Remove characters except digits from string using Python?

Question

How can I remove all characters except numbers from string?

Answer 1

Use re.sub, like so:

>>> import re
>>> re.sub('\D', '', 'aas30dsa20')
'3020'

\D matches any non-digit character so, the code above, is essentially replacing every non-digit character for the empty string.

Or you can use filter, like so (in Python 2):

>>> filter(str.isdigit, 'aas30dsa20')
'3020'

Since in Python 3, filter returns an iterator instead of a list, you can use the following instead:

>>> ''.join(filter(str.isdigit, 'aas30dsa20'))
'3020'

Answer 2

In Python 2.*, by far the fastest approach is the .translate method:

>>> x='aaa12333bb445bb54b5b52'
>>> import string
>>> all=string.maketrans('','')
>>> nodigs=all.translate(all, string.digits)
>>> x.translate(all, nodigs)
'1233344554552'
>>> 

string.maketrans makes a translation table (a string of length 256) which in this case is the same as ''.join(chr(x) for x in range(256)) (just faster to make;-). .translate applies the translation table (which here is irrelevant since all essentially means identity) AND deletes characters present in the second argument -- the key part.

.translate works very differently on Unicode strings (and strings in Python 3 -- I do wish questions specified which major-release of Python is of interest!) -- not quite this simple, not quite this fast, though still quite usable.

Back to 2.*, the performance difference is impressive...:

$ python -mtimeit -s'import string; all=string.maketrans("", ""); nodig=all.translate(all, string.digits); x="aaa12333bb445bb54b5b52"' 'x.translate(all, nodig)'
1000000 loops, best of 3: 1.04 usec per loop
$ python -mtimeit -s'import re;  x="aaa12333bb445bb54b5b52"' 're.sub(r"\D", "", x)'
100000 loops, best of 3: 7.9 usec per loop

Speeding things up by 7-8 times is hardly peanuts, so the translate method is well worth knowing and using. The other popular non-RE approach...:

$ python -mtimeit -s'x="aaa12333bb445bb54b5b52"' '"".join(i for i in x if i.isdigit())'
100000 loops, best of 3: 11.5 usec per loop

is 50% slower than RE, so the .translate approach beats it by over an order of magnitude.

In Python 3, or for Unicode, you need to pass .translate a mapping (with ordinals, not characters directly, as keys) that returns None for what you want to delete. Here's a convenient way to express this for deletion of "everything but" a few characters:

import string

class Del:
  def __init__(self, keep=string.digits):
    self.comp = dict((ord(c),c) for c in keep)
  def __getitem__(self, k):
    return self.comp.get(k)

DD = Del()

x='aaa12333bb445bb54b5b52'
x.translate(DD)

also emits '1233344554552'. However, putting this in xx.py we have...:

$ python3.1 -mtimeit -s'import re;  x="aaa12333bb445bb54b5b52"' 're.sub(r"\D", "", x)'
100000 loops, best of 3: 8.43 usec per loop
$ python3.1 -mtimeit -s'import xx; x="aaa12333bb445bb54b5b52"' 'x.translate(xx.DD)'
10000 loops, best of 3: 24.3 usec per loop

...which shows the performance advantage disappears, for this kind of "deletion" tasks, and becomes a performance decrease.

Answer 3

s=''.join(i for i in s if i.isdigit())

Another generator variant.

Answer 4

You can use filter:

filter(lambda x: x.isdigit(), "dasdasd2313dsa")

On python3.0 you have to join this (kinda ugly :( )

''.join(filter(lambda x: x.isdigit(), "dasdasd2313dsa"))

Answer 5

You can easily do it using Regex

>>> import re
>>> re.sub("\D","","£70,000")
70000

Answer 6

along the lines of bayer's answer:

''.join(i for i in s if i.isdigit())

Answer 7

The op mentions in the comments that he wants to keep the decimal place. This can be done with the re.sub method (as per the second and IMHO best answer) by explicitly listing the characters to keep e.g.

