On Tue, May 21, 2013 at 5:10 AM, JNV <[email protected]> wrote:
Hi there,
I've got this matrix D with, say 10 rows and 20 columns. For each
row I want
to sum the first 3 non zero elements and put them in a vector z.
So if the first row D[1,] is
0 3 5 0 8 9 3 2 4 0
then I want z
z<-D[1,2]+D[1,3]+D[1,5]
But if there are less than 3 non zero elements, those should be
summed. If
there are no non zero elements, the result must be zero.
So if the first row D[1,] is
0 0 3 0 1 0 0 0 0 0
then I want z
z<-D[1,3]+D[1,5]
Here is a matrix, D, with those two rows The t(apply(...)) replaces
the first non-zero element in each row with 1, the 2nd with 2, etc.
(It puts garbage into the elements that are 0.) We then convert
this to T/F according to whether each element less than or equal to 3
or not and multiply by the original data which both zaps the garbage
in the zero positions and zaps those positions which are a 4th or
more
non-zero in each row. This multiplication also inserts the correct
values into the good positions. Finally we sum the rows using what
is
left:
D <- matrix( c(0, 0, 3, 0, 5, 3, 0, 0, 8, 1, 9, 0, 3,
+ 0, 2, 0, 4, 0, 0, 0), 2)
D
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 3 5 0 8 9 3 2 4 0
[2,] 0 0 3 0 1 0 0 0 0 0
as.data.table(D)[, rowSums((t(apply(.SD > 0, 1, cumsum)) <= 3) *
.SD)]
[1] 16 4
Not sure if this really benefits from data.table as we could have
written this without data.table:
rowSums((t(apply(D > 0, 1, cumsum)) <= 3) * D)
[1] 16 4
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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