Help with ER

[avechuche]

Hi folks, I have a problem. I need that with regular expressions, you can replace "x^x" with "x" (where x is any letter) and also need to replace "xy" with "x" (where x, y are any letter)

For example: p^p = p, q^q = q, r^r=r, etc etc.
qq = q

I need to check that $1 and $2 are the same, in order to replace the expression for $1 or $2 (in this case is the same)

$inputExpresion = StringRegExpReplace($inputExpresion, "([a-z])^([a-z])", "??????")

and

How do I verify that the previous character is the same as that continues and so replace the two, with one.

StringRegExpReplace($inputExpresion, "([a-z])+", "?????")

 

any solution? Thanks

Edited March 13, 2014 by avechuche

[Melba23]

avechuche,

This seems to work:

$sList = "blah q^q blah" & @CRLF & _
    "blah pp blah" & @CRLF & _
    "blah a^b blah" & @CRLF & _
    "blah cd blah"

$sNewList = StringRegExpReplace($sList, "(.)(\^)?(\1)", "$1")

ConsoleWrite($sNewList & @CRLF)

SRER decode:

(.)      - Capture a character
(\^)?    - There might be an optional ^
(\1)     - Is it the same character again?

$1       - If it was, then replace the whole lot by the single character

Any use?

M23

[avechuche]

ooooooo Nice! vey vey nice. Thanks! this is the secret, did not know

(\1)     - Is it the same character again?

"1" refers to the first captured group, right?

but (there is always a but)

q^q = q its ok!!!

pp = p its ok!!!

a^b = a^b its ok!!!

cd = cd wrong, I need only the letter "c" is maintained

I'm doing a program to validate logical expressions, only that I needed to check certain things

[Melba23]

avechuche,

 

"1" refers to the first captured group, right?

Yes - look at the Help file for StringRegExp under "Backreferences and references to subroutines".

Do I understand correctly that you want any 2-character phrase reduced to the first character? If so then this should do the trick:

$sList = "blah q^q blah" & @CRLF & _
    "blah pp blah" & @CRLF & _
    "blah a^b blah" & @CRLF & _
    "blah cd blah"

$sTempList = StringRegExpReplace($sList, "(.)(\^)?(\1)", "$1")
$sNewList = StringRegExpReplace($sTempList, "\s(.)(.)\s", " $1 ")

ConsoleWrite($sNewList & @CRLF)

M23

[avechuche]

Hello, I have another problem. for example

"(A | B | C) | (D | E | F)" ===> For example

If you find A, B or C, replace by "X" and if it is D, E or F, the Replace with "Y". It is possible by a single expression? Thx!!!

[Melba23]

avechuche,

It is trivial to do it in 2 passes:

$sString = "(A | B | C) | (D | E | F)"

$sNewString = StringRegExpReplace(StringRegExpReplace($sString, "[ABC]", "X"), "[DEF]", "Y")

ConsoleWrite($sNewString & @CRLF)

Why the requirement for a single pass?

M23

P.S. And was the code in post #4 what you needed?

M23

[avechuche]

Oh yes, post #4 is correct.

 

I need it in one step, because, for example:

p ^ q | r =====> "|" operator is the logical "or" and "^" is the logical operator "and"

In expressions, except that finds parentheses are resolved from left to right (basic math ^ ^). Thus, the first operation would have to resolve "^" (p ^ q) and then the result of that, I apply the "or". In short "p ^ q = x", then "x | r"

With the example you give me the expression, solves first all "or" then all "and" and that's wrong, you have to resolve as it finds a logical operator.

This is a simple example that the result is the same, but not always so.

Edited March 15, 2014 by avechuche

[Melba23]

avechuche,

Judging from that explanation you need a proper expression parser, not a simple (or even complicated) RegEx.

M23

[jchd]

You still can parse in one pass the expressions you have shown so far by using pattern recursion, namely (?R) or (?0) or subpattern recursion. Lookup the References to subroutines paragraph in the regexp help.

If you need to deal with parenthesized sub-expressions (possibly nested) you'll have to define them as a top-level subpattern.

Edited March 17, 2014 by jchd