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

This was what I was searching online but unfortunately found not online but while experimenting on python interpreter.

>>> case  = "caseCamel"
>>> label = "Case Camel"
>>> list  = ["apple", "banana"]
>>>
>>> (case or label) in list
False
>>> list = ["apple", "caseCamel"]
>>> (case or label) in list
True
>>> (case and label) in list
False
>>> list = ["case", "caseCamel", "Case Camel"]
>>> (case and label) in list
True
>>>

and if you have a looong list of variables held in a sublist variable

>>>
>>> list  = ["case", "caseCamel", "Case Camel"]
>>> label = "Case Camel"
>>> case  = "caseCamel"
>>>
>>> sublist = ["unique banana", "very unique banana"]
>>>
>>> # example for if any (at least one) item contained in superset (or statement)
...
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
False
>>>
>>> sublist[0] = label
>>>
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
True
>>>
>>> # example for whether a subset (all items) contained in superset (and statement)
...
>>> # a bit of demorgan's law
...
>>> next((False for item in sublist if item not in list), True)
False
>>>
>>> sublist[1] = case
>>>
>>> next((False for item in sublist if item not in list), True)
True
>>>
>>> next((True for item in sublist if next((True for x in list if x == item), False)), False)
True
>>>
>>>

Answer 6

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

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