Is there a way to check for two different patterns in the same string in regex python?

Question

I want to extract certain digits from a string. The problem is that the string can contain the digits in two different patterns. How can I create a regex pattern in re.search such that I can have both patterns to search in a single string?

For e.g.,

## extract 65.45 from this string
string = '1112 (65.45%)'

So, if I do the following, it works

re.search('.*?$(.*)%$', string).group(1)

and I get the expected result 65.45.

Now, I have another kind of string in the same text that I need to look for.

## from this string, extract 4.00 which appears before [
string = '4.00 [3.00 - 4.50]'

re.search('^(\S+)\s\[.*', string).group(1)

gives me the desired result: 4.00

But if I combine them like the following, it only extracts the one that matches first.

re.search('^(\S+)\s\[.*|.*?$(.*)%$', string).group(1)

in which case, only the string that contains the square bracket extracts the value, not if the string has a % sign. How can I fix this?

For e.g., If I have a list of strings like the following:

['73 (1.40%)', '38 (1.55%)', '27 (2.17%)', '32 (1.46%)', '10 (1.46%)', '11 (1.04%)', '11 (1.41%)', '7 (1.34%)', '4 (1.24%)', '28 (1.27%)', '750 (14.41%)', '381 (15.54%)', '182 (14.60%)', '313 (14.27%)', '4.10 [3.73 - 4.45]', '4.08 [3.70 - 4.42]', '4.13 [3.77 - 4.47]', '4.13 [3.78 - 4.47]', '4.07 [3.70 - 4.42]', '4.07 [3.70 - 4.43]', '4.07 [3.70 - 4.40]', '4.09 [3.73 - 4.42]', '4.03 [3.63 - 4.40]', '4.10 [3.70 - 4.47]']

I want to do certain things with each value that is extracted and compare with a specific threshold value.

Using the for-loop, I did something like this:

for val in string: 
    match = re.search('^(\S+)\s\[.*|.*?$(.*)%$', val)
    print(match)

which results in the following:

But not sure how to extract the exact value.

I have to do the .group() to extract the value, but it requires me to know the exact location. And I'm struggling to figure out how to do that.

If I do match.group(2), then I get the following result:

1.40
1.55
2.17
1.46
1.46
1.04
1.41
1.34
1.24
1.27
14.41
15.54
14.60
14.27
None
None
None
None
None
None
None
None
None
None

user_1863229 · Answer

Here is an approach which works with your exact input, in which each list entry always would have one of the two matching patterns:

inp = ['73 (1.40%)', '38 (1.55%)', '27 (2.17%)', '32 (1.46%)', '10 (1.46%)', '11 (1.04%)', '11 (1.41%)', '7 (1.34%)', '4 (1.24%)', '28 (1.27%)', '750 (14.41%)', '381 (15.54%)', '182 (14.60%)', '313 (14.27%)', '4.10 [3.73 - 4.45]', '4.08 [3.70 - 4.42]', '4.13 [3.77 - 4.47]', '4.13 [3.78 - 4.47]', '4.07 [3.70 - 4.42]', '4.07 [3.70 - 4.43]', '4.07 [3.70 - 4.40]', '4.09 [3.73 - 4.42]', '4.03 [3.63 - 4.40]', '4.10 [3.70 - 4.47]']
matches = [re.findall(r'\b\d+ $(\d+(?:\.\d+)?%)$|(\d+(?:\.\d+)?) $$\d+(?:\.\d+)? - \d+(?:\.\d+)?$$', x) for x in inp]
matches = [x[0][0] + x[0][1] for x in matches]
print(matches)

This prints:

['1.40%', '1.55%', '2.17%', '1.46%', '1.46%', '1.04%', '1.41%', '1.34%',
 '1.24%', '1.27%', '14.41%', '15.54%', '14.60%', '14.27%', '4.10', '4.08',
 '4.13', '4.13', '4.07', '4.07', '4.07', '4.09', '4.03', '4.10']

The strategy used above is to match, in two separate groups, either the first digit in the percentage input, or the number outside the square brackets. Then, in a list comprehension, we concatenate the two capture groups together. Since one of the two groups is guaranteed to be empty, the concatenated result always corresponds to the desired match.

user_3280538 · Answer

I would just use a list of simple regexs and iterate through them for each string I want to test. The first regex that gets a hit will be used. I would also compile the regex upfront to save CPU cycles. This is easier to follow readability wise and easy to add new patterns to:

import re

regexs = [
    re.compile(r".*?$(.*)%$"), 
    re.compile(r"^(\S+)\s\[.*"),
]

data = [
    "73 (1.40%)",
    "38 (1.55%)",
    "27 (2.17%)",
    "750 (14.41%)",
    "381 (15.54%)",
    "4.10 [3.73 - 4.45]",
    "4.08 [3.70 - 4.42]",
    "4.13 [3.77 - 4.47]",
    "this shouldn't match"
]


for val in data:
    for regex in regexs:
        if match := regex.search(val):
            print("Matched: " + match.group(1))
            break
    else:
        print("No match: " + val)

Outputs:

Matched: 1.40
Matched: 1.55
Matched: 2.17
Matched: 14.41
Matched: 15.54
Matched: 4.10
Matched: 4.08
Matched: 4.13
No match: this shouldn't match

user_5958455 · Answer

.group returns captured groups, so .group(1) always returns the first captured group.

To get the other capture group, use .group(2)

user_5424988 · Answer

Another option is to use lookarounds to get a match only:

(?<=$)\d+(?:\.\d+)?(?=%$)|\d+(?:\.\d+)?(?=\s*$$[^][]*])

The pattern matches

(?<=\() Positive lookbehind, assert ( to the left
\d+(?:\.\d+)? Match 1+ digits with an optional decimal part
(?=%\)) Positive lookahead, assert ) to the right
| Or
\d+(?:\.\d+)? Match 1+ digits with an optional decimal part
(?=\s*\[[^][]*]) Positive lookahead, assert an opening till closing square bracket to the right (You could make it more specific by specifying the exact format between the square brackets)

Regex demo | Python demo

import re

pattern = r"(?<=\()\d+(?:\.\d+)?(?=%\))|\d+(?:\.\d+)?\b(?=\s*\[[^][]*$$)"
strings = ['73 (1.40%)', '38 (1.55%)', '27 (2.17%)', '32 (1.46%)', '10 (1.46%)', '11 (1.04%)', '11 (1.41%)', '7 (1.34%)', '4 (1.24%)', '28 (1.27%)', '750 (14.41%)', '381 (15.54%)', '182 (14.60%)', '313 (14.27%)', '4.10 [3.73 - 4.45]', '4.08 [3.70 - 4.42]', '4.13 [3.77 - 4.47]', '4.13 [3.78 - 4.47]', '4.07 [3.70 - 4.42]', '4.07 [3.70 - 4.43]', '4.07 [3.70 - 4.40]', '4.09 [3.73 - 4.42]', '4.03 [3.63 - 4.40]', '4.10 [3.70 - 4.47]']
for val in strings:
    match = re.search(pattern, val)
    print(match)

Output

Is there a way to check for two different patterns in the same string in regex python?

4 Answers