Message-ID: <Pine.A41.4.44.0306060928150.82226-100000@homer33.u.washington.edu>
Date: 2003-06-06T17:06:55Z
From: Thomas Lumley
Subject: little manipulation on data frame
In-Reply-To: <20030606154100.63292.qmail@web41313.mail.yahoo.com>
On Fri, 6 Jun 2003, N Dey wrote:
> Dear all,
>
> I have data like 3 coulmns and many rows. Each entry
> is less than 10.
>
> Example
> x y z
> 1 5 3 2
> 2 3 7 8
> 3 8 9 5
> 4 5 4 6
> --------------------------
> ---------------------------
>
> I have to sum entries of each coulmn (seperately) till
> it be 10. This i have to start for each row. And I
> want to assign no. of rows needed including that row
> too(it to be 10 or 10+, the moment it exceeds 10, i
> need to stop and count the no. of rows)in additional
> coulmns say N1 (corresponding to coulmn x), N2 (y) and
> N3 (z).
>
>
> I want my new table like
>
> x y z N1 N2 N3
> 1 5 3 2 3 2 2
> 2 3 7 8 2 2 2
> 3 8 9 5 2 2 2
> 4 5 4 6 depends upon next row
>
It depends a bit on how many is `many'.
You can get cumulative sums with cumsum, and the first entry in each
column is then
min(which(cumsum(x) > 10))
The i+1th entry is
min(which( cumsum(x)-cumsum(x)[-(1:i)] > 10))
If the number of rows is not very large I would do
sumx<-cumsum(x)
N1<-min(which(cumsum(x) > 10))
N1<-c(N1, sapply(1:(length(x)-1), function(i)
min(which(sumx[-(1:i)]-sumx[i]>10))))
or the equivalent for() loop.
If the number of rows is very large it would be more efficient to rely on
the fact that no more than 10 rows are needed (assuming that zeros aren't
possible)
n<-length(x)
sumx<-cumsum(x)
sumlags<-matrix(nrow=n,ncol=10)
for(i in 0:9)
sumlags[,i+1]<-sumx[ c((i+1):n, rep(n,i))]
N1<-rowSums(sumlags < c(0,sumx[1:(n-1)])+10)+1
-thomas