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Catherine
Catherine

Reputation: 5407

How does one reorder columns in a data frame?

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

Upvotes: 396

Views: 890766

Answers (12)

Pau
Pau

Reputation: 71

Dplyr has a function that allows you to move specific columns to before or after other columns. That is a critical tool when you work with big data frameworks (if it is 4 columns, it's faster to use select as mentioned before).

https://dplyr.tidyverse.org/reference/relocate.html

In your case, it would be:

df <- df %>% relocate(Out, .after = In)

Simple and elegant. It also allows you to move several columns together and move it to the beginning or to the end:

df <- df %>% relocate(any_of(c('ColX', 'ColY', 'ColZ')), .after = last_col())

Again: super powerful when you work with big dataframes :)

Upvotes: 4

Xavier Guardiola
Xavier Guardiola

Reputation: 2799

# reorder by column name
data <- data[, c("A", "B", "C")] # leave the row index blank to keep all rows

#reorder by column index
data <- data[, c(1,3,2)] # leave the row index blank to keep all rows

Upvotes: 220

usct01
usct01

Reputation: 898

You can use the data.table package:

How to reorder data.table columns (without copying)

require(data.table)
setcolorder(DT,myOrder)

Upvotes: 23

lroha
lroha

Reputation: 34291

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)

Upvotes: 45

Ben G
Ben G

Reputation: 4328

A dplyr solution (part of the tidyverse package set) is to use select:

select(table, "Time", "Out", "In", "Files") 

# or

select(table, Time, Out, In, Files)

Upvotes: 43

Hossein Noorazar
Hossein Noorazar

Reputation: 133

data.table::setcolorder(table, c("Out", "in", "files"))

Upvotes: 3

Vrokipal
Vrokipal

Reputation: 824

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.

Upvotes: 19

Cybernetic
Cybernetic

Reputation: 13334

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.

Upvotes: 1

landroni
landroni

Reputation: 2988

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

Upvotes: 50

richiemorrisroe
richiemorrisroe

Reputation: 9493

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")

Upvotes: 418

user3482899
user3482899

Reputation: 320

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.

Upvotes: 27

dalloliogm
dalloliogm

Reputation: 8940

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.

Upvotes: 125

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