Question

我有一个大型列表，看起来像这样：

entries = ["['stuff']...other stuff", "['stuff']...stuff", "['stuff']...more stuff", ...]

我想删除列表中不包含“其他”或“这些”字样的所有元素。

我尝试了这个，但它没有删除我需要它的所有元素（只有一些接近结尾）：

for e in entries:
    if 'other' or 'things' not in e:
        entries.remove(e)
print entries

我做错了什么？

Answer 1

在迭代列表时，您不应该从列表中删除项目。此外，您的条件语句并不符合您的意思：它会检查'other'的真实性，并仅'things'进行遏制。要解决此问题，请使用and和两个单独的in支票。

如果列表不是很大，你可以使用列表推导来重建它：

entries = [e for e in entries if "other" not in e and "things" not in e]

否则，从列表末尾循环到开头并按索引删除项目。

for i in range(len(entries)-1, -1, -1):
    if "other" in entries[i] and "things" in entries[i]:
        del entries[i]

Answer 2

正如其他人已经指出的那样，在你的版本中有三个主要问题：

for e in entries:
    if 'other' or 'things' not in e: #or returns first truthy value, and `if other` is always true.  Also, you need and, not or.
        entries.remove(e) #mutating the item you are iterating over is bad
print entries

以下是您修改上述问题的修订版本：

for e in words[:]: #words[:] is a copy of words, solves mutation issue while iterating
    if 'other' not in e and 'things' not in e: #want words that both don't contain 'other' AND dont contain 'things'
        print(e)
        words.remove(e)
print(words)

以下是一些替代方法：

import re

words = ['this doesnt contain chars you want so gone',
         'this contains other so will be included',
         'this is included bc stuff']

answer = list(filter(lambda x: re.search('other|stuff',x),words))
other_way = [sentence for sentence in words if re.search('other|stuff',sentence)]

print(answer)
print(other_way)

Answer 3

您可以使用all(..)使用列表理解表达式来检查子字符串：

>>> [entry for entry in entries if any(something in entry  for something in  ["other", "things"])]

这将返回包含＆＃34;其他＆＃34;的新单词列表。或＆＃34;事物＆＃34;。

从缺少某些字符串的列表中删除元素 - python

3 个答案: