CODEWITHSUNDEEP
pythonmodular-arithmetic
azerbajdzan
azerbajdzan

Reputation: 169

Is there a modulo function with offset in Python?

Is there a function that computes modulo but with offset like in second output?

Something like mod(23,13,-6)==-3 where -6 is offset. It is equivalent to (23+6)%13-6==-3.

I know that I can define my own function but I am not sure it would be as fast as built-in function.

d=13
off=int((d-1)/2)
print([x%d for x in range(d)])
print([(x+off)%d-off for x in range(d)])

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
[0, 1, 2, 3, 4, 5, 6, -6, -5, -4, -3, -2, -1]

Update:

The variant with offset equal (d-1)/2 can be defined also this way (see Wikipedia link in my comment below):

d=13
print([x-d*round(x/d) for x in range(d)])

[0, 1, 2, 3, 4, 5, 6, -6, -5, -4, -3, -2, -1]

Upvotes: 1

Views: 280

Answers (1)

azerbajdzan
azerbajdzan

Reputation: 169

I just found this (going to check whether I again any speed improvements over the definitions in OP or not):

Also the down side is that math.remainder(x,d) needs additional int to convert to integers.

import math
d=13
print([int(math.remainder(x,d)) for x in range(d)])

[0, 1, 2, 3, 4, 5, 6, -6, -5, -4, -3, -2, -1]

Upvotes: 0

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