How to cbind or rbind different lengths vectors without repeating the elements of the shorter vectors?

RVector

R Problem Overview


cbind(1:2, 1:10)  
     [,1] [,2]  
  [1,]    1    1  
  [2,]    2    2  
  [3,]    1    3  
  [4,]    2    4  
  [5,]    1    5  
  [6,]    2    6  
  [7,]    1    7  
  [8,]    2    8  
  [9,]    1    9  
 [10,]    2   10  

I want an output like below

[,1] [,2]  
[1,] 1 1  
[2,] 2 2  
[3,]   3  
[4,]   4  
[5,]   5  
[6,]   6  
[7,]   7  
[8,]   8  
[9,]   9  
[10,]  10  

R Solutions


Solution 1 - R

The trick is to make all your inputs the same length.

x <- 1:2
y <- 1:10
n <- max(length(x), length(y))
length(x) <- n                      
length(y) <- n

If you want you output to be an array, then cbind works, but you get additional NA values to pad out the rectangle.

cbind(x, y)
       x  y
 [1,]  1  1
 [2,]  2  2
 [3,] NA  3
 [4,] NA  4
 [5,] NA  5
 [6,] NA  6
 [7,] NA  7
 [8,] NA  8
 [9,] NA  9
[10,] NA 10

To get rid of the NAs, the output must be a list.

Map(function(...) 
   {
      ans <- c(...)
      ans[!is.na(ans)]
   }, as.list(x), as.list(y)
)
[[1]]
[1] 1 1

[[2]]
[1] 2 2

[[3]]
[1] 3

[[4]]
[1] 4

[[5]]
[1] 5

[[6]]
[1] 6

[[7]]
[1] 7

[[8]]
[1] 8

[[9]]
[1] 9

[[10]]
[1] 10

EDIT: I swapped mapply(..., SIMPLIFY = FALSE) for Map.

Solution 2 - R

I came across similar problem and I would like to suggest that additional solution that some, I hope, may find useful. The solution is fairly straightforward and makes use of the qpcR package and the provided cbind.na function.

Example

x <- 1:2
y <- 1:10
dta <- qpcR:::cbind.na(x, y)

Results

> head(dta)
      x y
[1,]  1 1
[2,]  2 2
[3,] NA 3
[4,] NA 4
[5,] NA 5
[6,] NA 6

Side comments

Following the OP's original example, column names can be easily removed:

colnames(dta) <- NULL

the operation would produce the desired output in full:

> head(dta)
     [,1] [,2]
[1,]    1    1
[2,]    2    2
[3,]   NA    3
[4,]   NA    4
[5,]   NA    5
[6,]   NA    6

Solution 3 - R

I would like to propose an alternate solution that makes use of the rowr package and their cbind.fill function.

> rowr::cbind.fill(1:2,1:10, fill = NA);

   object object
1       1      1
2       2      2
3      NA      3
4      NA      4
5      NA      5
6      NA      6
7      NA      7
8      NA      8
9      NA      9
10     NA     10

Or alternatively, to match the OP's desired output:

> rowr::cbind.fill(1:2,1:10, fill = '');

   object object
1       1      1
2       2      2
3              3
4              4
5              5
6              6
7              7
8              8
9              9
10            10

Solution 4 - R

Helper function...

bind.pad <- function(l, side="r", len=max(sapply(l,length)))
{
  if (side %in% c("b", "r")) {
    out <- sapply(l, 'length<-', value=len)
  } else {
    out <- sapply(sapply(sapply(l, rev), 'length<-', value=len, simplify=F), rev)}
  if (side %in% c("r", "l")) out <- t(out)
  out
}

Examples:

> l <- lapply(c(3,2,1,2,3),seq)
> lapply(c("t","l","b","r"), bind.pad, l=l, len=4)
[[1]]
     [,1] [,2] [,3] [,4] [,5]
[1,]   NA   NA   NA   NA   NA
[2,]    1   NA   NA   NA    1
[3,]    2    1   NA    1    2
[4,]    3    2    1    2    3

[[2]]
     [,1] [,2] [,3] [,4]
[1,]   NA    1    2    3
[2,]   NA   NA    1    2
[3,]   NA   NA   NA    1
[4,]   NA   NA    1    2
[5,]   NA    1    2    3

[[3]]
     [,1] [,2] [,3] [,4] [,5]
[1,]    1    1    1    1    1
[2,]    2    2   NA    2    2
[3,]    3   NA   NA   NA    3
[4,]   NA   NA   NA   NA   NA

[[4]]
     [,1] [,2] [,3] [,4]
[1,]    1    2    3   NA
[2,]    1    2   NA   NA
[3,]    1   NA   NA   NA
[4,]    1    2   NA   NA
[5,]    1    2    3   NA

Solution 5 - R

Given that some of the solutions above rely on packages that are no longer available, here a helper function that only uses dplyr.

bind_cols_fill <- function(df_list) {

  max_rows <- map_int(df_list, nrow) %>% max()
  
  map(df_list, function(df) {
    if(nrow(df) == max_rows) return(df)
    first <- names(df)[1] %>% sym()
    df %>% add_row(!!first := rep(NA, max_rows - nrow(df)))
  }) %>% bind_cols()
}

Note that this takes a list of data frames, so that it is slightly cumbersome if one only wants to combine two vectors:

x <- 1:2
y <- 1:10
bind_cols_fill(list(tibble(x), tibble(y)) 

Solution 6 - R

Another solution with no dependencies:

my_bind <- function(x, y){
if(length(x = x) > length(x = y)){
    len_diff <- length(x) - length(y)
    y <- c(y, rep(NA, len_diff))
}else if(length(x = x) < length(x = y)){
    len_diff <- length(y) - length(x)
    x <- c(x, rep(NA, len_diff))
}
cbind(x, y)
}
my_bind(x = letters[1:4], y = letters[1:2])

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionCharesView Question on Stackoverflow
Solution 1 - RRichie CottonView Answer on Stackoverflow
Solution 2 - RKonradView Answer on Stackoverflow
Solution 3 - RAlex WView Answer on Stackoverflow
Solution 4 - RThellView Answer on Stackoverflow
Solution 5 - RLukas WallrichView Answer on Stackoverflow
Solution 6 - RHelgi GuðmundssonView Answer on Stackoverflow