On Jan 11, 2010, at 9:38 PM, Márcio Resende wrote:
Hi,
I have a 100x15 matrix and in each row a set of 15 random numbers
out of 25.
for example:
b <- c(1:25)
a <-matrix(0,10,15)
for (i in 1:10){
a[i,] <- sample(b,15,replace = FALSE)
}
I would like to create another matrix (25x100), for example "d" with
the
probability of each number from the first matrix
Therefore I need to track, for example, if number 1 is present in
the first
row (d[1,1]) (which would give me an probability of 1 out of 1).
Then, track again if number 1 is present on the second row (d[2,1])
(And if
not, the probability would be 1 out of 2 = 0.5)...and so on for all
the 25
collumns (25 numbers) and all the 100 rows.
Could anybody help how to do it??
Maybe:
> apply(a, 2, table)[[1]]
1 5 6 7 10 16 21 22 25
1 2 1 1 1 1 1 1 1
> apply(a, 2, table)[[1]]/10
1 5 6 7 10 16 21 22 25
0.1 0.2 0.1 0.1 0.1 0.1 0.1 0.1 0.1
> lapply(apply(a, 2, table), "/", 10)
[[1]]
1 5 6 7 10 16 21 22 25
0.1 0.2 0.1 0.1 0.1 0.1 0.1 0.1 0.1
[[2]]
1 2 7 9 12 15 16 17 25
0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.2
snipped remaining 13 vectors in the 15 element list
> str(lapply(apply(a, 2, table), "/", 10)[[1]])
table [1:9(1d)] 0.1 0.2 0.1 0.1 0.1 0.1 0.1 0.1 0.1
- attr(*, "dimnames")=List of 1
..$ : chr [1:9] "1" "5" "6" "7" ...
> as.numeric(names(lapply(apply(a, 2, table), "/", 10)[[1]]))
[1] 1 5 6 7 10 16 21 22 25
> d <- matrix(0, nrow=25, ncol=15)
> d[as.numeric(names(lapply(apply(a, 2, table), "/", 10)[[1]])), 1]
<- lapply(apply(a, 2, table), "/", 10)[[1]]
> d
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[,13] [,14] [,15]
[1,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[2,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[3,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[4,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[5,] 0.2 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[6,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[7,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[8,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[9,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[10,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[11,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[12,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[13,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[14,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[15,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[16,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[17,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[18,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[19,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[20,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[21,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[22,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[23,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[24,] 0.0 0 0 0 0 0 0 0 0 0 0
0 0 0 0
[25,] 0.1 0 0 0 0 0 0 0 0 0 0
0 0 0 0
So that "worked for one column. Too bad the R syntax does not allow
slightly more flexible handling of indices or this might have worked:
> d[as.numeric(names(lapply(apply(a, 2, table), "/", 10)[[1:15]])),
1] <- lapply(apply(a, 2, table), "/", 10)[[1:15]]
Error in lapply(apply(a, 2, table), "/", 10)[[1:15]] :
recursive indexing failed at level 2
As it is a loop does the job:
> for(i in 1:15) {d[as.numeric(names(lapply(apply(a, 2, table), "/",
10)[[i]])), i] <- lapply(apply(a, 2, table), "/", 10)[[i]]}
> d
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[,13] [,14] [,15]
[1,] 0.1 0.1 0.1 0.1 0.0 0.0 0.0 0.1 0.0 0.0 0.1
0.1 0.1 0.0 0.0
[2,] 0.1 0.1 0.1 0.0 0.1 0.0 0.0 0.0 0.0 0.0 0.1
0.0 0.0 0.0 0.0
[3,] 0.1 0.0 0.0 0.0 0.1 0.0 0.0 0.0 0.1 0.2 0.0
0.1 0.0 0.0 0.0
[4,] 0.1 0.0 0.0 0.2 0.0 0.0 0.2 0.0 0.0 0.0 0.0
0.0 0.0 0.0 0.1
[5,] 0.2 0.0 0.1 0.0 0.1 0.0 0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.0 0.3
[6,] 0.1 0.0 0.0 0.0 0.0 0.0 0.1 0.1 0.0 0.0 0.0
0.0 0.2 0.1 0.0
[7,] 0.1 0.1 0.0 0.1 0.0 0.0 0.0 0.1 0.1 0.0 0.2
0.0 0.0 0.0 0.0
[8,] 0.1 0.0 0.0 0.1 0.0 0.1 0.0 0.0 0.1 0.1 0.0
0.1 0.0 0.1 0.0
[9,] 0.1 0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.2 0.0 0.0
0.0 0.0 0.1 0.0
[10,] 0.1 0.0 0.0 0.1 0.1 0.0 0.0 0.0 0.0 0.1 0.1 0.0
0.0 0.2 0.0
[11,] 0.1 0.0 0.0 0.1 0.1 0.0 0.0 0.1 0.0 0.0 0.2 0.0
0.1 0.0 0.0
[12,] 0.2 0.1 0.1 0.0 0.1 0.0 0.0 0.1 0.0 0.2 0.0 0.0
0.0 0.0 0.0
[13,] 0.1 0.0 0.1 0.0 0.0 0.1 0.0 0.1 0.0 0.0 0.1 0.1
0.0 0.1 0.0
[14,] 0.1 0.0 0.1 0.0 0.1 0.0 0.1 0.0 0.1 0.0 0.0 0.1
0.0 0.1 0.1
[15,] 0.1 0.1 0.0 0.1 0.0 0.0 0.0 0.1 0.0 0.1 0.0 0.0
0.0 0.0 0.0
[16,] 0.1 0.1 0.0 0.1 0.0 0.2 0.0 0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.0
[17,] 0.2 0.1 0.0 0.0 0.0 0.1 0.0 0.0 0.1 0.0 0.0 0.1
0.2 0.0 0.0
[18,] 0.1 0.0 0.0 0.0 0.0 0.1 0.2 0.0 0.0 0.1 0.0 0.2
0.1 0.1 0.0
[19,] 0.1 0.0 0.1 0.0 0.1 0.1 0.2 0.1 0.1 0.0 0.1 0.0
0.0 0.0 0.0
[20,] 0.1 0.0 0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.1 0.0 0.1
0.0 0.1 0.1
[21,] 0.1 0.0 0.0 0.0 0.2 0.1 0.0 0.0 0.0 0.0 0.1 0.0
0.0 0.0 0.1
[22,] 0.1 0.0 0.1 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.1
0.2 0.0 0.0
[23,] 0.1 0.0 0.0 0.0 0.0 0.0 0.1 0.1 0.2 0.1 0.0 0.0
0.1 0.0 0.0
[24,] 0.2 0.0 0.0 0.1 0.0 0.1 0.1 0.0 0.0 0.0 0.0 0.0
0.0 0.1 0.2
[25,] 0.1 0.2 0.1 0.0 0.0 0.1 0.0 0.1 0.0 0.0 0.0 0.0
0.0 0.0 0.1
>
Thanks in advance
Marcio
--
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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