value - r sort one column by another

How to sort a dataframe by multiple column(s)? (12)

I want to sort a data.frame by multiple columns. For example, with the data.frame below I would like to sort by column z (descending) then by column b (ascending):

dd <- data.frame(b = factor(c("Hi", "Med", "Hi", "Low"), 
      levels = c("Low", "Med", "Hi"), ordered = TRUE),
      x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),
      z = c(1, 1, 1, 2))
    b x y z
1  Hi A 8 1
2 Med D 3 1
3  Hi A 9 1
4 Low C 9 2

Your choices

  • order from base
  • arrange from dplyr
  • setorder and setorderv from data.table
  • arrange from plyr
  • sort from taRifx
  • orderBy from doBy
  • sortData from Deducer

Most of the time you should use the dplyr or data.table solutions, unless having no-dependencies is important, in which case use base::order.

I recently added to a CRAN package, making it class compatible as discussed here: Best way to create generic/method consistency for

Therefore, given the data.frame dd, you can sort as follows:

dd <- data.frame(b = factor(c("Hi", "Med", "Hi", "Low"), 
      levels = c("Low", "Med", "Hi"), ordered = TRUE),
      x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),
      z = c(1, 1, 1, 2))
sort(dd, f= ~ -z + b )

If you are one of the original authors of this function, please contact me. Discussion as to public domaininess is here:

You can also use the arrange() function from plyr as Hadley pointed out in the above thread:


Benchmarks: Note that I loaded each package in a new R session since there were a lot of conflicts. In particular loading the doBy package causes sort to return "The following object(s) are masked from 'x (position 17)': b, x, y, z", and loading the Deducer package overwrites from Kevin Wright or the taRifx package.

#Load each time
dd <- data.frame(b = factor(c("Hi", "Med", "Hi", "Low"), 
      levels = c("Low", "Med", "Hi"), ordered = TRUE),
      x = c("A", "D", "A", "C"), y = c(8, 3, 9, 9),
      z = c(1, 1, 1, 2))

# Reload R between benchmarks
microbenchmark(dd[with(dd, order(-z, b)), ] ,
    dd[order(-dd$z, dd$b),],

Median times:

dd[with(dd, order(-z, b)), ] 778

dd[order(-dd$z, dd$b),] 788

microbenchmark(sort(dd, f= ~-z+b ),times=1000)

Median time: 1,567


Median time: 862

microbenchmark(orderBy(~-z+b, data=dd),times=1000)

Median time: 1,694

Note that doBy takes a good bit of time to load the package.

microbenchmark(sortData(dd,c("z","b"),increasing= c(FALSE,TRUE)),times=1000)

Couldn't make Deducer load. Needs JGR console.

esort <- function(x, sortvar, ...) {
x <- x[with(x,order(sortvar,...)),]

microbenchmark(esort(dd, -z, b),times=1000)

Doesn't appear to be compatible with microbenchmark due to the attach/detach.

m <- microbenchmark(
  sort(dd, f= ~-z+b ),
  dd[with(dd, order(-z, b)), ] ,
  dd[order(-dd$z, dd$b),],

uq <- function(x) { fivenum(x)[4]}  
lq <- function(x) { fivenum(x)[2]}

y_min <- 0 # min(by(m$time,m$expr,lq))
y_max <- max(by(m$time,m$expr,uq)) * 1.05

p <- ggplot(m,aes(x=expr,y=time)) + coord_cartesian(ylim = c( y_min , y_max )) 
p + stat_summary(fun.y=median,fun.ymin = lq, fun.ymax = uq, aes(fill=expr))

(lines extend from lower quartile to upper quartile, dot is the median)

Given these results and weighing simplicity vs. speed, I'd have to give the nod to arrange in the plyr package. It has a simple syntax and yet is almost as speedy as the base R commands with their convoluted machinations. Typically brilliant Hadley Wickham work. My only gripe with it is that it breaks the standard R nomenclature where sorting objects get called by sort(object), but I understand why Hadley did it that way due to issues discussed in the question linked above.

Alternatively, using the package Deducer

dd<- sortData(dd,c("z","b"),increasing= c(FALSE,TRUE))

Dirk's answer is good but if you need the sort to persist you'll want to apply the sort back onto the name of that data frame. Using the example code:

dd <- dd[with(dd, order(-z, b)), ] 

Dirk's answer is great. It also highlights a key difference in the syntax used for indexing data.frames and data.tables:

## The data.frame way
dd[with(dd, order(-z, b)), ]

## The data.table way: (7 fewer characters, but that's not the important bit)
dd[order(-z, b)]

The difference between the two calls is small, but it can have important consequences. Especially if you write production code and/or are concerned with correctness in your research, it's best to avoid unnecessary repetition of variable names. data.table helps you do this.

Here's an example of how repetition of variable names might get you into trouble:

Let's change the context from Dirk's answer, and say this is part of a bigger project where there are a lot of object names and they are long and meaningful; instead of dd it's called quarterlyreport. It becomes :


Ok, fine. Nothing wrong with that. Next your boss asks you to include last quarter's report in the report. You go through your code, adding an object lastquarterlyreport in various places and somehow (how on earth?) you end up with this :


That isn't what you meant but you didn't spot it because you did it fast and it's nestled on a page of similar code. The code doesn't fall over (no warning and no error) because R thinks it is what you meant. You'd hope whoever reads your report spots it, but maybe they don't. If you work with programming languages a lot then this situation may be all to familiar. It was a "typo" you'll say. I'll fix the "typo" you'll say to your boss.

