Whenever I want to do something "map"py in R, I usually try to use a function in the apply family.
However, I've never quite understood the differences between them -- how {sapply, lapply, etc.} apply the function to the input/grouped input, what the output will look like, or even what the input can be -- so I often just go through them all until I get what I want.
Can someone explain how to use which one when?
My current (probably incorrect/incomplete) understanding is...
sapply(vec, f): input is a vector. output is a vector/matrix, where elementiisf(vec[i]), giving you a matrix iffhas a multi-element outputlapply(vec, f): same assapply, but output is a list?apply(matrix, 1/2, f): input is a matrix. output is a vector, where elementiis f(row/col i of the matrix)tapply(vector, grouping, f): output is a matrix/array, where an element in the matrix/array is the value offat a groupinggof the vector, andggets pushed to the row/col namesby(dataframe, grouping, f): letgbe a grouping. applyfto each column of the group/dataframe. pretty print the grouping and the value offat each column.aggregate(matrix, grouping, f): similar toby, but instead of pretty printing the output, aggregate sticks everything into a dataframe.
Side question: I still haven't learned plyr or reshape -- would plyr or reshape replace all of these entirely?
Best Answer
R has many *apply functions which are ably described in the help files (e.g. ?apply). There are enough of them, though, that beginning useRs may have difficulty deciding which one is appropriate for their situation or even remembering them all. They may have a general sense that "I should be using an *apply function here", but it can be tough to keep them all straight at first.
Despite the fact (noted in other answers) that much of the functionality of the *apply family is covered by the extremely popular plyr package, the base functions remain useful and worth knowing.
This answer is intended to act as a sort of signpost for new useRs to help direct them to the correct *apply function for their particular problem. Note, this is not intended to simply regurgitate or replace the R documentation! The hope is that this answer helps you to decide which *apply function suits your situation and then it is up to you to research it further. With one exception, performance differences will not be addressed.
apply - When you want to apply a function to the rows or columnsof a matrix (and higher-dimensional analogues); not generally advisable for data frames as it will coerce to a matrix first.
# Two dimensional matrixM <- matrix(seq(1,16), 4, 4)# apply min to rowsapply(M, 1, min)[1] 1 2 3 4# apply max to columnsapply(M, 2, max)[1] 4 8 12 16# 3 dimensional arrayM <- array( seq(32), dim = c(4,4,2))# Apply sum across each M[*, , ] - i.e Sum across 2nd and 3rd dimensionapply(M, 1, sum)# Result is one-dimensional[1] 120 128 136 144# Apply sum across each M[*, *, ] - i.e Sum across 3rd dimensionapply(M, c(1,2), sum)# Result is two-dimensional[,1] [,2] [,3] [,4][1,] 18 26 34 42[2,] 20 28 36 44[3,] 22 30 38 46[4,] 24 32 40 48If you want row/column means or sums for a 2D matrix, be sure toinvestigate the highly optimized, lightning-quick
colMeans,rowMeans,colSums,rowSums.lapply - When you want to apply a function to each element of alist in turn and get a list back.
This is the workhorse of many of the other *apply functions. Peelback their code and you will often find
lapplyunderneath.x <- list(a = 1, b = 1:3, c = 10:100) lapply(x, FUN = length) $a [1] 1$b [1] 3$c [1] 91lapply(x, FUN = sum) $a [1] 1$b [1] 6$c [1] 5005sapply - When you want to apply a function to each element of alist in turn, but you want a vector back, rather than a list.
