Question

假设我有一个列表

test = ["a","bb","ph","phi","phi_ph"]

其中test的成员可以包含字符串ph，字符串phi，这两者的组合，或者不包含任何字符串ph。如何过滤此列表以仅保留包含test_filtered = ["ph","phi_ph"]的元素，以便：

[x for x in test if 'ph' in x]

例如

>> ["ph","phi","phi_ph"]

返回

Calendar

Answer 1

使用re.compile()和re.search()函数的解决方案：

import re

test = ["a","bb","ph","phi","phi_ph", "phi_abc", "ph_a"]
search_str = 'ph'

pattern = re.compile(r'(^|[^a-z0-9])'+ search_str + '([^a-z0-9]|$)')
result = [i for i in test if re.search(pattern, i)]

print(result)

输出：

['ph', 'phi_ph', 'ph_a']

(^|[^a-z0-9]) - 交替组，确保搜索字符串（即ph）应出现在字符串的开头或以非字母数字字符开头

([^a-z0-9]|$) - 交替组，确保搜索字符串（即ph）应出现在字符串末尾或后跟非字母数字字符

Answer 2

这应该照顾它：

test = ["a","bb","ph","phi","phi_ph"]

result = [y for y in test for x in y.split('_') if x == "ph"]

这样您只需添加"ph"，而且不需要排除任何内容。

Answer 3

我发现以下列表理解完成了这项工作：

test_filtered = [x for x in test if ('ph' in x and x.count('ph')!=x.count('phi'))]
>> ['ph', 'phi_ph']

Answer 4

试试这个：

In [26]: new_filter=[ x.split('_')[0] for x in test_filtered]+test_filtered
In [27]: [x for x in test if x in new_filter]
Out[27]: ['ph', 'phi', 'phi_ph']

检查子字符串是否在字符串

4 个答案: