How would one change this input (with the sequence: time, in, out, files):
Time In Out Files
1 2 3 4
2 3 4 5
To this output (with the sequence: time, out, in, files)?
Time Out In Files
1 3 2 4
2 4 3 5
Here's the dummy R data:
table <- data.frame(Time=c(1,2), In=c(2,3), Out=c(3,4), Files=c(4,5))
table
## Time In Out Files
##1 1 2 3 4
##2 2 3 4 5
# reorder by column name
data <- data[c("A", "B", "C")]
#reorder by column index
data <- data[c(1,3,2)]
You can use the data.table package:
How to reorder data.table columns (without copying)
require(data.table)
setcolorder(DT,myOrder)
The only one I have seen work well is from here.
shuffle_columns <- function (invec, movecommand) {
movecommand <- lapply(strsplit(strsplit(movecommand, ";")[[1]],
",|\\s+"), function(x) x[x != ""])
movelist <- lapply(movecommand, function(x) {
Where <- x[which(x %in% c("before", "after", "first",
"last")):length(x)]
ToMove <- setdiff(x, Where)
list(ToMove, Where)
})
myVec <- invec
for (i in seq_along(movelist)) {
temp <- setdiff(myVec, movelist[[i]][[1]])
A <- movelist[[i]][[2]][1]
if (A %in% c("before", "after")) {
ba <- movelist[[i]][[2]][2]
if (A == "before") {
after <- match(ba, temp) - 1
}
else if (A == "after") {
after <- match(ba, temp)
}
}
else if (A == "first") {
after <- 0
}
else if (A == "last") {
after <- length(myVec)
}
myVec <- append(temp, values = movelist[[i]][[1]], after = after)
}
myVec
}
Use like this:
new_df <- iris[shuffle_columns(names(iris), "Sepal.Width before Sepal.Length")]
Works like a charm.
As mentioned in this comment, the standard suggestions for re-ordering columns in a data.frame are generally cumbersome and error-prone, especially if you have a lot of columns.
This function allows to re-arrange columns by position: specify a variable name and the desired position, and don't worry about the other columns.
##arrange df vars by position
##'vars' must be a named vector, e.g. c("var.name"=1)
arrange.vars <- function(data, vars){
##stop if not a data.frame (but should work for matrices as well)
stopifnot(is.data.frame(data))
##sort out inputs
data.nms <- names(data)
var.nr <- length(data.nms)
var.nms <- names(vars)
var.pos <- vars
##sanity checks
stopifnot( !any(duplicated(var.nms)),
!any(duplicated(var.pos)) )
stopifnot( is.character(var.nms),
is.numeric(var.pos) )
stopifnot( all(var.nms %in% data.nms) )
stopifnot( all(var.pos > 0),
all(var.pos <= var.nr) )
##prepare output
out.vec <- character(var.nr)
out.vec[var.pos] <- var.nms
out.vec[-var.pos] <- data.nms[ !(data.nms %in% var.nms) ]
stopifnot( length(out.vec)==var.nr )
##re-arrange vars by position
data <- data[ , out.vec]
return(data)
}
Now the OP's request becomes as simple as this:
table <- data.frame(Time=c(1,2), In=c(2,3), Out=c(3,4), Files=c(4,5))
table
## Time In Out Files
##1 1 2 3 4
##2 2 3 4 5
arrange.vars(table, c("Out"=2))
## Time Out In Files
##1 1 3 2 4
##2 2 4 3 5
To additionally swap Time and Files columns you can do this:
arrange.vars(table, c("Out"=2, "Files"=1, "Time"=4))
## Files Out In Time
##1 4 3 2 1
##2 5 4 3 2
Maybe it's a coincidence that the column order you want happens to have column names in descending alphabetical order. Since that's the case you could just do:
df<-df[,order(colnames(df),decreasing=TRUE)]
That's what I use when I have large files with many columns.
dplyr version 1.0.0 includes the relocate() function to easily reorder columns:
dat <- data.frame(Time=c(1,2), In=c(2,3), Out=c(3,4), Files=c(4,5))
library(dplyr) # from version 1.0.0 only
dat %>%
relocate(Out, .before = In)
or
dat %>%
relocate(Out, .after = Time)
You can also use the subset function:
data <- subset(data, select=c(3,2,1))
You should better use the [] operator as in the other answers, but it may be useful to know that you can do a subset and a column reorder operation in a single command.
Update:
You can also use the select function from the dplyr package:
data = data %>% select(Time, out, In, Files)
I am not sure about the efficiency, but thanks to dplyr's syntax this solution should be more flexible, specially if you have a lot of columns. For example, the following will reorder the columns of the mtcars dataset in the opposite order:
mtcars %>% select(carb:mpg)
And the following will reorder only some columns, and discard others:
mtcars %>% select(mpg:disp, hp, wt, gear:qsec, starts_with('carb'))
Read more about dplyr's select syntax.
Your dataframe has four columns like so df[,c(1,2,3,4)].
Note the first comma means keep all the rows, and the 1,2,3,4 refers to the columns.
To change the order as in the above question do df2[,c(1,3,2,4)]
If you want to output this file as a csv, do write.csv(df2, file="somedf.csv")
The three top-rated answers have a weakness.
If your dataframe looks like this
df <- data.frame(Time=c(1,2), In=c(2,3), Out=c(3,4), Files=c(4,5))
> df
Time In Out Files
1 1 2 3 4
2 2 3 4 5
then it's a poor solution to use
> df2[,c(1,3,2,4)]
It does the job, but you have just introduced a dependence on the order of the columns in your input.
This style of brittle programming is to be avoided.
The explicit naming of the columns is a better solution
data[,c("Time", "Out", "In", "Files")]
Plus, if you intend to reuse your code in a more general setting, you can simply
out.column.name <- "Out"
in.column.name <- "In"
data[,c("Time", out.column.name, in.column.name, "Files")]
which is also quite nice because it fully isolates literals. By contrast, if you use dplyr's select
data <- data %>% select(Time, out, In, Files)
then you'd be setting up those who will read your code later, yourself included, for a bit of a deception. The column names are being used as literals without appearing in the code as such.
data.table::setcolorder(table, c("Out", "in", "files"))
Source: Stackoverflow.com