If you find yourself typing
unlist(lapply(...)), stop and considersapply.x <- list(a = 1, b = 1:3, c = 10:100)# Compare with above; a named vector, not a list sapply(x, FUN = length) a b c 1 3 91sapply(x, FUN = sum) a b c 1 6 5005In more advanced uses of
sapplyit will attempt to coerce theresult to a multi-dimensional array, if appropriate. For example, if our function returns vectors of the same length,sapplywill use them as columns of a matrix:sapply(1:5,function(x) rnorm(3,x))If our function returns a 2 dimensional matrix,
sapplywill do essentially the same thing, treating each returned matrix as a single long vector:sapply(1:5,function(x) matrix(x,2,2))Unless we specify
simplify = "array", in which case it will use the individual matrices to build a multi-dimensional array:sapply(1:5,function(x) matrix(x,2,2), simplify = "array")Each of these behaviors is of course contingent on our function returning vectors or matrices of the same length or dimension.
vapply - When you want to use
sapplybut perhaps need tosqueeze some more speed out of your code or want more type safety.For
vapply, you basically give R an example of what sort of thingyour function will return, which can save some time coercing returnedvalues to fit in a single atomic vector.x <- list(a = 1, b = 1:3, c = 10:100)#Note that since the advantage here is mainly speed, this# example is only for illustration. We're telling R that# everything returned by length() should be an integer of # length 1. vapply(x, FUN = length, FUN.VALUE = 0L) a b c 1 3 91mapply - For when you have several data structures (e.g.vectors, lists) and you want to apply a function to the 1st elementsof each, and then the 2nd elements of each, etc., coercing the resultto a vector/array as in
sapply.This is multivariate in the sense that your function must acceptmultiple arguments.
#Sums the 1st elements, the 2nd elements, etc. mapply(sum, 1:5, 1:5, 1:5) [1] 3 6 9 12 15#To do rep(1,4), rep(2,3), etc.mapply(rep, 1:4, 4:1) [[1]][1] 1 1 1 1[[2]][1] 2 2 2[[3]][1] 3 3[[4]][1] 4Map - A wrapper to
mapplywithSIMPLIFY = FALSE, so it is guaranteed to return a list.Map(sum, 1:5, 1:5, 1:5)[[1]][1] 3[[2]][1] 6[[3]][1] 9[[4]][1] 12[[5]][1] 15rapply - For when you want to apply a function to each element of a nested list structure, recursively.
To give you some idea of how uncommon
rapplyis, I forgot about it when first posting this answer! Obviously, I'm sure many people use it, but YMMV.rapplyis best illustrated with a user-defined function to apply:# Append ! to string, otherwise incrementmyFun <- function(x){if(is.character(x)){return(paste(x,"!",sep=""))}else{return(x + 1)}}#A nested list structurel <- list(a = list(a1 = "Boo", b1 = 2, c1 = "Eeek"), b = 3, c = "Yikes", d = list(a2 = 1, b2 = list(a3 = "Hey", b3 = 5)))# Result is named vector, coerced to character rapply(l, myFun)# Result is a nested list like l, with values alteredrapply(l, myFun, how="replace")tapply - For when you want to apply a function to subsets of avector and the subsets are defined by some other vector, usually afactor.
The black sheep of the *apply family, of sorts. The help file's use ofthe phrase "ragged array" can be a bit confusing, but it is actuallyquite simple.
A vector:
x <- 1:20A factor (of the same length!) defining groups:
y <- factor(rep(letters[1:5], each = 4))Add up the values in
xwithin each subgroup defined byy:tapply(x, y, sum) a b c d e 10 26 42 58 74More complex examples can be handled where the subgroups are definedby the unique combinations of a list of several factors.
tapplyissimilar in spirit to the split-apply-combine functions that arecommon in R (aggregate,by,ave,ddply, etc.) Hence itsblack sheep status.
On the side note, here is how the various plyr functions correspond to the base *apply functions (from the intro to plyr document from the plyr webpage http://had.co.nz/plyr/)
Base function Input Output plyr function ---------------------------------------aggregate d d ddply + colwise apply a a/l aaply / alply by d l dlply lapply l l llply mapply a a/l maply / mlply replicate r a/l raply / rlply sapply l a laply One of the goals of plyr is to provide consistent naming conventions for each of the functions, encoding the input and output data types in the function name. It also provides consistency in output, in that output from dlply() is easily passable to ldply() to produce useful output, etc.
