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  8176 
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.
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']
Ah...the picture is becoming a little more clear:
>>r = re.compile(r'(?:\([^\)]*\)|\[[^\]]*\]|\S)+')
r.findall(s)
['(a c)b(c d)', 'e']
It also works on my original test data, and is a cleaner regexp
than the original.
The clearer the problem, the clearer the answer. :)
-tkc
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)" ?
Simon Forman wrote:
What are the desired results in cases like this:
"(a b)[c d]" or "(a b)(c d)" ?
["(a b)[c d]"], ["(a b)(c d)"]
Tim Chase wrote:
Ah...the picture is becoming a little more clear:
>>r = re.compile(r'(?:\([^\)]*\)|\[[^\]]*\]|\S)+')
>>r.findall(s)
['(a c)b(c d)', 'e']
It also works on my original test data, and is a cleaner regexp
than the original.
The clearer the problem, the clearer the answer. :)
Ah, it's exactly what I want! I thought the left and right sides of
"|" are equal, but it is not true. I think I must sleep right now,
lacking of sleep makes me a dull :-p. Thank you and Simon for your
kindly help!
"Tim Chase" <py*********@tim.thechases.comwrote in message
news:ma***************************************@pyt hon.org...
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
:) Cute! (but how come no pyparsing on your machine?)
Ok, I confess I looked at the pyparsing list parser to see how it compares.
Pyparsing's examples include a list parser that comprehends nested lists
within lists, but this is a bit different, and really more straightforward.
Here's my test program for this modified case:
wrd = Word(alphas)
parenList = Combine( Optional(wrd) + "(" + SkipTo(")") + ")" +
Optional(wrd) )
brackList = Combine( Optional(wrd) + "[" + SkipTo("]") + "]" +
Optional(wrd) )
listExpr = ZeroOrMore( parenList | brackList | wrd )
txt = "a (b c) d [e f g] h i(j k) l [m n o]p q"
print listExpr.parseString(txt)
Gives:
['a', '(b c)', 'd', '[e f g]', 'h', 'i(j k)', 'l', '[m n o]p', 'q']
Comparitive timing of pyparsing vs. re comes in at about 2ms for pyparsing,
vs. 0.13 for re's, so about 15x faster for re's. If psyco is used (and we
skip the first call, which incurs all the compiling overhead), the speed
difference drops to about 7-10x. I did try compiling the re, but this
didn't appear to make any difference - probably user error.
Since the OP indicates a concern for speed (he must be compiling a lot of
strings, I guess), it would be tough to recommend pyparsing - especially in
the face of a working re that so neatly does the trick. But if at some
point it became necessary to add support for {}'s and <>'s, or quoted
strings, I'd rather be working with a pyparsing grammar than that crazy re
gibberish!
-- Paul
Ah, I had just made the same change!
from pyparsing import *
wrd = Word(alphas)
parenList = "(" + SkipTo(")") + ")"
brackList = "[" + SkipTo("]") + "]"
listExpr = ZeroOrMore( Combine( OneOrMore( parenList | brackList | wrd ) ) )
t = "a (b c) d [e f g] h i(j k) l [m n o]p q r[s] (t u)v(w) (x)(y)z"
print listExpr.parseString(t)
Gives:
['a', '(b c)', 'd', '[e f g]', 'h', 'i(j k)', 'l', '[m n o]p', 'q', 'r[s]',
'(t u)v(w)', '(x)(y)z']
>>r = re.compile(r'(?:\([^\)]*\)|\[[^\]]*\]|\S)+')
>>r.findall(s)
['(a c)b(c d)', 'e']
Ah, it's exactly what I want! I thought the left and right
sides of "|" are equal, but it is not true.
In theory, they *should* be equal. I was baffled by the nonparity
of the situation. You *should" be able to swap the two sides of
the "|" and have it treated the same. Yet, when I tried it with
the above regexp, putting the \S first, it seemed to choke and
give different results. I'd love to know why.
Thank you and Simon for your kindly help!
My pleasure. A nice diversion from swatting spammers and getting
our network back up and running today. I had hoped to actually
get something productive done (i.e. writing some python code)
rather than putting out fires. Sigh.
-tkc
Paul McGuire wrote:
Comparitive timing of pyparsing vs. re comes in at about 2ms for pyparsing,
vs. 0.13 for re's, so about 15x faster for re's. If psyco is used (and we
skip the first call, which incurs all the compiling overhead), the speed
difference drops to about 7-10x. I did try compiling the re, but this
didn't appear to make any difference - probably user error.
That is because of how the methods in the sre module are implemented...
Compiling a regex really just saves you a dictionary lookup.
def findall(pattern, string, flags=0):
"""snip"""
return _compile(pattern, flags).findall(string)
def compile(pattern, flags=0):
"""snip"""
return _compile(pattern, flags)
def _compile(*key):
# internal: compile pattern
cachekey = (type(key[0]),) + key
p = _cache.get(cachekey)
if p is not None:
return p
#snip
--
- Justin
"Tim Chase" <py*********@tim.thechases.comwrote in message
news:ma***************************************@pyt hon.org...
>r = re.compile(r'(?:\([^\)]*\)|\[[^\]]*\]|\S)+')
r.findall(s)
['(a c)b(c d)', 'e']
Ah, it's exactly what I want! I thought the left and right
sides of "|" are equal, but it is not true.
In theory, they *should* be equal. I was baffled by the nonparity
of the situation. You *should" be able to swap the two sides of
the "|" and have it treated the same. Yet, when I tried it with
the above regexp, putting the \S first, it seemed to choke and
give different results. I'd love to know why.
Does the re do left-to-right matching? If so, then the \S will eat the
opening parens/brackets, and never get into the other alternative patterns.
\S is the most "matchable" pattern, so if it comes ahead of the other
alternatives, then it will always be the one matched. My guess is that if
you put \S first, you will only get the contiguous character groups,
regardless of ()'s and []'s. The expression might as well just be \S+.
Or I could be completely wrong...
-- Paul This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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