Interesting. I didn't know about ≡. That's helpful, but the issue you raise with 'a' 'b' leaves us back to where I started. There should be a simple way for me to determine whether to do ⊂ or not. I just haven't figured out how.
I read that there was a big debate about what ⊂ does on a scalar. I suppose this is part of what I am running up against. I'm sure the problem can be solved with a bunch of loops and branches, but I was hoping there is a clean APL way. Thanks! Blake On Fri, May 9, 2014 at 10:05 AM, Elias Mårtenson <loke...@gmail.com> wrote: > 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 >>>>> >>>>> >>>>> >>>> >>> >> >
