Python divmod() — A Simple Guide with Video

Python’s built-in divmod(a, b) function takes two integer or float numbers a and b as input arguments and returns a tuple (a // b, a % b). The first tuple value is the result of the integer division a//b. The second tuple is the result of the remainder, also called modulo operation a % b. In case of float inputs, divmod() still returns the division without remainder by rounding down to the next round number.

Python divmod() visual explanation

Usage

Learn by example! Here are some examples of how to use the divmod() built-in function with integer arguments:

# divmod() with integers
>>> divmod(10, 2)
(5, 0)
>>> divmod(10, 3)
(3, 1)
>>> divmod(10, 4)
(2, 2)
>>> divmod(10, 5)
(2, 0)
>>> divmod(10, 10)
(1, 0)

You can also use float arguments as follows:

# divmod() with floats
>>> divmod(10.0, 2.0)
(5.0, 0.0)
>>> divmod(10.0, 3.0)
(3.0, 1.0)
>>> divmod(10.0, 4.0)
(2.0, 2.0)
>>> divmod(10.0, 5.0)
(2.0, 0.0)
>>> divmod(10.0, 10.0)
(1.0, 0.0)

Video divmod()

Syntax divmod()

Syntax: 
divmod(a, b) -> returns a tuple of two numbers. The first is the result of the division without remainder a/b. The second is the remainder (modulo) a%b.  
ArgumentsintegerThe dividend of the division operation.
integerThe divisor of the division operation.
Return ValuetupleReturns a tuple of two numbers. The first is the result of the division without remainder. The second is the remainder (modulo).

Interactive Shell Exercise: Understanding divmod()

Consider the following interactive code:

Exercise: Guess the output before running the code.


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Exact Mathematical Definition divmod()

You can generally use the divmod(a, b) function with two integers, one integer and one float, or two floats.

Two integers. Say you call divmod(a, b) with two integers a and b. In this case, the exact mathematical definition of the return value is (a // b, a % b).

a = 5
b = 2
print((a // b, a % b))
print(divmod(a, b))
# OUTPUT:
# (2, 1)
# (2, 1)

One integer and one float. Say you call divmod(a, b) with an integer a and a float b. In this case, the exact mathematical definition of the return value is the return value of converting the integer to a float and calling divmod(a, float(b)).

a = 5.0
b = 2
print((a // b, a % b))
print(divmod(a, b))
# OUTPUT:
# (2.0, 1.0)
# (2.0, 1.0)

Two floats. Say you call divmod(a, b) with two floats a and b. In this case, the exact mathematical definition of the return value is (float(math.floor(a / b)), a % b).

import math

a = 5.0
b = 2.0
print((float(math.floor(a / b)), a % b))
print(divmod(a, b))
# OUTPUT:
# (2.0, 1.0)
# (2.0, 1.0)

Note that because of the imprecision of floating point arithmetic, the result may have a small floating point error in one of the lower decimal positions. You can read more about the floating point trap on the Finxter blog.

Related Tutorial: Floating Point Error Explained

Python divmod() Negative Numbers

Can you use the divmod() method on negative numbers for the dividend or the divisor?

You can use divmod(a, b) for negative input arguments a, b, or both. In any case, if both arguments are integers, Python performs integer division a // b to obtain the first element and modulo division a % b to obtain the second element of the returned tuple. Both operations allow negative inputs a or b. The returned tuple (x, y) is calculated so that x * b + y = a.

Here’s an example of all three cases:

>>> divmod(-10, -3)
(3, -1)
>>> divmod(-10, 3)
(-4, 2)
>>> divmod(10, -3)
(-4, -2)

Python divmod() Performance — Is It Faster Than Integer Division // and Modulo % Operators?

There are two semantically identical ways to create a tuple where the first element is the result of the integer division and the second is the result of the modulo operation:

  • Use the divmod(a, b) function.
  • Use the (a // b, a % b) explicit operation with Python built-in operators.

Next, we measure the performance of calculating the elapsed runtime in milliseconds when performing 10 million computations for relatively small integers. Let’s start with divmod():

import time
import random

# Small Operands
operands = zip([random.randint(1, 100) for i in range(10**7)],
               [random.randint(1, 100) for i in range(10**7)])

start = time.time()

for i, j in operands:
    divmod(i, j)

stop = time.time()
print('divmod() elapsed time: ', (stop-start), 'milliseconds')
# divmod() elapsed time:  1.7654337882995605 milliseconds

Compare this to integer division and modulo:

import time
import random

# Small Operands
operands = zip([random.randint(1, 100) for i in range(10**7)],
               [random.randint(1, 100) for i in range(10**7)])

start = time.time()

for i, j in operands:
    (i // j, i % j)

stop = time.time()
print('(i // j, i % j) elapsed time: ', (stop-start), 'milliseconds')
# (i // j, i % j) elapsed time:  1.9048900604248047 milliseconds

The result of this performance benchmark is that divmod() requires 1.76 milliseconds and the explicit way of using integer division and modulo requires 1.90 milliseconds for 10,000,000 operations. Thus, divmod() is 8% faster. The reason is that the explicit way performs many duplicate operations to calculate the result of the integer division and the modulo operation which internally uses integer division again. This effect becomes even more pronounced if you use larger integers.

Performance difference divmod() vs Integer Division and Modulo

Python divmod() Implementation

For integer input arguments, here’s a semantically equivalent divmod() implementation:

>>> def divmod_own(x, y):
	return (x // y, x % y)

>>> divmod_own(10, 3)
(3, 1)
>>> divmod(10, 3)
(3, 1)

But note that this implementation still performs redundant computations (e.g., integer division) and, therefore, is less efficient than divmod().

Summary

Python’s built-in divmod(a, b) function takes two integer or float numbers a and b as input arguments and returns a tuple (a // b, a % b).

In case of float inputs, divmod() still returns the division without remainder by rounding down to the next round number.


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