I am putting some values in a matrix through a binomial probability expression. The problem is that I have some NA values that I need to keep in my matrix, but rbinom() returns an error when NA is inputted.
My algorithm consists of the following:
Here is a reproducible example.
set.seed(10)
xn <- matrix(c(NA, 100, 100, 100, 100, NA, NA, 100, 100, NA), byrow=TRUE, ncol=2)
dup <- xn * 2
z <- matrix(rbinom(n=rep(1,length(dup)), size = as.vector(dup), prob = 0.5), nrow = nrow(dup))
Warning message:
In rbinom(n = rep(1, length(dup)), size = as.vector(dup), prob = 0.5) :
NAs produced
I thought about only selecting the values in the matrix that have actual values.
set.seed(6)
xn_bin <- rbinom(n=rep(1,length(dup[-which(is.na(dup))])), size = as.vector(dup[-which(is.na(dup))]),prob = 0.5)
I don't know how to get the matrix back together since I need take the results from xn_bin and put them back into the same position in a new matrix as the dup values were located inputted into rbinom().
If that does not make any sense. xn_bin will give the values: xn_bin [1] 101 115 112 98 103 103
dup will give:
dup
[,1] [,2]
[1,] NA 200
[2,] 200 200
[3,] 200 NA
[4,] NA 200
[5,] 200 NA
I want the final matrix to have the values from xn_bin and the NA values from xn:
[,1] [,2]
[1,] NA 101
[2,] 115 112
[3,] 98 NA
[4,] NA 103
[5,] 103 NA
Any idea how to do this effectively?
Notice that the order is not the same but position in the matrix structure is as desired and since this is a random process that should not be determined by positon I do not see why this result is not as valid as your ordering (R matrices being filled by column rather than by row):
xn_bin <- z
set.seed(6)
xn_bin[ !is.na(z) ] <- rbinom(n=rep(1,length(dup[-which(is.na(dup))])),
size = as.vector(dup[-which(is.na(dup))]),prob = 0.5)
xn_bin
[,1] [,2]
[1,] NaN 98
[2,] 101 103
[3,] 115 NaN
[4,] NaN 103
[5,] 112 NaN
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