Python: See if one set contains another entirely?

Question

Is there a fast way to check if one set entirely contains another?

Something like:

>>>[1, 2, 3].containsAll([2, 1])
True

>>>[1, 2,         


        

Answer 1

If you suspect a set to be a subset of another, and intersect those two sets together, the result is equal to itself if it is a subset.

a = [2,1,3,3]
b = [5,4,3,2,1]
set(a).intersection(set(b)) == set(a)
>>True

Answer 2

For completeness: this is equivalent to issubset (although arguably a bit less explicit/readable):

>>> set([1,2,3]) >= set([2,1])
True
>>> set([1,2,3]) >= set([3,5,9])
False

Answer 3

One option is left untouched -- subtraction:

>>> {1, 2} - {1, 2, 3}
set([])
>>> {1, 2, 3} - {1, 2}
set([3])

Basically you check what elements in first list are not in second list.

I found it very handy since you could show what values are missing:

>>> def check_contains(a, b):
...     diff = a - b
...     if not diff:
...         # All elements from a are present in b
...         return True
...     print('Some elements are missing: {}'.format(diff))
...     return False
...
>>> check_contains({1, 2}, {1, 2, 3})
True
>>> check_contains({1, 2, 3}, {1, 2})
Some elements are missing: set([3])
False

Answer 4

>>> set([1,2,3]).issuperset(set([2,1]))
True 
>>>    
>>> set([1,2,3]).issuperset(set([3,5,9]))
False

Answer 5

Below function return 0 if mainlist doesn't contains sublist fully and 1 if contains fully.

def islistsubset(sublist,mainlist):
     for item in sublist:
             if item in mainlist:
                     contains = 1
             else:
                     contains = 0
                     break;
     return contains

Answer 6

You can use either set.issubset() or set.issuperset() (or their operator based counterparts: <= and >=). Note that the methods will accept any iterable as an argument, not just a set:

>>> {1, 2}.issubset([1, 2, 3])
True
>>> {1, 2, 3}.issuperset([1, 2])
True

However, if you use operators, both arguments must be sets:

>>> {1, 2} <= {1, 2, 3}
True
>>> {1, 2, 3} >= {1, 2}
True