Sort Frequency of observations as decreasing in R [duplicate] - r

Question

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

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

Answer 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 sort.data.frame to a CRAN package, making it class compatible as discussed here:
Best way to create generic/method consistency for sort.data.frame?
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))
library(taRifx)
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: https://chat.stackoverflow.com/transcript/message/1094290#1094290
You can also use the arrange() function from plyr as Hadley pointed out in the above thread:
library(plyr)
arrange(dd,desc(z),b)
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 sort.data.frame 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))
library(microbenchmark)
# Reload R between benchmarks
microbenchmark(dd[with(dd, order(-z, b)), ] ,
dd[order(-dd$z, dd$b),],
times=1000
)
Median times:
dd[with(dd, order(-z, b)), ] 778
dd[order(-dd$z, dd$b),] 788
library(taRifx)
microbenchmark(sort(dd, f= ~-z+b ),times=1000)
Median time: 1,567
library(plyr)
microbenchmark(arrange(dd,desc(z),b),times=1000)
Median time: 862
library(doBy)
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.
library(Deducer)
microbenchmark(sortData(dd,c("z","b"),increasing= c(FALSE,TRUE)),times=1000)
Couldn't make Deducer load. Needs JGR console.
esort <- function(x, sortvar, ...) {
attach(x)
x <- x[with(x,order(sortvar,...)),]
return(x)
detach(x)
}
microbenchmark(esort(dd, -z, b),times=1000)
Doesn't appear to be compatible with microbenchmark due to the attach/detach.
m <- microbenchmark(
arrange(dd,desc(z),b),
sort(dd, f= ~-z+b ),
dd[with(dd, order(-z, b)), ] ,
dd[order(-dd$z, dd$b),],
times=1000
)
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.

Answer 3

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 :
quarterlyreport[with(quarterlyreport,order(-z,b)),]
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 :
quarterlyreport[with(lastquarterlyreport,order(-z,b)),]
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
quarterlyreport[with(lastquarterlyreport,order(-z,b)),]
it's just
quarterlyreport[order(-z,b)]
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.

Answer 4

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):
library(dplyr)
# 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

Answer 5

The R package data.table provides both fast and memory efficient ordering of data.tables with a straightforward syntax (a part of which Matt has highlighted quite nicely in his answer). There has been quite a lot of improvements and also a new function setorder() since then. From v1.9.5+, setorder() also works with data.frames.
First, we'll create a dataset big enough and benchmark the different methods mentioned from other answers and then list the features of data.table.
Data:
require(plyr)
require(doBy)
require(data.table)
require(dplyr)
require(taRifx)
set.seed(45L)
dat = data.frame(b = as.factor(sample(c("Hi", "Med", "Low"), 1e8, TRUE)),
x = sample(c("A", "D", "C"), 1e8, TRUE),
y = sample(100, 1e8, TRUE),
z = sample(5, 1e8, TRUE),
stringsAsFactors = FALSE)
Benchmarks:
The timings reported are from running system.time(...) on these functions shown below. The timings are tabulated below (in the order of slowest to fastest).
orderBy( ~ -z + b, data = dat) ## doBy
plyr::arrange(dat, desc(z), b) ## plyr
arrange(dat, desc(z), b) ## dplyr
sort(dat, f = ~ -z + b) ## taRifx
dat[with(dat, order(-z, b)), ] ## base R
# convert to data.table, by reference
setDT(dat)
dat[order(-z, b)] ## data.table, base R like syntax
setorder(dat, -z, b) ## data.table, using setorder()
## setorder() now also works with data.frames
# R-session memory usage (BEFORE) = ~2GB (size of 'dat')
# ------------------------------------------------------------
# Package function Time (s) Peak memory Memory used
# ------------------------------------------------------------
# doBy orderBy 409.7 6.7 GB 4.7 GB
# taRifx sort 400.8 6.7 GB 4.7 GB
# plyr arrange 318.8 5.6 GB 3.6 GB
# base R order 299.0 5.6 GB 3.6 GB
# dplyr arrange 62.7 4.2 GB 2.2 GB
# ------------------------------------------------------------
# data.table order 6.2 4.2 GB 2.2 GB
# data.table setorder 4.5 2.4 GB 0.4 GB
# ------------------------------------------------------------
data.table's DT[order(...)] syntax was ~10x faster than the fastest of other methods (dplyr), while consuming the same amount of memory as dplyr.
data.table's setorder() was ~14x faster than the fastest of other methods (dplyr), while taking just 0.4GB extra memory. dat is now in the order we require (as it is updated by reference).
data.table features:
Speed:
data.table's ordering is extremely fast because it implements radix ordering.
The syntax DT[order(...)] is optimised internally to use data.table's fast ordering as well. You can keep using the familiar base R syntax but speed up the process (and use less memory).
Memory:
Most of the times, we don't require the original data.frame or data.table after reordering. That is, we usually assign the result back to the same object, for example:
DF <- DF[order(...)]
The issue is that this requires at least twice (2x) the memory of the original object. To be memory efficient, data.table therefore also provides a function setorder().
setorder() reorders data.tables by reference (in-place), without making any additional copies. It only uses extra memory equal to the size of one column.
Other features:
It supports integer, logical, numeric, character and even bit64::integer64 types.
Note that factor, Date, POSIXct etc.. classes are all integer/numeric types underneath with additional attributes and are therefore supported as well.
In base R, we can not use - on a character vector to sort by that column in decreasing order. Instead we have to use -xtfrm(.).
However, in data.table, we can just do, for example, dat[order(-x)] or setorder(dat, -x).

