Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
662 views
in Technique[技术] by (71.8m points)

python - Test if set is a subset, considering the number (multiplicity) of each element in the set

I know I can test if set1 is a subset of set2 with:

{'a','b','c'} <= {'a','b','c','d','e'} # True

But the following is also True:

{'a','a','b','c'} <= {'a','b','c','d','e'} # True

How do I have it consider the number of times an element in the set occurs so that:

{'a','b','c'}     <= {'a','b','c','d','e'}      # True
{'a','a','b','c'} <= {'a','b','c','d','e'}      # False since 'a' is in set1 twice but set2 only once
{'a','a','b','c'} <= {'a','a','b','c','d','e'}  # True because both sets have two 'a' elements

I know I could do something like:

A, B, C = ['a','a','b','c'], ['a','b','c','d','e'], ['a','a','b','c','d','e']
all([A.count(i) == B.count(i) for i in A]) # False
all([A.count(i) == C.count(i) for i in A]) # True

But I was wondering if there was something more succinct like set(A).issubset(B,count=True) or a way to stay from list comprehensions. Thanks!

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

As stated in the comments, a possible solution using Counter:

from collections import Counter

def issubset(X, Y):
    return len(Counter(X)-Counter(Y)) == 0

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...