Vector to sliding matrix in R - r

Question

I am trying to create a function that takes a vector and creates two sliding matrix, like bellow:
Input, Output
[d01, d02, d03, d04, d05, d06, d07], [d08, d09, d10, d11, d12, d13, d14]
[d02, d03, d04, d05, d06, d07, d08], [d09, d10, d11, d12, d13, d14, d15]
...
I tried to adapt a Python code to R but I am having some problems and I cannot find the error (I am not used to R)
This is the R code:
create_dataset = function(data, n_input, n_out){
dataX = c()
dataY = c()
in_start = 0
for (i in 1:range(length(data))) {
#define the end of the input sequence
in_end = in_start + n_input
out_end = in_end + n_out
if(out_end <= length(data)){
x_input = data[in_start:in_end, 1]
X = append(x_input)
y = append(data[in_end:out_end], 1)
}
#move along one time step
in_start = in_start + 1
}
X; Y
}
I got this error when calling this function
> create_dataset(data, n_input = 5, n_out = 5)
Error in data[in_start:in_end, 1] : incorrect number of dimensions
In addition: Warning message:
In 1:range(length(data)) :
numerical expression has 2 elements: only the first used
EDIT:
Adding the Python code I trying to adapt to R
# convert history into inputs and outputs
def to_supervised(train, n_input, n_out):
X, y = list(), list()
in_start = 0
# step over the entire history one time step at a time
for _ in range(len(data)):
# define the end of the input sequence
in_end = in_start + n_input
out_end = in_end + n_out
# ensure we have enough data for this instance
if out_end <= len(data):
x_input = data[in_start:in_end, 0]
x_input = x_input.reshape((len(x_input), 1))
X.append(x_input)
y.append(data[in_end:out_end, 0])
# move along one time step
in_start += 1
return array(X), array(y)

Answer 1

Here are two approaches. Also see Lagging time series data
1) Normally in R one takes the whole object approach rather than iterating over indexes. Now, assuming inputs v, k1 and k2 we compute e as the sliding matrix with k1+k2 columns. Then first k1 columns is the first matrix and the remaining columns is the second.
# inputs
v <- 1:12 # 1, 2, ..., 12
k1 <- k2 <- 3
k <- k1 + k2
e <- embed(v, k)[, k:1]
ik1 <- 1:k1
e[, ik1]
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 2 3 4
## [3,] 3 4 5
## [4,] 4 5 6
## [5,] 5 6 7
## [6,] 6 7 8
## [7,] 7 8 9
e[, -ik1]
## [,1] [,2] [,3]
## [1,] 4 5 6
## [2,] 5 6 7
## [3,] 6 7 8
## [4,] 7 8 9
## [5,] 8 9 10
## [6,] 9 10 11
## [7,] 10 11 12
2) Regarding the R code in the question:
in R the range function takes a vector input and returns a 2 element vector of the minimum and maximum so it is not what is wanted in the for loop, use seq_along instead
indexes in R start at 1 rather than 0
the return value of a function must be a single object. We return a two element list of matrices.
iteratively appending to an object is inefficient in R. This can be addressed by preallocating the result or not using a loop; however, we don't address this problem below as we already have a better implementation above in (1).
there was inconsistent naming of variables in the question's code
Although this entire approach is not how one would normally write R software, in order to make the minimal changes to get it to work we can write the following.
# data is plain vector, n_input and n_out are scalars
# result is 2 element list of matrices
create_dataset = function(data, n_input, n_out){
X <- matrix(nrow = 0, ncol = n_input)
Y <- matrix(nrow = 0, ncol = n_out)
in_start <- 0
for (i in seq_along(data)) {
#define the end of the input sequence
in_end <- in_start + n_input
out_end <- in_end + n_out
if(out_end <= length(data)){
X <- rbind(X, data[(in_start+1):in_end])
Y <- rbind(Y, data[(in_end+1):out_end])
}
#move along one time step
in_start = in_start + 1
}
list(X, Y)
}
# inputs defined in (1)
create_dataset(v, k1, k2)
giving this two element list of matrices:
[[1]]
[,1] [,2] [,3]
[1,] 1 2 3
[2,] 2 3 4
[3,] 3 4 5
[4,] 4 5 6
[5,] 5 6 7
[6,] 6 7 8
[7,] 7 8 9
[[2]]
[,1] [,2] [,3]
[1,] 4 5 6
[2,] 5 6 7
[3,] 6 7 8
[4,] 7 8 9
[5,] 8 9 10
[6,] 9 10 11
[7,] 10 11 12