Re: False positives in editing data

On Nov 26, 8:36 pm, Raul Parolari <[e-mail]> wrote:
> RichardOnRails wrote:
>
> > I forgot to tell you that I finally understand your second example.
>
> >> md =3D s.match( /^ (.*) [a-zA-Z] /x )
> >> md[1]  # =3D> "2.1Topi"
>
> > Without the question mark, in principal, the ".* initially consumes
> > all the characters,  but then it sees the match fails, because there's
> > no match for the "[a-zA-Z]".  So the ".*" sort of "backs off" and
> > satisfies it self with "2.1Topi", leaving the "c" to satisfy "[a-zA-
> > Z]".
>
> Very good, Richard!
> It is a question on when one is 'content'; if you need a metaphor to
> remember it, think of a WallStreet banker (.*, .+) vs a Franciscan monk
> (.*?, .+?)
> :-)
>
> > The one I like settled on is:
>
> > s=3D"2.1Topic 2.1"
> > md =3D s.match( /^ ([\.\d]*) [^\.\d] /x )
> >   #md[0]=3D2.1T
> >   #md[1]=3D2.1
>
> I see that you have solved the problem in your previous post (that I
> could not reply to), when you wrote (removing all other code):
>
> s =3D "2.002.1Topic 2.2.1"
>
> s =3D~ /^ (\d+[.]?)+  [^\.\d] /x
>
> I must confess: I was stunned myself that it did not work; foolish of
> us, in fact it was working, but you failed to collect the bounty! you
> needed parenthesis to include the '+'!
>
> s =3D~ /^ ((\d+[.]?)+)  [^\.\d] /x
>
> p $1, 2 # =3D> "2.002.1", "1"
>
> However it is better to avoid collecting also the inner results as they
> overwrite each other in $2 and then confuse us (that's the reason that
> you saw the last digit captured above..); so let's use the '?:' trick,
> to avoid writing in $2, where instead we will capture the 'non
> digits/dots' that come after:
>
> s =3D~ /^ ((?:\d+[.]?)*)  ([^\.\d]+) /x
>
> p $1, $2  # =3D> "2.002.1", "Topic "
>
> Do you see it? I think you do. Now, to finish, let's examine how you
> solved the problem in this post:
>
> > s=3D"2.1Topic 2.1"
> > md =3D s.match( /^ ([.\d]*) [^\.\d] /x )
>
> Ah, you resorted to 'pragmatism'.. you said: "the bloody '\d+[.]?)+'
> does not work, so I will change it". This was ok, but do you see the
> difference between:
>
>    ((?:\d+[.]?)*)   # I changed + -> * to compare
>
>    ([\d[.]]*)
>
> aside that the second one is easier to read? (you may want to stop
> reading and think about this as this is your test to graduate from
> "intermediate level regexp" :-)
>
> Ok: if they could speak, they would say respectively:
> 1) I want 0 or more sequences of (digits followed optionally by a dot)
> 2) I want 0 or more combinations of digits and dots as they come
>
> Do you see?
> both would match: "2.002.1" but the 2nd would also match "...1..37"!
>
> The last question you had was: how do I pick up the digits once I
> collected the "2.002.1"? Study scan in Pickaxe and then do:
>
> str =3D "2.002.1"
>
> str.scan(/ (\d+) /x) # =3D> [["2"], ["002"], ["1"]]
>
> All right, let's call it a Regexp day,
>
> Raul
>
> --
> Posted viahttp://www.ruby-forum.com/.

Hi Raul,

Thank you for your further  support of my obstinacy =83=BA Your help has
guided me to the solution I wanted.  Your original one is succinct,
perhaps even elegant in that it decomposes the problem into two sub-
problems which admit of essentially one-line solutions.  While I truly
appreciate that approach,  I wanted to find a "natural" solution,
which is the one included below.  It has one caveat:  it's aimed at
processing files of only a few megabytes.  That said,  I'd be pleased
to hear of any downsides you may foresee.

> > I forgot to tell you that I finally understand your second example.
>  [snip]
> Very good, Richard!
> It is a question on when one is 'content'; if you need a metaphor ...

Thanks.  I've got that stuff wired into brain now.

> I must confess: I was stunned myself that it did not work; foolish of
> us, in fact it was working, but you failed to collect the bounty! you
> needed parenthesis to include the '+'!

That approach is old news, now that I've conceived of my "natural"
approach

> However it is better to avoid collecting also the inner results as they
> overwrite each other in $2 and then confuse us

Understood!  As you'll see,  I avoided that pitfall below.

[snip]
> Do you see it? I think you do.
Quit so.

[snip]
This was ok, but do you see the
> difference between:
>
>    ((?:\d+[.]?)*)   # I changed + -> * to compare
>
>    ([\d[.]]*)
[snip]
> Do you see?
For sure!

> All right, let's call it a Regexp day,

I'll drink to that!

With Thanks and Best Wishes, I remain
Yours truly,
Richard

# "Natural" Solution
input =3D <<DATA
05Topic 05
1.0Topic 1.0
2.002.1Topic 2.2.1
3.15.26.37Topic 3.15.26.37
DATA

MaxDepth =3D 5
sRE =3D "^"
(1..MaxDepth).each { |i|
	sRE << '  (\d*)(?:\.?)'
}
sRE +=3D '  ([^\.\d].*)'
re =3D Regexp.new(sRE, Regexp::EXTENDED)

input.each { |line|
	puts '=3D'*10
	puts line
	puts '=3D'*10

	# puts re.to_s	# Debug
	md =3D line.match( re )
	(0..MaxDepth+1).each { |i|
		puts "md[#{i}] =3D " + md[i]  if md[i]
	}
	puts
}