python检查单词是否在列表的某些元素中 [英] python check if word is in certain elements of a list

问题描述

我想知道是否有更好的放置方式:

I was wondering if there was a better way to put:

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

推荐答案

非常简单的任务，以及多种处理方式.令人兴奋！这是我的想法:

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

如果您确定wordList很小(否则可能效率太低)，那么我建议您使用此列表:

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:])

---

否则，我会大概去做(仍然要视情况而定！):

---

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

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

例如，如果您要忽略多个索引:

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)

这是pythonic，并且可以缩放.

This is pythonic and it scales.

但是，还有一些值得注意的选择:

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

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