Greetings,
I wrote the following APL1 code a long time ago.
∇
[0] r←d Parse v
[1] ⍝ Convert vector v into a matrix breaking at delimiter d
[2] r←(((0≠⍴v)×⍴r),⌈/r)⍴(,r∘.≥⍳⌈/r←¯1+(r,1+⍴v)-0,r←r/⍳⍴v)\(~r←v∈d)/v←,v
∇
It worked in 1981 and it works now.
I believe at some point someone o
Hi,
Parse2 works for me. Albeit a rather simple test.
What was your data you parsed??
To explain:
r←(1↑d),(1↑d),v ⍝⍝ add 2 time the delimiter in front of vector. If d
is a vector, take the first element. If d is a matrix, fail.
⍝ normally use a different name than the return var. Just improve
On Mon, Mar 21, 2022 at 3:34 PM Hans-Peter Sorge <
hanspeterso...@netscape.net> wrote:
> Hi,
>
> Parse2 works for me. Albeit a rather simple test.
> What was your data you parsed??
>
',' Parse 'One,Two,Three,Four'
One
Two
Three
Four
',' Parse2 'One,Two,Three,Four'
One Two Three Four
