Count digits in a given number using recursion

Question

Here is my code that finds the number of digits in a given integer (either positive or negative). The code works and results in expected output.

'''
Program that returns number of digits in a given integer (either positive or negative)
'''
def ndigits(x):
 # Assume for the input 0 the output is 0
 if(x == 0):
 return 0
 if(abs(x) / 10 == 0):
 return 1
 else:
 return 1 + ndigits(x / 10)

• N3buchadnezzar: def ndigits(x): return sum(int(i) for i in str(x)) Here are a few other solutions stackoverflow.com/questions/14939953/….
• MrSmith42: @N3buchadnezzar: He is not interested in the sum of the digits only in the number of digits. But the solutions can be adapted to this difference.
• N3buchadnezzar: @MrSmith42 def ndigits(x): return 1 + int(log(x, 10) Something I am still missing?
• MrSmith42: @N3buchadnezzar: The link to the 'few other solutions' is a link to where the sum of the digits is calculated.

Answer 1

pro:

• code is clean and easy to understand

con:

• a recursive solution may not be optimized to a loop by your interpreter so there might be a lot of memory (stack) waste because of the recursion overhead. (So you could implement it as a loop instead of a recursion)
• You do not need the if(abs(x) / 10 == 0) branch. So to simplify your code you could remove it.

Simplified recursive code:

def ndigits(x):
 # Assume for the input 0 the output is 0
 if(x == 0):
 return 0
 else:
 return 1 + ndigits(abs(x) / 10)

Simplified tail-recursive code: end-recursive methods are likely to be detected by the interpreter and transformed to a loop. So this code might be faster and may nor waste memory.
For more information see wikipedia Tail call

def ndigits(x):
 return ndigits_(0, abs(x))
def ndigits_(s,x):
 if(x == 0):
 return s
 else:
 return ndigits(s+1, x / 10)

• TheBlackCat: The python interpreter doesn't do that sort of optimization period. There is some sort of work going on in that area, but that is the case right now.
• MrSmith42: @TheBlackCat: tail recursion is a very common optimization, so we can expect it also from the python interpreter in the near future. But good to know that it is not the case yet.
• TheBlackCat: Not going to happen. Guido has explicitly rejected adding tail recursion to Python.