I am using python to pattern match a string multiple times in a string. Problem:
string = 'The value = 1 The value = 2 The value = 3'
I want to grep only value but my output should be like:
['value = 1 value = 2 value = 3']
I am doing like this:
pattern = re.compile('[value = (\d+)]*')
values = pattern.search(string)
values.group(0)
Output:
''
i.e NULL (no match)
Please help me give the right regular expression to grep the required output.
>>> [' '.join(re.findall(r'value = \d+', string))]
['value = 1 value = 2 value = 3']
You are using a character class grouping. You should just use a normal grouping with (.
import re
string = 'The value = 1 The value = 2 The value = 3'
pattern = re.compile(r'(value = \d+)')
pattern.findall(string)
# OUT: ['value = 1', 'value = 2', 'value = 3']
" ".join(pattern.findall(string))
# OUT: 'value = 1 value = 2 value = 3'
Your use of square brackets ([]) in the RE source is very odd. Those form character sets.
You should use something like:
>>> pattern = re.compile(r'([^=]+)\s*=\s*(\d+)')
>>> pattern.findall(string)
[('The value ', '1'), (' The value ', '2'), (' The value ', '3')]
Note use of findall() to get all the matches, and grouping to get the value names too.
=equals signs and the numbers in your input example, but your regular expression expects a space there. The output example is a single string in a list, did you mean to make that a list of 3 strings instead?*after yourbracket?string.replace('The ', '')works in this case ;)