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? 17 8206
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?
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?
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 "(".
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
"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
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
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']
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.
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)" ? This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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