Python: Using while loop only to return string without the leading zeros

Question

The function is below. I'd like to use a while loop for this and no methods, as I want to get more practice using the while loop and if-statements.

def remove_leading_zeros(s):
'''(str) -> str
Return s, but without any extra leading zeros
e.g. given "007", return "7"
Precondition: Each character of s is a number.
'''
i = 0
r = ''
while i < len(s):
    if s[i] != '0':
        r = r + s[i]
    i += 1
return r

So, when I type remove_leading_zeros('0001950'), I'd like the output to be '1950'.
However, with the function above, I seem to be omitting all the zeros. What should I modify to create a function that only omits the leading zeros?

I also tried r = r + s[i:] , so that as soon as a non zero is encountered, it would return everything afterwards, but I'm not sure how to end the loop at just that.

Resolved. Thanks to all that helped!

Mark Tolonen · Accepted Answer · 2013-06-17 01:58:06Z

2

Just return the rest of the string when you find the non-zero:

i = 0
r = ''
while i < len(s):
    if s[i] != '0':
        return s[i:]
    i += 1
return r

Or even simpler:

i = 0
while i < len(s) and s[i] == '0':
    i += 1
return s[i:]

And for completeness, there is a function for it:

s.lstrip('0')

edited Jun 17, 2013 at 1:58

answered Jun 17, 2013 at 1:49

Mark Tolonen

181k26 gold badges182 silver badges278 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Engineero · Accepted Answer · 2013-06-17 01:54:07Z

1

You could update it so that you stop checking whether the character is a zero after you encounter your first non-zero (currently you are still checking even after you find a non-zero number). For instance, you could do:

def remove_leading_zeros(s):
    i = 0
    while i < len(s):
        if s[i] == '0':
            # do nothing
            i += 1
        else:
            return s[i:]

I tested this and it seemed to work. Something along these lines should do the trick. The idea is that after you hit your first non-zero number you stop checking anything and just return whatever is left.

EDIT: Fixed a couple of bugs in my original answer.

edited Jun 17, 2013 at 1:54

answered Jun 17, 2013 at 1:44

Engineero

13k5 gold badges56 silver badges79 bronze badges

3 Comments

user2489861 Over a year ago

Do you know exactly what to put if you wanted it to do nothing? This is good to know! It seems I need an indented block there with something to run this code. Thanks!

Engineero Over a year ago

Yeah, sorry. I updated that block just a second ago with what you should put there. You need to increment your counter in the "do nothing" case.

Engineero Over a year ago

No prob, glad I could help.

Nhan Nguyen · Accepted Answer · 2013-06-17 01:44:40Z

1

You could add a boolean variable to indicate whether the first non-zero digit has been encountered yet, and only append to r from there:

def remove_leading_zeros(s):
    i = 0
    r = ''
    started = False
    while i < len(s):
        if started or s[i] != '0' or i + 1 == len(s):
            started = True
            r = r + s[i]
        i += 1
    return r

The i + 1 == len(s) is the extra check for the edge case of remove_leading_zeros('000') to make sure it returns '0' instead of ''.

answered Jun 17, 2013 at 1:44

Nhan Nguyen

4102 silver badges12 bronze badges

Comments

Ivan Chub · Accepted Answer · 2013-06-17 01:55:56Z

1

What your loop is doing is skipping all the zeroes. You have no logic that remembers whether or not you already had a non-zero number. You can fix it by replacing your loop with this:

nonZeroEncountered = False

while i < len(s):
    if nonZeroEncountered:
        r = r + s[i]
    elif s[i] != '0':
        nonZeroEncountered = True;
        r = r + s[i]
    i += 1
return r

Basically, you want all the numbers after the first non-zero number. This loop will skip zeroes, until it reaches a non-zero number, after which point it will no longer skip any characters.

edited Jun 17, 2013 at 1:55

answered Jun 17, 2013 at 1:39

Ivan Chub

4423 silver badges13 bronze badges

Comments

rmunn · Accepted Answer · 2013-06-17 04:46:20Z

One thing that nobody's mentioned yet is how to end a loop early when you're not yet ready to return from your function: use the break statement. Something like this:

while True:  # A common way to write an "infinite" loop
    print("Doing some work...")
    result = do_something()
    if result is None:  # Or whatever other signal means "we're finished here"
        break  # The next thing printed will be "We're done!"
    print("Going through the loop again...")
print("We're done!")

See how that works? The break statement interrupts the loop and resumes program flow at the next statement after the loop, which in this case is the print("We're done!") function call.

There's also the continue statement, which jumps straight to the top of the loop again, skipping over any remaining statements. If that break in the above loop had been a continue instead, then getting a None result would lead to skipping the print("Going through the loop again...") call and going straight back to print("Doing some work..."). Of course, then there would be no way to exit the loop and it really would be an infinite loop, which would be a Bad Thing™.

One other thing about continue: if you're using a for loop, it will (as the name suggests) continue on to the next item in the loop:

for i in range(10):
    if (i % 2) == 0:
        continue
    print(i)  # Will print 1, 3, 5, 7 and 9

These two tools, continue and break, will make a lot of loop processing easier.

Collectives™ on Stack Overflow

Python: Using while loop only to return string without the leading zeros

5 Answers 5

Comments

3 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

3 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related