Python - find digits in a string

Question

def get_digits(str1):
    c = ""

    for i in str1:
        if i.isdigit():
            c += i
            return c

Above is the code that I used and the problem is that it only returns only the first digit of strings. For this, I have to keep both for loop and return statement. Anyone knows how to fix?

Thanks.

Answer 1

As the others said, you have a semantic problem on your indentation, but you don't have to write such function to do that, a more pythonic way to do that is:

def get_digits(text):
    return filter(str.isdigit, text)

On the interpreter:

>>> filter(str.isdigit, "lol123")
'123'

Some advice

Always test things yourself when people shows 'faster' methods:

from timeit import Timer

def get_digits1(text):
    c = ""
    for i in text:
        if i.isdigit():
            c += i
    return c

def get_digits2(text):
    return filter(str.isdigit, text)

def get_digits3(text):
    return ''.join(c for c in text if c.isdigit())

if __name__ == '__main__':
    count = 5000000
    t = Timer("get_digits1('abcdef123456789ghijklmnopq123456789')", "from __main__ import get_digits1")
    print t.timeit(number=count)

    t = Timer("get_digits2('abcdef123456789ghijklmnopq123456789')", "from __main__ import get_digits2")
    print t.timeit(number=count)

    t = Timer("get_digits3('abcdef123456789ghijklmnopq123456789')", "from __main__ import get_digits3")
    print t.timeit(number=count)



~# python tit.py
19.990989106  # Your original solution
16.7035926379 # My solution
24.8638381019 # Accepted solution

Answer 2

Your indentation is a bit borked (indentation in Python is quite important). Better:

def get_digits(str1):
    c = ""
    for i in str1:
        if i.isdigit():
            c += i
    return c

A shorter and faster solution using generator expressions:

''.join(c for c in my_string if c.isdigit())

Answer 3

it's because your return statement is inside the for loop, so it returns after the first true if condition and stops.

  def get_digits(str1):
      c = ""
      for i in str1:
        if i.isdigit():
            c += i
      return c

Answer 4

there is an indentation problem which returns when it finds the first digit, as with current indentation, it is intepreted as a statement inside if statement, it needs to be parallel to the for statement to be considered outside for statement.

def get_digits(str1):
    c = ""

    for i in str1:
        if i.isdigit():
            c += i

    return c

digits = get_digits("abd1m4m3m22mmmbb4")
print(digits)

A curly braces equivalent of your incorrect code is :

def get_digits(str1){
    c = ""

    for i in str1 {
        if i.isdigit(){
            c += i
            return c    # Notice the error here
        }
    }
}

And when the code is corrected to move the return statement in alignment with for, the equivalent is :

def get_digits(str1){
        c = ""

        for i in str1 {
            if i.isdigit(){
                c += i               
            }
        }
        return c    # Correct as required
    }

Answer 5

Your code was almost ok, except the return statement needed to be moved to the level of your for-loop.

def get_digits(str1):
    c = ""
    for i in str1:
        if i.isdigit():
            c += i
    return c   ## <--- moved to correct level

so, now:

get_digits('this35ad77asd5')

yields:

'35775'

Explanation:

Previously, your function was returning only the first digit because when it found one the if statement was executed, as was the return (causing you to return from the function which meant you didn't continue looking through the string).

Whitespace/indentation really matters in Python as you can see (unlike many other languages).

Answer 6

Of course it returns only the first digit, you explicitly tell Python to return as soon as you have a digit.

Change the indentation of the return statement and it should work:

def get_digits(str1):
    c = ""

    for i in str1:
        if i.isdigit():
            c += i

    # Note the indentation here
    return c