正则表达式匹配字符串Python

Question

I wanted to match the numeric values of a string:

1,000 metric tonnes per contract month
Five cents ($0.05) per tonne
Five cents ($0.05) per tonne
1,000 metric tonnes per contract month

My current approach:

size = re.findall(r'(\d+(,?\d*).*?)', my_string)

What I get with my approach:

print size
[(u'1,000', u',000')]

As you can see, the number 1 was being cut out from the second element of the list, why is that? Also, could I get a hint as to how I can match the $0.05 terms?

Answer 1

Something like this:

>>> import re
>>>  strs = """1,000 metric tonnes per contract month
Five cents ($0.05) per tonne
Five cents ($0.05) per tonne
1,000 metric tonnes per contract month"""
>>> [m.group(0) for m in re.finditer(r'\$?\d+([,.]\d+)?', strs)]
['1,000', '$0.05', '$0.05', '1,000']

Demo : http://rubular.com/r/UomzIY3SD3

Answer 2

re,findall() returns a tuple of all the capturing groups for each match, and each set of normal parentheses generates one such group. Write your regex like this:

size = re.findall(r'\d{1,3}(?:,\d{3})*(?:\.\d+)?', my_string)

Explanation:

\d{1,3}      # One to three digits
(?:,\d{3})*  # Optional thousands groups
(?:\.\d+)?   # Optional decimal part

This assumes that all numbers have commas as thousands separators, ie no numbers like 1000000 . If you need to match those too, use

size = re.findall(r'\d+(?:,\d{3})*(?:\.\d+)?', my_string)

Answer 3

Try this regex:

(\$?\d+(?:[,.]?\d*(?:\.\d+)?)).*?

Live demo

Answer 4

Why are you grouping your regex? Try this r'\\$?\\d+,?\\d*\\.?\\d*'

Answer 5

I would try this regex:

r'[0-9]+(?:,[0-9]+) (?:.[0-9] )?'

Add \\$? at the beginning to optionally catch the $