How to exclude characters when counting digits

[juliabrushett]

Hello
I am trying to write a program that will count only numeric digits within a string, excluding all letters. What I have currently counts EVERYTHING. How do I fix this?

def countDigits(digits) :
    digits = 0
    x = len(s)

    for i in range(0, x, 1) :
        ((ord(s[i]) >= 65) and (ord(s[i]) <= 57))
        digits += 1
    return digits


s = input("Enter a string\n")

print("There are", countDigits(s), "digits in your string.")

[j.crater]

Try to check if string (character) is a digit with string's isdigit() method.

[juliabrushett]

I tried that, but I am returning 0 now (because isdigit() looks for a strings with ALL characters).. How can I fix this?

def countDigits(s) :
    digits = 0
    x = len(s)

    for i in range(0, x, 1) :
        ((ord(s[i]) >= 65) and (ord(s[i]) <= 57))
        digits += 1
        if s.isdigit() == False :
            digits -= 1
            break
        else:
            continue
    return digits


s = input("Enter a string\n")

print("There are", countDigits(s), "digits in your string.")

[gruntfutuk]

As you are dealing with a string, you can iterate through it with just:

for character in s:

character takes each character in turn in the loop.

As isdigit() returns True or False, you only need:

if character.isdigit():
    digit += 1

inside the loop.

For future information, you could replace it all with just a return line:

return sum(c.isdigit() for c in s)

[buran]

more efficient approach

import string

def smart_count_digits(my_string):
    orig_length = len(my_string)
    for digit in string.digits:
        my_string = my_string.replace(digit, '')
    return orig_length - len(my_string) 

[woooee]

You can also use the string version of a digit (there is no such thing as "only numeric digits within a string", they are all strings)

def countDigits(digits) :
    digits = 0
    for ch in digits :
        ## assumes zero is not a digit
        ## and there are no negative numbers
        if "1" <= ch <= "9":
            digits += 1
    return digits
 
s = input("Enter a string\n")
 
print("There are", countDigits(s), "digits in your string.")  

[volcano63]

Since you were given a lot of advice, just some details about this code

def countDigits(s) :
    digits = 0
    x = len(s)

    for i in range(0, x, 1) :
        ((ord(s[i]) >= 65) and (ord(s[i]) <= 57)) <----- This line does nothing
        digits += 1
        if s.isdigit() == False :
            digits -= 1
            break
        else:   <--- This is parasitic
            continue 
    return digits


s = input("Enter a string\n")

print("There are", countDigits(s), "digits in your string.")

[nilamo]

        ((ord(s[i]) >= 65) and (ord(s[i]) <= 57))
        digits += 1

What do you think that does? There's no if there, so counting every character in the string as a digit.

[volcano63]

Couple of more things

• you never compare Boolean to True or False - because the result will be ... Boolean. For negating Boolean, use operator not
  
• You may check that a value lies between to boundaries in an easier way
  

  57 <= ord(c) <= 65

[gruntfutuk]

more efficient approach

import string

def smart_count_digits(my_string):
    orig_length = len(my_string)
    for digit in string.digits:
        my_string = my_string.replace(digit, '')
    return orig_length - len(my_string) 

Just checking, as I thought my preceding:

def count_func(s):
    return sum(c.isdigit() for c in s)

would be pretty efficient, but I didn't run time it.