Question

我有一个列表“ abc”（字符串），我正在尝试从列表“ abc”中删除列表“ stop”中存在的某些单词以及abc中存在的所有数字。

abc=[ 'issues in performance 421',
 'how are you doing',
 'hey my name is abc, 143 what is your name',
 'attention pleased',
 'compliance installed 234']
stop=['attention', 'installed']

我正在使用列表推导将其删除，但是下面的这段代码无法删除该单词。

new_word=[word for word in abc if word not in stop ]

结果：（注意词仍然存在。）

['issues in performance',
 'how are you doing',
 'hey my name is abc, what is your name',
 'attention pleased',
 'compliance installed']

所需的输出：

 ['issues in performance',
     'how are you doing',
     'hey my name is abc, what is your name',
     'pleased',
     'compliance']

谢谢

Answer 1

在过滤掉stop中的单词后，您需要将每个短语分解为单词，然后将单词重新组合为短语。

[' '.join(w for w in p.split() if w not in stop) for p in abc]

这将输出：

['issues in performance', 'how are you doing', 'hey my name is abc, what is your name', 'pleased', 'compliance installed']

Answer 2

只需使用set就可以解决这个问题。因为每个项目可能有多个单词，所以不能使用in。您应该将set与&一起使用以获取公共词。如果您的stop集已存在，则将返回True。因为您只关心其余部分，所以我们可以在这里使用if not。

new_word=[word for word in abc if  not set(word.split(' ')) & set(stop)]

更新

如果您还想删除所有包含数字的项目，只需执行以下操作即可：

new_word=[word for word in abc if  not (set(word.split(' ')) & set(stop) or any([i.strip().isdigit() for i in word.split(' ')]))]

Answer 3

这是一个解决方案，将简单的正则表达式与re.sub方法一起使用。 此解决方案还删除了数字。

import re

abc=[ 'issues in performance 421',
 'how are you doing',
 'hey my name is abc, 143 what is your name',
 'attention pleased',
 'compliance installed 234']
stop=['attention\s+', 'installed\s+', '[0-9]']

[(lambda x: re.sub(r'|'.join(stop), '', x))(x) for x in abc]


'Output':
['issues in performance ',
'how are you doing',
 'hey my name is abc,  what is your name',
 'pleased',
 'compliance ']

Answer 4

list1 = []
for word in abc:
    word1 = ''
    for remove_word in stop:
        word1 = remove_word
        word1 = word.replace(word1, '')
    list1.append(word1)

Answer 5

这至少是我要做的：

abc=[ 'issues in performance 421',
    'how are you doing',
    'hey my name is abc, 143 what is your name',
    'attention pleased',
    'compliance installed 234'
]
stop=['attention', 'installed']
for x, elem in enumerate(abc):
    abc[x] = " ".join(filter(lambda x: x not in stop and not x.isdigit(), elem.split()))
print(abc)

结果：

['issues in performance',
    'how are you doing',
    'hey my name is abc, what is your name',
    'pleased',
    'compliance']

希望对您有帮助：）

从Python列表中删除单词

5 个答案: