Python regex search range of numbers

Question

I couldn't seem to find a thread on this one, but it seems like something that should be pretty simple. I'm trying to use regex to search a line in an output for a number 0-99, and do one action, but if the number is 100 then it'll do a different action. Here's what I have tried(simplified version):

OUTPUT = #Some command that will store the output in variable OUTPUT
OUTPUT = OUTPUT.split('\n')
for line in OUTPUT:
    if (re.search(r"Rebuild status:  percentage_complete", line)): #searches for the line, regardless of number
        if (re.search("\d[0-99]", line)): #if any number between 0 and 99 is found
            print("error")
        if (re.search("100", line)): #if number 100 is found
            print("complete")

I've tried this and it still picks up the 100 and prints error.

Answer 1

This: \d[0-99] means a digit (\d), followed by a number (0-9) or 9. If you are after the numeric range of [0-99], you would need to use something akin to \b\d{1,2}\b. This will match any numeric value which is made of 1 or 2 digits.

Answer 2

You can simplify your regexes by re-ordering your number tests, and using elif instead of if on the test for 2 digit numbers.

for line in output:
    if re.search("Rebuild status:  percentage_complete", line): 
        if re.search("100", line):
            print "complete"
        elif re.search(r"\d{1,2}", line): 
            print "error"

The test for a 2 digit number is performed only if the test for "100" fails.

Using a raw string isn't strictly necessary with r"\d{1,2}" but it's a good habit to use a raw string for any regex that contains a backslash.

Note that you don't need parentheses around conditions in Python, so using them just adds unnecessary clutter.

---

As dawg mentions in the comments, the test for "100" can be tightened to re.search(r"\b100\b", line), but that's not needed if we can guarantee that we're only testing integer percentages in the range 0 - 100.

Answer 3

0 - 99:

>>> s='\n'.join(["line {} text".format(i) for i in range(-2,101) ])
>>> import re
>>> re.findall(r'(?<!\-)\b(\d\d|\d)\b', s)
['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23', '24', '25', '26', '27', '28', '29', '30', '31', '32', '33', '34', '35', '36', '37', '38', '39', '40', '41', '42', '43', '44', '45', '46', '47', '48', '49', '50', '51', '52', '53', '54', '55', '56', '57', '58', '59', '60', '61', '62', '63', '64', '65', '66', '67', '68', '69', '70', '71', '72', '73', '74', '75', '76', '77', '78', '79', '80', '81', '82', '83', '84', '85', '86', '87', '88', '89', '90', '91', '92', '93', '94', '95', '96', '97', '98', '99']

The regex '(?<!\-)\b(\d\d|\d)\b' matches 2 digits 0-99 and does not match negative numbers such as -9

Demo

100 is easy: '(?<!\-)\b100\b'

If you do not want to match floats: \b(?<![-.])(\d\d|\d)(?!\.)\b

Demo