Conceptually, learning plyr is no more difficult than understanding the base *apply functions.
plyr and reshape functions have replaced almost all of these functions in my every day use. But, also from the Intro to Plyr document:
Related functions
tapplyandsweephave no corresponding function inplyr, and remain useful.mergeis useful for combining summaries with the original data.
From slide 21 of http://www.slideshare.net/hadley/plyr-one-data-analytic-strategy:
(Hopefully it's clear that apply corresponds to @Hadley's aaply and aggregate corresponds to @Hadley's ddply etc. Slide 20 of the same slideshare will clarify if you don't get it from this image.)
(on the left is input, on the top is output)
First start with Joran's excellent answer -- doubtful anything can better that.
Then the following mnemonics may help to remember the distinctions between each. Whilst some are obvious, others may be less so --- for these you'll find justification in Joran's discussions.
Mnemonics
lapplyis a list apply which acts on a list or vector and returns a list.sapplyis a simplelapply(function defaults to returning a vector or matrix when possible)vapplyis a verified apply (allows the return object type to be prespecified)rapplyis a recursive apply for nested lists, i.e. lists within liststapplyis a tagged apply where the tags identify the subsetsapplyis generic: applies a function to a matrix's rows or columns (or, more generally, to dimensions of an array)
Building the Right Background
If using the apply family still feels a bit alien to you, then it might be that you're missing a key point of view.
These two articles can help. They provide the necessary background to motivate the functional programming techniques that are being provided by the apply family of functions.
Users of Lisp will recognise the paradigm immediately. If you're not familiar with Lisp, once you get your head around FP, you'll have gained a powerful point of view for use in R -- and apply will make a lot more sense.
- Advanced R: Functional Programming, by Hadley Wickham
- Simple Functional Programming in R, by Michael Barton
Since I realized that (the very excellent) answers of this post lack of by and aggregate explanations. Here is my contribution.
BY
The by function, as stated in the documentation can be though, as a "wrapper" for tapply. The power of by arises when we want to compute a task that tapply can't handle. One example is this code:
ct <- tapply(iris$Sepal.Width , iris$Species , summary )cb <- by(iris$Sepal.Width , iris$Species , summary )cbiris$Species: setosaMin. 1st Qu. Median Mean 3rd Qu. Max. 2.300 3.200 3.400 3.428 3.675 4.400 -------------------------------------------------------------- iris$Species: versicolorMin. 1st Qu. Median Mean 3rd Qu. Max. 2.000 2.525 2.800 2.770 3.000 3.400 -------------------------------------------------------------- iris$Species: virginicaMin. 1st Qu. Median Mean 3rd Qu. Max. 2.200 2.800 3.000 2.974 3.175 3.800 ct$setosaMin. 1st Qu. Median Mean 3rd Qu. Max. 2.300 3.200 3.400 3.428 3.675 4.400 $versicolorMin. 1st Qu. Median Mean 3rd Qu. Max. 2.000 2.525 2.800 2.770 3.000 3.400 $virginicaMin. 1st Qu. Median Mean 3rd Qu. Max. 2.200 2.800 3.000 2.974 3.175 3.800 If we print these two objects, ct and cb, we "essentially" have the same results and the only differences are in how they are shown and the different class attributes, respectively by for cb and array for ct.
As I've said, the power of by arises when we can't use tapply; the following code is one example:
tapply(iris, iris$Species, summary )Error in tapply(iris, iris$Species, summary) : arguments must have same lengthR says that arguments must have the same lengths, say "we want to calculate the summary of all variable in iris along the factor Species": but R just can't do that because it does not know how to handle.
