更新时间:2023-12-05 15:40:46
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().