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Asked: 2026-06-11T02:47:51+00:00

Can anyone tell me why the regex pattern in Python works most of the

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Can anyone tell me why the regex pattern in Python works most of the time but have difficulty with the following text.

PATTERN:

patternd = re.compile(r"""\(VFSCAN\)[^=]*=\s*    # first line of a section: (VFSCAN) AT TIME =  1.1800 UP    TO  100 BUSES WITH LOW VOLTAGE DEVIATION BELOW -0.200 
(\d*(?:\.\d+)?)                 # group 1 - first number of first line: 1.1800
\D+ 
\d+                             # second number of first line: 100 
\s+BUSES\s+WITH\s+LOW\s+VOLTAGE\s+(DEVIATION)\s+BELOW.*? 
\D+                             # skip second line 
(?:                             # a data line: 18436 [LENZIE 618.0] -0.245 18433 [LENZIE 318.0] -0.245
(?:\d+\s+\[(.+?)\]\s+\S+\s*)+   # Component of data line
(?=[\r\n\s]+|$) 
)+                              # This search ends with an empty line
""", re.VERBOSE)

The text I am having problems with is:

test3 = r'''(VFSCAN) AT TIME =  1.1800 UP TO  100 BUSES WITH LOW VOLTAGE DEVIATION BELOW -0.200:

X ----- BUS ------ X    VDEV       X ----- BUS ------ X    VDEV
18436 [LENZIE 618.0]   -0.245      18433 [LENZIE 318.0]   -0.245     
18431 [LENZIE 118.0]   -0.214      18435 [LENZIE 518.0]   -0.214     
18434 [LENZIE 418.0]   -0.214      18432 [LENZIE 218.0]   -0.214     

(VFSCAN) AT TIME =  1.5167 UP TO  100 BUSES WITH LOW VOLTAGE DEVIATION BELOW -0.200:

X ----- BUS ------ X    VDEV       X ----- BUS ------ X    VDEV
69036 [DNLP2G21.575]   -0.414      69038 [DNLP2G22.575]   -0.414     
69040 [DNLP2G23.575]   -0.414      69032 [DNLP1_G1.575]   -0.402     
65460 [DIFICULT 230]   -0.384      69027 [7MIHL G1.575]   -0.355     
69076 [HORIZ_G .575]   -0.303      67237 [MEDBOWCO 115]   -0.301     
67940 [STNDPSVC 230]   -0.300      65976 [MINERS  34.5]   -0.294     
65585 [FT CRK1 34.5]   -0.261      65584 [FT CRK2 34.5]   -0.261     
69073 [HIPLN_G .575]   -0.214     

(VFSCAN) AT TIME =  1.1800 UP TO  100 BUSES WITH LOW VOLTAGE DEVIATION BELOW -0.200:

X ----- BUS ------ X    VDEV       X ----- BUS ------ X    VDEV
65191 [BONANZA 24.0]   -0.572      65192 [BONANZA  138]   -0.434     
65194 [CHAPITA  138]   -0.433      66278 [RANGELY  138]   -0.320     
65371 [COVE TP  138]   -0.302      79265 [CALAMRDG 138]   -0.286     
79400 [DES.MINE 138]   -0.285      65086 [ASHLEY  69.0]   -0.284     
79067 [VERNAL   138]   -0.277      67257 [MOONLAK269.0]   -0.268     
67256 [MOONLAK169.0]   -0.266      79264 [W.RV.CTY 138]   -0.206     

'''

I am using “findall” to match the text using the pattern above and expect to extract a list of tuples. When I run this I am getting:

[('1.1800', 'DEVIATION', 'LENZIE 218.0'), ('1.5167', 'DEVIATION', 'HIPLN_G .575'),     ('1.1800', 'DEVIATION', 'W.RV.CTY 138')]

But I should be getting a lot more matched tuples in my list. I am really only interested in extracting the time on the first line of each match along with all the items in the square brackets “[]”. I should be getting 6+13+12 = 31 tuples in the list. Any help would be appreciated.

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1 Answer

  1. Editorial Team

    What you want cannot be done using regular expressions. (At least, not in a single regular expression). The number of subgroups inside a group or a match cannot be variable. Compare this simplified version of your regular expression.

    p = re.compile('(a(b)+)')
test3 = 'ababbabbbabbbb'
print p.findall(test3)

    You want it to return something like this:

    [('ab', 'b'), ('abb', 'b', 'b'), ('abbb', 'b', 'b', 'b'), ('abbbb', 'b')]

    Instead it returns this:

    [('ab', 'b'), ('abb', 'b'), ('abbb', 'b'), ('abbbb', 'b')]

    Based on the example data, you could use a two pass algorithm. You could first use your original regular expression (without the grouping that makes it fail) to extract the big blocks of text. Then, using a second regular expression, you can extract the subpatterns from the blocks you found in the first pass.

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