If I have two lists that look like this:
list1 = ['a', 'a', 'b', 'c']
list2 = ['a', 'a', 'a', 'b']
How can I check if the second list is contained in the first? I assume sets would not work, because the letters can be repeated.
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Does order matter, or simply the number of occurrences of values? – eestrada Dec 12 '15 at 00:50
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@eestrada the order does not matter – Dec 12 '15 at 00:51
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Possible duplicate of [Python - Comparing two lists](http://stackoverflow.com/questions/16138015/python-comparing-two-lists) – Achayan Dec 12 '15 at 00:51
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So in this case, the expect outpu should be False or True? – Remi Guan Dec 12 '15 at 01:04
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@KevinGuan In this case, False. Only two a's in the first list. – Dec 12 '15 at 01:05
2 Answers
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Sets won't work if the frequency matter but counting the frequency will:
from collections import Counter
list1 = ['a', 'a', 'b', 'c']
list2 = ['a', 'a', 'a', 'b']
cn1, cn2 = Counter(list1), Counter(list2)
print(all(cn2[k] <= v for k, v in cn1.items()))
If the count of each string in list2 is <= to the amount of times it appears in list1 you have all the strings from list2 in list1, which for the lists in your question would return False but for,
list1 = ['a', 'a', 'b', 'c', 'a']
list2 = ['a', 'a', 'a', 'b']
would return True as you have the same amount of a's and a b.
Padraic Cunningham
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@shomz, I edited, so you want to basically see if you could make up the first list by using elements in the second? – Padraic Cunningham Dec 12 '15 at 00:56
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Similar to @Padraic_Cunningham's answer:
from collections import Counter
def contained(l1, l2):
cntr1 = Counter(list1)
cntr2 = Counter(list2)
for k in cntr2:
if cntr2[k] != cntr1.get(k):
return False
return True
list1 = ['a', 'a', 'b', 'c']
list2 = ['a', 'a', 'a', 'b']
contained(list1, list2)
eestrada
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