what is equivalent to do.call(rbind, list)?

Question

In my real data I got warning number of columns of result is not a multiple of vector length (arg 1) ,though my list has a unique number of cols, when I used do.call("rbind"

I want to try something else that produce the same output as do.call(rbind, list) to check whither the problem is in my list or not.

example

       n = c(2, 3, 5,4) 
       n1 = c(2, 7, 4,6) 
       n2 = c(NA, NA, NA,NA) 
       x = list(n, n1, n2)
      dat <- do.call("rbind", x)

I tried this:

      df=matrix(as.numeric(unlist(x)), nrow= 3)

but

      identical(dat,df)
     > identical(dat,df)
      [1] FALSE

PS: I do not want to change the class or str of my list

Answer 1

Just a note: Looking at

> dat
     [,1] [,2] [,3] [,4]
[1,]    2    3    5    4
[2,]    2    7    4    6
[3,]   NA   NA   NA   NA
> df
     [,1] [,2] [,3] [,4]
[1,]    2    4    4   NA
[2,]    3    2    6   NA
[3,]    5    7   NA   NA

I am not surprised about the result that

identical(dat,df)
[1] FALSE

However, look at

df=matrix(as.numeric(unlist(x)), nrow= 3, byrow = T)
identical(dat,df)
[1] TRUE

Alternatives to do.call(rbind, list)

If you are looking for an alternative to do.call, look at dplyr::bind_rows (which feeds on dataframes and is quite efficient). A second alternative may be Reduce, as in:

Reduce(rbind, x)
     [,1] [,2] [,3] [,4]
init    2    3    5    4
        2    7    4    6
       NA   NA   NA   NA

A further alternative is data.table::rbindlist, which feeds on a list (!). Note though that the output will be transposed, meaning that each row will appear as one column. See

library(data.table)
rbindlist(list(x))
   V1 V2 V3
1:  2  2 NA
2:  3  7 NA
3:  5  4 NA
4:  4  6 NA

You can wrap it into t() to transpose the result, though.

More generally speaking, though, if you want to check if all elements in your list have the same length, you might be more successful in finding the error with something like

sapply(x, length)
[1] 4 4 4

or depending on how your actual data looks like, a variation such as ncol or so instead of length. If the list elements have column names, then names might be a valid alternative as well.

Answer 2

You need to add byrow = TRUE:

df=matrix(as.numeric(unlist(x)), nrow= 3, byrow = TRUE)