On Apr 20, 2012, at 6:57 AM, Adrian Duşa wrote:
One solution among many:
seq(970, 970 - 30*14, by=-30)
If you want to be fancy, a function can be easily written:
seqf <- function(startp, lengthv, byv) {
seq(startp, startp + (lengthv - 1)*byv, by=byv)
}
seqf(970, 15, -30)
[1] 970 940 910 880 850 820 790 760 730 700 670 640 610 580 550
A slightly more verbose but potentially more generalizable solution:
> Reduce("-", rep(30, 15), init=1000, accumulate=TRUE)
[1] 1000 970 940 910 880 850 820 790 760 730 700 670
640 610 580 550
An application to construct a geometrically declining sequence of
specific length:
> Reduce("*", rep(.75, 15), init=1000, accumulate=TRUE)
[1] 1000.00000 750.00000 562.50000 421.87500 316.40625
237.30469 177.97852 133.48389
[9] 100.11292 75.08469 56.31351 42.23514 31.67635
23.75726 17.81795 13.36346
--
David.
Hth,
Adrian
On Fri, Apr 20, 2012 at 13:45, <r-help-boun...@r-project.org> wrote:
---------- Forwarded message ----------
From: uday <uday_143...@hotmail.com>
To: r-help@r-project.org
Cc:
Date: Fri, 20 Apr 2012 02:14:03 -0700 (PDT)
Subject: vector subtraction
I would like to calculate vector from existing value
e.g
v <- 1000
s <- 30
d1 <- v-s
d1 <- 970
d2 <- d1 -s
d2 <- 940
d 3 <- d2-s
d3 <- 910
:
:
d15 <- .....
so how I should get vector of length 15 d < - 970,940 ,
910 ,
.......
--
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--
Adrian Dusa
Romanian Social Data Archive
1, Schitu Magureanu Bd.
050025 Bucharest sector 5
Romania
Tel.:+40 21 3126618 \
+40 21 3120210 / int.101
Fax: +40 21 3158391
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David Winsemius, MD
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