Reputation: 2396
I want to create 27 matrix with 2 columns and a variable number of rows. I could just write 27 lines of code, like this:
x1 = cbind(rep(1,34), rnorm(34))
x2 = cbind(rep(1,36), rnorm(36))
....
x27 = cbind(rep(1,k), rnorm(k))
But it must have a better way to do that. I thought of a loop, something like this:
aux = c(34, 36, ..., k) # auxiliar variable with number of rows for each matrix
for (i in 1:27) paste("x",i, sep="") = cbind(rep(1,aux[i]), rnorm(aux[i]))
However, it doesn't work. I feel like this is a simple task, but I am out of ideas.
Any help?
ps.: I thought of an array, but I wasn't able to use it. Maybe a list can do the job, I don't know.
Upvotes: 3
Views: 13877
Reputation: 13123
The code below shows how I used Joshua Ulrich's approach to create slightly more complex matrices. Hopefully this answer is helpful is showing some of the flexibility possible in creating objects. If not, I can delete my answer.
I suspect this approach can be modified easily to create matrices that differ in size, for example, by setting nrow
or ncol
equal to a variable and using rep(q, z)
with some variable z
to duplicate elements in the vector inside the matrix
or rbind
statement:
p1.c1 <- 0.10
p2.c1 <- 0.20
p3.c1 <- 0.30
p4.c1 <- 0.40
s1.c1 <- matrix(c(p1.c1, p1.c1, (1 - p1.c1),
p1.c1, p1.c1, (1 - p1.c1),
0, 0, 1), nrow=3, ncol=3, byrow = TRUE)
s2.c1 <- matrix(c(p2.c1, p2.c1, (1 - p2.c1),
p2.c1, p2.c1, (1 - p2.c1),
0, 0, 1), nrow=3, ncol=3, byrow = TRUE)
s3.c1 <- matrix(c(p3.c1, p3.c1, (1 - p3.c1),
p3.c1, p3.c1, (1 - p3.c1),
0, 0, 1), nrow=3, ncol=3, byrow = TRUE)
s4.c1 <- matrix(c(p4.c1, p4.c1, (1 - p4.c1),
p4.c1, p4.c1, (1 - p4.c1),
0, 0, 1), nrow=3, ncol=3, byrow = TRUE)
n <- 5
p.c1 <- c(p1.c1, p2.c1, p3.c1, p4.c1)
for (i in 1: (n - 1)) {
assign(paste('xs', i, '.c1', sep=""), matrix(c(p.c1[i], p.c1[i], (1-p.c1[i]),
p.c1[i], p.c1[i], (1-p.c1[i]),
0, 0, 1 ), nrow=3, ncol=3, byrow = TRUE))
}
identical(xs1.c1, s1.c1)
identical(xs2.c1, s2.c1)
identical(xs3.c1, s3.c1)
identical(xs4.c1, s4.c1)
for (i in 1: (n - 1)) {
assign(paste('ys', i, '.c1', sep=""), rbind(c(p.c1[i], p.c1[i], (1-p.c1[i])),
c(p.c1[i], p.c1[i], (1-p.c1[i])),
c(0, 0, 1)))
}
identical(ys1.c1, s1.c1)
identical(ys2.c1, s2.c1)
identical(ys3.c1, s3.c1)
identical(ys4.c1, s4.c1)
Upvotes: 1
Reputation: 70653
If you have nothing against lists, here's another shot at your question:
aux <- 30:40
manoel <- sapply(aux, function(x) {
matrix(NA, ncol = 2, nrow = x)
}, simplify = FALSE)
> str(manoel)
List of 11
$ : logi [1:30, 1:2] NA NA NA NA NA NA ...
$ : logi [1:31, 1:2] NA NA NA NA NA NA ...
$ : logi [1:32, 1:2] NA NA NA NA NA NA ...
$ : logi [1:33, 1:2] NA NA NA NA NA NA ...
$ : logi [1:34, 1:2] NA NA NA NA NA NA ...
$ : logi [1:35, 1:2] NA NA NA NA NA NA ...
$ : logi [1:36, 1:2] NA NA NA NA NA NA ...
$ : logi [1:37, 1:2] NA NA NA NA NA NA ...
$ : logi [1:38, 1:2] NA NA NA NA NA NA ...
$ : logi [1:39, 1:2] NA NA NA NA NA NA ...
$ : logi [1:40, 1:2] NA NA NA NA NA NA ...
Upvotes: 2
Reputation: 28672
If you need different number of rows then you definitely need to use a list. See the example below:
First, declaring your aux
variable which holds the number of rows to generate:
aux <- 50:77
And let your loop spin:
x <- vector("list",27)
for (i in 1:27) {
x[[i]] <- cbind(rep(1,aux[i]), rnorm(aux[i]))
}
Which will give back the list of your data frames. See e.g. str(x)
for the list and str(x[[1]])
for the first matrixe's structure. The latter would return:
num [1:50, 1:2] 1 1 1 1 1 1 1 1 1 1 ...
And you are right: it could be written a lot nicer than this proposal, look for lapply
and so, but others will help you out with that tricky part :)
After editing: added lapply
example
Try the code below if you are familiar with the above loop (and of course look for ?lapply
):
aux <- 50:77
fun <- function(x) cbind(rep(1,x), rnorm(x))
x <- lapply(aux, fun)
Upvotes: 3
Reputation: 176718
You need assign
:
for (i in 1:27) {
assign(paste("x",i,sep=""), cbind(rep(1,aux[i]), rnorm(aux[i])))
}
This will create 27 matrix objects in your global environment.
Upvotes: 7