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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