Im trying to write a function which counts the number of digits in a number - with no string operation. here is my code:
def count_digits(n): count_list=[] while (n>0): n=n%10 i=n count_list.append(i) n=n/10 return len(count_list) n=12345 print count_digits(n) By using % i get the last digits - in order to add it to a list. By using / i throw the digit from the number.
The script does not work. For each n i put, the script just prints 1.
Thanks!
There are several problems to your code:
return statement should be outside of the loop.n = n % 10 statement modifies n, making it impossible to get the other digits.//. In Python 3, n / 10 would give a floating point number.0 as having 0 digits. You need to check whether n is 0.Here is a corrected version of your code:
def count_digits(n): if n == 0: return 1 count_list = [] while n > 0: count_list.append(n % 10) n = n // 10 return len(count_list) Also, as it was said in the comments, since your goal is just to count the digits, you don't need to maintain a list:
def count_digits(n): if n == 0: return 1 count = 0 while n > 0: n = n // 10 count += 1 return count 
0 as having zero digits, when usually, it should still count as having one digit. It can be handled by either making the while condition while True with an if n <= 0: break at the end of the loop (so 0 gets counted once), or by adding an if n == 0: return 1 to the top of the function. Obviously, still doesn't handle negative numbers (but handling them involves defining what digit count means; simplest would be to assign n = abs(n) at the top of the function and ignore the sign).Perhaps you can try this, a much simpler approach. No lists involved :)
def dcount(num): count = 0 if num == 0: return 1 while (num != 0): num /= 10 count += 1 return count count_list stores the digits.
def count_digits(n): count_list=[] while (n>0): count_list.append(n%10) n-=n%10 n = n/10 return count_list n=12345 print len(count_digits(n)) Without using a list
def count_digits(n): count_list=0 while (n>0): count_list+=1 n-=n%10 n = n/10 return count_list n=12345 print count_digits(n) 
n -= n % 10? It makes count_list as list of 0s, which doesn't really matter (you just get the len of it anyway), but it costs more work for zero gain.
n //= 10 (or n = n // 10 if you prefer) and in both Py2 and Py3 it will do floor division; subtracting the remainder becomes unnecessary (and you remove the opportunity for floating point errors to bite you).eskaev solution, thank you for your comment.
returnstatement is not properly indented. It should be outside thewhileloop.count_listwhen you could just use an incrementing counter seems kind of silly (only trick is that you handle0specially, or use a somewhat kludgy Python equivalent ofdo/whileto avoid the need to handle it specially).import math, you can get the number of digits usingint(math.log10(n)) + 1