In the 3 Dimensional array bellow :
ar <- array(someData, c(5, 5, 5));
rownames(ar) <- ...; #to set up row names
colnames(ar) <- ...; #to set up col names
How can i set the third dimension names ?
You can either set the dimnames argument when defining the array:
ar <- array(data = 1:27,
dim = c(3, 3, 3),
dimnames = list(c("a", "b", "c"),
c("d", "e", "f"),
c("g", "h", "i")))
and/or you can set the dimnames of the third dimension like so:
dimnames(ar)[[3]] <- c("G", "H", "I")
Still starting in R but I found this way that may be useful for large multidimensional array.
Instead of naming each of the indexes ('a','b','c','d',....), you can use provideDimnames() function to automatic generate the index names following the pattern you choose.
Creating data
ar <- array (data = 1:(4*3*2) , dim=c(4,3,2))
> ar
, , 1
[,1] [,2] [,3]
[1,] 1 5 9
[2,] 2 6 10
[3,] 3 7 11
[4,] 4 8 12
, , 2
[,1] [,2] [,3]
[1,] 13 17 21
[2,] 14 18 22
[3,] 15 19 23
[4,] 16 20 24
Labelling dimensions
ar <- provideDimnames(ar , sep = "_", base = list('row','col','lev'))
And you get
> ar
, , lev
col col_1 col_2
row 1 5 9
row_1 2 6 10
row_2 3 7 11
row_3 4 8 12
, , lev_1
col col_1 col_2
row 13 17 21
row_1 14 18 22
row_2 15 19 23
row_3 16 20 24
Related
Suppose
t=c(0,0.5,0.7,0.9,1,1.2) and
v=matrix(1:40, nrow=5, ncol=8)
v
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,] 1 6 11 16 21 26 31 36
[2,] 2 7 12 17 22 27 32 37
[3,] 3 8 13 18 23 28 33 38
[4,] 4 9 14 19 24 29 34 39
[5,] 5 10 15 20 25 30 35 40
I would like to make an array of order 5x6x4.
array <- array(NA, dim = c(5, 6, 4))
5 is the number of rows, 6 is the length of t and 4 is the number of arrays. to create the first array, I would like to consider only the first two columns of v, and for each time point, the first element of the first matrix is filled by v[1,1]+v[1,2]*t[] which is the value of the first element of the first matrix of the first array which is the value of array[1,1,1]
, similarly, array[2,1,1]e v[2,1]+v[2,2]*t[1]
For the second array consider only 3rd and 4th column of v. For the third array consider only 5th and 6th column of v, and Finally, for the fourth array, consider the last two columns of v.
I would appreciate if anyone can help me using a for loop or alternative ways?
Thanks
Try these nested loops:
ary <- array(NA, dim = c(5, 6, 4))
for (i in 1:5) {
for (j in 1:6) {
for (k in 1:4) {
ary[i, j, k] <- v[i, k * 2 - 1] + v[i, k * 2] * t[j]
}
}
}
I need to create a 3D array sorted by row, from left to right and descendent.
x <- 100
I have tried with this:
b <- array(1:96, dim= c(8,4,3))
but it sorts firstly descendently. Using apperm(b) doesn't work as well
The result I want is this:
, , 1
1 2 3 4 5
6 7 8 9 10
11 12 13 14
15 16 17 18
19 20 21 22
array by default fill values along 1st dimension, then 2nd dimension, then 3rd; What you are looking for is fill it in the order of (2nd, 1st, 3rd), you can initialize the array with the shape of 1st dimension and 2nd dimension switched and then use aperm on it:
b <- aperm(array(1:96, dim= c(4,8,3)), c(2,1,3))
# ^ ^ ^ ^ switch the dimension twice here
b
, , 1
[,1] [,2] [,3] [,4]
[1,] 1 2 3 4
[2,] 5 6 7 8
[3,] 9 10 11 12
Edit: My first try, but #Psidom's answer is the right way to do this.
You need to make it as a combination of 3 matrices and then combine them into an array. In the code below I used 96*i/3 to make it flexible for more than 3 matrices to be combined.
b <- array( c( aperm(array(1:(96*1/3), dim = c(4,8))),
aperm(array(33:(96*2/3), dim = c(4,8))),
aperm(array(65:(96*3/3), dim = c(4,8))) ) ,
dim = c(8, 4, 3))
This will be the output:
b[, , 1]
# [,1] [,2] [,3] [,4]
# [1,] 1 2 3 4
# [2,] 5 6 7 8
# [3,] 9 10 11 12
# [4,] 13 14 15 16
# [5,] 17 18 19 20
# [6,] 21 22 23 24
# [7,] 25 26 27 28
# [8,] 29 30 31 32
I would like to obtain the indices of a 3D array in R (ie. arr[x,y,z]) based on multiple values. Specifically, using the first z dimension to subset values in the second z dimension. Here is a example:
# create example array
> m1 <- matrix(c(rep("a",5), rep("b",5), rep("c",5)), nr = 5)
> m2 <- matrix(c(rep(100, 5), rep(10, 5), rep(10, 5)), nr = 5)
> arr <- array(c(m1, m2), dim = c(dim(m1), 2))
#use which() to return the indices in m2 that correspond to indices with
#"a" and "c" values in m1. This does not work as expected.
