Count the digits in a number in Python

The number 987654321 has 9 digits. This page counts them with a loop, and then with a string.

The program

n = int(input("Enter a whole number: "))

number = abs(n)
count = 0
while True:
    count += 1
    number //= 10
    if number == 0:
        break

print(f"{n} has {count} digits")

Typed in this example: -98765. Press Run to type your own.

Enter a whole number: -98765
-98765 has 5 digits

How it works

• number //= 10 removes the last digit. The program counts how many times it can do this before the number becomes 0.
• The digit is counted before the check, so the loop always runs at least once. This makes 0 count as 1 digit. A plain while number > 0: loop would give 0 digits for the number 0, which is wrong.
• abs(n) removes the minus sign, because the sign is not a digit.

Another way: len(str())

for n in [987654321, -98765, 0, 7]:
    print(n, len(str(abs(n))))

987654321 9
-98765 5
0 1
7 1

str() turns the number into text and len() counts the characters. Without abs(), len(str(-98765)) would be 6, because the minus sign is a character too.