How to remove leading and trailing zeros in a stri

Question

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-n00000','alphanumeric0000', '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)

Answer 1

You can simply do this with a bool:

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

   number = int(number)

else:

   number = float(number)

Answer 2

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']

Answer 3

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 ]

Answer 4

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

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

Answer 5

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]