如何从 Python 中的列表中过滤项目？
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How can I filter items from a list in Python?
提问by Allen S. Rout
I have data naively collected from package dependency lists.
我从包依赖列表中天真地收集了数据。
Depends: foo bar baz >= 5.2
取决于： foo bar baz >= 5.2
I end up with
我结束了
d = set(['foo','bar','baz','>=','5.2'])
I don't want the numerics and the operands.
我不想要数字和操作数。
In Perl I would
在 Perl 中，我会
@new = grep {/^[a-z]+$/} @old
but I can't find a way to e.g. pass remove() a lambda, or something.
但我找不到一种方法，例如通过 remove() 一个 lambda 或其他东西。
The closest I've come is ugly:
我最接近的是丑陋的：
[ item != None for item in [ re.search("^[a-zA-Z]+$",atom) for atom in d] ]
which gets me a map of which values out of the set I want...ifthe order of the set is repeatable? I know that's not the case in Perl hashes.
这让我得到了我想要的集合中哪些值的映射......如果集合的顺序是可重复的？我知道在 Perl 哈希中情况并非如此。
I know how to iterate. :) I'm trying to do it the pythonesque Right Way
我知道如何迭代。:) 我正在尝试以正确的方式做到这一点
回答by Triptych
No need for regular expressions here. Use str.isalpha. With and without list comprehensions:
这里不需要正则表达式。使用str.isalpha. 有和没有列表推导式：
my_list = ['foo','bar','baz','>=','5.2']
# With
only_words = [token for token in my_list if token.isalpha()]
# Without
only_words = filter(str.isalpha, my_list)
PersonallyI don't think you have to use a list comprehension for everything in Python, but I always get frowny-faced when I suggest mapor filteranswers.
就我个人而言，我认为您不必对 Python 中的所有内容都使用列表理解，但是当我提出建议map或filter回答时，我总是皱眉。
回答by Ian Clelland
How about
怎么样
d = set([item for item in d if re.match("^[a-zA-Z]+$",item)])
that gives you just the values you want, back in d (the order may be different, but that's the price you pay for using sets.
这给了你你想要的值，回到 d （顺序可能不同，但这是你使用集合支付的价格。
