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splitting words with brackets

I've got some strings to split. They are main words, but some words
are inside a pair of brackets and should be considered as one unit. I
prefer to use re.split, but haven't written a working one after hours
of work.

Example:

"a (b c) d [e f g] h i"
should be splitted to
["a", "(b c)", "d", "[e f g]", "h", "i"]

As speed is a factor to consider, it's best if there is a single line
regular expression can handle this. I tried this but failed:
re.split(r"(?![\(\[].*?)\s+(?!.*?[\)\]])", s). It work for "(a b) c"
but not work "a (b c)" :(

Any hint?

Jul 26 '06 #1
17 8201
re.findall('\([^\)]*\)|\[[^\]]*|\S+', s)

Qiangning Hong wrote:
I've got some strings to split. They are main words, but some words
are inside a pair of brackets and should be considered as one unit. I
prefer to use re.split, but haven't written a working one after hours
of work.

Example:

"a (b c) d [e f g] h i"
should be splitted to
["a", "(b c)", "d", "[e f g]", "h", "i"]

As speed is a factor to consider, it's best if there is a single line
regular expression can handle this. I tried this but failed:
re.split(r"(?![\(\[].*?)\s+(?!.*?[\)\]])", s). It work for "(a b) c"
but not work "a (b c)" :(

Any hint?
Jul 26 '06 #2
er,
....|\[[^\]]*\]|...
^_^

faulkner wrote:
re.findall('\([^\)]*\)|\[[^\]]*|\S+', s)

Qiangning Hong wrote:
I've got some strings to split. They are main words, but some words
are inside a pair of brackets and should be considered as one unit. I
prefer to use re.split, but haven't written a working one after hours
of work.

Example:

"a (b c) d [e f g] h i"
should be splitted to
["a", "(b c)", "d", "[e f g]", "h", "i"]

As speed is a factor to consider, it's best if there is a single line
regular expression can handle this. I tried this but failed:
re.split(r"(?![\(\[].*?)\s+(?!.*?[\)\]])", s). It work for "(a b) c"
but not work "a (b c)" :(

Any hint?
Jul 26 '06 #3
faulkner wrote:
re.findall('\([^\)]*\)|\[[^\]]*|\S+', s)
sorry i forgot to give a limitation: if a letter is next to a bracket,
they should be considered as one word. i.e.:
"a(b c) d" becomes ["a(b c)", "d"]
because there is no blank between "a" and "(".

Jul 26 '06 #4
faulkner wrote:
er,
...|\[[^\]]*\]|...
^_^
That's why it is nice to use re.VERBOSE:

def splitup(s):
return re.findall('''
\( [^\)]* \) |
\[ [^\]]* \] |
\S+
''', s, re.VERBOSE)

Much less error prone this way

--
- Justin

Jul 26 '06 #5
"a (b c) d [e f g] h i"
should be splitted to
["a", "(b c)", "d", "[e f g]", "h", "i"]

As speed is a factor to consider, it's best if there is a
single line regular expression can handle this. I tried
this but failed:
re.split(r"(?![\(\[].*?)\s+(?!.*?[\)\]])", s). It work
for "(a b) c" but not work "a (b c)" :(

Any hint?
[and later added]
sorry i forgot to give a limitation: if a letter is next
to a bracket, they should be considered as one word. i.e.:
"a(b c) d" becomes ["a(b c)", "d"] because there is no
blank between "a" and "(".
>>import re
s ='a (b c) d [e f g] h ia abcd(b c)xyz d [e f g] h i'
r = re.compile(r'(? :\S*(?:\([^\)]*\)|\[[^\]]*\])\S*)|\S+')
r.findall(s )
['a', '(b c)', 'd', '[e f g]', 'h', 'ia', 'abcd(b c)xyz', 'd',
'[e f g]', 'h', 'i']

I'm sure there's a *much* more elegant pyparsing solution to
this, but I don't have the pyparsing module on this machine.
It's much better/clearer and will be far more readable when
you come back to it later.

However, the above monstrosity passes the tests I threw at
it.

-tkc


Jul 26 '06 #6
Qiangning Hong wrote:
faulkner wrote:
re.findall('\([^\)]*\)|\[[^\]]*|\S+', s)

sorry i forgot to give a limitation: if a letter is next to a bracket,
they should be considered as one word. i.e.:
"a(b c) d" becomes ["a(b c)", "d"]
because there is no blank between "a" and "(".
This variation seems to do it:

import re

s = "a (b c) d [e f g] h i(j k) l [m n o]p q"

def splitup(s):
return re.findall('''
\S*\( [^\)]* \)\S* |
\S*\[ [^\]]* \]\S* |
\S+
''', s, re.VERBOSE)

print splitup(s)

# Prints

['a', '(b c)', 'd', '[e f g]', 'h', 'i(j k)', 'l', '[m n o]p', 'q']
Peace,
~Simon

Jul 26 '06 #7
Tim Chase wrote:
>>import re
>>s ='a (b c) d [e f g] h ia abcd(b c)xyz d [e f g] h i'
>>r = re.compile(r'(? :\S*(?:\([^\)]*\)|\[[^\]]*\])\S*)|\S+')
>>r.findall(s )
['a', '(b c)', 'd', '[e f g]', 'h', 'ia', 'abcd(b c)xyz', 'd',
'[e f g]', 'h', 'i']
[...]
However, the above monstrosity passes the tests I threw at
it.
but it can't pass this one: "(a c)b(c d) e"
the above regex gives out ['(a c)b(c', 'd)', 'e'], but the correct one
should be ['(a c)b(c d)', 'e']

Jul 26 '06 #8
Simon Forman wrote:
def splitup(s):
return re.findall('''
\S*\( [^\)]* \)\S* |
\S*\[ [^\]]* \]\S* |
\S+
''', s, re.VERBOSE)
Yours is the same as Tim's, it can't handle a word with two or more
brackets pairs, too.

I tried to change the "\S*\([^\)]*\)\S*" part to "(\S|\([^\)]*\))*",
but it turns out to a mess.

Jul 26 '06 #9

Qiangning Hong wrote:
Tim Chase wrote:
>>import re
>>s ='a (b c) d [e f g] h ia abcd(b c)xyz d [e f g] h i'
>>r = re.compile(r'(? :\S*(?:\([^\)]*\)|\[[^\]]*\])\S*)|\S+')
>>r.findall(s )
['a', '(b c)', 'd', '[e f g]', 'h', 'ia', 'abcd(b c)xyz', 'd',
'[e f g]', 'h', 'i']
[...]
However, the above monstrosity passes the tests I threw at
it.

but it can't pass this one: "(a c)b(c d) e"
the above regex gives out ['(a c)b(c', 'd)', 'e'], but the correct one
should be ['(a c)b(c d)', 'e']
What are the desired results in cases like this:

"(a b)[c d]" or "(a b)(c d)" ?

Jul 26 '06 #10

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