Right-align NA values in 4d array-CodePudding

Context: I'm working with survey data organized as a 4D array with this structure: m[n_sites, n_surveys, n_years, n_species].

Question: There are data missing randomly, though, and I want to move the missing data to the end of each row.

Example: Here's the original data:

, , 1, 1

   1  2  3  4  5
1 NA  2 NA  2  3
2 NA  3  1 NA NA
3  4 NA NA  4  6
4  2 NA NA  2  1

... and I want to rearrange this to be:

, , 1, 1

   1  2  3  4  5
1  2  2  3 NA NA
2  3  1 NA NA NA
3  4  4  6 NA NA
4  2  2  1 NA NA

Note: The data are very large, though, so I need something efficient and fairly simple.

Reproducible code:

library(magrittr)  ## for %>% pipe
library(reshape2)  ## for acast
set.seed(1)

# Simulate survey data
df <- expand.grid(
  species = c(1,2),
  year = c(1,2,3),
  site = c(1,2,3,4),
  survey = c(1,2,3,4,5))
df$counts <- rpois(n = nrow(df), lambda = 3)

# Add random NAs (missing data)
posNA <- sample(x = 1:nrow(df), size = 0.5 * nrow(df), replace = FALSE)
df$counts[posNA] <- NA

# Cast to 4d array
m <- df %>% acast(site ~ survey ~ year ~ species)

CodePudding user response：

We could use

m1 <- m
for(i in seq_len(dim(m)[3]))
    for(j in seq_len(dim(m)[4]))
      m1[,, i, j]  <- t(apply(m1[,, i, j], 1, 
              function(x) x[order(is.na(x))]))

-output

> m1[,, 1, 1]
  1 2  3  4  5
1 2 2  3 NA NA
2 3 1 NA NA NA
3 4 4  6 NA NA
4 2 2  1 NA NA

CodePudding user response：

You may use na.omit on rows and the 3rd and 4th dimension, correct the length by cumber of columns and transpose result using aperm.

## 4D-array
apply(m, c(1, 3, 4), \(x) `length<-`(na.omit(x), dim(m)[2])) |> aperm(c(2, 1, 3, 4))
# , , 1, 1
# 
#   [,1] [,2] [,3] [,4] [,5]
# 1    2    2    3   NA   NA
# 2    3    1   NA   NA   NA
# 3    4    4    6   NA   NA
# 4    2    2    1   NA   NA
# 
# , , 2, 1
# 
#   [,1] [,2] [,3] [,4] [,5]
# 1    3    0    3   NA   NA
# 2    3    3    2    3   NA
# 3    3    4   NA   NA   NA
# 4    6    3    1   NA   NA
# 
# , , 3, 1
# 
#   [,1] [,2] [,3] [,4] [,5]
# 1    2   NA   NA   NA   NA
# 2    2    1   NA   NA   NA
# 3    4    2   NA   NA   NA
# 4    4    0    4    3   NA
# 
# , , 1, 2
# 
#   [,1] [,2] [,3] [,4] [,5]
# 1    2    2   NA   NA   NA
# 2    4    3    8   NA   NA
# 3    2    1    2    3   NA
# 4    4   NA   NA   NA   NA
# 
# , , 2, 2
# 
#   [,1] [,2] [,3] [,4] [,5]
# 1    5    5   NA   NA   NA
# 2    4   NA   NA   NA   NA
# 3    2    1   NA   NA   NA
# 4    5   NA   NA   NA   NA
# 
# , , 3, 2
# 
#   [,1] [,2] [,3] [,4] [,5]
# 1    5    2    2    2   NA
# 2    1    4    2    3   NA
# 3    8    2    3   NA   NA
# 4    5   NA   NA   NA   NA

Here additionally at a simpler array to demonstrate the logic:

## 3D-array
a
# , , 1
# 
#      [,1] [,2] [,3] [,4]
# [1,]   NA   NA    1    3
# [2,]    4    1    1   NA
# [3,]   NA    3   NA    4
# 
# , , 2
# 
#      [,1] [,2] [,3] [,4]
# [1,]    3    2    2    4
# [2,]    1   NA    3    4
# [3,]    1   NA    4    3

apply(a, c(1, 3), \(x) `length<-`(na.omit(x), dim(a)[2])) |> aperm(c(2, 1, 3))
# , , 1
# 
#      [,1] [,2] [,3] [,4]
# [1,]    1    3   NA   NA
# [2,]    4    1    1   NA
# [3,]    3    4   NA   NA
# 
# , , 2
# 
#      [,1] [,2] [,3] [,4]
# [1,]    3    2    2    4
# [2,]    1    3    4   NA
# [3,]    1    4    3   NA

Data:

m <- structure(c(NA, NA, 4L, 2L, 2L, 3L, NA, NA, NA, 1L, NA, NA, 2L, 
NA, 4L, 2L, 3L, NA, 6L, 1L, 3L, 3L, NA, 6L, 0L, 3L, NA, 3L, NA, 
2L, 3L, 1L, 3L, 3L, 4L, NA, NA, NA, NA, NA, 2L, 2L, 4L, 4L, NA, 
NA, NA, 0L, NA, NA, NA, NA, NA, NA, NA, 4L, NA, 1L, 2L, 3L, NA, 
4L, 2L, NA, NA, 3L, 1L, NA, NA, NA, 2L, 4L, 2L, NA, NA, NA, 2L, 
8L, 3L, NA, 5L, NA, NA, NA, NA, NA, NA, NA, 5L, NA, 2L, NA, NA, 
4L, 1L, 5L, NA, NA, NA, NA, 5L, 1L, 8L, NA, 2L, 4L, NA, NA, 2L, 
NA, 2L, 5L, NA, 2L, NA, NA, 2L, 3L, 3L, NA), dim = c(4L, 5L, 
3L, 2L), dimnames = list(c("1", "2", "3", "4"), c("1", "2", "3", 
"4", "5"), c("1", "2", "3"), c("1", "2")))


a <- structure(c(NA, 4L, NA, NA, 1L, 3L, 1L, 1L, NA, 3L, NA, 4L, 3L, 
1L, 1L, 2L, NA, NA, 2L, 3L, 4L, 4L, 4L, 3L), dim = c(3L, 4L, 
2L))