从缺少某些字符串的列表中删除元素 - python

Question

我有一个看起来像这样的大列表：

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

我想删除列表中不包含单词"other" 或 [=26]的所有元素=].

我试过了，但它并没有删除我需要的所有元素（只有一些接近尾声的元素）：

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

我做错了什么？

Answer 1

您不应该在遍历列表时从列表中删除项目。此外，您的条件语句不符合您的意思：它检查 'other' 是否真实，只检查 'things' 是否包含。要修复它，请使用 and 和两个单独的 in 检查。

如果列表不是很大，您可以使用列表理解来重建它：

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

否则，从列表的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，在您的版本中存在三个主要问题：

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

这是您的版本，已修改以解决上述问题：

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)

这里有一些替代方法：

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 列表理解表达式 使用 all(..) 检查子字符串为：

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

这将为您 return 包含 "other" 或 "things" 的新单词列表。

从缺少某些字符串的列表中删除元素 - python

Removing elements from a list that lack certain strings - python

python

list

elements