Hows about this?
b=.1 0 1 0 1 0 1
x0=.i.7
x1=.7+i.7
(b*x0)+(-.b)*x1
0 8 2 10 4 12 6
Leigh J. Halliwell wrote:
Thank you, Roger and Cliff. On my first question I was indeed thinking of
the monadic amend. And Cliff's use of it for my second question is nice,
too. But I'd still like to know how to make the identity/left operator work
one-to one, as per my second question:
"2) X0 and X1 are numeric vectors, and B is a Boolean vector. They all have
the same length. I'd like to select from X0 where B is 0, and from X1 where
B is 1. I try the expression: X1 [^:B X0. But the adverb ^:B wants to get
two-dimensional, i.e., to apply each B to every pair of X1 and X2. How can
I make J to apply the adverb one-to-one with the arguments?"
Sincerely,
Leigh
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