With the by function R dispatch a specific method for data frame class and then let the summary function works even if the length of the first argument (and the type too) are different.
bywork <- by(iris, iris$Species, summary )byworkiris$Species: setosaSepal.Length Sepal.Width Petal.Length Petal.Width Species Min. :4.300 Min. :2.300 Min. :1.000 Min. :0.100 setosa :50 1st Qu.:4.800 1st Qu.:3.200 1st Qu.:1.400 1st Qu.:0.200 versicolor: 0 Median :5.000 Median :3.400 Median :1.500 Median :0.200 virginica : 0 Mean :5.006 Mean :3.428 Mean :1.462 Mean :0.246 3rd Qu.:5.200 3rd Qu.:3.675 3rd Qu.:1.575 3rd Qu.:0.300 Max. :5.800 Max. :4.400 Max. :1.900 Max. :0.600 -------------------------------------------------------------- iris$Species: versicolorSepal.Length Sepal.Width Petal.Length Petal.Width Species Min. :4.900 Min. :2.000 Min. :3.00 Min. :1.000 setosa : 0 1st Qu.:5.600 1st Qu.:2.525 1st Qu.:4.00 1st Qu.:1.200 versicolor:50 Median :5.900 Median :2.800 Median :4.35 Median :1.300 virginica : 0 Mean :5.936 Mean :2.770 Mean :4.26 Mean :1.326 3rd Qu.:6.300 3rd Qu.:3.000 3rd Qu.:4.60 3rd Qu.:1.500 Max. :7.000 Max. :3.400 Max. :5.10 Max. :1.800 -------------------------------------------------------------- iris$Species: virginicaSepal.Length Sepal.Width Petal.Length Petal.Width Species Min. :4.900 Min. :2.200 Min. :4.500 Min. :1.400 setosa : 0 1st Qu.:6.225 1st Qu.:2.800 1st Qu.:5.100 1st Qu.:1.800 versicolor: 0 Median :6.500 Median :3.000 Median :5.550 Median :2.000 virginica :50 Mean :6.588 Mean :2.974 Mean :5.552 Mean :2.026 3rd Qu.:6.900 3rd Qu.:3.175 3rd Qu.:5.875 3rd Qu.:2.300 Max. :7.900 Max. :3.800 Max. :6.900 Max. :2.500 it works indeed and the result is very surprising. It is an object of class by that along Species (say, for each of them) computes the summary of each variable.
Note that if the first argument is a data frame, the dispatched function must have a method for that class of objects. For example is we use this code with the mean function we will have this code that has no sense at all:
by(iris, iris$Species, mean)iris$Species: setosa[1] NA------------------------------------------- iris$Species: versicolor[1] NA------------------------------------------- iris$Species: virginica[1] NAWarning messages:1: In mean.default(data[x, , drop = FALSE], ...) :argument is not numeric or logical: returning NA2: In mean.default(data[x, , drop = FALSE], ...) :argument is not numeric or logical: returning NA3: In mean.default(data[x, , drop = FALSE], ...) :argument is not numeric or logical: returning NAAGGREGATE
aggregate can be seen as another a different way of use tapply if we use it in such a way.
at <- tapply(iris$Sepal.Length , iris$Species , mean)ag <- aggregate(iris$Sepal.Length , list(iris$Species), mean)atsetosa versicolor virginica 5.006 5.936 6.588 agGroup.1 x1 setosa 5.0062 versicolor 5.9363 virginica 6.588The two immediate differences are that the second argument of aggregate must be a list while tapply can (not mandatory) be a list and that the output of aggregate is a data frame while the one of tapply is an array.
The power of aggregate is that it can handle easily subsets of the data with subset argument and that it has methods for ts objects and formula as well.
