Removing numbers from string


How can I remove digits from a string?

Asked By: user1739954



I’d love to use regex to accomplish this, but since you can only use lists, loops, functions, etc..

here’s what I came up with:

stringWithNumbers="I have 10 bananas for my 5 monkeys!"
stringWithoutNumbers=''.join(c if c not in map(str,range(0,10)) else "" for c in stringWithNumbers)
print(stringWithoutNumbers) #I have  bananas for my  monkeys!
Answered By: Sean Johnson

Would this work for your situation?

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

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():

# 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
Answered By: RocketDonkey

If i understand your question right, one way to do is break down the string in chars and then check each char in that string using a loop whether it’s a string or a number and then if string save it in a variable and then once the loop is finished, display that to the user

Answered By: Baahubali

What about this:

out_string = filter(lambda c: not c.isdigit(), in_string)
Answered By: Pavel Paulau

Say st is your unformatted string, then run

st_nodigits=''.join(i for i in st if i.isalpha())

as mentioned above.
But my guess that you need something very simple
so say s is your string
and st_res is a string without digits, then here is your code

l = ['0','1','2','3','4','5','6','7','8','9']
for ch in s:
 if ch not in l:
Answered By: iddqd

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

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



Much more efficient than looping…


for i in range(10):
Answered By: Alain Nisam

Just a few (others have suggested some of these)

Method 1:

''.join(i for i in myStr if not i.isdigit())

Method 2:

def removeDigits(s):
    answer = []
    for char in s:
        if not char.isdigit():
    return ''.join(answer)

Method 3:

''.join(filter(lambda x: not x.isdigit(), mystr))

Method 4:

nums = set(map(int, range(10)))
''.join(i for i in mystr if i not in nums)

Method 5:

''.join(i for i in mystr if ord(i) not in range(48, 58))
Answered By: inspectorG4dget

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'
Answered By: Jon Clements
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