Merge Multiple Data Frames by Row Names

Question

I\'m trying to merge multiple data frames by row names.

I know how to do it with two:

x = data.frame(a = c(1,2,3), row.names = letters[1:3])
y = data.fra         


        

Answer 1

Merging by row.names does weird things - it creates a column called Row.names, which makes subsequent merges hard.

To avoid that issue you can instead create a column with the row names (which is generally a better idea anyway - row names are very limited and hard to manipulate). One way of doing that with the data as given in OP (not the most optimal way, for more optimal and easier ways of dealing with rectangular data I recommend getting to know data.table instead):

Reduce(merge, lapply(l, function(x) data.frame(x, rn = row.names(x))))

Answer 2

As an alternative to Reduce and merge:

If you put all the data frames into a list, you can then use grep and cbind to get the data frames with the desired row names.

## set up the data
> x <- data.frame(x1 = c(2,4,6), row.names = letters[1:3])
> y <- data.frame(x2 = c(3,6,9), row.names = letters[1:3])
> z <- data.frame(x3 = c(1,2,3), row.names = letters[1:3])
> a <- data.frame(x4 = c(4,6,8), row.names = letters[4:6])
> lst <- list(a, x, y, z)

## combine all the data frames with row names = letters[1:3]
> gg <- grep(paste(letters[1:3], collapse = ""), 
             sapply(lapply(lst, rownames), paste, collapse = ""))
> do.call(cbind, lst[gg])
##   x1 x2 x3
## a  2  3  1
## b  4  6  2
## c  6  9  3

Answer 3

maybe there exists a faster version using do.call or *apply, but this works in your case:

x = data.frame(X = c(1,2,3), row.names = letters[1:3])
y = data.frame(Y = c(1,2,3), row.names = letters[1:3])
z = data.frame(Z = c(1,2,3), row.names = letters[1:3])

merge.all <- function(x, ..., by = "row.names") {
  L <- list(...)
  for (i in seq_along(L)) {
    x <- merge(x, L[[i]], by = by)
    rownames(x) <- x$Row.names
    x$Row.names <- NULL
  }
  return(x)
}

merge.all(x,y,z)

important may be to define all the parameters (like by) in the function merge.all you want to forward to merge since the whole ... arguments are used in the list of objects to merge.