Answer 6

With this (very helpful) function by Kevin Wright, posted in the tips section of the R wiki, this is easily achieved.
sort(dd,by = ~ -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

Answer 7

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.

Answer 8

or you can use package doBy
library(doBy)
dd <- orderBy(~-z+b, data=dd)

Answer 9

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

Answer 10

Alternatively, using the package Deducer
library(Deducer)
dd<- sortData(dd,c("z","b"),increasing= c(FALSE,TRUE))

Answer 11

In response to a comment added in the OP for how to sort programmatically:
Using dplyr and data.table
library(dplyr)
library(data.table)
dplyr
Just use arrange_, which is the Standard Evaluation version for arrange.
df1 <- tbl_df(iris)
#using strings or formula
arrange_(df1, c('Petal.Length', 'Petal.Width'))
arrange_(df1, ~Petal.Length, ~Petal.Width)
Source: local data frame [150 x 5]
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
(dbl) (dbl) (dbl) (dbl) (fctr)
1 4.6 3.6 1.0 0.2 setosa
2 4.3 3.0 1.1 0.1 setosa
3 5.8 4.0 1.2 0.2 setosa
4 5.0 3.2 1.2 0.2 setosa
5 4.7 3.2 1.3 0.2 setosa
6 5.4 3.9 1.3 0.4 setosa
7 5.5 3.5 1.3 0.2 setosa
8 4.4 3.0 1.3 0.2 setosa
9 5.0 3.5 1.3 0.3 setosa
10 4.5 2.3 1.3 0.3 setosa
.. ... ... ... ... ...
#Or using a variable
sortBy <- c('Petal.Length', 'Petal.Width')
arrange_(df1, .dots = sortBy)
Source: local data frame [150 x 5]
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
(dbl) (dbl) (dbl) (dbl) (fctr)
1 4.6 3.6 1.0 0.2 setosa
2 4.3 3.0 1.1 0.1 setosa
3 5.8 4.0 1.2 0.2 setosa
4 5.0 3.2 1.2 0.2 setosa
5 4.7 3.2 1.3 0.2 setosa
6 5.5 3.5 1.3 0.2 setosa
7 4.4 3.0 1.3 0.2 setosa
8 4.4 3.2 1.3 0.2 setosa
9 5.0 3.5 1.3 0.3 setosa
10 4.5 2.3 1.3 0.3 setosa
.. ... ... ... ... ...
#Doing the same operation except sorting Petal.Length in descending order
sortByDesc <- c('desc(Petal.Length)', 'Petal.Width')
arrange_(df1, .dots = sortByDesc)
more info here: https://cran.r-project.org/web/packages/dplyr/vignettes/nse.html
It is better to use formula as it also captures the environment to evaluate an expression in
data.table
dt1 <- data.table(iris) #not really required, as you can work directly on your data.frame
sortBy <- c('Petal.Length', 'Petal.Width')
sortType <- c(-1, 1)
setorderv(dt1, sortBy, sortType)
dt1
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1: 7.7 2.6 6.9 2.3 virginica
2: 7.7 2.8 6.7 2.0 virginica
3: 7.7 3.8 6.7 2.2 virginica
4: 7.6 3.0 6.6 2.1 virginica
5: 7.9 3.8 6.4 2.0 virginica
---
146: 5.4 3.9 1.3 0.4 setosa
147: 5.8 4.0 1.2 0.2 setosa
148: 5.0 3.2 1.2 0.2 setosa
149: 4.3 3.0 1.1 0.1 setosa
150: 4.6 3.6 1.0 0.2 setosa

