I'm not comfortable with array manipulation (and English writing, sorry ..)
I have this data frame (aa):
aa<-data.frame(replicate(10,sample(0:17,30,rep=TRUE)))
> aa
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 17 7 9 2 3 7 17 0 15 1
2 12 5 10 10 8 17 13 7 2 2
3 14 14 7 7 16 1 13 0 14 6
4 12 10 10 15 7 2 7 11 4 0
5 1 9 5 5 8 15 15 11 8 17
6 8 0 9 6 7 11 9 12 4 17
7 17 1 17 5 11 8 16 0 2 15
8 10 7 15 6 17 3 0 16 16 15
9 8 3 14 13 16 5 15 8 14 10
10 11 13 15 3 17 13 13 4 11 12
11 9 13 0 7 4 13 15 1 2 0
12 1 3 17 13 10 4 12 5 4 15
13 5 8 9 8 0 6 14 13 0 8
14 17 11 10 4 15 10 7 1 1 7
15 2 0 16 7 13 10 13 3 10 7
16 5 5 15 7 0 17 10 14 11 4
17 10 17 9 11 0 9 9 17 0 4
18 12 8 8 16 11 4 10 16 4 7
19 5 7 13 12 17 17 17 17 6 8
20 13 17 1 2 0 1 8 4 17 17
21 15 15 5 13 6 16 5 5 14 13
22 12 4 5 1 2 7 17 2 9 9
23 12 5 13 16 6 6 15 2 13 10
24 8 6 12 4 5 11 7 12 14 10
25 5 11 15 1 17 3 8 10 4 4
26 3 10 8 14 1 13 16 1 16 11
27 10 2 14 11 6 8 13 3 8 10
28 14 5 7 12 8 14 16 9 16 14
29 5 17 16 17 12 1 3 8 2 0
30 5 17 12 2 8 9 3 1 14 15
I would like to create an array which is the result of a cutting of dataframe (aa) like this (by seq of x elements, here for example 10): an array with 3 dimensions
, , 1
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
1 17 7 9 2 3 7 17 0 15 1
2 12 5 10 10 8 17 13 7 2 2
3 14 14 7 7 16 1 13 0 14 6
4 12 10 10 15 7 2 7 11 4 0
5 1 9 5 5 8 15 15 11 8 17
6 8 0 9 6 7 11 9 12 4 17
7 17 1 17 5 11 8 16 0 2 15
8 10 7 15 6 17 3 0 16 16 15
9 8 3 14 13 16 5 15 8 14 10
10 11 13 15 3 17 13 13 4 11 12
, , 2
X1 X2 X3 X4 X5 X6 X7 X8 X9 X10
11 9 13 0 7 4 13 15 1 2 0
12 1 3 17 13 10 4 12 5 4 15
13 5 8 9 8 0 6 14 13 0 8
14 17 11 10 4 15 10 7 1 1 7
15 2 0 16 7 13 10 13 3 10 7
16 5 5 15 7 0 17 10 14 11 4
17 10 17 9 11 0 9 9 17 0 4
18 12 8 8 16 11 4 10 16 4 7
19 5 7 13 12 17 17 17 17 6 8
20 13 17 1 2 0 1 8 4 17 17
etc...
I've already tried this...
aa_lag <-array(aa[1:10,],dim=c(dim(aa),3))
thank you so much for answer...
You can try
aa1 <- t(aa)
dim(aa1) <- c(10,10,3)
aa2 <- aperm(aa1, c(2,1,3))
Checking the results
m1 <- as.matrix(aa[1:10,])
dimnames(m1) <- NULL
identical(m1, aa2[,,1])
#[1] TRUE
Or using seq
lst <- lapply(seq(1,30, by=10), function(i) aa[i:(i+9),])
aa3 <- array(unlist(lst), dim=c(10,10,3))
identical(aa2, aa3)
#[1] TRUE
set.seed(24)
aa<-data.frame(replicate(10,sample(0:17,30,rep=TRUE)))