You did not specify exactly what you mean by the inverse of cumsum.
If you want (in pseudocode)
cumsumfromright(a)(k):=(sum(a(i),i=k..length(a))
giving you the sums over the tails of the sequence, then
cumsumfromright <- function(x) rev(cumsum(rev(x)))
is probably what you want.
If you want to find the sequence which has a given sequence a as its
cumsum, you want
firstdiff <- function(x) {
shifted <- c(0,x[1:(length(x)-1)])
x-shifted
}
With this function,
cumsum(firstdiff(x)) == x and
firstdiff(cumsum(x)) == x
for every numeric vector x.
(Ted Harding) wrote:
On 18-Jun-08 10:17:09, Alfredo Alessandrini wrote:
I've a matrix like this:
1985 1.38 1.27 1.84 2.10 0.59 3.47
1986 1.05 1.13 1.21 1.54 0.21 2.14
1987 1.33 1.21 1.77 1.44 0.27 2.85
1988 1.86 1.06 2.33 2.14 0.55 1.40
1989 2.10 0.65 2.74 2.43 1.19 1.45
1990 1.55 0.00 1.59 1.94 0.99 2.14
1991 0.92 0.72 0.50 1.29 0.54 1.22
1992 2.15 1.28 1.23 2.26 1.22 3.17
1993 1.50 0.87 1.68 1.97 0.83 2.55
1994 0.69 0.00 0.76 1.89 0.60 0.87
1995 1.13 1.04 1.19 1.52 1.13 1.78
Can I utilise a cumsum inverse? from 1995 to 1985?
Thank you,
Alfredo
Try on the lines of:
cumsum(c(1,2,3,4,5))
# [1] 1 3 6 10 15
cumsum(rev(c(1,2,3,4,5)))
# [1] 5 9 12 14 15
Hoping this helps,
Ted.
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E-Mail: (Ted Harding) <[EMAIL PROTECTED]>
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Date: 18-Jun-08 Time: 11:50:40
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