In data.table we're concerned about tiny details like this. So we've done something simple to avoid typing variable names twice. Something very simple. i is evaluated within the frame of dd already, automatically. You don't need with() at all.

Instead of

dd[with(dd, order(-z, b)), ]

it's just

dd[order(-z, b)]

And instead of


it's just


It's a very small difference, but it might just save your neck one day. When weighing up the different answers to this question, consider counting the repetitions of variable names as one of your criteria in deciding. Some answers have quite a few repeats, others have none.

I learned about order with the following example which then confused me for a long time:


ID        = 1:10
Age       = round(rnorm(10, 50, 1))
diag      = c("Depression", "Bipolar")
Diagnosis = sample(diag, 10, replace=TRUE)

data = data.frame(ID, Age, Diagnosis)

databyAge = data[order(Age),]

The only reason this example works is because order is sorting by the vector Age, not by the column named Age in the data frame data.

To see this create an identical data frame using read.table with slightly different column names and without making use of any of the above vectors: <- read.table(text = '

  id age  diagnosis
   1  49 Depression
   2  50 Depression
   3  51 Depression
   4  48 Depression
   5  50 Depression
   6  51    Bipolar
   7  49    Bipolar
   8  49    Bipolar
   9  49    Bipolar
  10  49 Depression

', header = TRUE)

The above line structure for order no longer works because there is no vector named age:

databyage =[order(age),]

The following line works because order sorts on the column age in

databyage =[order($age),]

I thought this was worth posting given how confused I was by this example for so long. If this post is not deemed appropriate for the thread I can remove it.

EDIT: May 13, 2014

Below is a generalized way of sorting a data frame by every column without specifying column names. The code below shows how to sort from left to right or by right to left. This works if every column is numeric. I have not tried with a character column added.

I found the code a month or two ago in an old post on a different site, but only after extensive and difficult searching. I am not sure I could relocate that post now. The present thread is the first hit for ordering a data.frame in R. So, I thought my expanded version of that original code might be useful.


v1  <- c(0,0,0,0, 0,0,0,0, 1,1,1,1, 1,1,1,1)
v2  <- c(0,0,0,0, 1,1,1,1, 0,0,0,0, 1,1,1,1)
v3  <- c(0,0,1,1, 0,0,1,1, 0,0,1,1, 0,0,1,1)
v4  <- c(0,1,0,1, 0,1,0,1, 0,1,0,1, 0,1,0,1)

df.1 <- data.frame(v1, v2, v3, v4) 

rdf.1 <- df.1[sample(nrow(df.1), nrow(df.1), replace = FALSE),]

order.rdf.1 <- rdf.1[, as.list(rdf.1)),]

order.rdf.2 <- rdf.1[, rev(as.list(rdf.1))),]

rdf.3 <- data.frame(rdf.1$v2, rdf.1$v4, rdf.1$v3, rdf.1$v1) 

order.rdf.3 <- rdf.1[, as.list(rdf.3)),]

I was struggling with the above solutions when I wanted to automate my ordering process for n columns, whose column names could be different each time. I found a super helpful function from the psych package to do this in a straightforward manner:

dfOrder(myDf, columnIndices)

where columnIndices are indices of one or more columns, in the order in which you want to sort them. More information here:

dfOrder function from 'psych' package

Just like the mechanical card sorters of long ago, first sort by the least significant key, then the next most significant, etc. No library required, works with any number of keys and any combination of ascending and descending keys.

 dd <- dd[order(dd$b, decreasing = FALSE),]

Now we're ready to do the most significant key. The sort is stable, and any ties in the most significant key have already been resolved.

dd <- dd[order(dd$z, decreasing = TRUE),]

This may not be the fastest, but it is certainly simple and reliable

Suppose you have a data.frame A and you want to sort it using column called x descending order. Call the sorted data.frame newdata

newdata <- A[order(-A$x),]

If you want ascending order then replace "-" with nothing. You can have something like

newdata <- A[order(-A$x, A$y, -A$z),]

where x and z are some columns in data.frame A. This means sort data.frame A by x descending, y ascending and z descending.

The arrange() in dplyer is my favorite option. Use the pipe operator and go from least important to most important aspect

dd1 <- dd %>%
    arrange(z) %>%

There are a lot of excellent answers here, but dplyr gives the only syntax that I can quickly and easily remember (and so now use very often):

# sort mtcars by mpg, ascending... use desc(mpg) for descending
arrange(mtcars, mpg)
# sort mtcars first by mpg, then by cyl, then by wt)
arrange(mtcars , mpg, cyl, wt)

For the OP's problem:

arrange(dd, desc(z),  b)

    b x y z
1 Low C 9 2
2 Med D 3 1
3  Hi A 8 1
4  Hi A 9 1

You can use the order() function directly without resorting to add-on tools -- see this simpler answer which uses a trick right from the top of the example(order) code:

R> dd[with(dd, order(-z, b)), ]
    b x y z
4 Low C 9 2
2 Med D 3 1
1  Hi A 8 1
3  Hi A 9 1

Edit some 2+ years later: It was just asked how to do this by column index. The answer is to simply pass the desired sorting column(s) to the order() function:

R> dd[order(-dd[,4], dd[,1]), ]
    b x y z
4 Low C 9 2
2 Med D 3 1
1  Hi A 8 1
3  Hi A 9 1

rather than using the name of the column (and with() for easier/more direct access).

if SQL comes naturally to you, sqldf handles ORDER BY as Codd intended.