What about a basic

your_string.strip("0")

to remove both trailing and leading zeros ? If you're only interested in removing trailing zeros, use .rstrip instead (and .lstrip for only the leading ones).

More info in the doc.

You could use some list comprehension to get the sequences you want like so:

trailing_removed = [s.rstrip("0") for s in listOfNum]
leading_removed = [s.lstrip("0") for s in listOfNum]
both_removed = [s.strip("0") for s in listOfNum]

Use lstrip:

>>> '00000010'.lstrip('0')
'10'

(strip removes both leading and trailing zeros.)

This messes up '0' (turning it into an empty string). There are several ways to fix this:

#1:

>>> re.sub(r'0+(.+)', r'\1', '000010')
'10'
>>> re.sub(r'0+(.+)', r'\1', '0')
'0'

#2:

>>> str(int('0000010'))
'10'

#3:

>>> s = '000010'
>>> s[:-1].lstrip('0') + s[-1]

You can use

re.sub(r'^\D*0*', '', text)

See the regex demo. Details

• ^ - start of string
• \D* - any zero or more non-digit chars
• 0* - zero or more zeros.

See Python demo:

import re
text = "U012034"
print( re.sub(r'^\D*0*', '', text) )
# => 12034

If there is more text after the first number, use

print( re.sub(r'^\D*0*(\d+).*', r'\1', text) )

See this regex demo. Details:

• ^ - start of string
• \D* - zero or more non-digits
• 0* - zero or more zeros
• (\d+) - Group 1: one or more digits (use (\d+(?:\.\d+)?) to match float or int values)
• `.* - the rest of the string.

The replacement is the Group 1 value.

You can use the strip method:

x = 0.13
res = str(x).strip('0')
print('This is my formatted number: {}'.format(res))

Here is another way:

>>> ("%.4f" % k).lstrip('0')
'.1337'

It is slightly more general than [1:] in that it also works with numbers >=1.

Neither method correctly handles negative numbers, however. The following is better in this respect:

>>> re.sub('0(?=[.])', '', ("%0.4f" % -k))
'-.1337'

Not particularly elegant, but right now I can't think of a better method.