棘手的问题:)


尝试了一会儿在放弃之前我虽然使用了2个相同的列表。

Tricky problem :)

After trying a while, before giving up I though about using 2 identical lists.

不喜欢过滤器？你毕竟是过滤了;）


两种可能性：


x = [要过滤的列表]

y = [要过滤的术语列表]


过滤器（lambda x：x不在y，x）


或


[如果p不在y，则p为p in x]


et voila。


list comp is可能更快（我还没有测试过）

no love for filter? you are filtering after all ;)

two possibilities:

x =[list to be filtered]
y =[list of terms to be filtered out]

filter(lambda x: x not in y,x)

or

[p for p in x if p not in y]

et voila.

list comp is probably faster (i haven''t tested)

非常好的benjumanji。 filter（）解决方案很好，但由于速度和可读性，我更喜欢列表理解。使用 timeit（）进行测试时，列表理解速度提高了两倍。

Very good benjumanji. The filter() solution is fine, but I prefer the list comprehension due to its speed and readability. The list comprehension was twice as fast when tested with timeit().