Counting number of digits of input using python

Question

I am trying to count the number of digits of an input. However, whenever I input 10 or 11 or any two digit number, the output is 325.

Answer 1

num = int(input())
count = 0
while num > 0:
  count += 1
  num = num // 10
print(count)

Answer 2

You must be using Python3, logically your function is right. You just have to change

countnumber = countnumber // 10

because Python3, // is floor division, meanwhile / is true division.

>>>print(1 / 10)
0.1
>>>print(1 // 10)
0

Btw, as @chrisz said above, you can just simply using the len() function to get the number of digits of the input

>>>print(len(input())

Answer 3

Your error mainly happened here:

countnumber=countnumber/10

Note that you are intending to do integer division. Single-slash division in Python 3 is always "float" or "real" division, which yields a float value and a decimal part if necessary.

Replace it with double-slash division, which is integer division: countnumber = countnumber // 10. Each time integer division is performed in this case, the rightmost digit is cut.

You also have to watch out if your input is 0. The number 0 is considered to be one digit, not zero.

Answer 4

I would not convert that beautiful input to int to be honest.

print(len(input())

would be sufficient.

An easily understandable one liner that no one can complain about.

• But of course, if negative sign bothers you like wisty said,

  len(str(abs(int (v))))
  

  will be safer for sure.

• Again, if you are worried about the non numeric inputs like mulliganaceous said, you better cover that case.

  str = input()
  if str.isnumeric():
      print(len(str(abs(v))))
  else:
      print("bad input")
  

Answer 5

The reason is that in python 3 the division of two integers yields a floating point number. It can be fixed using the // operator:

number = int(input())
digits_count = 0
while number > 0:
    digits_count += 1
    number = number // 10