# Python divmod() — A Simple Guide with Video

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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.

## 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)```

## 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.  ````

## 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.

## 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)`.

• 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.

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