These elements make aggregate easier to work with that tapply in some situations.Here are some examples (available in documentation):
ag <- aggregate(len ~ ., data = ToothGrowth, mean)agsupp dose len1 OJ 0.5 13.232 VC 0.5 7.983 OJ 1.0 22.704 VC 1.0 16.775 OJ 2.0 26.066 VC 2.0 26.14We can achieve the same with tapply but the syntax is slightly harder and the output (in some circumstances) less readable:
att <- tapply(ToothGrowth$len, list(ToothGrowth$dose, ToothGrowth$supp), mean)attOJ VC0.5 13.23 7.981 22.70 16.772 26.06 26.14There are other times when we can't use by or tapply and we have to use aggregate.
ag1 <- aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, mean)ag1Month Ozone Temp1 5 23.61538 66.730772 6 29.44444 78.222223 7 59.11538 83.884624 8 59.96154 83.961545 9 31.44828 76.89655We cannot obtain the previous result with tapply in one call but we have to calculate the mean along Month for each elements and then combine them (also note that we have to call the na.rm = TRUE, because the formula methods of the aggregate function has by default the na.action = na.omit):
ta1 <- tapply(airquality$Ozone, airquality$Month, mean, na.rm = TRUE)ta2 <- tapply(airquality$Temp, airquality$Month, mean, na.rm = TRUE)cbind(ta1, ta2)ta1 ta25 23.61538 65.548396 29.44444 79.100007 59.11538 83.903238 59.96154 83.967749 31.44828 76.90000while with by we just can't achieve that in fact the following function call returns an error (but most likely it is related to the supplied function, mean):
by(airquality[c("Ozone", "Temp")], airquality$Month, mean, na.rm = TRUE)Other times the results are the same and the differences are just in the class (and then how it is shown/printed and not only -- example, how to subset it) object:
byagg <- by(airquality[c("Ozone", "Temp")], airquality$Month, summary)aggagg <- aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, summary)The previous code achieve the same goal and results, at some points what tool to use is just a matter of personal tastes and needs; the previous two objects have very different needs in terms of subsetting.
There are lots of great answers which discuss differences in the use cases for each function. None of the answer discuss the differences in performance. That is reasonable cause various functions expects various input and produces various output, yet most of them have a general common objective to evaluate by series/groups. My answer is going to focus on performance. Due to above the input creation from the vectors is included in the timing, also the apply function is not measured.
I have tested two different functions sum and length at once. Volume tested is 50M on input and 50K on output. I have also included two currently popular packages which were not widely used at the time when question was asked, data.table and dplyr. Both are definitely worth to look if you are aiming for good performance.
library(dplyr)library(data.table)set.seed(123)n = 5e7k = 5e5x = runif(n)grp = sample(k, n, TRUE)timing = list()# sapplytiming[["sapply"]] = system.time({lt = split(x, grp)r.sapply = sapply(lt, function(x) list(sum(x), length(x)), simplify = FALSE)})# lapplytiming[["lapply"]] = system.time({lt = split(x, grp)r.lapply = lapply(lt, function(x) list(sum(x), length(x)))})# tapplytiming[["tapply"]] = system.time(r.tapply <- tapply(x, list(grp), function(x) list(sum(x), length(x))))# bytiming[["by"]] = system.time(r.by <- by(x, list(grp), function(x) list(sum(x), length(x)), simplify = FALSE))# aggregatetiming[["aggregate"]] = system.time(r.aggregate <- aggregate(x, list(grp), function(x) list(sum(x), length(x)), simplify = FALSE))# dplyrtiming[["dplyr"]] = system.time({df = data_frame(x, grp)r.dplyr = summarise(group_by(df, grp), sum(x), n())})# data.tabletiming[["data.table"]] = system.time({dt = setnames(setDT(list(x, grp)), c("x","grp"))r.data.table = dt[, .