(Using Python3)
I have a list of tuples, (of strings)
have = [ ('a', 'b', 'c', 'd'), ('a', 'b', 'c', 'e'), ('a', 'b', 'c', 'f'), ('a', 'b', 'c', 'g'), ('a', 'b', 'd', 'e'), ('a', 'b', 'd', 'f'), ('a', 'b', 'd', 'g'), ('a', 'b', 'e', 'f'), ('a', 'b', 'e', 'g'), ('a', 'b', 'f', 'g'), ('a', 'c', 'd', 'e'), ('a', 'c', 'd', 'f'), ('a', 'c', 'd', 'g'), ('a', 'c', 'e', 'f'), ('a', 'c', 'e', 'g'), ('a', 'c', 'f', 'g'), ('a', 'd', 'e', 'f'), ('a', 'd', 'e', 'g'), ('a', 'd', 'f', 'g'), ('a', 'e', 'f', 'g'), ('b', 'c', 'd', 'e'), ('b', 'c', 'd', 'f'), ('b', 'c', 'd', 'g'), ('b', 'c', 'e', 'f'), ('b', 'c', 'e', 'g'), ('b', 'c', 'f', 'g'), ('b', 'd', 'e', 'f'), ('b', 'd', 'e', 'g'), ('b', 'd', 'f', 'g'), ('b', 'e', 'f', 'g'), ('c', 'd', 'e', 'f'), ('c', 'd', 'e', 'g'), ('c', 'd', 'f', 'g'), ('c', 'e', 'f', 'g'), ('d', 'e', 'f', 'g') ]
I also have a list of tuples (also strings) which I want to “exclude”
exclude = [('a', 'd'), ('b', 'c')]
I’m trying to find an efficient way to remove any element in have
that contains both the elements in each exclude
tuple. Ordering does not matter.
My goal is to return something like this:
[ ('a', 'b', 'e', 'f'), ('a', 'b', 'e', 'g'), ('a', 'b', 'f', 'g'), ('a', 'c', 'e', 'f'), ('a', 'c', 'e', 'g'), ('a', 'c', 'f', 'g'), ('a', 'e', 'f', 'g'), ('b', 'd', 'e', 'f'), ('b', 'd', 'e', 'g'), ('b', 'd', 'f', 'g'), ('b', 'e', 'f', 'g'), ('c', 'd', 'e', 'f'), ('c', 'd', 'e', 'g'), ('c', 'd', 'f', 'g'), ('c', 'e', 'f', 'g'), ('d', 'e', 'f', 'g') ]
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Answer
You could convert the exclude
tuples to set
s and then check for each element of have
is the excluded set isn’t a subset of it:
excludeSet = [set(e) for e in exclude] filteredHave = [h for h in have if not any(e for e in excludeSet if e.issubset(h))]