Compare/find a list in another list's nested lists

Question

I have a few lists called find. I want to know if these find are part of full_list. Lists find 1-4 are part of full_list while lists find 5-7 are not.

The example below returns "Yes".

find1 = [[1, 1]]
find2 = [[1, 1], [2, 2]]
find3 = [[1, 1], [3, 3]]
find4 = [[4, 4], [2, 2], [3, 3]]

find5 = [[1, 0]]
find6 = [[1, 1], [2, 0]]
find7 = [[1, 1], [3, 0]]

full_list = [[1, 1], [2, 2], [3, 3], [4, 4]]

if find2[0] in full_list and find2[1] in full_list:
    print("Yes")
else:
    print("No")

Because len(find2) != len(find4), the current if statement is very clumsy and almost useless.

How to make this work in a more universal way?

Austin · Accepted Answer · 2018-10-06 17:21:33Z

3

You could use all() with a generator which returns a True if everything is a truthy or False:

if all(x in full_list for x in find2):
    print("Yes")
else:
    print("No")

# => prints Yes

This one is generic, just need to change find2 to any list you need to check with full_list.

edited Oct 6, 2018 at 17:21

answered Oct 6, 2018 at 17:15

Austin

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Comments

Romain Reboulleau · Accepted Answer · 2018-10-06 17:20:24Z

1

Try this:

set1 = set(find1)
full_set = set(full_list)
if set1.issubset(full_set):
    print("Yes")
else:
    print("No")

edited Oct 6, 2018 at 17:20

answered Oct 6, 2018 at 17:18

Romain Reboulleau

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1 Comment

Artem Fedosov Over a year ago

We have to convert list to tuples first, as they aren't hashable.

vash_the_stampede · Accepted Answer · 2018-10-10 15:51:41Z

1

You could set up a function to process all of these, all the mentioned methods will work, just displaying other options, you could use filter and compare the lens as well

def is_in_full(k):
    l = list(filter(lambda x: x in k, full_list))
    if len(l) == len(k):
        return 'Yes'
    else:
        return 'No'

print(is_in_full(find1))
# Yes

edited Oct 10, 2018 at 15:51

answered Oct 6, 2018 at 17:32

vash_the_stampede

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