Answer 12

The arrange() in dplyr is my favorite option. Use the pipe operator and go from least important to most important aspect
dd1 <- dd %>%
arrange(z) %>%
arrange(desc(x))

Answer 13

I learned about order with the following example which then confused me for a long time:
set.seed(1234)
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),]
databyAge
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:
my.data <- 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 = my.data[order(age),]
The following line works because order sorts on the column age in my.data.
databyage = my.data[order(my.data$age),]
I thought this was worth posting given how confused I was by this example for so long. If this post is not deemed appropriate for the thread I can remove it.
EDIT: May 13, 2014
Below is a generalized way of sorting a data frame by every column without specifying column names. The code below shows how to sort from left to right or by right to left. This works if every column is numeric. I have not tried with a character column added.
I found the do.call code a month or two ago in an old post on a different site, but only after extensive and difficult searching. I am not sure I could relocate that post now. The present thread is the first hit for ordering a data.frame in R. So, I thought my expanded version of that original do.call code might be useful.
set.seed(1234)
v1 <- c(0,0,0,0, 0,0,0,0, 1,1,1,1, 1,1,1,1)
v2 <- c(0,0,0,0, 1,1,1,1, 0,0,0,0, 1,1,1,1)
v3 <- c(0,0,1,1, 0,0,1,1, 0,0,1,1, 0,0,1,1)
v4 <- c(0,1,0,1, 0,1,0,1, 0,1,0,1, 0,1,0,1)
df.1 <- data.frame(v1, v2, v3, v4)
df.1
rdf.1 <- df.1[sample(nrow(df.1), nrow(df.1), replace = FALSE),]
rdf.1
order.rdf.1 <- rdf.1[do.call(order, as.list(rdf.1)),]
order.rdf.1
order.rdf.2 <- rdf.1[do.call(order, rev(as.list(rdf.1))),]
order.rdf.2
rdf.3 <- data.frame(rdf.1$v2, rdf.1$v4, rdf.1$v3, rdf.1$v1)
rdf.3
order.rdf.3 <- rdf.1[do.call(order, as.list(rdf.3)),]
order.rdf.3

Answer 14

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)), ]

Answer 15

Just for the sake of completeness, since not much has been said about sorting by column numbers... It can surely be argued that it is often not desirable (because the order of the columns could change, paving the way to errors), but in some specific situations (when for instance you need a quick job done and there is no such risk of columns changing orders), it might be the most sensible thing to do, especially when dealing with large numbers of columns.
In that case, do.call() comes to the rescue:
ind <- do.call(what = "order", args = iris[,c(5,1,2,3)])
iris[ind, ]
## Sepal.Length Sepal.Width Petal.Length Petal.Width Species
## 14 4.3 3.0 1.1 0.1 setosa
## 9 4.4 2.9 1.4 0.2 setosa
## 39 4.4 3.0 1.3 0.2 setosa
## 43 4.4 3.2 1.3 0.2 setosa
## 42 4.5 2.3 1.3 0.3 setosa
## 4 4.6 3.1 1.5 0.2 setosa
## 48 4.6 3.2 1.4 0.2 setosa
## 7 4.6 3.4 1.4 0.3 setosa
## (...)

Answer 16

For the sake of completeness: you can also use the sortByCol() function from the BBmisc package:
library(BBmisc)
sortByCol(dd, c("z", "b"), asc = c(FALSE, TRUE))
b x y z
4 Low C 9 2
2 Med D 3 1
1 Hi A 8 1
3 Hi A 9 1
Performance comparison:
library(microbenchmark)
microbenchmark(sortByCol(dd, c("z", "b"), asc = c(FALSE, TRUE)), times = 100000)
median 202.878
library(plyr)
microbenchmark(arrange(dd,desc(z),b),times=100000)
median 148.758
microbenchmark(dd[with(dd, order(-z, b)), ], times = 100000)
median 115.872

Answer 17

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

Answer 18

Another alternative, using the rgr package:
> library(rgr)
> gx.sort.df(dd, ~ -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

Answer 19

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