Removing numbers from string [closed]

How can I remove digits from a string?

Solution 1:

Would this work for your situation?

>>> s = '12abcd405'
>>> result = ''.join([i for i in s if not i.isdigit()])
>>> result
'abcd'

This makes use of a list comprehension, and what is happening here is similar to this structure:

no_digits = []
# Iterate through the string, adding non-numbers to the no_digits list
for i in s:
    if not i.isdigit():
        no_digits.append(i)

# Now join all elements of the list with '', 
# which puts all of the characters together.
result = ''.join(no_digits)

As @AshwiniChaudhary and @KirkStrauser point out, you actually do not need to use the brackets in the one-liner, making the piece inside the parentheses a generator expression (more efficient than a list comprehension). Even if this doesn't fit the requirements for your assignment, it is something you should read about eventually :) :

>>> s = '12abcd405'
>>> result = ''.join(i for i in s if not i.isdigit())
>>> result
'abcd'

Solution 2:

And, just to throw it in the mix, is the oft-forgotten str.translate which will work a lot faster than looping/regular expressions:

For Python 2:

from string import digits

s = 'abc123def456ghi789zero0'
res = s.translate(None, digits)
# 'abcdefghizero'

For Python 3:

from string import digits

s = 'abc123def456ghi789zero0'
remove_digits = str.maketrans('', '', digits)
res = s.translate(remove_digits)
# 'abcdefghizero'

Solution 3:

Not sure if your teacher allows you to use filters but...

filter(lambda x: x.isalpha(), "a1a2a3s3d4f5fg6h")

returns-

'aaasdffgh'

Much more efficient than looping...

Example:

for i in range(10):
  a.replace(str(i),'')

Solution 4:

What about this:

out_string = filter(lambda c: not c.isdigit(), in_string)