Python:在 any() 函数中扩展迭代器变量的范围 [英] Python: Expanding the scope of the iterator variable in the any() function

问题描述

我写了一些结构上等效的现实世界代码，我预计 firstAdjective 的结果将是 quick.但结果显示 word 超出范围.有什么巧妙的解决方案可以克服这个问题，但仍然保留我想做的语言"风格?

<预><代码>>>>text = '敏捷的棕色狐狸跳过懒狗'>>>形容词 = ['慢'、'快'、'棕色'、'懒惰']>>>如果有的话(文本中的单词对形容词中的单词):... firstAdjective = word...回溯(最近一次调用最后一次):文件<交互式输入>"，第 2 行，在 <module> 中.NameError: 名称 'word' 未定义

解决方案

您可以在生成器表达式上使用next:

firstAdjective = next((如果单词在文本中，则在形容词中逐字逐句)，无)if first形容词:...

当找不到单词时返回默认值 None (credit @Bakuriu)

试用:

<预><代码>>>>firstadjective = next((如果单词在文本中，则形容词中的单词一个单词)，无)>>>第一个形容词'快的'

I wrote some structurally equivalent real world code, where I anticipated the result for firstAdjective would be quick. But the result shows that word is out of scope. What is a neat solution that will overcome this but still retain the 'linguistic' style of what I want to do?

>>> 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

解决方案

You can use next on a generator expression:

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

if firstAdjective:
    ... 

A default value of None is returned when the word is not found (credit @Bakuriu)

Trial:

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

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