didn't look totally at what you are trying to do, but
(1+(i.4)),(1.25+(i.4))
1 2 3 4 1.25 2.25 3.25 4.25
5 ({: - {.)\ 1 2 3 4 1.25 2.25 3.25 4.25
0.25 0.25 0.25 0.25
\ used dyadically using overlapping groups of length 5 in this case.
----- Original Message -----
From: Joe Bogner <joebog...@gmail.com>
To: programm...@jsoftware.com
Cc:
Sent: Friday, February 21, 2014 5:58:36 PM
Subject: [Jprogramming] simple time series (lag)
is there a better way to do this? Calculate change over the 4th or Nth
previous cell?
d=.(1+(i.4)),(1.25+(i.4))
d %~ ((# d) {. _4 {. d)
1.25 1.125 1.08333 1.0625 0 0 0 0
Here's how it's done in R
> d<-ts(c(1:4, c(1:4)+0.25))
> d/lag(d,k=-4)
Time Series:
Start = 5
End = 8
Frequency = 1
[1] 1.250000 1.125000 1.083333 1.062500
>
The parameter isn't very intuitive:
d %~ ((# d) {. _6 {. d)
3 2 0.416667 0.5625 2.6 1.88889 0 0
> ts(c(1:4, c(1:4)+0.25)) / lag(ts(1:4),k=-2)
Time Series:
Start = 3
End = 6
Frequency = 1
[1] 3.0000000 2.0000000 0.4166667 0.5625000
If this is the right way, I'd probably define a dyad, or if I knew better
an conjunction(?)
lag=: 4 : 'y %~ ((# y) {. (_1*( x -~ # y)) {. y)'
2 lag d
3 2 0.416667 0.5625 2.6 1.88889 0 0
I naively thought I could just replace %~ with u
lag=: 2 : 'y u ((# y) {. (_1*( x -~ # y)) {. y)'
4 %~ lag d
|syntax error
| 4%~lag d
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