> ac.ind <- which(arr[,,1] %in% c("a", "c"), arr.ind = T)
> ac.ind
[1] 1 2 3 4 5 11 12 13 14 15
which() returns an vector of positions in m1 that correspond to "a" and "c", not the matrix indices (the (x,y) positions). I would like ac.ind to return:
row col
[1,] 1 1
[2,] 2 1
[3,] 3 1
[4,] 4 1
[5,] 5 1
[1,] 1 3
[2,] 2 3
[3,] 3 3
[4,] 4 3
[5,] 5 3
If I do a more simple which() subset, it does return the indices:
#use which to return indices in m2 that correspond to only "a" in m1
>a.ind <- which(arr[,,1] == c("a"), arr.ind = T)
>a.ind
row col
[1,] 1 1
[2,] 2 1
[3,] 3 1
[4,] 4 1
[5,] 5 1
I am using %in% since I want to subset based on two values in m1 ("a" and "c" values). Is there a way to return the indices of an array based on two values in R?
The issue is that arr[,,1] %in% c("a", "c") returns a vector. One way is to cast this as a matrix with the number of rows equaling the first dimension of arr:
ac.ind <- which(matrix(arr[,,1] %in% c("a", "c"), nrow=dim(arr)[1]), arr.ind = T)
## row col
## [1,] 1 1
## [2,] 2 1
## [3,] 3 1
## [4,] 4 1
## [5,] 5 1
## [6,] 1 3
## [7,] 2 3
## [8,] 3 3
## [9,] 4 3
##[10,] 5 3
Something like this but it is not very efficient because it has to go twice through the data:
ac.ind <- which(arr[,,1] == "c" | arr[,,1] == "a" , arr.ind = T)
ac.ind
row col
[1,] 1 1
[2,] 2 1
[3,] 3 1
[4,] 4 1
[5,] 5 1
[6,] 1 3
[7,] 2 3
[8,] 3 3
[9,] 4 3
[10,] 5 3
at first i have a matrix like this :
x <- matrix(rnorm(1e3),260)
and then an Array
lst <- lapply(seq(1,length(x[,1]), by=52), function(i) x[i:(i+51),])
Data_array <- array(unlist(lst), dim=c(52,length(x[1,]),(length(x[,1])/52)))
This array is a sequence of the Dataframe by 52 (weeks).
It's a temporal analysis (weekly)
I would like to compute an ecdf function on this array.
, , 1
[,1] [,2] [,3]
[1,] **0.66319631** 0.01004290 0.02133477
[2,] -1.64273648 0.23105503 1.02862145
[3,] 1.17083363 -0.49700717 -0.01119745
, , 2
[,1] [,2] [,3]
[1,] **-0.79365987** 1.28394049 -0.547763434
[2,] -0.09221301 1.07676841 0.570294731
[3,] 0.20293308 1.00182888 0.247373981
, , 3
[,1] [,2] [,3]
[1,] **1.03862172** -0.961678683 1.25334651
[2,] 0.58476540 0.745250484 -0.06183788
[3,] 0.24057690 1.226575038 0.23363005
compute ecdf function for each cell. It's for a weekly seasonal analysis.
i.e. calcul quantile for this time series (**): 0.66319631;-0.79365987;1.03862172
for MEAN it's works :
array_lag_sum<-apply(Data_array,c(1,2),FUN=function(x){mean(x,na.rm=TRUE)})
i tried a similar function whith ecdf, but it doesn't work.
percent_array<-apply(Data_array,c(1,2),FUN=function(u){ecdf(u)(u)})
Then...it is not finish, i would like to reformat this array like the original format of the data dataframe (x). (like a rbind but on an array.)
Thank you so much for your help.
edit :
sorry, but i don't know if i was so clear. It's sur that array is complicated for me;
but with your method, if i have this simple data frame :
B <- matrix(seq(1,20), 20, 3)
> B
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 4 4
[5,] 5 5 5
[6,] 6 6 6
[7,] 7 7 7
[8,] 8 8 8
[9,] 9 9 9
[10,] 10 10 10
[11,] 11 11 11
[12,] 12 12 12
[13,] 13 13 13
[14,] 14 14 14
[15,] 15 15 15
[16,] 16 16 16
[17,] 17 17 17
[18,] 18 18 18
[19,] 19 19 19
[20,] 20 20 20
Your function gives :
Data_array <- array( B, dim=c(10,3,5))
, , 1
[,1] [,2] [,3]
[1,] 1 11 1
[2,] 2 12 2
[3,] 3 13 3
[4,] 4 14 4
[5,] 5 15 5
[6,] 6 16 6
[7,] 7 17 7
[8,] 8 18 8
[9,] 9 19 9
[10,] 10 20 10
, , 2
[,1] [,2] [,3]
[1,] 11 1 11
[2,] 12 2 12
[3,] 13 3 13
[4,] 14 4 14
[5,] 15 5 15
[6,] 16 6 16
[7,] 17 7 17
[8,] 18 8 18
[9,] 19 9 19
[10,] 20 10 20
or i would more something like this :
,,1
[,1] [,2] [,3]
[1,] 1 1 1
[2,] 2 2 2
[3,] 3 3 3
[4,] 4 4 4
[5,] 5 5 5
[6,] 6 6 6
[7,] 7 7 7
[8,] 8 8 8
[9,] 9 9 9
[10,] 10 10 10
,,2
[,1] [,2] [,3]
[1,] 11 11 11
[2,] 12 12 12
[3,] 13 13 13
[4,] 14 14 14
[5,] 15 15 15
[6,] 16 16 16
[7,] 17 17 17
[8,] 18 18 18
[9,] 19 19 19
[10,] 20 20 20
and get in result a table which is the percentile value of the time series.