(sum(x), .N), grp]})# all output size match to group countsapply(list(sapply=r.sapply, lapply=r.lapply, tapply=r.tapply, by=r.by, aggregate=r.aggregate, dplyr=r.dplyr, data.table=r.data.table), function(x) (if(is.data.frame(x)) nrow else length)(x)==k)# sapply lapply tapply by aggregate dplyr data.table # TRUE TRUE TRUE TRUE TRUE TRUE TRUE # print timingsas.data.table(sapply(timing, `[[`, "elapsed"), keep.rownames = TRUE)[,.(fun = V1, elapsed = V2)][order(-elapsed)]# fun elapsed#1: aggregate 109.139#2: by 25.738#3: dplyr 18.978#4: tapply 17.006#5: lapply 11.524#6: sapply 11.326#7: data.table 2.686Despite all the great answers here, there are 2 more base functions that deserve to be mentioned, the useful outer function and the obscure eapply function
outer
outer is a very useful function hidden as a more mundane one. If you read the help for outer its description says:
The outer product of the arrays X and Y is the array A with dimension c(dim(X), dim(Y)) where element A[c(arrayindex.x, arrayindex.y)] = FUN(X[arrayindex.x], Y[arrayindex.y], ...).which makes it seem like this is only useful for linear algebra type things. However, it can be used much like mapply to apply a function to two vectors of inputs. The difference is that mapply will apply the function to the first two elements and then the second two etc, whereas outer will apply the function to every combination of one element from the first vector and one from the second. For example:
A<-c(1,3,5,7,9)B<-c(0,3,6,9,12)mapply(FUN=pmax, A, B)> mapply(FUN=pmax, A, B)[1] 1 3 6 9 12outer(A,B, pmax)> outer(A,B, pmax)[,1] [,2] [,3] [,4] [,5][1,] 1 3 6 9 12[2,] 3 3 6 9 12[3,] 5 5 6 9 12[4,] 7 7 7 9 12[5,] 9 9 9 9 12I have personally used this when I have a vector of values and a vector of conditions and wish to see which values meet which conditions.
eapply
eapply is like lapply except that rather than applying a function to every element in a list, it applies a function to every element in an environment. For example if you want to find a list of user defined functions in the global environment:
A<-c(1,3,5,7,9)B<-c(0,3,6,9,12)C<-list(x=1, y=2)D<-function(x){x+1}> eapply(.GlobalEnv, is.function)$A[1] FALSE$B[1] FALSE$C[1] FALSE$D[1] TRUE Frankly I don't use this very much but if you are building a lot of packages or create a lot of environments it may come in handy.
It is maybe worth mentioning ave. ave is tapply's friendly cousin. It returns results in a form that you can plug straight back into your data frame.
dfr <- data.frame(a=1:20, f=rep(LETTERS[1:5], each=4))means <- tapply(dfr$a, dfr$f, mean)## A B C D E ## 2.5 6.5 10.5 14.5 18.5 ## great, but putting it back in the data frame is another line:dfr$m <- means[dfr$f]dfr$m2 <- ave(dfr$a, dfr$f, FUN=mean) # NB argument name FUN is needed!dfr## a f m m2## 1 A 2.5 2.5## 2 A 2.5 2.5## 3 A 2.5 2.5## 4 A 2.5 2.5## 5 B 6.5 6.5## 6 B 6.5 6.5## 7 B 6.5 6.5## ...There is nothing in the base package that works like ave for whole data frames (as by is like tapply for data frames). But you can fudge it:
dfr$foo <- ave(1:nrow(dfr), dfr$f, FUN=function(x) {x <- dfr[x,]sum(x$m*x$m2)})dfr## a f m m2 foo## 1 1 A 2.5 2.5 25## 2 2 A 2.5 2.5 25## 3 3 A 2.5 2.5 25## ...I recently discovered the rather useful sweep function and add it here for the sake of completeness:
