I have the following case:
Test (2.00001) Test (2.000) Test 2.1 Test (2,0001) Test 2,000 Test 2,1000 test 2
I try to use regex to find only the integers:
2.0002,0002but not the other float numbers.
I tried different things:
re.search('(?<![0-9.])2(?![.,]?[1-9])(?=[.,]*[0]*)(?![1-9]),...)
but this returns true for:
2.000012.0002,0002,00012What have I to do?
UPDATE
I have updated the question and it should also find an integer without any comma and point, too (2).
I would use:
import re
text = 'Test (2.00001) Test (2.000) Test 2.1 Test (2,0001) Test 2,000 Test 2,1000'
re.findall(r'(\d+[.,]0+)(?!\d)', text)
Output:
['2.000', '2,000']
Regex:
( # start capturing
\d+ # match digit(s)
[.,] # match . or ,
0+ # match one or more zeros
) # stop capturing
(?!\d) # ensure the last zero is not followed by a digit
If you also want to match "intergers" alone, surrounded by spaces or parentheses/brackets:
import re
text = 'Test (2.00001) Test (2.000) Test 2.1 Test (2,0001) Test 2,000 Test 2,1000 2'
re.findall(r'(?:^|[(\s[])(\d+(?:[.,]0+(?!\d))?)(?=[]\s)]|$)', text)
Regex:
(?:^|[(\s[]) # match the start of string or [ or ( or space
( # start capturing
\d+ # match digit(s)
(?:[.,]0+(?!\d))? # optionally match . or , with only zeros
) # stop capturing
(?=[]\s)]|$) # match the end of string or ] or ) or space
2 alone without , and ..
abc123 or 127.0.0.0?2 2.00 .You can use
re.findall(r'\b(?<!\d[.,])\d+(?:[.,]0+)?\b(?![,.]\d)', text)
See the regex demo. Details:
\b - a word boundary(?<!\d[.,]) - no digit followed with . or , immediately on the left\d+ - one or more digits(?:[.,]0+)? - an optional sequence of . or , and then one or more zeros\b - a word boundary(?![,.]\d) - no , or . and a digit allowed immediately to the right.If you need to support thousand separators:
pattern = r'\b(?<!\d[.,])(?:\d{1,3}(?:(?=([.,]))(?:\1\d{3})+)?|\d{4,})(?:(?!\1)[.,]0+)?\b(?![,.]\d)'
matches = [x.group() for x in re.finditer(pattern, text)]
See this regex demo.
GB2L 114, it will also return the 2 inside the string.(?<!\d[.,])\b\d+(?:[.,]0+)?\b(?![,.]\d), see this regex demo.2,102.120 i.e. a decimal with one zero at the end.Without the need for regex, you can also consider using is_integer() after trying to conver the values into their respective numeric formats. While a little bit harder to read, it removes the need for regex and should be robust for further use cases given the string structure you provide:
[x for x in string.split() if float((pd.to_numeric(x.replace(r'(','').replace(r')','').replace(r',','.'),errors='coerce'))).is_integer()]
Returning the former values in the list:
['(2.000)', '2,000', '2']
Or if you'd like them cleaned:
[x for x in string.replace(r'(','').replace(r')','').replace(r',','.').split() if float((pd.to_numeric(x,errors='coerce'))).is_integer()]
Returning:
['2.000', '2.000', '2']
This should be easy - just get a number and check "is this an int value?". Meaby something like this...
import re
text = 'Test (2.00001) Test (2.000) Test 2.1 Test (2,0001) Test 2,000 Test 2,1000 test 2'
out_ints = []
for x in re.findall(r'([0-9.,]+)', text):
possible_int = x.replace(',', '.')
is_int = int(float(possible_int)) == float(possible_int)
if is_int:
out_ints.append(int(float(possible_int)))
print(out_ints)
Output:
[2, 2, 2]
Or am i missing something?
(?<!\d)(?<!\d[.,])\d{1,3}(?:[.,]\d{3})*(?![,.]?\d), see demo.test 2 2.002.00001, why do you want to match2.00? How can you formulate the pattern requirements regarding differentiation between valid and non-valid floats?(?<!\d)(?<!\d[.,])(?:\d{1,3}(?:([.,])\d{3})*|\d{4,})(?:(?!\1)[.,]0+)?(?![,.]?\d)? See regex101.com/r/qrG8hg/2(?<!\d)(?<!\d[.,])\d+(?:[.,]0+)?(?![,.]?\d)- see this demo.