[R] using lapply()
dxc13
dxc13 at health.state.ny.us
Tue Jan 8 18:11:23 CET 2008
useR's,
I am trying to find a quick way to change some values in a list that are
subject to a condition to be NA. Consider the 3x1 matrix:
delta <- matrix(c(2.5,2.5,1), nrow = 1)
And consider the list named v that has 3 elements
> v
v[[1]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[,13] [,14]
[1,] 4.25 3.25 2.25 1.25 0.25 0.75 1.75 2.75 3.75 4.25 3.25 2.25 1.25
0.25
[2,] 1.25 0.25 0.75 1.75 2.75 3.75 4.75 5.75 6.75 1.25 0.25 0.75 1.75
2.75
[3,] 7.25 6.25 5.25 4.25 3.25 2.25 1.25 0.25 0.75 7.25 6.25 5.25 4.25
3.25
[4,] 3.25 2.25 1.25 0.25 0.75 1.75 2.75 3.75 4.75 3.25 2.25 1.25 0.25
0.75
[5,] 6.25 5.25 4.25 3.25 2.25 1.25 0.25 0.75 1.75 6.25 5.25 4.25 3.25
2.25
v[[2]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[,13] [,14]
[1,] 9.25 9.25 9.25 9.25 9.25 9.25 9.25 9.25 9.25 8.25 8.25 8.25 8.25
8.25
[2,] 7.25 7.25 7.25 7.25 7.25 7.25 7.25 7.25 7.25 6.25 6.25 6.25 6.25
6.25
[3,] 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25 1.25 0.25 0.25 0.25 0.25
0.25
[4,] 4.25 4.25 4.25 4.25 4.25 4.25 4.25 4.25 4.25 3.25 3.25 3.25 3.25
3.25
[5,] 6.25 6.25 6.25 6.25 6.25 6.25 6.25 6.25 6.25 5.25 5.25 5.25 5.25
5.25
v[[3]]
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13]
[,14]
[1,] 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5
0.5
[2,] 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5 1.5
1.5
[3,] 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5
2.5
[4,] 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5 3.5
3.5
[5,] 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5 4.5
4.5
The first element of the matrix delta (2.5) corresponds to the first list
element v[[1]], and so on...
What I want to do is to set any values in this list that are greater than
its corresponding delta to be NA. For example, if any value in v[[1]] is
greater than 2.5 than set it to NA, and similarly for v[[2]]. If any value
in v[[3]] is > 1, then set it NA.
Can this be done using lapply or some other quick, efficient way? Hope this
makes sense.
Thanks,
Derek
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