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I have a character string 'aabaacaba'. Starting from left, I am trying to get substrings of all sizes >=2, which appear later in the string. For instance, aa appears again in the string and so is the case with ab.
I wrote following regex code:
re.findall(r'([a-z]{2,})(?:[a-z]*)(?:\1)', 'aabaacaba')
and I get ['aa'] as answer. Regular expression misses ab pattern. I think this is because of overlapping characters. Please suggest a solution, so that the expression could be fixed. Thank you.
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You can simply use DEF as your regexp. To identify strings that don't contain it, simply return the strings that don't match the above expression.
Use re.search() to extract a substring matching a regular expression pattern. Specify the regular expression pattern as the first parameter and the target string as the second parameter. \d matches a digit character, and + matches one or more repetitions of the preceding pattern.
Use the substring() method to get the substring before a specific character, e.g. const before = str. substring(0, str. indexOf('_')); . The substring method will return a new string containing the part of the string before the specified character.
You can use look-ahead assertion which does not consume matched string:
>>> re.findall(r'(?=([a-z]{2,})(?=.*\1))', 'aabaacaba')
['aa', 'aba', 'ba']
NOTE: aba matched instead of ab. (trying to match as long as possible)
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