,. is equivalent to ,"_1 :
,'abc' ,"_1"1 (6 3 $ 'defghijklmnopqrstu')
adbecfagbhciajbkclambncoapbqcrasbtcu
Did you try two ranks?
Henry Rich
On 1/8/2018 11:26 AM, Nick S wrote:
I want to shuffle arrays, such that I pair them, a member of 1 with a
member of the next, as:
'abc' ,"_1 'def'
ad
be
cf
,'abc' ,"_1 'def'
adbecf
But what I want to do is to:
so the same sort of pairing with this as a right argument:
6 3 $ 'defghijklmnopqrstu'
Now, honestly, I had worked on this for a very long time for such a simple
thing. I had tried every variation I could think of. Then I thought about
it for a couple of days, read the language doc again, and finally decided
to ask the question here. So I was formulating the question and, of
course, this is part of another problem, the details of which are
irrelevant here, so I was trying to create an isolated example, and I typed
this expecting it to fail. I thiought I had tried this before, but I had
mostly worked with ravel, not ravel items, because the statement in the
book implied (to me, anyway) that the two verbs were equivalent except with
respect to rank.
,'abc' ,."1 (6 3 $ 'defghijklmnopqrstu')
adbecfagbhciajbkclambncoapbqcrasbtcu
Exactly what I wanted. Of course, in practice, the dimensions will be
variable but the length of the vector will always be equal to the second
dimension of the y table.
So the book says that ,. is the same as ,"_1 - but I can't figure out how
to do what I did with , (ravel) no matter what I do with the ranks. I
think this only works because of what it says in the description, that ,.
(ravel items) joins items of x to items of y.
Am I wrong about that? Is there a way to do this item by item lacing with
ravel? Or can it only be done with ravel items? If, in fact, the verbs
are not equivalent except with respect to rank, is it reasonably to reword
the statement to indicate that somehow so that others don't chase down the
blind alley? I can certainly understand that an unranked ravel items is
the same as a ravel items with the rank specified as _1.
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