python检查单词是否在列表的某些元素中

Question

我想知道是否有更好的方法：

if word==wordList[0] or word==wordList[2] or word==wordList[3] or word==worldList[4]

Answer 1

word in wordList

或者，如果你想先检查4，

word in wordList[:4]

Answer 2

Very simple task, and so many ways to deal with it. Exciting! Here is what I think:

If you know for sure that wordList is small (else it might be too inefficient), then I recommend using this one:

b = word in (wordList[:1] + wordList[2:])

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Otherwise I would probably go for this (still, it depends!):

b = word in (w for i, w in enumerate(wordList) if i != 1)

For example, if you want to ignore several indexes:

ignore = frozenset([5, 17])
b = word in (w for i, w in enumerate(wordList) if i not in ignore)

This is pythonic and it scales.

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However, there are noteworthy alternatives:

### Constructing a tuple ad-hoc. Easy to read/understand, but doesn't scale.
# Note lack of index 1.
b = word in (wordList[0], wordList[2], wordList[3], wordList[4])

### Playing around with iterators. Scales, but rather hard to understand.
from itertools import chain, islice
b = word in chain(islice(wordList, None, 1), islice(wordList, 2, None))

### More efficient, if condition is to be evaluated many times in a loop.
from itertools import chain
words = frozenset(chain(wordList[:1], wordList[2:]))
b = word in words

Answer 3

Have indexList be a list of the indicies you want to check (ie, [0,2,3] ) and have wordList be all the words you want to check. Then, the following command will return the 0th, 2nd, and 3rd elements of wordList, as a list:

[wordList[i] for i in indexList]

This will return [wordList[0], wordList[2], wordList[3]] .