Removing all non-numeric characters from string in Python

How do we remove all non-numeric characters from a string in Python?

>>> import re
>>> re.sub("[^0-9]", "", "sdkjh987978asd098as0980a98sd")
'987978098098098'

Not sure if this is the most efficient way, but:

>>> ''.join(c for c in "abc123def456" if c.isdigit())
'123456'

The ''.join part means to combine all the resulting characters together without any characters in between. Then the rest of it is a list comprehension, where (as you can probably guess) we only take the parts of the string that match the condition isdigit.

This should work for both strings and unicode objects in Python2, and both strings and bytes in Python3:

# python <3.0
def only_numerics(seq):
    return filter(type(seq).isdigit, seq)

# python ≥3.0
def only_numerics(seq):
    seq_type= type(seq)
    return seq_type().join(filter(seq_type.isdigit, seq))

@Ned Batchelder and @newacct provided the right answer, but ...

Just in case if you have comma(,) decimal(.) in your string:

import re
re.sub("[^\d\.]", "", "$1,999,888.77")
'1999888.77'

Just to add another option to the mix, there are several useful constants within the string module. While more useful in other cases, they can be used here.

>>> from string import digits
>>> ''.join(c for c in "abc123def456" if c in digits)
'123456'

There are several constants in the module, including:

• ascii_letters (abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ)
• hexdigits (0123456789abcdefABCDEF)

If you are using these constants heavily, it can be worthwhile to covert them to a frozenset. That enables O(1) lookups, rather than O(n), where n is the length of the constant for the original strings.

>>> digits = frozenset(digits)
>>> ''.join(c for c in "abc123def456" if c in digits)
'123456'

Many right answers but in case you want it in a float, directly, without using regex:

x= '$123.45M'

float(''.join(c for c in x if (c.isdigit() or c =='.'))

123.45

You can change the point for a comma depending on your needs.

change for this if you know your number is an integer

x='$1123'    
int(''.join(c for c in x if c.isdigit())

1123

Fastest approach, if you need to perform more than just one or two such removal operations (or even just one, but on a very long string!-), is to rely on the translate method of strings, even though it does need some prep:

>>> import string
>>> allchars = ''.join(chr(i) for i in xrange(256))
>>> identity = string.maketrans('', '')
>>> nondigits = allchars.translate(identity, string.digits)
>>> s = 'abc123def456'
>>> s.translate(identity, nondigits)
'123456'

The translate method is different, and maybe a tad simpler simpler to use, on Unicode strings than it is on byte strings, btw:

>>> unondig = dict.fromkeys(xrange(65536))
>>> for x in string.digits: del unondig[ord(x)]
... 
>>> s = u'abc123def456'
>>> s.translate(unondig)
u'123456'

You might want to use a mapping class rather than an actual dict, especially if your Unicode string may potentially contain characters with very high ord values (that would make the dict excessively large;-). For example:

>>> class keeponly(object):
...   def __init__(self, keep): 
...     self.keep = set(ord(c) for c in keep)
...   def __getitem__(self, key):
...     if key in self.keep:
...       return key
...     return None
... 
>>> s.translate(keeponly(string.digits))
u'123456'
>>>