Python itertools.count() Function

The Python itertools.count() function is used to create an iterator that generates an infinite sequence of numbers, starting from a specified value and incrementing by a given step. This function is commonly used for generating counters in loops or for creating sequences of numbers dynamically.

By default, the function starts from 0 and increments by 1 if no arguments are provided.

Syntax

Following is the syntax of the Python itertools.count() function −

itertools.count(start=0, step=1)

Parameters

This function accepts the following parameters −

• start (optional): The starting value of the sequence (default is 0).
• step (optional): The difference between consecutive values (default is 1).

Return Value

This function returns an iterator that produces an infinite sequence of numbers.

Example 1

Following is an example of the Python itertools.count() function. Here, we are generating an infinite sequence of numbers starting from 5 −

import itertools

counter = itertools.count(5)
for _ in range(5):
   print(next(counter))

Following is the output of the above code −

5
6
7
8
9

Example 2

Here, we specify a step value of 2, generating an arithmetic sequence with a difference of 2 between each value −

import itertools

counter = itertools.count(10, 2)
for _ in range(5):
   print(next(counter))

Output of the above code is as follows −

10
12
14
16
18

Example 3

Now, we use the itertools.count() function with a floating-point step value to generate a sequence of decimal numbers −

import itertools

counter = itertools.count(1.5, 0.5)
for _ in range(5):
   print(next(counter))

The result obtained is as shown below −

1.5
2.0
2.5
3.0
3.5

Example 4

If you use the itertools.count() function without limiting its iteration, it will run indefinitely. To prevent infinite loops, you can use conditions or the islice() function from itertools.

Here, we generate a sequence of numbers but limit it using itertools.islice() function −

import itertools

counter = itertools.count(0, 3)
limited_counter = itertools.islice(counter, 5)
for num in limited_counter:
   print(num)

The result produced is as follows −

0
3
6
9
12