sweep
The basic idea is to sweep through an array row- or column-wise and return a modified array. An example will make this clear (source: datacamp):
Let's say you have a matrix and want to standardize it column-wise:
dataPoints <- matrix(4:15, nrow = 4)# Find means per column with `apply()`dataPoints_means <- apply(dataPoints, 2, mean)# Find standard deviation with `apply()`dataPoints_sdev <- apply(dataPoints, 2, sd)# Center the points dataPoints_Trans1 <- sweep(dataPoints, 2, dataPoints_means,"-")# Return the resultdataPoints_Trans1## [,1] [,2] [,3]## [1,] -1.5 -1.5 -1.5## [2,] -0.5 -0.5 -0.5## [3,] 0.5 0.5 0.5## [4,] 1.5 1.5 1.5# NormalizedataPoints_Trans2 <- sweep(dataPoints_Trans1, 2, dataPoints_sdev, "/")# Return the resultdataPoints_Trans2## [,1] [,2] [,3]## [1,] -1.1618950 -1.1618950 -1.1618950## [2,] -0.3872983 -0.3872983 -0.3872983## [3,] 0.3872983 0.3872983 0.3872983## [4,] 1.1618950 1.1618950 1.1618950NB: for this simple example the same result can of course be achieved more easily by
apply(dataPoints, 2, scale)
In the collapse package recently released on CRAN, I have attempted to compress most of the common apply functionality into just 2 functions:
dapply(Data-Apply) applies functions to rows or (default) columns of matrices and data.frames and (default) returns an object of the same type and with the same attributes (unless the result of each computation is atomic anddrop = TRUE). The performance is comparable tolapplyfor data.frame columns, and about 2x faster thanapplyfor matrix rows or columns. Parallelism is available viamclapply(only for MAC).
Syntax:
dapply(X, FUN, ..., MARGIN = 2, parallel = FALSE, mc.cores = 1L, return = c("same", "matrix", "data.frame"), drop = TRUE)Examples:
# Apply to columns:dapply(mtcars, log)dapply(mtcars, sum)dapply(mtcars, quantile)# Apply to rows:dapply(mtcars, sum, MARGIN = 1)dapply(mtcars, quantile, MARGIN = 1)# Return as matrix:dapply(mtcars, quantile, return = "matrix")dapply(mtcars, quantile, MARGIN = 1, return = "matrix")# Same for matrices ...BYis a S3 generic for split-apply-combine computing with vector, matrix and data.frame method. It is significantly faster thantapply,byandaggregate(an also faster thanplyr, on large datadplyris faster though).
Syntax:
BY(X, g, FUN, ..., use.g.names = TRUE, sort = TRUE,expand.wide = FALSE, parallel = FALSE, mc.cores = 1L,return = c("same", "matrix", "data.frame", "list"))Examples:
# Vectors:BY(iris$Sepal.Length, iris$Species, sum)BY(iris$Sepal.Length, iris$Species, quantile)BY(iris$Sepal.Length, iris$Species, quantile, expand.wide = TRUE) # This returns a matrix # Data.framesBY(iris[-5], iris$Species, sum)BY(iris[-5], iris$Species, quantile)BY(iris[-5], iris$Species, quantile, expand.wide = TRUE) # This returns a wider data.frameBY(iris[-5], iris$Species, quantile, return = "matrix") # This returns a matrix# Same for matrices ...Lists of grouping variables can also be supplied to g.
Talking about performance: A main goal of collapse is to foster high-performance programming in R and to move beyond split-apply-combine alltogether. For this purpose the package has a full set of C++ based fast generic functions: fmean, fmedian, fmode, fsum, fprod, fsd, fvar, fmin, fmax, ffirst, flast, fNobs, fNdistinct, fscale, fbetween, fwithin, fHDbetween, fHDwithin, flag, fdiff and fgrowth. They perform grouped computations in a single pass through the data (i.e. no splitting and recombining).
