Hello !
I am trying to understand pyparsing. Here is a little test program
to check Optional subclass:
from pyparsing import Word,nums,Liter al,Optional
lbrack=Literal( "[").suppress ()
rbrack=Literal( "]").suppress ()
ddot=Literal(": ").suppress ()
start = Word(nums+".")
step = Word(nums+".")
end = Word(nums+".")
sequence=lbrack +start+Optional (ddot+step)+ddo t+end+rbrack
tokens = sequence.parseS tring("[0:0.1:1]")
print tokens
tokens1 = sequence.parseS tring("[1:2]")
print tokens1
It works on tokens, but the error message is showed on
the second string ("[1:2]"). I don't get it. I did use
Optional for ddot and step so I guess they are optional.
Any hints what I am doing wrong?
The versions are pyparsing 1.1.2 and Python 2.3.3.
Thanks,
B. 4 2216
On Thu, 13 May 2004 08:05:32 +0200, bo***********@m f.uni-lj.si
(Bo¹tjan Jerko) wrote: Hello !
I am trying to understand pyparsing. Here is a little test program to check Optional subclass:
from pyparsing import Word,nums,Liter al,Optional
lbrack=Literal ("[").suppress () rbrack=Literal ("]").suppress () ddot=Literal(" :").suppress () start = Word(nums+".") step = Word(nums+".") end = Word(nums+".")
sequence=lbrac k+start+Optiona l(ddot+step)+dd ot+end+rbrack
tokens = sequence.parseS tring("[0:0.1:1]") print tokens
tokens1 = sequence.parseS tring("[1:2]") print tokens1
It works on tokens, but the error message is showed on the second string ("[1:2]"). I don't get it. I did use Optional for ddot and step so I guess they are optional.
Any hints what I am doing wrong?
The versions are pyparsing 1.1.2 and Python 2.3.3.
Thanks,
B.
I don't see anything "obviously" wrong to me, but changing it thusly
seems to resolve the problem (I added a few intermediate rules to
make it more obvious):
pref = lbrack + start
midf = ddot + step
suff = ddot + end + rbrack
sequence = pref + midf + suff | pref + suff
I've run into "this kind of thing" now and again, and have always
been able to resolve it by reorganizing my rules.
--dang
"Bo¹tjan Jerko" <bo***********@ mf.uni-lj.si> wrote in message
news:87******** ****@bostjan-pc.mf.uni-lj.si... Hello !
I am trying to understand pyparsing. Here is a little test program to check Optional subclass:
from pyparsing import Word,nums,Liter al,Optional
lbrack=Literal( "[").suppress () rbrack=Literal( "]").suppress () ddot=Literal(": ").suppress () start = Word(nums+".") step = Word(nums+".") end = Word(nums+".")
sequence=lbrack +start+Optional (ddot+step)+ddo t+end+rbrack
tokens = sequence.parseS tring("[0:0.1:1]") print tokens
tokens1 = sequence.parseS tring("[1:2]") print tokens1
It works on tokens, but the error message is showed on the second string ("[1:2]"). I don't get it. I did use Optional for ddot and step so I guess they are optional.
Any hints what I am doing wrong?
The versions are pyparsing 1.1.2 and Python 2.3.3.
Thanks,
B.
Bostjan -
Here's how pyparsing is processing your input strings:
[0:0.1:1]
[ = lbrack
0 = start
:0.1 = ddot + step (Optional match)
: = ddot
1 = end
] = rbrack
[1:2]
[ = lbrack
1 = start
:2 = ddot + step (Optional match)
] = oops! expected ddot -> failure
Dang Griffith proposed one alternative construct, here's another, perhaps
more explicit:
lbrack + ( ( ddot + step + ddot + end ) | (ddot + end) ) + rbrack
Note that the order of the inner construct is important, so as to not match
ddot+end before trying ddot+step+ddot+ end; '|' is a greedy matching
operator, creating a MatchFirst object from pyparsing's class library. You
could avoid this confusion by using '^', which generates an Or object:
lbrack + ( (ddot + end) ^ ( ddot + step + ddot + end ) ) + rbrack
This will evaluate both subconstructs, and choose the longer of the two.
Or you can use another pyparsing helper, the delimited list
lbrack + delimitedlist( Word(nums+"."), delim=":") + rbrack
This implicitly suppresses delimiters, so that all you will get back are
["1","0.1"," 1"] in the first case and ["1","2"] in the second.
Happy pyparsing!
-- Paul
> Dang Griffith proposed one alternative construct, here's another, perhaps more explicit: lbrack + ( ( ddot + step + ddot + end ) | (ddot + end) ) + rbrack
should be:
lbrack + start + ( ( ddot + step + ddot + end ) | (ddot + end) ) +
rbrack
Note that the order of the inner construct is important, so as to not
match ddot+end before trying ddot+step+ddot+ end; '|' is a greedy matching operator, creating a MatchFirst object from pyparsing's class library.
You could avoid this confusion by using '^', which generates an Or object: lbrack + ( (ddot + end) ^ ( ddot + step + ddot + end ) ) + rbrack
should be:
lbrack + start + ( (ddot + end) ^ ( ddot + step + ddot + end ) ) +
rbrack
This will evaluate both subconstructs, and choose the longer of the two.
Or you can use another pyparsing helper, the delimited list lbrack + delimitedlist( Word(nums+"."), delim=":") + rbrack
at least this one is correct! No, wait, I mis-cased delimitedList!
should be:
lbrack + delimitedList( Word(nums+"."), delim=":") + rbrack
This implicitly suppresses delimiters, so that all you will get back are ["1","0.1"," 1"] in the first case and ["1","2"] in the second.
Happy pyparsing! -- Paul
Sorry for the sloppiness,
-- Paul
Paul,
thanks for the explanation.
Bo¹tjan
On Fri, 14 May 2004, pt***@austin.rr ._bogus_.com spake: Dang Griffith proposed one alternative construct, here's another, perhaps more explicit: lbrack + ( ( ddot + step + ddot + end ) | (ddot + end) ) + rbrack
should be: lbrack + start + ( ( ddot + step + ddot + end ) | (ddot + end) ) + rbrack
Note that the order of the inner construct is important, so as to not match ddot+end before trying ddot+step+ddot+ end; '|' is a greedy matching operator, creating a MatchFirst object from pyparsing's class library. You could avoid this confusion by using '^', which generates an Or object: lbrack + ( (ddot + end) ^ ( ddot + step + ddot + end ) ) + rbrack
should be: lbrack + start + ( (ddot + end) ^ ( ddot + step + ddot + end ) ) + rbrack
This will evaluate both subconstructs, and choose the longer of the two.
Or you can use another pyparsing helper, the delimited list lbrack + delimitedlist( Word(nums+"."), delim=":") + rbrack
at least this one is correct! No, wait, I mis-cased delimitedList! should be: lbrack + delimitedList( Word(nums+"."), delim=":") + rbrack
This implicitly suppresses delimiters, so that all you will get back are ["1","0.1"," 1"] in the first case and ["1","2"] in the second.
Happy pyparsing! -- Paul
Sorry for the sloppiness, -- Paul This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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