percentile values of 1 and 11, 2 and 12 for each column and each row (i know it's not pertinent but it's just for exemple)
Sorry if my last question was not understandable
The answer is:
ecdf_mat <- apply( Data_array, 1:2, ecdf)
This passes values from each combination of the first two indices to the the function, ecdf. Each of those passes will return a function into a matrix location. You are getting something most people will not be able to use without a bit of coaching: one 52 x 4 matrix of functions. The functions are contained in lists which are valid matrix or array elements:
> dim(apply( Data_array, 1:2, ecdf) )
[1] 52 4
To access them you need to first pull them out of the matrix with standard "[" indexing but then pull them out of the list container with a call to "[[1]]":
> str(apply( Data_array, 1:2, ecdf)[1,1] )
List of 1
$ :function (v)
..- attr(*, "class")= chr [1:3] "ecdf" "stepfun" "function"
..- attr(*, "call")= language FUN(newX[, i], ...)
> apply( Data_array, 1:2, ecdf)[1,1][[1]]
Empirical CDF
Call: FUN(newX[, i], ...)
x[1:5] = -0.92217, -0.37471, 0.058284, 0.28502, 0.44391
> apply( Data_array, 1:2, ecdf)[1,1][[1]](0)
[1] 0.4
Edit:------
It appears you don't want the ecdf's themselves (despite getting no response to my efforts at getting you to recognize the distinction), but rather want an identically shaped array with the percentile values for the i-j positions considered as individual length k-sequences. I can think of two ways to do this. The first one would use that matrix of ecdf functions I built and demonstrated, but I believe that is the more baroque method and it would be easier to give you a more direct route. I've take the liberty of making this more manageable by making the long first dimension only 10-long.
x <- matrix(rnorm(1e3),260)
lst <- lapply(seq(1,length(x[,1]), by=10), function(i) x[i:(i+51),])
Data_array <- array(unlist(lst), dim=c(10,length(x[1,]),(length(x[,1])/52
pctiles2 <- apply( Data_array, 1:2, function(x) ecdf(x)(x) )
> str(pctiles2)
num [1:5, 1:10, 1:4] 0.8 0.4 0.6 0.2 1 0.4 1 0.2 0.6 0.8 ...
They aren't actually percentiles, but that could be easily remedied by slipping a 100* in from of the ecdf call (or multiplying the result by 100.. You will notice that the structure has been permuted so that the quantile/percentiles sequences run down the first column. That because apply always delivers its result in column major order. There is a function aperm which would allow you to re-arrange these in the original order:
re_pctiles <- aperm(pctiles, c(2,3,1) )
Within R I would like to transform an array (dimensions: i, j, k) into a matrix such that the observations (i.e. rows) of the new matrix are each element from the array pulled k "layers" at a time. Essentially, again, the rows of the new matrix will be composed of each element of the previous array with the columns of the matrix being equivalent to the k dimension of the array. Thus, the new matrix should be composed of i*j rows with k columns.
Please let me know if I can clarify or provide an example of input / output!
Thanks!
Edit:
This code works (but is not optimized) —
m = array(1:27,dim = c(3,3,3))
m
dim = dim(m)
mparam = dim[3]
listm = list()
for (i in 1:mparam){
listm[[i]] = as.vector(m[,,i])
}
untran = do.call(rbind,listm)
transposed = t(untran)
transposed
Like this?
m <- array(1:27,dim = c(3,3,3))
> m
, , 1
[,1] [,2] [,3]
[1,] 1 4 7
[2,] 2 5 8
[3,] 3 6 9
, , 2
[,1] [,2] [,3]
[1,] 10 13 16
[2,] 11 14 17
[3,] 12 15 18
, , 3
[,1] [,2] [,3]
[1,] 19 22 25
[2,] 20 23 26
[3,] 21 24 27
> matrix(m,9,3)
[,1] [,2] [,3]
[1,] 1 10 19
[2,] 2 11 20
[3,] 3 12 21
[4,] 4 13 22
[5,] 5 14 23
[6,] 6 15 24
[7,] 7 16 25
[8,] 8 17 26
[9,] 9 18 27