Python - Creating multiple optional groups in regex

Question

I have two strings, where some part of the string is optional. So, I tried to create an optional pattern by using ? after every group which I want to make optional. But, it's giving None as output for those groups.

Text-1:

text1 = '95031B2	R	C01 N1 P93 R-- 12:39:18.540 08/05/20 0000002802 R  -                               No_barcode  FLC   F  LR    7.673353 sccm   Pt   25.288202 psig   FL  536.651917 sccm   EDC   0.000000 sccm   PQ    7.668324 sccm   QF  536.289246 sccm   QP   25.287605 psig   LLQ  -0.109524 sccm   HLQ   4.440174 sccm   CLF   1.429953 sccm   MF    0.000000 sccm   LF  100.000015 sccm   MQF   0.000000 sccm   LQF 100.000015 sccm   FPR  25.290846 psig 
'

Text-2:

text2 = '5102060	R	C01 N1 P93 R-- 12:38:52.140 08/05/20 0000002801 FO -                               No_barcode 
'

Working pattern for text1:

pattern1 = ['(?P\d\d:\d\d:\d\d.\d{3})\s',
           '(?P\d\d/\d\d/\d\d)\s',
           '(?P\d{10})\s',
           '(?P\w{1,2}).*?-',
           '\s*',
           '(?PNo_barcode|\W{20})',
           '\s{2}',
           '(?P\w{3})',
           '.*?',
           '(?PPt.*?\d*[.]?\d*\s[a-z]+)'
           '\s{1,3}',
           '(?PFL.*?\d*[.]?\d*\s[a-z]+)'
           ]

Tried to make optional part in above pattern to work with both the strings:

>>> pattern2 = ['(?P\d\d:\d\d:\d\d.\d{3})\s', # time pattern
           '(?P\d\d/\d\d/\d\d)\s',           # date pattern
           '(?P\d{10})\s',                    # 10 digits
           '(?P\w{1,2}).*?-',              # 1 or 2 alphabets follows with anything and then hyphen('-')
           '\s*',                                  # zero or more spaces
           '(?PNo_barcode|\W{20})',         # No_barcode or any alphanumeric with 20 length
                                       
                     # OPTIONAL PART STARTS (Not working)
           '(\s{1,2}|',                            # 1 or 2 spaces or
           '(?P\w{3})|',                     # 3 alphabets or
           '.*?|',                                 # anything getting ignored or
           '(?PPt.*?\d*[.]?\d*\s[a-z]+)|'      # Pt digits optional decimal followed with digits, 1 space, 1 or more a-z alphabets or
           '\s{1,3}|',                             # 1 to 3 spaces or
           '(?PFL.*?\d*[.]?\d*\s[a-z]+))?'     # FL digits optional decimal followed with digits, 1 space, 1 or more a-z alphabets 
           ]

Output:

>>> res = re.search(r''.join(pattern1), text) # pattern1
>>> res.groups()
('12:39:18.540', '08/05/20', '0000002802', 'R', 'No_barcode', 'FLC', 'Pt   25.288202 psig', 'FL  536.651917 sccm')

>>> res = re.search(r''.join(pattern2), text) # pattern2, trying to get same output as pattern1
>>> res.groups()
('12:39:18.540', '08/05/20', '0000002802', 'R', 'No_barcode', '  ', None, None, None)

Expected output:

For pattern2(after adding an optional part to pattern1), I should get the same output as the output of pattern1.

>>> res = re.search(r''.join(pattern2), text) # pattern2
>>> res.groups()
('12:39:18.540', '08/05/20', '0000002802', 'R', 'No_barcode', 'FLC', 'Pt   25.288202 psig', 'FL  536.651917 sccm')

user_3832970 · Answer

You wrapped all the part after obligatory part with an optional group, and inserted | alternation operator in between each line in the pattern, which is a wrong way to make optional groups.

If the parts are all optional and the subsequent subpatterns do not depend on whether the preceding subpattern is found or not, you need to either wrap each part with an optional group: (?:...(optional_1)...)?(?:...(optional_2)...)?(?:...(optional_n)...)?.

Or, make nested optional groups if each subsequent pattern cannot appear if the preceding subpattern is missing: (?:...(optional_1)...(?:...(optional_2)...(?:...(optional_n)...)?)?)?.

So, you may use either this regex:

(?P\d\d:\d\d:\d\d.\d{3})\s(?P\d\d/\d\d/\d\d)\s(?P\d{10})\s(?P\w{1,2}).*?-\s*(?PNo_barcode|\W{20})(?:\s+(?P\w{3}))?(?:.*?(?PPt.*?\d*[.]?\d*\s[a-z]+))?(?:\s{1,3}(?PFL.*?\d*[.]?\d*\s[a-z]+))?

Or this regex:

(?P\d\d:\d\d:\d\d.\d{3})\s(?P\d\d/\d\d/\d\d)\s(?P\d{10})\s(?P\w{1,2}).*?-\s*(?PNo_barcode|\W{20})(?:\s+(?P\w{3})(?:.*?(?PPt.*?\d*[.]?\d*\s[a-z]+)(?:\s{1,3}(?PFL.*?\d*[.]?\d*\s[a-z]+))?)?)?

Since you say you need Scheme 1, you may use

pattern = [r'(?P\d\d:\d\d:\d\d.\d{3})\s',
           r'(?P\d\d/\d\d/\d\d)\s',
           r'(?P\d{10})\s',
           r'(?P\w{1,2}).*?-',
           r'\s*',
           r'(?PNo_barcode|\W{20})',
           r'(?:\s+',                     # (?: starts a non-capturing group...
           r'(?P\w{3}))?',          # ...)? closes the group, 
           r'(?:.*?',
           r'(?PPt.*?\d*[.]?\d*\s[a-z]+))?'
           r'(?:\s{1,3}',
           r'(?PFL.*?\d*[.]?\d*\s[a-z]+))?'
          ]
print(r''.join(pattern))

See Python demo.

Python - Creating multiple optional groups in regex

1 Answers