如何删除字符串中的前导和尾随零？Python

I have several alphanumeric strings like these

listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', '000alphanumeric']

The desired output for removing trailing zeros would be:

listOfNum = ['000231512-n','1209123100000-n','alphanumeric', '000alphanumeric']

The desired output for leading trailing zeros would be:

listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']

The desire output for removing both leading and trailing zeros would be:

listOfNum = ['231512-n','1209123100000-n', 'alphanumeric', 'alphanumeric']

For now i’ve been doing it the following way, please suggest a better way if there is:

listOfNum = ['000231512-n','1209123100000-n00000','alphanumeric0000', \
'000alphanumeric']
trailingremoved = []
leadingremoved = []
bothremoved = []

# Remove trailing
for i in listOfNum:
  while i[-1] == "0":
    i = i[:-1]
  trailingremoved.append(i)

# Remove leading
for i in listOfNum:
  while i[0] == "0":
    i = i[1:]
  leadingremoved.append(i)

# Remove both
for i in listOfNum:
  while i[0] == "0":
    i = i[1:]
  while i[-1] == "0":
    i = i[:-1]
  bothremoved.append(i)

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]

Remove leading + trailing ‘0’:

list = [i.strip('0') for i in listOfNum ]

Remove leading ‘0’:

list = [ i.lstrip('0') for i in listOfNum ]

Remove trailing ‘0’:

list = [ i.rstrip('0') for i in listOfNum ]

You can simply do this with a bool:

if int(number) == float(number):

   number = int(number)

else:

   number = float(number)

Did you try with strip() :

listOfNum = ['231512-n','1209123100000-n00000','alphanumeric0000', 'alphanumeric']
print [item.strip('0') for item in listOfNum]

>>> ['231512-n', '1209123100000-n', 'alphanumeric', 'alphanumeric']

str.strip is the best approach for this situation, but more_itertools.strip is also a general solution that strips both leading and trailing elements from an iterable:

Code

import more_itertools as mit


iterables = ["231512-n\n","  12091231000-n00000","alphanum0000", "00alphanum"]
pred = lambda x: x in {"0", "\n", " "}
list("".join(mit.strip(i, pred)) for i in iterables)
# ['231512-n', '12091231000-n', 'alphanum', 'alphanum']

Details

Notice, here we strip both leading and trailing "0"s among other elements that satisfy a predicate. This tool is not limited to strings.

See also docs for more examples of

• more_itertools.strip: strip both ends
• more_itertools.lstrip: strip the left end
• more_itertools.rstrip: strip the right end

is a third-party library installable via > pip install more_itertools.

Assuming you have other data types (and not only string) in your list try this. This removes trailing and leading zeros from strings and leaves other data types untouched. This also handles the special case s = ‘0’

e.g

a = ['001', '200', 'akdl00', 200, 100, '0']

b = [(lambda x: x.strip('0') if isinstance(x,str) and len(x) != 1 else x)(x) for x in a]

b
>>>['1', '2', 'akdl', 200, 100, '0']