Syntax:
fFUN(x, g = NULL, [w = NULL,] TRA = NULL, [na.rm = TRUE,] use.g.names = TRUE, drop = TRUE)Examples:
v <- iris$Sepal.Lengthf <- iris$Species# Vectorsfmean(v) # meanfmean(v, f) # grouped meanfsd(v, f) # grouped standard deviationfsd(v, f, TRA = "/") # grouped scalingfscale(v, f) # grouped standardizing (scaling and centering)fwithin(v, f) # grouped demeaningw <- abs(rnorm(nrow(iris)))fmean(v, w = w) # Weighted meanfmean(v, f, w) # Weighted grouped meanfsd(v, f, w) # Weighted grouped standard-deviationfsd(v, f, w, "/") # Weighted grouped scalingfscale(v, f, w) # Weighted grouped standardizingfwithin(v, f, w) # Weighted grouped demeaning# Same using data.frames...fmean(iris[-5], f) # grouped meanfscale(iris[-5], f) # grouped standardizingfwithin(iris[-5], f) # grouped demeaning# Same with matrices ...In the package vignettes I provide benchmarks. Programming with the fast functions is significantly faster than programming with dplyr or data.table, especially on smaller data, but also on large data.
Starting in R 4.3.0, tapply will support data frames and both tapply and by will support grouping data frame rows with a formula.
> R.version.string[1] "R version 4.3.0 beta (2023-04-07 r84200)"> dd <- data.frame(x = 1:10, f = gl(5L, 2L), g = gl(2L, 5L))x f g1 1 1 12 2 1 13 3 2 14 4 2 15 5 3 16 6 3 27 7 4 28 8 4 29 9 5 210 10 5 2> tapply(dd, ~f + g, nrow)gf 1 21 2 02 2 03 1 14 0 25 0 2> by(dd, ~g, identity)g: 1x f g1 1 1 12 2 1 13 3 2 14 4 2 15 5 3 1------------------------------------------------------------ g: 2x f g6 6 3 27 7 4 28 8 4 29 9 5 210 10 5 2There are some alternatives from some packages as well which are not discussed above.
The parApply() function in the parallels package provides an alternative to the apply family of functions for executing parallel computations on a cluster. Other alternatives for parallel computation in R include the foreach package and the doParallel package, which allow for parallel execution of loops and functions. The future package provides a simple and consistent API for using futures, which are a way to evaluate expressions asynchronously, either in parallel or sequentially. Additionally, the purrr package provides a functional programming approach to iteration and mapping, and supports parallelization through the future package.
Here are some examples
parApply() example:
library(parallel)# Create a matrixm <- matrix(1:20, nrow = 5)# Define a function to apply to each column of the matrixmy_fun <- function(x) {x^2}# Apply the function to each column of the matrix in parallelresult <- parApply(cl = makeCluster(2), X = m, MARGIN = 2, FUN = my_fun)# View the resultresultforeach example:
library(foreach)library(doParallel)# Register a parallel backendregisterDoParallel(cores = 2)# Create a list of numbersmy_list <- list(1, 2, 3, 4, 5)# Define a function to apply to each element of the listmy_fun <- function(x) {x^2}# Apply the function to each element of the list in parallelresult <- foreach(i = my_list) %dopar% my_fun(i)# View the resultresultfuture example:
library(future)# Plan to use a parallel backendplan(multisession, workers = 2)# Create a list of numbersmy_list <- list(1, 2, 3, 4, 5)# Define a function to apply to each element of the listmy_fun <- function(x) {x^2}# Apply the function to each element of the list in parallel using futuresresult <- future_map(my_list, my_fun)# View the resultresultpurrr example:
library(purrr)library(future)# Plan to use a parallel backendplan(multisession, workers = 2)# Create a list of numbersmy_list <- list(1, 2, 3, 4, 5)# Define a function to apply to each element of the listmy_fun <- function(x) {x^2}# Apply the function to each element of the list in parallel using purrrresult <- future_map(my_list, my_fun)# View the resultresultEDIT 2023-07-02 (by future author): Replaced deprecated and no-longer existing multiprocess future backend with multisession.