>>> re.sub("[^0123456789\.]","","poo123.4and5fish")
'123.45'

Answer 8

x.translate(None, string.digits)

will delete all digits from string. To delete letters and keep the digits, do this:

x.translate(None, string.letters)

Answer 9

Use a generator expression:

>>> s = "foo200bar"
>>> new_s = "".join(i for i in s if i in "0123456789")

Answer 10

A fast version for Python 3:

# xx3.py
from collections import defaultdict
import string
_NoneType = type(None)

def keeper(keep):
    table = defaultdict(_NoneType)
    table.update({ord(c): c for c in keep})
    return table

digit_keeper = keeper(string.digits)

Here's a performance comparison vs. regex:

$ python3.3 -mtimeit -s'import xx3; x="aaa12333bb445bb54b5b52"' 'x.translate(xx3.digit_keeper)'
1000000 loops, best of 3: 1.02 usec per loop
$ python3.3 -mtimeit -s'import re; r = re.compile(r"\D"); x="aaa12333bb445bb54b5b52"' 'r.sub("", x)'
100000 loops, best of 3: 3.43 usec per loop

So it's a little bit more than 3 times faster than regex, for me. It's also faster than class Del above, because defaultdict does all its lookups in C, rather than (slow) Python. Here's that version on my same system, for comparison.

$ python3.3 -mtimeit -s'import xx; x="aaa12333bb445bb54b5b52"' 'x.translate(xx.DD)'
100000 loops, best of 3: 13.6 usec per loop

Answer 11

Try:

import re

string = '1abcd2XYZ3'
string_without_letters = re.sub(r'[a-z]', '', string.lower())

this should give:

123

Answer 12

Ugly but works:

>>> s
'aaa12333bb445bb54b5b52'
>>> a = ''.join(filter(lambda x : x.isdigit(), s))
>>> a
'1233344554552'
>>>

Answer 13

$ python -mtimeit -s'import re;  x="aaa12333bb445bb54b5b52"' 're.sub(r"\D", "", x)'

100000 loops, best of 3: 2.48 usec per loop

$ python -mtimeit -s'import re; x="aaa12333bab445bb54b5b52"' '"".join(re.findall("[a-z]+",x))'

100000 loops, best of 3: 2.02 usec per loop

$ python -mtimeit -s'import re;  x="aaa12333bb445bb54b5b52"' 're.sub(r"\D", "", x)'

100000 loops, best of 3: 2.37 usec per loop

$ python -mtimeit -s'import re; x="aaa12333bab445bb54b5b52"' '"".join(re.findall("[a-z]+",x))'

100000 loops, best of 3: 1.97 usec per loop

I had observed that join is faster than sub.

Answer 14

You can read each character. If it is digit, then include it in the answer. The str.isdigit() method is a way to know if a character is digit.

your_input = '12kjkh2nnk34l34'
your_output = ''.join(c for c in your_input if c.isdigit())
print(your_output) # '1223434'

Answer 15

You can use join + filter + lambda:

''.join(filter(lambda s: s.isdigit(), "20 years ago, 2 months ago, 2 days ago"))

Output: '2022'

Answer 16

Not a one liner but very simple:

buffer = ""
some_str = "aas30dsa20"

for char in some_str:
    if not char.isdigit():
        buffer += char

print( buffer )

Answer 17

I used this. 'letters' should contain all the letters that you want to get rid of:

Output = Input.translate({ord(i): None for i in 'letters'}))

Example:

Input = "I would like 20 dollars for that suit" Output = Input.translate({ord(i): None for i in 'abcdefghijklmnopqrstuvwxzy'})) print(Output)

Output: 20

Answer 18

my_string="sdfsdfsdfsfsdf353dsg345435sdfs525436654.dgg(" 
my_string=''.join((ch if ch in '0123456789' else '') for ch in my_string)
print(output:+my_string)

output: 353345435525436654

Answer 19

Another one:

import re

re.sub('[^0-9]', '', 'ABC123 456')

Result:

'123456'