如何在列表中每次连续的零运行中删除除x中的所有零以外的所有零? [英] How can I delete all zeros except for x of them in every run of consecutive zeros within a list?

问题描述

对于每次运行x或Python列表中的多个连续零，我想在运行中删除所有零，但x除外.如果为x = 0，则删除所有零.

For every run of x or more consecutive zeros in a list in Python, I would like to del all zeros in the run except for x of them. If x = 0, then delete all zeros.

我在想一个Python函数，它将一个列表L和一个数字x作为输入.

I was thinking of a Python function that took a list, L, and a number, x, as inputs.

例如，让L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 0, 8].

• 如果x = 0，则返回L = [7, 12, 2, 27, 10, 8]
• 如果x = 1，则返回L = [7, 0, 12, 0, 2, 0, 27, 10, 0, 8]
• 如果x = 2，则返回L = [7, 0, 12, 0, 0, 2, 0, 0, 27, 10, 0, 0, 8]
• 如果x = 3，则返回L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 8]
• 如果为x = 4，则返回L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 0, 8](与原始的L相同)
• 如果为x >= 5，则返回原始L，因为没有5个或更多连续零的游程.

• If x = 0, then return L = [7, 12, 2, 27, 10, 8]
• If x = 1, then return L = [7, 0, 12, 0, 2, 0, 27, 10, 0, 8]
• If x = 2, then return L = [7, 0, 12, 0, 0, 2, 0, 0, 27, 10, 0, 0, 8]
• If x = 3, then return L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 8]
• If x = 4, then return L = [7, 0, 12, 0, 0, 2, 0, 0, 0, 27, 10, 0, 0, 0, 0, 8] (Same as original L)
• If x >= 5, then return original L as there are no runs of 5 or more consecutive zeros.

衷心感谢您的帮助.

推荐答案

作为生成器很容易做到.如果要删除零游程的新列表，请在list构造函数中将对它的调用包装起来.

This is easy to do as a generator. Wrap your call to it in a list constructor if you want a fresh list with the zero-runs removed.

def compact_zero_runs(iterable, max_zeros):
    zeros = 0
    for i in iterable:
        if i == 0:
            zeros += 1
            if zeros <= max_zeros:
                yield i
        else:
            zeros = 0
            yield i

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