CODEWITHSUNDEEP
pythonlogic
Utopion
Utopion

Reputation: 1128

How to remove all zeros before the number in a string (and keep the last 0 if it's equals to 0)

INPUT: "00012" / "00453403" / "0000000" / "123223"

OUTPUT: "12" / "453403" / "0" / "123223"

I can do it in 3 or more lines of code and now I feel dumb because I think there must be a way to do it in one line (lets's say 2 to be safe).

Can someone help to rise the number of beautiful code in the world ? :)

    def remove_unnecessary_zero_in_the_beginning(nbr_as_str):
        while len(nbr_as_str) > 1 and nbr_as_str[0] == "0":
            nbr_as_str = nbr_as_str[1:]
        return nbr_as_str
    print(remove_unnecessary_zero_in_the_beginning("00012"))

Upvotes: 1

Views: 79

Answers (5)

Atul Kewat
Atul Kewat

Reputation: 1

Say you have a list of the inputs ie : l1 = [ "00012" , "00453403" , "0000000" , "123223"]

Code can be :

l1 = [ "00012" , "00453403" , "0000000" , "123223"]
test = list(map(lambda x :  x.strip('0') if int(x)!=0 else int(x),l1))
print(test)

Upvotes: 0

phuycke
phuycke

Reputation: 331

You could use this

i = ["00012","00453403","0000000","123223"]
o = ["{0:d}".format(int(e)) for e in l]

Output:

o
Out[8]: ['12', '453403', '0', '123223']

Upvotes: 0

Kelly Bundy
Kelly Bundy

Reputation: 27588

str(int(mystring))

(too few chars)

Upvotes: 3

tdelaney
tdelaney

Reputation: 77337

You could use a regular expression with a match group. This one matches all leading zeros then captures all digits after that. A \d*? non-greedy match followed by \d ensures that at least the last digit is always present, even if its 0.

import re

def remove_unnecessary_zero_in_the_beginning(nbr_as_str):
    return re.match(r"0*(\d*?\d)$", t).group(1)

test = ["00012", "00453403", "0000000", "123223"]
for t in test:
    print(remove_unnecessary_zero_in_the_beginning(t))

Upvotes: 0

Lachlan
Lachlan

Reputation: 175

import re
re.sub("^0+", "", n) or 0

Upvotes: 0

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