python re - split a string before a character

Question

how to split a string at positions before a character?

• split a string before 'a'
• input: "fffagggahhh"
• output: ["fff", "aggg", "ahhh"]

the obvious way doesn't work:

>>> h=re.compile("(?=a)")
>>> h.split("fffagggahhh")
['fffagggahhh']
>>>

• kennytm: What do you expect when you split "aaa" — ['', 'a', 'a', 'a'] or ['a', 'a', 'a']?
• kakarukeys: "aaa" -> "a", "a", "a" or "", "a", "a", "a"
• S.Lott: -1: "aaa" -> ["a", "a", "a"] or ["", "a", "a", "a"]. That's the least helpful thing I've ever seen. Both are right? In that case, no pattern can ever work. Close this question.
• kakarukeys: either one of them will do. if you have coded in python before, you would know a simple filter(bool, L) will filter out the empty element.
• fireattack: Did something change in Python? Now your "the obvious way" works flawlessly.

Answer 1

Ok, not exactly the solution you want but I thought it will be a useful addition to problem here.

Solution without re

Without re:

>>> x = "fffagggahhh"
>>> k = x.split('a')
>>> j = [k[0]] + ['a'+l for l in k[1:]]
>>> j
['fff', 'aggg', 'ahhh']
>>> 

• pyfunc: @knitti: Thanks. I understand it is not the re based solution and I wanted to write it first before I write re solution. By the time, I finished writing this, the re based solution had come.
• knitti: yeah, why use a hammer on a single nail if you've got a nail shooter.

Answer 2

>>> rx = re.compile("(?:a|^)[^a]*")
>>> rx.findall("fffagggahhh")
['fff', 'aggg', 'ahhh']
>>> rx.findall("aaa")
['a', 'a', 'a']
>>> rx.findall("fgh")
['fgh']
>>> rx.findall("")
['']

• John Machin: -1 re.findall("(?:^|a)[^a]*", "aaa") produces ['', 'a', 'a']

Answer 3

>>> r=re.compile("(a?[^a]+)")
>>> r.findall("fffagggahhh")
['fff', 'aggg', 'ahhh']

EDIT:

This won't handle correctly double as in the string:

>>> r.findall("fffagggaahhh")
['fff', 'aggg', 'ahhh']

KennyTM's re seems better suited.

• Jeff Mercado: I wonder if the OP would want to keep the empty string from the split if it started with an 'a'.
• John Machin: -1 Uncool. Fails on repeated a's ... e.g. "aaa" -> empty list

Answer 4

import re
def split_before(pattern,text):
 prev = 0
 for m in re.finditer(pattern,text):
 yield text[prev:m.start()]
 prev = m.start()
 yield text[prev:]
if __name__ == '__main__':
 print list(split_before("a","fffagggahhh"))

re.split treats the pattern as a delimiter.

>>> print list(split_before("a","afffagggahhhaab"))
['', 'afff', 'aggg', 'ahhh', 'a', 'ab']
>>> print list(split_before("a","ffaabcaaa"))
['ff', 'a', 'abc', 'a', 'a', 'a']
>>> print list(split_before("a","aaaaa"))
['', 'a', 'a', 'a', 'a', 'a']
>>> print list(split_before("a","bbbb"))
['bbbb']
>>> print list(split_before("a",""))
['']

Answer 5

This one works on repeated a's

 >>> re.findall("a[^a]*|^[^a]*", "aaaaa")
 ['a', 'a', 'a', 'a', 'a']
 >>> re.findall("a[^a]*|[^a]+", "ffaabcaaa")
 ['ff', 'a', 'abc', 'a', 'a', 'a']

Approach: the main chunks that you are looking for are an a followed by zero or more not-a. That covers all possibilities except for zero or more not-a. That can happen only at the start of the input string.

Answer 6

>>> foo = "abbcaaaabbbbcaaab"
>>> bar = foo.split("c")
>>> baz = [bar[0]] + ["c"+x for x in bar[1:]]
>>> baz
['abb', 'caaaabbbb', 'caaab']

Due to how slicing works, this will work properly even if there are no occurrences of c in foo.

Answer 7

split() takes an argument for the character to split on:

>>> "fffagggahhh".split('a')
['fff', 'ggg', 'hhh']