transpose: Efficient transpose of list

Description

transpose is an efficient way to transpose lists, data.frames or data.tables.

Usage

transpose(l, fill=NA, ignore.empty=FALSE, keep.names=NULL,
          make.names=NULL, list.cols=FALSE)

Value

A transposed list, data.frame or data.table.

list outputs will only be named according to make.names.

Arguments

l: A list, data.frame or data.table.
fill: Default is NA. It is used to fill shorter list elements so as to return each element of the transposed result of equal lengths.
ignore.empty: Default is FALSE. TRUE will ignore length-0 list elements.
keep.names: The name of the first column in the result containing the names of the input; e.g. keep.names="rn". By default NULL and the names of the input are discarded.
make.names: The name or number of a column in the input to use as names of the output; e.g. make.names="rn". By default NULL and default names are given to the output columns.
list.cols: Default is FALSE. TRUE will avoid promoting types and return columns of type list instead. factor will always be cast to character.

Details

The list elements (or columns of data.frame/data.table) should be all atomic. If list elements are of unequal lengths, the value provided in fill will be used so that the resulting list always has all elements of identical lengths. The class of input object is also preserved in the transposed result.

The ignore.empty argument can be used to skip or include length-0 elements.

This is particularly useful in tasks that require splitting a character column and assigning each part to a separate column. This operation is quite common enough that a function tstrsplit is exported.

factor columns are converted to character type. Attributes are not preserved at the moment. This may change in the future.

Examples

Run this code

ll = list(1:5, 6:8)
transpose(ll)
setDT(transpose(ll, fill=0))[]

DT = data.table(x=1:5, y=6:10)
transpose(DT)

DT = data.table(x=1:3, y=c("a","b","c"))
transpose(DT, list.cols=TRUE)

# base R equivalent of transpose
l = list(1:3, c("a", "b", "c"))
lapply(seq(length(l[[1]])), function(x) lapply(l, `[[`, x))
transpose(l, list.cols=TRUE)

ll = list(nm=c('x', 'y'), 1:2, 3:4)
transpose(ll, make.names="nm")