如何 return 给定一个字符串列表和一个列表列表的字数？

Question

假设我有一长串包含标点符号、空格等的列表，如下所示：

list_1 = [[the guy was plaguy but unable to play football, but he was able to play tennis],[That was absolute cool],...,[This is an implicit living.]]

我还有一个像这样的长列表：

list_2 =['unable', 'unquestioning', 'implicit',...,'living', 'relative', 'comparative']

如何为 list_1 的每个子列表提取出现在 list_2 中的所有单词的计数或频率？例如上面的列表：

list_2 =['unable', 'unquestioning', 'implicit',...,'living', 'relative', 'comparative']

[the guy was unable to play football, but he was able to play tennis]

由于 unable 出现在 list_2 的前一个子列表中，因此此列表的计数为 1。

list_2 =['unable', 'unquestioning', 'implicit',...,'living', 'relative', 'comparative']

[That was absolute cool]

由于list_2的词没有出现在上一个子列表中，因此计数为0。

list_2 =['unable', 'unquestioning', 'implicit',...,'living', 'relative', 'comparative']

[This is an implicit living.]

由于 implicit 和 living 出现在 list_2 的前一个子列表中，因此此列表的计数为 2。

所需的输出是 [1,0,2]。

知道如何处理此任务以 return 计数列表吗？提前谢谢大家。

例如：

>>> [sum(1 for word in list_2 if word in sentence) for sublist in list_1 for sentence in sublist]

是错误的，因为混淆了两个词 guy 和 playguy。知道如何解决这个问题吗？

Answer 1

使用内置函数 sum 和列表理解

>>> list_1 = [['the guy was unable to play football, but he was able to play tennis'],['That was absolute cool'],['This is implicit living.']]
>>> list_2 =['unable', 'unquestioning', 'implicit','living', 'relative', 'comparative']   
>>> [sum(1 for word in list_2 if word in sentence) for sublist in list_1 for sentence in sublist]

[1, 0, 2]

Answer 2

诀窍是使用 split() 方法和列表理解。如果只用空格分隔：

list_1 = ["the guy was unable to play football but he was able to play tennis", "That was absolute cool", "This is implicit living"]

list_2 =['unable', 'unquestioning', 'implicit','living', 'relative', 'comparative']

print([sum(sum(1 for j in list_2 if j in i.split()) for i in k for k) inlist_1])

但是，如果你想使用所有非字母数字来分词，你应该使用re:

import re

list_1 = ["the guy was unable to play football,but he was able to play tennis", "That was absolute cool", "This is implicit living"]
list_2 =['unable', 'unquestioning', 'implicit','living', 'relative', 'comparative']

print(sum([sum(1 for j in list_2 if re.split("\W",i)) for i in k) for k in list_1])

\W 字符集全部为非字母数字。

Answer 3

我宁愿使用正则表达式。首先，因为你需要匹配一个完整的单词，这与其他字符串搜索方法相比很复杂。而且，即使它看起来像火箭筒，它通常也非常有效。

您首先从 list_2 生成一个正则表达式，然后使用它搜索 list_1 的句子。正则表达式的构造如下："(\bword1\b|\bword2\b|...)" 表示 "either whole word1 or whole word2 or..."。 \b表示在单词的开头或结尾匹配。

我假设您想要 list_1 的每个子列表的结果，而不是每个子列表的每个句子的结果。

_regex = re.compile(r"(\b{}\b)".format(r"\b|\b".join(list_2)))
word_counts = [ 
    sum(
        sum(1 for occurence in _regex.findall(sentence))
        for sentence in sublist
    ) for sublist in list_1
]

Here you can find a whole sample code 与普通字符串搜索的性能比较，知道匹配整个单词需要更多的工作，因此效率会更低。

如何 return 给定一个字符串列表和一个列表列表的字数？

How to return the word count given one list of strings and one list of lists?

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