尝试使用 lambda 过滤列表

Question

I'm trying to filter this list or real and fake Python keywords. I'm using the keyword module to obtain the list. While I got it to work using list comprehension, I'm stuck with lambda as I keep getting the same list in return.

Here is the code:

import keyword

keyword_list = keyword.kwlist
iskeyword = ['False','Fake','else','however','or', 'True','Real','attempt','try','and','moreover']

# using list comprehension
new_keyword_list = [f'{k} is keyword' if k in keyword_list else f'{k} not keyword' for k in iskeyword]
new_keyword_list

# user-defined function
def check_keyword(c):
    return new_keyword_list

res = list(filter(check_keyword, new_keyword_list))
print(res)

And this is the output, which is correct:

['False is keyword', 'Fake not keyword', 'else is keyword', 'however not keyword', 'or is keyword', 'True is keyword', 'Real not keyword', 'attempt not keyword', 'try is keyword', 'and is keyword', 'moreover not keyword']

Now, if I use lambda, I get the same result from the original iskeyword list.

# using lambda
new_keyword_list = filter(lambda k : (f'{k} is keyword' if k in keyword_list else f'{k} not keyword'), iskeyword)
new_keyword_list

I know I'm messing up with lambda, but where? Thanks for helping!

Answer 1

That's how filter is supposed to work. filter returns the elements of the original list for which the lambda returns True, which all of yours do. I suspect you were really looking for map , but the list comprehension is a better solution.

Answer 2

filter用于获取所有True值，修复方法是使用map ：

new_keyword_list = map(lambda k: f'{k} is keyword' if k in keyword_list else f'{k} not keyword', iskeyword)