Question

我在python中得到以下示例的误报。我试图找出字符串中是否存在关键字。问题是字符串的单词通常用下划线或连字符连接，所以如果关键字在不在单词中时存在，我只想要肯定的结果。它可以用连字符，下划线或任何不是字母的东西来考虑真实的结果。通常它应该用下划线或连字符包围。它也不区分大小写。

test_list = ['server_test', 'server_dev', 'server_uat', 'server_dr', 'server-dr-NA', 'server-DR', 'dress_prod', 'testosterone','uatae','devacurl', 'dev_server']

结果应输出此True / False列表

[True, True, True, True, True, True, False, False, False, False, True]

实现：

key_words = ['uat','dr','test','qa','dev']
for name in test_list:
     if any(x in name.lower() for x in key_words):
         print('True')
     else:
         print('False')

结果：

True
True
True
True
True
True
True
True
True
True

在python中有更好的方法吗？

如果不是我如何在python中使用正则表达式？

请记住，这是在一个性能很重要的大型数据集上循环。

Answer 1

假设：

>>> test_list = ['server_test', 'server_dev', 'server_uat', 'server_dr', 'server-dr-NA', 'server-DR', 'dress_prod', 'testosterone','uatae','devacurl', 'dev_server']
>>> key_words = ['uat','dr','test','qa','dev']

您可以使用re.split和any：

>>> [any(word.lower() in key_words for word in re.split(r'[^a-zA-Z]', s))
...     for s in test_list]
[True, True, True, True, True, True, False, False, False, False, True]

与目标相同：

>>> tgt=[True, True, True, True, True, True, False, False, False, False, True]
>>> [any(word.lower() in key_words for word in re.split(r'[^a-zA-Z]', s))
...     for s in test_list]==tgt
True

Answer 2

使用基于负面lookbehind的正则表达式。

>>> test_list = ['server_test', 'server_dev', 'server_uat', 'server_dr', 'server-dr-NA', 'server-DR', 'dress_prod', 'testosterone','uatae','devacurl', 'dev_server']
>>> key_words = ['uat','dr','test','qa','dev']
>>> [True if re.search(r'(?i)(?<![a-z])(?:' + '|'.join(key_words) + ')(?![a-z])', i) else False for i in test_list]
[True, True, True, True, True, True, False, False, False, False, True]
>>>

Answer 3

另一种方法是使用\b来检测字边界。不幸， _被视为单词字符，因此我们需要检测\b 或 _。

不像Avinash的解决方案那样简洁或高效，但可能更具可读性。

import re

test_list = ['server_test', 'server_dev', 'server_uat', 'server_dr',
             'server-dr-NA', 'server-DR', 'dress_prod', 'testosterone',
             'uatae', 'devacurl', 'dev_server']

key_words = ['uat','dr','test','qa','dev']

for name in test_list:
    for kw in key_words:
        regex = r'(\b|_)'+kw+r'(\b|_)'
        if re.search(regex, name, re.IGNORECASE):
            print('True')
            break  # exit "for kw" loop
    else:  # only executed if "for kw" loop exits via exhaustion, not via break
        print('False')

Answer 4

我认为这种模式很容易理解和修改：

import re

pattern = r'.*(^|[^a-z])({names})([^a-z]|$).*'.format(names='|'.join(key_words))

# .*(^|[^a-z])(uat|dr|test|qa|dev)([^a-z]|$).*

for name in test_list:
    print(bool(re.search(pattern, name, re.IGNORECASE)))

Answer 5

import re

key_words = ['uat','dr','test','qa','dev']
test_list = ['server_test', 'server_dev', 'server_uat', 'server_dr', 'server-dr-NA', 
             'server-DR', 'dress_prod', 'testosterone','uatae','devacurl', 'dev_server']



def check(word):
    parts = re.split('[^a-z]', word.lower())
    return any(part in key_words for part in parts)

print([check(item) for item in test_list])

Python Regex - 在单词

5 个答案: