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by eswoo
2901 days ago
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It needs to be [bar(foo(y)) for y in (z for z in stuff if foo(z))]
(etc.) though, since `bar` takes as input the output of `foo`. This leads to the objectionable duplicate calls to `foo`, hence the new assignment expressions.I like Dunnorandom's [bar(x) for x in map(foo, stuff) if x]
best for a correct result using existing syntax, or [bar(x) for y in (foo(x) for x in stuff) if y]
if you don't like `map`. |
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