Regular Expression to extract specific text from string

Question

I am new to Regex and try to extract a 16x character piece of text from a list of strings.

Sample list:

myString = ['  pon-3-1    |    UnReg 5A594F4380661123           1234567890               Active',
            '  pon-3-1    |    UnReg 5A594F43805FA456           1234567890               Active',
            '  pon-3-1    |    UnReg 4244434D73B24789           1234567890               Active', 
            '  pon-3-1    |    UnReg 5A594F43805FB000           1234567890               Active',
            'sw-frombananaramatoyourmama-01'
           ]

I cannot use a simple regex like (\w{16}) as this will include all text with 16 characters. I also tried (\w+A) which, depending on the characters in the string, don't return the correct results.

newArry = []
for i in myString:
   number = re.search('(\w{16})', i)
   newArr.append(number[0])

print(newArr)

Returns:

['5A594F4380661123', '5A594F43805FA456', '4244434D73B24789', '5A594F43805FB000', 'frombananaramato']

1. I want to extract only:
   • 5A594F4380661123
   • 5A594F43805FA456
   • 4244434D73B24789
   • 5A594F43805FB000

Any ideas?

Many thanks in advance

Answer 1

If you want to make sure the 16 characters are surrounded by non-alphabetics, try

re.search(r'\b([0-9A-F]{16})\b', i)

The \b "word boundary" operator matches on a position which is surrounded by alphabetics on one side and nonalphabetics on the other.

(If you want to be more specific about which nonalphabetics exactly, you can use lookarounds:

re.search(r'(?<![0-9A-F])([0-9A-F]{16})(?![0-9A-F])', i)

where (?<!...) says "cannot be preceded by ..." and (?!...) says "cannot be followed by ...".)

You'll also notice that I tightened up the character class to only match hex digits, which by itself is already sufficient to solve your example problem, and used r'...' raw strings for the regexes, which you should probably always do (at least until you completely understand how backslashes in Python non-raw strings are mangled).

Answer 2

use a regex set

number = re.search("([\dABCDEF]{16})", i)

this will search for any 16 length string with any digit (\d), 'A', 'B', 'C', 'D', 'E' or 'F'

Answer 3

Be more specific in you regex: tell it what you know!

If you obviously realize that the actual results differ from the expected ones in some specific way, try to gain an advantage from it.

\w matches letters ([A-Za-z]), numbers ([0-9]) and _, you seem to are searching for 16 hexadecimal digits. Build a specific character class.

Another observation is that you want the 16-hex-digit blocks enclosed in spaces, this can be expressed by the character regex around the capturing part befor(capt)after or by adding anchors/boundaries.

Answer 4

You can try this, assuming the HEX code is always preceded by UnReg

re.findall(r'UnReg\s+([\dA-F]{16})',';'.join(myString))

Answer 5

Use re.findall to avoid for-loop. I'd specify UnReg in the pattern (if there is one in your real data), so that regex doesn't include other 16-character pieces of text.

>>> import re
>>> newArr = re.findall(r'UnReg\s(.{16})', ' '.join(myString))
>>> print(newArr)
['5A594F4380661123', '5A594F43805FA456', '4244434D73B24789', '5A594F43805FB000']