How it iterate over a matrix using a function in R? - r

Question

I have created a function to order a vector of length 2, using the following code
x = (c(6,2))
orders = function(x){
for(i in 1:(length(x)-1)){
if(x[i+1] < x[i]){
return(c(x[i+1], x[i]))} else{
(return(x))
}}}
orders(x)
I have been asked to use this function to process a dataset with 2 columns as follows. Iterate over the rows of the
data set, and if the element in the 2nd column of row i is less than the element in the first
column of row i, switch the order of the two entries in the row by making a suitable call to
the function you just wrote.
I've tried using the following code
set.seed(1128719)
data=matrix(rnorm(20),byrow=T,ncol=2)
df = for (i in 1:2) {
for(j in 1:10){
data = orders(c(x[i], x[j]))
return(data)
}
}
The output is null. I'm not quite sure where I'm going wrong.
Any suggestions?

Answer 1

I modified your code a bit but tried to keep the 'style' the same
Ther is no need for a loop
i in 1:(length(x)-1) always evaluates to
for i in 1:1 and i will only take the value of 1.
orders = function(x){
# Since the function will only work on vectors of length 2
# its good practice to raise an error right at the start
#
if (length(x) != 2) {
stop("x must be vector of lenght 2")
}
if (x[2] < x[1]) {
return(c(x[2], x[1]))
} else {
return(x)
}
}
orders(c(6, 2))
set.seed(1128719)
data <- matrix(rnorm(20),byrow=T,ncol=2)
The for loop itself cant be assigned to a variable
But we use the loop to mutate the matrix 'data'
in place
for (row in 1:nrow(data)) {
data[row, ] <- orders(data[row,])
}
data
Edit:
This is the input:
[,1] [,2]
[1,] -0.04142965 0.2377140
[2,] -0.76237866 -0.8004284
[3,] 0.18700893 -0.6800310
[4,] 0.76499646 0.4430643
[5,] 0.09193440 -0.2592316
[6,] 1.17478053 -0.4044760
[7,] -1.62262500 0.1652850
[8,] -1.54848857 0.7475451
[9,] -0.05907252 -0.8324074
[10,] -1.11064318 -0.1148806
This is the output i get:
[,1] [,2]
[1,] -0.04142965 0.23771403
[2,] -0.80042842 -0.76237866
[3,] -0.68003104 0.18700893
[4,] 0.44306433 0.76499646
[5,] -0.25923164 0.09193440
[6,] -0.40447603 1.17478053
[7,] -1.62262500 0.16528496
[8,] -1.54848857 0.74754509
[9,] -0.83240742 -0.05907252
[10,] -1.11064318 -0.11488062

Answer 2

Here are two ways of ordering the 2 columns matrix.
This is the test matrix posted in the question.
set.seed(1128719)
data <- matrix(rnorm(20), byrow = TRUE, ncol = 2)
1. With a function orders.
The function expects as input a 2 element vector. If they are out of order, return the vector with its elements reversed, else return the vector as is.
orders <- function(x){
stopifnot(length(x) == 2)
if(x[2] < x[1]){
x[2:1]
}else{
x
}
}
Test the function.
x <- c(6,2)
orders(x)
#[1] 2 6
Now with the matrix data.
df1 <- t(apply(data, 1, orders))
2. Vectorized code.
Creates a logical index with TRUE whenever the elements are out of order and reverse only those elements.
df2 <- data
inx <- data[,2] < data[,1]
df2[inx, ] <- data[inx, 2:1]
The results are the same.
identical(df1, df2)
#[1] TRUE