match numbers if line starts with keyword

Question

I've got a file that looks like this:

foo: 11.00 12.00  bar 13.00
bar: 11.00 12.00 bar
foo: 11.00 12.00

and would like to extract all numbers in lines beginning with the keyword "foo:". Expected result:

['11.00', '12.00', '13.00']
['11.00', '12.00']

Now, this is easy, if I use two regexes, like this:

    if re.match('^foo:', line):
        re.findall('\d+\.\d+', line)

but I was wondering, if it is possible to combine these into a single regex?

Thanks for your help, MD

Answer 1

Not exactly what you asked for, but since it's recommended to use standard Python tools instead of regexes where possible, I'd do something like this:

import re

with open('numbers.txt', 'r') as f:
    [re.findall(r'\d+\.\d+', line) for line in f if line.startswith('foo')]

UPDATE

And this will return the numbers after 'foo' even if it's anywhere in the string rather than just in the beginning:

with open('numbers.txt', 'r') as f:
    [re.findall(r'\d+\.\d+', line.partition('foo')[2]) for line in f]

Answer 2

If all lines in the file always have the same number of numbers, you can use the following regex:

"^foo:[^\d]*(\d*\.\d*)[^\d]*(\d*\.\d*)[^\d]*(\d*\.\d*)"

Example:

>>> import re
>>> line = "foo: 11.00 12.00 bar 13.00"
>>> re.match("^foo:[^\d]*(\d*\.\d*)[^\d]*(\d*\.\d*)[^\d]*(\d*\.\d*)", line).groups()
('11.00', '12.00', '13.00')
>>> 

Using parentheses around a part of the regular expression makes it into a group that can be extracted from the match object. See the Python documentation for more information.

Answer 3

You can do without the first regexp and instead filter lines in a list comprehension by comparing the first four characters of the line, and compile the inner regexp:

import re

with open("input.txt", "r") as inp:
    prog=re.compile("\d+\.\d+")
    results=[prog.findall(line) for line in inp if line[:4]=="foo:"]