[Bug-apl] West way to remove substrings?

[Elias Mårtenson]

In a discussion on the #lisp IRC channel recently, someone wanted to remove
a certain substring from a larger string. In Lisp, the solution is to
simply use cl-ppcre and remove by regex.

But, I started thinking about how to do this in APL, and this is what I
came up with:

RemoveSubstrings ← {,/ ⍵ ⊂⍹ ↑~√/ (-⍳⍎⍺) °.⌜ ⊂⍺⍷⍵}

Is this a good way of solving the problem? I realise that I have a bug when
the resulting string is empty:

* "foo" RemoveSubstrings "foo"*
DOMAIN ERROR
λ1[1] λ←,/⍵⊂⍚↑∌√/(-⍳⍎⍺)°.⌜⊂⍺⋞⍵
^ ^

Does anyone have a better solution?

Regards,
Elias

[Jay Foad]

This is a bit simpler and fixes the empty result problem, and returns a
non-enclosed result:

RemoveSubstrings ← {⍵/⍚ ~⊃∹/ (-⍳⍎⍺) ∘.⌜ ⊂⍺⍷⍵}

Like your solution this only works with ⎕IO←0.

Post by Elias MÃ¥rtenson
In a discussion on the #lisp IRC channel recently, someone wanted to
remove a certain substring from a larger string. In Lisp, the solution is
to simply use cl-ppcre and remove by regex.
But, I started thinking about how to do this in APL, and this is what I
RemoveSubstrings ← {,/ ⍵ ⊂⍹ ↑~√/ (-⍳⍎⍺) °.⌜ ⊂⍺⍷⍵}
Is this a good way of solving the problem? I realise that I have a bug
* "foo" RemoveSubstrings "foo"*
DOMAIN ERROR
λ1[1] λ←,/⍵⊂⍚↑∌√/(-⍳⍎⍺)°.⌜⊂⍺⋞⍵
^ ^
Does anyone have a better solution?
Regards,
Elias

[mwgamera]

How about
{⊃,/(¯1+↑⍴⍺)↓¨(~⍺⍷⍵)⊂⍵←⍺,⍵}
It doesn't rely on ⎕IO setting.

-k

Post by Elias MÃ¥rtenson
In a discussion on the #lisp IRC channel recently, someone wanted to remove
a certain substring from a larger string. In Lisp, the solution is to simply
use cl-ppcre and remove by regex.
But, I started thinking about how to do this in APL, and this is what I came
RemoveSubstrings ← {,/ ⍵ ⊂⍨ ↑~∨/ (-⍳⍴⍺) °.⌽ ⊂⍺⍷⍵}
Is this a good way of solving the problem? I realise that I have a bug when
"foo" RemoveSubstrings "foo"
DOMAIN ERROR
λ1[1] λ←,/⍵⊂⍨↑∼∨/(-⍳⍴⍺)°.⌽⊂⍺⋸⍵
^ ^
Does anyone have a better solution?
Regards,
Elias

[e***@gmx.com]

something i had from a while back


#! rmsubstr

∇rs←substr rmsubstr string ⍝ " " quotes required for all strings and substrings

⍝ test variables - names for ease in testing
s1←string1←"abcdefghijkcdecef"
s2←string2←"abcdefghijkcdecefa"
s3←string3←"a" ⍝ single element string
ss1←substr1←"cde" ⍝ present in string 2x returns string with substring removed
ss2←substr2←"cee" ⍝ substring not there returns full string
ss3←substr3←"efa" ⍝ substring there in s2/string2 but not s1/string1
ss4←substr4←"a" ⍝ single element substr
ss4←substr5←"" ⍝ substring not there returns full string (special test for 0=⍴,substring)

r←(Rs←⍴string)-¯1+Rss←⍴substr

⍎((0=Rs)∨(0=Rss))/'''either string and/or substring are not given'' ◊ rs←string ◊ →0'

⍎(Rss>Rs)/'''substring is bigger then string'' ◊ rs←string ◊ →0'
⍝ 'r : ', r, ' Rs : ', Rs, ' Rss : ', Rss

ssa←((r,Rss)⍴substr)
⍝ 'ssa : '
⍝ ssa

sa←(string[((r,Rss)⍴¯1+⍳Rss)+⍉(Rss,r)⍴⍳r])
⍝ 'sa : '
⍝ sa

'string : ', string ⍝ comment for use as working function
'substring to remove from string : ', substr ⍝ comment for use as working function

di←(Rss=+/ssa=sa)/⍳r
⍝ di←(Rss=ssa+.=sa)/⍳r ⍝ inner product fails here but should work ?

⍎(0=⍴di)/'rs←string ◊ →0' ⍝ substring not found - intact string returned

o←Rs⍴1
o[,(((⍴di), Rss)⍴¯1+⍳Rss)+⍉(Rss,⍴di)⍴di]←0

rs←o/string

∇




On Tue, 14 Mar 2017 22:20:42 +0800

Post by Elias MÃ¥rtenson
In a discussion on the #lisp IRC channel recently, someone wanted to remove
a certain substring from a larger string. In Lisp, the solution is to
simply use cl-ppcre and remove by regex.
But, I started thinking about how to do this in APL, and this is what I
RemoveSubstrings ← {,/ ⍵ ⊂⍨ ↑~∨/ (-⍳⍴⍺) °.⌽ ⊂⍺⍷⍵}
Is this a good way of solving the problem? I realise that I have a bug when
* "foo" RemoveSubstrings "foo"*
DOMAIN ERROR
λ1[1] λ←,/⍵⊂⍨↑∼∨/(-⍳⍴⍺)°.⌽⊂⍺⋸⍵
^ ^
Does anyone have a better solution?
Regards,
Elias