基于模数/余数计算填充的简单方法 [英] Simple way to calculate padding based on modulo/remainder

问题描述

如果我有一个长度为L = 77的字符串，我希望将其填充为N = 10的倍数的长度.我对仅计算所需的填充量感兴趣.使用N - (L % N)可以很容易地做到这一点，除了L % N为零的情况.

If I have a string of length L=77 that I want to pad to a length that is a multiple of N=10. I am interested in computing just the amount of padding required. This can be easily done with N - (L % N), except for cases where L % N is zero.

我现在一直在使用以下内容:

I have been using the following for now:

pad = (N - (L % N)) % N

这似乎不是特别清晰，所以有时我会用

This does not seem particularly legible, so sometimes I use

pad = N - (L % N)
if pad == N:
    pad = 0

对于这么简单的事情，使用三行代码似乎是矫kill过正.

It seems overkill to use three lines of code for something so simple.

或者，我可以找到k * N >= L的k，但是使用math.ceil似乎也过分了.

Alternatively, I can find the k for which k * N >= L, but using math.ceil seems like overkill as well.

我还有更好的选择吗?也许某个地方有一个简单的功能?

Is there a better alternative that I am missing? Perhaps a simple function somewhere?

推荐答案

负L的模量可以做到这一点.

The modulus of negative L will do it.

pad = -L % N

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