删除与Python列表中的条件匹配的前N个项目 [英] Remove the first N items that match a condition in a Python list

问题描述

如果我具有函数matchCondition(x)，如何删除与该条件匹配的Python列表中的前n个项目?

If I have a function matchCondition(x), how can I remove the first n items in a Python list that match that condition?

一种解决方案是遍历每个项目，将其标记为删除(例如，将其设置为None)，然后使用理解过滤列表.这需要遍历列表两次，并对数据进行突变.有没有更惯用或有效的方法来做到这一点?

One solution is to iterate over each item, mark it for deletion (e.g., by setting it to None), and then filter the list with a comprehension. This requires iterating over the list twice and mutates the data. Is there a more idiomatic or efficient way to do this?

n = 3

def condition(x):
    return x < 5

data = [1, 10, 2, 9, 3, 8, 4, 7]
out = do_remove(data, n, condition)
print(out)  # [10, 9, 8, 4, 7] (1, 2, and 3 are removed, 4 remains)

推荐答案

使用 和 itertools.count :

from itertools import count, filterfalse

data = [1, 10, 2, 9, 3, 8, 4, 7]
output = filterfalse(lambda L, c=count(): L < 5 and next(c) < 3, data)

然后list(output)，给您:

[10, 9, 8, 4, 7]

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