Python:在any()函数中扩展迭代器变量的范围

Question

我写了一些结构上等效的真实代码，我预计 firstAdjective 的结果会是 quick。但是结果显示 word 超出了范围。有什么巧妙的解决方案可以克服这个问题，但仍保留我想做的事情的“语言”风格？

>>> text = 'the quick brown fox jumps over the lazy dog'
>>> adjectives = ['slow', 'quick', 'brown', 'lazy']
>>> if any(word in text for word in adjectives):
...     firstAdjective = word
...     
Traceback (most recent call last):
  File "<interactive input>", line 2, in <module>
NameError: name 'word' is not defined

Answer 1

您可以在生成器表达式 上使用next:

firstAdjective = next((word for word in adjectives if word in text), None)

if firstAdjective:
    ... 

当找不到单词时返回默认值 None(信用@Bakuriu)

试用:

>>> firstadjective = next((word for word in adjectives if word in text), None)
>>> firstadjective
'quick'