The most efficient and easy way to get the length of an integer is the “string conversion” approach.

The len() function returns the length of a string, not an integer, so we have to convert an input integer into a string using the str() function and then find its length using the len() function.

integer = 329484

print(len(str(integer)))

# Output: 6

And we got the exact total number of digits, which is 3.

Negative integer

If we have a negative integer, we can use the abs() function before using the str() function to get the absolute value, convert it into a string, and then use the len() function to determine its length.

def get_length_str(n: int) -> int:
    return len(str(abs(n))) if n != 0 else 1


integer = -329484

print(get_length_str(integer))

# Output: 6

Floating-Point Numbers

For floating-point numbers, determining the “length” (number of digits) is more nuanced because of the decimal point and potential fractional parts.

Convert the float to a string using the str() function and split it into integer and fractional parts.

def get_length_float(n: float) -> tuple[int, int]:
    integer_part, fractional_part = str(abs(n)).split('.')
    return len(integer_part), len(fractional_part)


number = 329.4845

print(get_length_float(number))

# Output: (3, 4)

You can see that the integer part has a length of 3, and the fractional part has a length of 4.

Here are two other ways:

1. Using math.log10()
2. Using Iterative Division

Approach 1: Using math.log10()

Another less-known approach is logarithm base 10. The main formula is: floor(log10(n)) + 1, which only works for positive integers.

For zero, log10 is undefined, and for numbers less than 1 (but integers can’t be negative here except zero).

So handling edge cases like n=0 is necessary. Wait, log10(0) would be a math error, so we need to handle n=0 separately.

import math


def get_length_log(n: int) -> int:
    if n == 0:
        return 1
    return math.floor(math.log10(abs(n))) + 1


integer = 329484

print(get_length_log(integer))

# Output: 6

Approach 2: Using Iterative Division

You can repeatedly divide the number by 10 until it becomes 0, counting iterations. That way, you will have an exact counting of digits.

def get_length_loop(n: int) -> int:
    if n == 0:
        return 1
    n = abs(n)
    count = 0
    while n > 0:
        n //= 10
        count += 1
    return count


integer = 329484

print(get_length_loop(integer))

# Output: 6

It is a very less efficient approach since it can take long iterations if the input is too big.