how to match any sequence of characters in ss/ssr function

Community Support
1 
Hi All,

I would like to use **\* pattern** in **ss** function ([https://code.kx.com/v2/kb/regex/]("https://code.kx.com/v2/kb/regex/")and[https://code.kx.com/v2/ref/ss/]("https://code.kx.com/v2/ref/ss/")) but i can't . it works fine for **"like"** but not for ss/ssr

please advice




"test\_p.xxx" ss "?\_p.?" / works fine  
"test\_p.xxx" like "\*\_p.\*" / works fine  
  
BUT
"test\_p.xxx" ss "\*\_p.\*" / doesn't work  
  
q)"test\_p.xxx" ss "\*\_p.\*"  
'length
[0]"test\_p.xxx" ss "\*\_p.\*"
2

This is by design most likely. Since ss gives the index/indices of where your substring begins, e.g.

q)“test_p.xxx” ss  “?_p.?”

,3

q)“test_p.xxx” ss  “??_p.?”

,2

q)“test_p.xxx” ss  “???_p.?”

,1

then it wouldn’t know what the starting index is if you supply a open-ended wildcard such as “*_p.*”

Terry

3

Hi Terry, 

then it wouldn’t know what the starting index is if you supply a open-ended wildcard such as “*_p.*”

Hm… why  not  0  for  my first example ( “test_p.xxx”   , and wild card is “*_p.*”)

or why not 4 for the following example

“test_p.xxx” ss  “_p.*”

 

This is by design most likely. Since ss gives the index/indices of where your substring begins, e.g.

q)“test_p.xxx” ss  “?_p.?”

,3

q)“test_p.xxx” ss  “??_p.?”

,2

q)“test_p.xxx” ss  “???_p.?”

,1

then it wouldn’t know what the starting index is if you supply a open-ended wildcard such as “*_p.*”

Terry

4

You can always create a custom ss function!

q)ss1:{$["*"=first y;(long$();0)x like y;““=last y;x ss ssr[y;””;“?”];x ss y]};q)q)ss1[“test_p.xxx”;“_p.”]0q)ss1[“test_p.xxx”;“_p.*”],4`

I haven’t tested this a whole lot but it might achieve what you’re looking for.

Terry

5

Yes, i know… Also i can use c++ extension  but i asked to be sure that i didn’t miss something

Thanks a lot

13:44, October 1, 2019, TerryLynch <tlyncher@gmail.com>:

You can always create a custom ss function!

q)ss1:{$["*"=first y;(long$();0)x like y;““=last y;x ss ssr[y;””;“?”];x ss y]};q)q)ss1[“test_p.xxx”;“_p.”]0q)ss1[“test_p.xxx”;“_p.*”],4`

I haven’t tested this a whole lot but it might achieve what you’re looking for.

Terry

On Monday, September 30, 2019 at 8:23:44 PM UTC+1, Igor M wrote:

Hi Terry, 

then it wouldn’t know what the starting index is if you supply a open-ended wildcard such as “*_p.*”

Hm… why  not  0  for  my first example ( “test_p.xxx”   , and wild card is “*_p.*”)

or why not 4 for the following example

“test_p.xxx” ss  “_p.*”

 

This is by design most likely. Since ss gives the index/indices of where your substring begins, e.g.

q)“test_p.xxx” ss  “?_p.?”

,3

q)“test_p.xxx” ss  “??_p.?”

,2

q)“test_p.xxx” ss  “???_p.?”

,1

then it wouldn’t know what the starting index is if you supply a open-ended wildcard such as “*_p.*”

Terry

On Sunday, September 29, 2019 at 1:24:37 PM UTC+1, Igor M wrote:

Hi All,

I would like to use * pattern in ss function (https://code.kx.com/v2/kb/regex/andhttps://code.kx.com/v2/ref/ss/) but i can’t . it works fine for “like” but not for ss/ssr

please advice

“test_p.xxx” ss “?_p.?” / works fine
“test_p.xxx” like “*_p.*” / works fine

BUT
“test_p.xxx” ss “*_p.*” / doesn’t work

q)“test_p.xxx” ss “*_p.*”
'length
[0]“test_p.xxx” ss “*_p.*”

–
Submitted via Google Groups