抓取在字符串中找到的列表中的第一个单词. ( Python ) [英] Grab the first word in a list that is found in a string. ( Python )

问题描述

所以，我有一个像这样的单词列表:

So, I have a list of words like so:

activationWords = ['cactus', 'cacti', 'rofl']

我想找到这些单词中的任何一个，并返回出现在随机字符串中的所有这些单词中的第一个单词.我将使用此字符串作为示例:

And I want to find any of those words and return the first word of any of those words appearing in a random string. I'll use this string as an example:

str = "Wow, rofl I found a cactus in a cacti pile."

如上面的示例字符串所示，列表中单词的第一个实例是"rofl".我希望能够检测到该问题并将该单词返回一个字符串，我可以根据自己的判断使用该字符串.我该怎么办?

As you can see with the above example string, the first instance of a word in the list is "rofl". I want to be able to detect that and return the word into a string that I can use to my discretion. How would I do this?

请记住，该字符串只是一个示例.每次运行此命令时，它将使用不同的字符串.

Keep in mind that that string is just an example. Each time I run this it will be using a different string.

推荐答案

您可以使用

You can use str.find() here:

>>> activationWords = ['cactus', 'cacti', 'rofl']
>>> s = "Wow, rofl I found a cactus in a cacti pile."
>>> temp = [(s.find(i), i) for i in activationWords if i in s]
>>> print temp
[(20, 'cactus'), (32, 'cacti'), (5, 'rofl')]
>>> print min(temp)[1]
rofl

str.find()查找该字符串在另一个字符串中的位置.它返回第一个字母的索引.

str.find() finds where the string is in the other string. It returns the index of the first letter.

[(s.find(i), i) for i in activationWords]是一个列表推导，返回一个元组列表，第一项是str.find()的结果，第二项是单词.

[(s.find(i), i) for i in activationWords] is a list comprehension that returns a list of tuples, the first item being the result of str.find(), the second being the word.

然后我们在此处使用 min() 来获得最低的数字，然后按[1]即可获得单词.

Then we use min() here to get the lowest number, and [1] to get the word.

请注意，您不应将字符串命名为str.它会覆盖内置类型.

Note that you should not name a string str. It overrides the built-in type.

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