1] = 2
>>> next((False for item in sub_list if item not in my_list), True)
True
>>> next((True for item in sub_list if next((True for x in my_list if x == item), False)), False)
True