python regex: extracting number from string, unknown number format

Question

I need to extract the first number from a string, but I don't know the exact format of the number.

The number could be one of the following formats... 1.224 some decimal... 3,455,000 some number with unknown number of commas... 45% a percentage ... or just an integer 5

it would be something like blah blah $ 2,400 or blah blah 45% or blah blah $1.23 or blah blah 7

would be interesting if it was intelligent enough to do word numbers too like blah blah seven

I don't need the dollar sign, just the number

Answer 1

While this problem has many cases, here is a solution which solves most of them using some regex and the re module:

import re

def extractVal(s):
    return re.sub(r'^[^0-9$\-]*| .*$', '', s)

(1) It removes all leading string characters that are not 0-9, or $

(2) It removes all ending characters up to and including the first space (after (1))

Here's some data in action:

>>> data = ['blah $50,000 10', 'blah -1.224 blah', 'blah 3,455,000 blah', 'blah 45% 10 10 blah', '5 6 4']
>>> print(list(map(extractVal,data)))
['$50,000', '-1.224', '3,455,000', '45%', '5']

This solution assumes that the first number ends in a space.

We can go further as others have stated by converting these strings into numbers :

def valToInt(s):
    if '%' in s:
        a = float(s[:-1])/100
    else:
        a =  float(re.sub(r'[,$]','',s))
    return int(a) if a == int(a) else a

Resulting to (with the map() function again):

[50000, -1.224, 3455000, 0.45, 5]

Answer 2

If you insist on a regex, then this should work (only limited to cases you mentioned):

rgx = re.compile(r'\d+(,|\.)?\d*')
assert rgx.search("blah blah $ 2,400")
assert rgx.search("blah blah 45%")
assert rgx.search("blah blah $1.23")
assert rgx.search("blah blah 7")

As for the blah blah seven I do not thing a regex would cut it (at least not for anything more complex than a single digit).

Answer 3

For extracting the first number from a string, with different formats, you could use re.findall() :

 import re

strings = ['45% blah 43%', '1.224 blah 3.2', '3,455,000 blah 4,3', '$1.2 blah blah $ 2,400', '3 blah blah 7']

for string in strings:
    first_match = re.findall(r'[0-9$,.%]+\d*', string)[0]
    print(first_match)

Which Outputs:

45%
1.224
3,455,000
$1.2
3

Answer 4

Assuming you want an actual number, and that percents should be converted to a decimal:

str_ = "blah blah $ 2,400"
number, is_percent = re.search(r"([0-9,.]+)\s*(%?)", str_).groups() or (None, None)
if number is not None:
    number = float(number.replace(",", ""))
    if is_percent:
        number /= 100