The modulo operation on negative numbers in Python

Question

I\'ve found some strange behaviour in Python regarding negative numbers:

>>> -5 % 4
3

Could anyone explain what\'s going on?

Answer 1

Modulo, equivalence classes for 4:

• 0: 0, 4, 8, 12... and -4, -8, -12...
• 1: 1, 5, 9, 13... and -3, -7, -11...
• 2: 2, 6, 10... and -2, -6, -10...
• 3: 3, 7, 11... and -1, -5, -9...

Here's a link to modulo's behavior with negative numbers. (Yes, I googled it)

Answer 2

Unlike C or C++, Python's modulo operator (%) always return a number having the same sign as the denominator (divisor). Your expression yields 3 because

(-5) / 4 = -1.25 --> floor(-1.25) = -2

(-5) % 4 = (-2 × 4 + 3) % 4 = 3.

It is chosen over the C behavior because a nonnegative result is often more useful. An example is to compute week days. If today is Tuesday (day #2), what is the week day N days before? In Python we can compute with

return (2 - N) % 7

but in C, if N ≥ 3, we get a negative number which is an invalid number, and we need to manually fix it up by adding 7:

int result = (2 - N) % 7;
return result < 0 ? result + 7 : result;

(See http://en.wikipedia.org/wiki/Modulo_operator for how the sign of result is determined for different languages.)

Answer 3

It's also worth to mention that also the division in python is different from C: Consider

>>> x = -10
>>> y = 37

in C you expect the result

0

what is x/y in python?

>>> print x/y
-1

and % is modulo - not the remainder! While x%y in C yields

-10

python yields.

>>> print x%y
27

You can get both as in C

The division:

>>> from math import trunc
>>> d = trunc(float(x)/y)
>>> print d
0

And the remainder (using the division from above):

>>> r = x - d*y
>>> print r
-10

This calculation is maybe not the fastest but it's working for any sign combinations of x and y to achieve the same results as in C plus it avoids conditional statements.

Answer 4

As pointed out, Python modulo makes a well-reasoned exception to the conventions of other languages.

This gives negative numbers a seamless behavior, especially when used in combination with the // integer-divide operator, as % modulo often is (as in math.divmod):

for n in range(-8,8):
    print n, n//4, n%4

Produces:

 -8 -2 0
 -7 -2 1
 -6 -2 2
 -5 -2 3

 -4 -1 0
 -3 -1 1
 -2 -1 2
 -1 -1 3

  0  0 0
  1  0 1
  2  0 2
  3  0 3

  4  1 0
  5  1 1
  6  1 2
  7  1 3

• Python % always outputs zero or positive*
• Python // always rounds toward negative infinity

_{* ... as long as the right operand is positive. On the other hand 11 % -10 == -9}

Answer 5

There is no one best way to handle integer division and mods with negative numbers. It would be nice if a/b was the same magnitude and opposite sign of (-a)/b. It would be nice if a % b was indeed a modulo b. Since we really want a == (a/b)*b + a%b, the first two are incompatible.

Which one to keep is a difficult question, and there are arguments for both sides. C and C++ round integer division towards zero (so a/b == -((-a)/b)), and apparently Python doesn't.

Answer 6

Here's an explanation from Guido van Rossum:

http://python-history.blogspot.com/2010/08/why-pythons-integer-division-floors.html

Essentially, it's so that a/b = q with remainder r preserves the relationships b*q + r = a and 0 <= r < b.