It seems like this is what you want to do, although there is probably a better
way to do it.
A.DT <- data.table(a1 = A.DT[,a1],
a2=sort(ifelse(B.DT[,b2] <= N/2 & B.DT[,b1] <
A.DT[nrow(A.DT):1,a1],
B.DT[nrow(A.DT):1,b1],
A.DT[nrow(A.DT):1, a2]), na.last=FALSE))
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Matteo Richiardi
Sent: Dienstag, 19. März 2013 08:24
To: r-help@r-project.org
Subject: [R] How can I eliminate a loop over a data.table?
I've two data.tables as shown below:
***
N = 10
A.DT <- data.table(a1 = c(rnorm(N,0,1)), a2 = NA)) B.DT <- data.table(b1 =
c(rnorm(N,0,1)), b2 = 1:N)
setkey(A.DT,a1)
setkey(B.DT,b1)
***
I tried to change my previous data.frame implementation to a data.table
implementation by changing the for-loop as shown below:
***
for (i in 1:nrow(B.DT)) {
for (j in nrow(A.DT):1) {
if (B.DT[i,b2] <= N/2
&& B.DT[i,b1] < A.DT[j,a1]) {
A.DT[j,]$a2 <- B.DT[i,]$b1
break
}
}
}
***
I get the following error message:
***
Error in `[<-.data.table`(`*tmp*`, j, a2, value = -0.391987468746123) :
object "a2" not found
***
I think the way I access data.table is not quite right. I am new to it. I guess
there is a quicker way of doing it than cycling up and down the two datatables.
I'd like to know if the loop shown above could be simplified/vectorised.
The data.table data for copy/paste reads:
***
# A.DT
a1 a2
1 -1.4917779 NA
2 -1.0731161 NA
3 -0.7533091 NA
4 -0.3673273 NA
5 -0.159569 NA
6 -0.1551948 NA
7 -0.0430574 NA
8 0.1783496 NA
9 0.4276034 NA
10 1.0697412 NA
# B.DT
b1 b2
1 0.64229018 1
2 1.00527902 2
3 0.24746294 3
4 -0.50288835 4
5 0.34447791 5
6 -0.22205129 6
7 0.60099079 7
8 -0.70242284 8
9 0.6298599 9
10 0.08917988 10
***
The output I expect is:
***
# OUTPUT
a1 a2
1 -1.4917779 NA
2 -1.0731161 NA
3 -0.7533091 NA
4 -0.3673273 NA
5 -0.159569 NA
6 -0.1551948 NA
7 -0.0430574 NA
8 0.1783496 -0.50288835
9 0.4276034 0.24746294
10 1.0697412 0.64229018
***
The algorithm goes down one table, and for each row go up the other table,
check some conditions and modify values accordingly. More specifically, it goes
down B.DT, and for each row in B.DT goes up A.DT and assigns to a2 the first
value of b1 such that b1 is smaller than a1. An additional condition is checked
before assignment (b2 being equal or smaller than 5 in this example).
0.64229018 is the first value in B.DT, and it is assigned to the last unit of
A.DT.
1.00527902 is the second value in B.DT, but it is left unassigned because it is
bigger than all other values in A.DT.
0.24746294 is the third value in B.DT, and it is assigned to the second last
unit in A.DT.
-0.50288835 is the fourth value in B.DT, and it is assigned to unit #8 in A.DT
0.34447791 is the fifth value in B.DT, and it is left unassigned because it is
too big.
This is of course a simplified problem (and therefore may not make much sense).
Thanks for your time and input.
Matteo
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and provide commented, minimal, self-contained, reproducible code.
