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What is the pythonic way to split a string before the occurrences of a given set of characters?
For example, I want to split 'TheLongAndWindingRoad' at any occurrence of an uppercase letter (possibly except the first), and obtain ['The', 'Long', 'And', 'Winding', 'Road'].
Edit: It should also split single occurrences, i.e. from 'ABC' I'd like to obtain ['A', 'B', 'C'].
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To split a string on capital letters, call the split() method with the following regular expression - /(? =[A-Z])/ . The regular expression uses a positive lookahead assertion to split the string on each capital letter and return an array of the substrings.
findall() method to split a string on uppercase letters, e.g. re. findall('[a-zA-Z][^A-Z]*', my_str) . The re. findall() method will split the string on uppercase letters and will return a list containing the results.
In Python, lower() is a built-in method used for string handling. The lower() methods returns the lowercased string from the given string. It converts all uppercase characters to lowercase. If no uppercase characters exist, it returns the original string.
=\\p{Upper}) matches an empty sequence followed by a uppercase letter, and split uses it as a delimiter. See javadoc for more info on Java regexp syntax.
Unfortunately it's not possible to split on a zero-width match in Python. But you can use re.findall instead:
>>> import re >>> re.findall('[A-Z][^A-Z]*', 'TheLongAndWindingRoad') ['The', 'Long', 'And', 'Winding', 'Road'] >>> re.findall('[A-Z][^A-Z]*', 'ABC') ['A', 'B', 'C'] Here is an alternative regex solution. The problem can be reprased as "how do I insert a space before each uppercase letter, before doing the split":
>>> s = "TheLongAndWindingRoad ABC A123B45" >>> re.sub( r"([A-Z])", r" \1", s).split() ['The', 'Long', 'And', 'Winding', 'Road', 'A', 'B', 'C', 'A123', 'B45'] This has the advantage of preserving all non-whitespace characters, which most other solutions do not.
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