清单的理解和条件? [英] List Comprehensions and Conditions?

问题描述

我正在尝试看看是否可以使用列表理解来使此代码更好.
可以说我有以下列表:

I am trying to see if I can make this code better using list comprehensions.
Lets say that I have the following lists:

a_list = [
        'HELLO',
        'FOO',
        'FO1BAR',
        'ROOBAR',
        'SHOEBAR'
        ]

regex_list =   [lambda x: re.search(r'FOO', x, re.IGNORECASE),
                lambda x: re.search(r'RO', x, re.IGNORECASE)]

我基本上想将regex_list中没有任何匹配项的所有元素添加到另一个列表中.

I basically want to add all the elements that do not have any matches in the regex_list into another list.

例如==>

newlist = []
for each in a_list:
    for regex in regex_list:
        if(regex(each) == None):
            newlist.append(each)

如何使用列表推导来做到这一点?甚至有可能吗?

How can I do this using list comprehensions? Is it even possible?

推荐答案

当然，我认为应该这样做

Sure, I think this should do it

newlist = [s for s in a_list if not any(r(s) for r in regex_list)]

EDIT :经过仔细检查，我注意到您的示例代码实际上将a_list中与正则表达式 all 不匹配的每个字符串添加到新列表中-而且，它为每个不匹配的正则表达式添加一次每个字符串.我的列表理解功能符合我的意思，即为每个与正则表达式任何不匹配的字符串添加一个副本.

EDIT: on closer inspection, I notice that your example code actually adds to the new list each string in a_list that doesn't match all the regexes - and what's more, it adds each string once for each regex that it doesn't match. My list comprehension does what I think you meant, which is add only one copy of each string that doesn't match any of the regexes.

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