Check if all elements in list1 exist in list2 python

Question

I want to be able to find whether all elements in list1 exist in list2, both lists containing chars. The number of times the same element appears in each list is important.

I tried using subsets however that didn't take note of the number of times an item appeared in a list

e.g.

list1 = [a, b, c]
list2 = [a, b, c, b]

It would find list2 a subset of list1, whereas I would only want my function to execute if:

list1 = [a, b, c, b, i]
list2 = [a, c, b, b]

as this means all items in list2 appear in list1.

If anyone was interested, using a counter was inefficient for large strings so I ended up adding all elements of list1 to a dictionary, with the values being number of occurances of each element and subtracted all elements of list2. If any values ended up negative not all items in list2 appear in list1

Get the count in both lists and then check if the count of all items in `list2` is exactly same in `list1`: https://docs.python.org/3/library/collections.html#collections.Counter — Ashwini Chaudhary, Mar 10 '16 at 08:58
This is very unclear, please give a better example of input/output — Idos, Mar 10 '16 at 09:03

Eugene Yarmash · Accepted Answer · 2016-03-10T09:14:14.687

4

You could count the items in both lists with collections.Counter and check that the counts for the corresponding items in the first list are >= than those in the second:

>>> from collections import Counter
>>> c = Counter("abcbi")
>>> d = Counter("acbb")
>>> c.subtract(d)
>>> if all(v >= 0 for k, v in c.items()):
...     print("All items exist")
... 
All items exist

edited Mar 10 '16 at 09:14

answered Mar 10 '16 at 09:07

Eugene Yarmash

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1

This is very nice, I didn't know of `subtract` before! I think we have a winner. – Frerich Raabe Mar 10 '16 at 09:15
This works however is very inefficient if the counters have large strings – user2201609 Mar 10 '16 at 09:50

score 1 · Answer 2 · answered Mar 10 '16 at 09:05

You can use collections.Counter

from collections import Counter
list1 = ['a', 'b', 'c']
list2 = ['a', 'b', 'c', 'b']    
c1 = Counter(list1)
c2 = Counter(list2)

for key, count in c1.iteritems():
    if c2[key] < count:
        break
else:
    print 'list2 has all the chracters in list1'

Idos · Answer 3 · 2016-03-10T09:05:08.923

0

You can use issuperset:

>>> list1 = ['a', 'b', 'c', 'b', 'i']
>>> list2 = ['a', 'c', 'b', 'b']
>>> set(list1).issuperset(set(list2)) # all items of list2 exist in list1
True
>>> set(list2).issuperset(set(list1)) # all items of list1 does not exist in list 2
False

edited Mar 10 '16 at 09:05

answered Mar 10 '16 at 08:57

Idos

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*I tried using subsets however that didn't take note of the **number of times** an item appeared in a list* – Ashwini Chaudhary Mar 10 '16 at 08:59
From the question statement, this seems *to me* as the correct answer – Idos Mar 10 '16 at 09:05
This does nothing different from what I can see. It doesn't acknowledge item frequency in the lists – user2201609 Mar 10 '16 at 09:13
Again, from what you state in your question this is what you "want". If you would be more clear we could give you a better answer. – Idos Mar 10 '16 at 09:14

score 0 · Answer 4 · answered Mar 10 '16 at 09:13

For your problem, you'll need to

Verify that all elements in list2 appear in list1 but
No element in list2 appears more often than it does in list2

This can be done quite easily using collections.Counter:

import collections

def f(list1, list2):
    d1 = collections.Counter(list1)
    d2 = collections.Counter(list2)
    return set(d1.keys()).issuperset(set(d2.keys())) \
       and all(d2[k] <= d1[k] for k in d1.keys())

This first checks (using set.issuperset) that all the different elements of list2 are contained in list1. If that's true, it checks how often each element occurs in list2 and verifies that it appears less often (or equally often) in list1.

This is a more verbose version of what "eugene y" wrote, if possible I'd rather go with his version than mine. — Frerich Raabe, Mar 10 '16 at 09:16

Check if all elements in list1 exist in list2 python

4 Answers4