Simple way to calculate padding based on modulo/remainder

Question

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.

I have been using the following for now:

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

This does not seem particularly legible, so sometimes I use

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

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

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

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

@alec_djinn. Because the fact that we are talking about strings is irrelevant. For that matter, what string formatting options? — Mad Physicist, Apr 18 '18 at 15:24
@MadPhysicist Thanks for the clarification. Check https://stackoverflow.com/questions/20309255/how-to-pad-a-string-to-a-fixed-length-with-spaces-in-python for padding solutions using string formatting — alec_djinn, Apr 18 '18 at 15:25
Is this being computed in a very hot loop or something? You seem rather worried about an extra subtraction or a call to math.ceil — Grant Williams, Apr 18 '18 at 15:38
@GrantWilliams. Purely a matter of aesthetics. Functionally, I am perfectly happy with my double modulus. Or `ceil`, or really even the `if` statement. — Mad Physicist, Apr 18 '18 at 15:40
@alec_djinn. All the options the answers suggest imply a prior knowledge of the total width. I am trying to avoid computing that. — Mad Physicist, Apr 18 '18 at 15:43

Bert Kellerman · Accepted Answer · 2019-11-21T14:38:54.300

4

The modulus of negative L will do it.

pad = -L % N

edited Nov 21 '19 at 14:38

answered Apr 18 '18 at 15:34

Bert Kellerman

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1

Why no just `-L % N`? – Mad Physicist Apr 18 '18 at 15:41

score 2 · Answer 2 · answered Apr 18 '18 at 15:38

2

Isn't this one enough?

pad = (N - L) % N

answered Apr 18 '18 at 15:38

Austin

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That actually reduces to the accepted answer. Clever. – Mad Physicist Apr 18 '18 at 15:49
Because I didn't see the obvious :) I don't use modulo with negative numbers much. – Mad Physicist Apr 18 '18 at 16:03

score 1 · Answer 3 · answered Apr 18 '18 at 15:25

1

With math.ceil:

from math import ceil

def pad(l, n):
    return n*ceil(l/n) - l

assert pad(59, 10) == 1
assert pad(60, 10) == 0
assert pad(61, 10) == 9

answered Apr 18 '18 at 15:25

Thierry Lathuille

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That does not seem any simpler than taking the modulo twice, but may be easier to understand. – Mad Physicist Apr 18 '18 at 15:26

score 0 · Answer 4 · answered Apr 18 '18 at 15:20

0

I mean couldn't you simply do

pad = N-L if N-L > 0 else 0

answered Apr 18 '18 at 15:20

Grant Williams

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Sorry, but my question was unclear. I updated it. I wanted to find the padding for cases where the final length needs to be some arbitrary multiple of `N`, where I don't know the multiple. Unfortunately, I've basically invalidated your answer. My apologies. – Mad Physicist Apr 18 '18 at 15:25
Also, I object to computing the difference twice for cases where it is nonzero, which reduces this solution to using the `if` statement that makes me unhappy. – Mad Physicist Apr 18 '18 at 15:28
fair enough. You could have also used `pad = max(N-L, 0)` if you wanted. Although i cant imagine an extra subtraction is going to matter all that much – Grant Williams Apr 18 '18 at 15:39

score 0 · Answer 5 · answered Apr 18 '18 at 15:39

0

Use the tenary conditional operator:

0 if (L % N) == 0 else N - (L % N)

answered Apr 18 '18 at 15:39

naroslife

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Simple way to calculate padding based on modulo/remainder

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