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)
$ 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.
Ugly but works:
>>> s
'aaa12333bb445bb54b5b52'
>>> a = ''.join(filter(lambda x : x.isdigit(), s))
>>> a
'1233344554552'
>>>
Not a one liner but very simple:
buffer = ""
some_str = "aas30dsa20"
for char in some_str:
if not char.isdigit():
buffer += char
print( buffer )
s=''.join(i for i in s if i.isdigit())
Another generator variant.
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"))
my_string="sdfsdfsdfsfsdf353dsg345435sdfs525436654.dgg("
my_string=''.join((ch if ch in '0123456789' else '') for ch in my_string)
print(output:+my_string)
output: 353345435525436654
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'
along the lines of bayer's answer:
''.join(i for i in s if i.isdigit())
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'
You can easily do it using Regex
>>> import re
>>> re.sub("\D","","£70,000")
70000
import re
string = '1abcd2XYZ3'
string_without_letters = re.sub(r'[a-z]', '', string.lower())
result = '123'
Use a generator expression:
>>> s = "foo200bar"
>>> new_s = "".join(i for i in s if i in "0123456789")
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
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
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'
Source: Stackoverflow.com