Python quotient vs remainder

Question

The python 2.6 docs state that x % y is defined as the remainder of x / y (http://docs.python.org/library/stdtypes.html#numeric-types-int-float-long-complex). I am not clear on what is really occurring though, as:

for i in range(2, 11):
    print 1.0 % i

prints "1.0" ten times, rather than "0.5, 0.333333, 0.25" etc. as I expected (1/2 = 0.5, etc).

Answer 1

I think you can get the result you want by doing something like this:

for i in range(2, 11):
    print 1.0*(1 % i) / i

This computes the (integer) remainder as explained by others. Then you divide by the denominator again, to produce the fractional part of the quotient.

Note that I multiply the result of the modulo operation by 1.0 to ensure that a floating point division operation is done (rather than integer division, which will result in 0).

Answer 2

Modulo is performed in the integer context, not fractional (remainders are integers). Therefore:

1 % 1  = 0  (1 times 1 plus 0)
1 % 2  = 1  (2 times 0 plus 1)
1 % 3  = 1  (3 times 0 plus 1)

6 % 3 = 0  (3 times 2 plus 0)
7 % 3 = 1  (3 times 2 plus 1)
8 % 3 = 2  (3 times 2 plus 2)

etc

How do I get the actual remainder of x / y?

By that I presume you mean doing a regular floating point division?

for i in range(2, 11):
    print 1.0 / i

Answer 3

You've confused division and modulus.

"0.5, 0.333333, 0.25" etc. as I expected (1/2 = 0.5, etc)."

That's the result of division.

Not modulus.

Modulus (%) is the remainder left over after integer division.

Your sample values are simple division, which is the / operator. Not the % operator.

Answer 4

Wouldn't dividing 1 by an number larger than it result in 0 with remainder 1?

The number theorists in the crowd may correct me, but I think modulus/remainder is defined only on integers.

Answer 5

We can have 2 types of division, that we can define through the return types:

Float: a/b. For example: 3/2=1.5

def division(a,b):
    return a/b

Int: a//b and a%b. For example: 3//2=1 and 3%2=1

def quotient(a,b):
    return a//b
def remainder(a,b):
    return a%b