How would you do that? The problem here is that you as a person see 'foo' and think of it as a string in this context and that you want to match on it. However, the ∊ function sees an array of elements (those elements being characters) so it acts upon them.
I suppose the best you can do would be to conditionally enclose the array if the depth is equal to 1. Something like: →((≡X)≠1)/skip X←⊂X skip: But I find that to be quite horrible for various reasons. One of those reasons being the fact that it would fail if you try to pass it the two strings: 'a' 'b'. This, of course, because 'a' is not a string at all, but a character. I certainly am not overly happy about the fact that a single-character string is in fact a character. I would find APL to be much more logical if 'a' was equivalent to ,'a'. But, it's not, so we'd better embrace it. :-) Regards, Elias On 9 May 2014 22:54, Blake McBride <blake1...@gmail.com> wrote: > Thanks for the tutorial. In my case, however, I already knew that. More > specifically, my problem is: > > x←'hello' 'there' 'how' 'are' 'you' > > ∇r←test v > r←(⊂v)∊x > ∇ > > This works for: > > test 'there' > > But doesn't work for: > > test 'there' 'are' > > I know I can remove the ⊂ for the second case. I need a single line that > works in both cases, or a test for v so I can branch to the correct code. > > Thanks! > > Blake > > > > On Fri, May 9, 2014 at 9:19 AM, Elias Mårtenson <loke...@gmail.com> wrote: > >> I was very confused about this when I started learning APL too (well >> documented in this very mailing list's archive). >> >> What happens can be illustrated by boxing the output. Let's look at a >> string: >> >> * 8⎕CR 'foo'* >> ┌→──┐ >> │foo│ >> └───┘ >> * ⍴'foo'* >> 3 >> >> In order words,this is a three-element array of characters. The third >> element of this array is the character 'o'. >> >> Here's another example: >> >> * 8⎕CR 'foo' 'bar' 'testing'* >> ┌→────────────────────┐ >> │┌→──┐ ┌→──┐ ┌→──────┐│ >> ││foo│ │bar│ │testing││ >> │└───┘ └───┘ └───────┘│ >> └∊────────────────────┘ >> * ⍴'foo' 'bar' 'testing'* >> 3 >> >> This is a three-element array of arrays. The subarrays themselves has the >> dimensions 3, 3 and 7 respectively. The third element is the string >> 'testing'. >> >> Since the ∊ function matches each individual element, when passed the >> string 'foo' it will match each individual one. I.e, the characters 'f', >> 'o' and 'o'. >> >> Finally, let's look at what the enclose function does: >> >> * 8⎕CR ⊂'foo'* >> ┌─────┐ >> │┌→──┐│ >> ││foo││ >> │└───┘│ >> └∊────┘ >> * 8⎕CR ⍴⊂'foo'* >> ┌⊖┐ >> │0│ >> └─┘ >> >> What the function does is to encapsulate the argument into a single >> scalar. A scalar has no dimensions, as seen by the fact that ⍴ returned ⍬. >> >> Of course, you could also put the string inside a single-element array. >> Such array can be constructed as such: >> >> * 8⎕CR ,⊂'foo'* >> ┌→────┐ >> │┌→──┐│ >> ││foo││ >> │└───┘│ >> └∊────┘ >> * ⍴,⊂'foo'* >> 1 >> >> The ∊ function will give identical results for that, since it interprets >> a scalar and a single-element array the same. >> >> I'm a beginner myself, so perhaps I made a mistake in my explanation. >> I'll leave it to others to fill in any information I have missed. >> >> Regards, >> Elias >> >> >> On 9 May 2014 20:07, Blake McBride <blake1...@gmail.com> wrote: >> >>> Greetings, >>> >>> >>> On Fri, May 9, 2014 at 12:10 AM, Daniel H. Leidisch >>> <li...@leidisch.net>wrote: >>> >>>> Hello! >>>> >>>> Blake McBride <blake1...@gmail.com> >>>> writes: >>>> >>>> > x←'abcd' 'efg' 'hijkl' >>>> > >>>> > Now, if I have: >>>> > >>>> > y←'hijkl' >>>> > >>>> > z←'hhh' >>>> > >>>> > How can I tell if y is in x? How can I tell if z is in x? >>>> >>>> Or for both at once: >>>> >>>> (y z)∊x >>>> 1 0 >>>> >>> >>> I like this best, except: >>> >>> u←'abcd' >>> g←'ghjk' 'dsaw' >>> >>> g∊x works. but I have to do: >>> >>> (⊂u)∊x >>> >>> but: >>> >>> u∊x doesn't work. >>> >>> the left side is being past as an argument to a function. I don't know >>> if it is going to be a string array or a general array of strings. I need >>> a way to work in either case. (Sorry for the stupid questions. I'm just >>> not straight with APL2 yest.) >>> >>> Thanks. >>> >>> Blake >>> >>> >>> >>> >>> >>>> >>>> >>>> Regards, >>>> >>>> Daniel >>>> >>>> >>>> >>> >> >
