附注:以下是各种plyr函数如何对应于基本*apply函数(来自plyr网页http://had.co.nz/plyr/的plyr介绍文档)

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 

plyr的目标之一是为每个函数提供一致的命名约定，在函数名中编码输入和输出数据类型。它还提供了输出的一致性，因为来自dlply()的输出很容易传递给ldply()以产生有用的输出，等等。

从概念上讲，学习plyr并不比理解基本的apply函数更难。

在我的日常使用中，Plyr和重塑函数几乎取代了所有这些函数。但是，同样来自Plyr文档的介绍:

相关函数tapply和sweep在plyr中没有相应的函数，仍然有用。Merge对于合并摘要和原始数据非常有用。

请看http://www.slideshare.net/hadley/plyr-one-data-analytic-strategy:的第21页

(希望这是清楚的，apply对应于@Hadley的apply和aggregate对应于@Hadley的ddply等。如果你没有从这张图片中得到它，同一幻灯片的第20张将会说明。)

(左边是输入，上面是输出)

R有许多在帮助文件中有巧妙描述的*apply函数(例如?apply)。但是，它们太多了，初学者可能很难决定哪一个适合他们的情况，甚至很难记住它们。他们可能有一个普遍的感觉，“我应该在这里使用一个*apply函数”，但一开始很难把它们都说清楚。

尽管事实上(在其他回答中提到)*apply系列的大部分功能都由非常流行的plyr包覆盖，但基本函数仍然有用，值得了解。

这个答案旨在作为新用户的一个路标，帮助他们针对特定的问题找到正确的*apply函数。注意，这不是为了简单地复制或替换R文档!希望这个答案能帮助您决定哪个*apply函数适合您的情况，然后由您进一步研究。除了一个例外，性能差异将不予处理。

apply - When you want to apply a function to the rows or columns of a matrix (and higher-dimensional analogues); not generally advisable for data frames as it will coerce to a matrix first. # Two dimensional matrix M <- matrix(seq(1,16), 4, 4) # apply min to rows apply(M, 1, min) [1] 1 2 3 4 # apply max to columns apply(M, 2, max) [1] 4 8 12 16 # 3 dimensional array M <- array( seq(32), dim = c(4,4,2)) # Apply sum across each M[*, , ] - i.e Sum across 2nd and 3rd dimension apply(M, 1, sum) # Result is one-dimensional [1] 120 128 136 144 # Apply sum across each M[*, *, ] - i.e Sum across 3rd dimension apply(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 48 If you want row/column means or sums for a 2D matrix, be sure to investigate the highly optimized, lightning-quick colMeans, rowMeans, colSums, rowSums. lapply - When you want to apply a function to each element of a list in turn and get a list back. This is the workhorse of many of the other *apply functions. Peel back their code and you will often find lapply underneath. x <- list(a = 1, b = 1:3, c = 10:100) lapply(x, FUN = length) $a [1] 1 $b [1] 3 $c [1] 91 lapply(x, FUN = sum) $a [1] 1 $b [1] 6 $c [1] 5005 sapply - When you want to apply a function to each element of a list in turn, but you want a vector back, rather than a list. If you find yourself typing unlist(lapply(...)), stop and consider sapply. 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 91 sapply(x, FUN = sum) a b c 1 6 5005 In more advanced uses of sapply it will attempt to coerce the result to a multi-dimensional array, if appropriate. For example, if our function returns vectors of the same length, sapply will use them as columns of a matrix: sapply(1:5,function(x) rnorm(3,x)) If our function returns a 2 dimensional matrix, sapply will 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 sapply but perhaps need to squeeze some more speed out of your code or want more type safety. For vapply, you basically give R an example of what sort of thing your function will return, which can save some time coercing returned values 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 91 mapply - For when you have several data structures (e.g. vectors, lists) and you want to apply a function to the 1st elements of each, and then the 2nd elements of each, etc., coercing the result to a vector/array as in sapply. This is multivariate in the sense that your function must accept multiple 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] 4 Map - A wrapper to mapply with SIMPLIFY = 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] 15 rapply - 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 rapply is, I forgot about it when first posting this answer! Obviously, I'm sure many people use it, but YMMV. rapply is best illustrated with a user-defined function to apply: # Append ! to string, otherwise increment myFun <- function(x){ if(is.character(x)){ return(paste(x,"!",sep="")) } else{ return(x + 1) } } #A nested list structure l <- 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 altered rapply(l, myFun, how="replace") tapply - For when you want to apply a function to subsets of a vector and the subsets are defined by some other vector, usually a factor. The black sheep of the *apply family, of sorts. The help file's use of the phrase "ragged array" can be a bit confusing, but it is actually quite simple. A vector: x <- 1:20 A factor (of the same length!) defining groups: y <- factor(rep(letters[1:5], each = 4)) Add up the values in x within each subgroup defined by y: tapply(x, y, sum) a b c d e 10 26 42 58 74 More complex examples can be handled where the subgroups are defined by the unique combinations of a list of several factors. tapply is similar in spirit to the split-apply-combine functions that are common in R (aggregate, by, ave, ddply, etc.) Hence its black sheep status.

附注:以下是各种plyr函数如何对应于基本*apply函数(来自plyr网页http://had.co.nz/plyr/的plyr介绍文档)

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 

plyr的目标之一是为每个函数提供一致的命名约定，在函数名中编码输入和输出数据类型。它还提供了输出的一致性，因为来自dlply()的输出很容易传递给ldply()以产生有用的输出，等等。

从概念上讲，学习plyr并不比理解基本的apply函数更难。

在我的日常使用中，Plyr和重塑函数几乎取代了所有这些函数。但是，同样来自Plyr文档的介绍:

相关函数tapply和sweep在plyr中没有相应的函数，仍然有用。Merge对于合并摘要和原始数据非常有用。

我最近发现了一个相当有用的扫描函数，为了完整起见，我将它添加到这里:

扫描

基本思想是逐行或逐列遍历数组并返回修改后的数组。下面的例子将说明这一点(来源:datacamp):

假设你有一个矩阵，想要按列对它进行标准化:

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 result
dataPoints_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

# Normalize
dataPoints_Trans2 <- sweep(dataPoints_Trans1, 2, dataPoints_sdev, "/")

# Return the result
dataPoints_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  1.1618950

注意:对于这个简单的例子，同样的结果当然可以通过应用(dataPoints, 2, scale)更容易实现。

因为我意识到这篇文章(非常优秀)的答案缺乏全面的解释。以下是我的贡献。

BY

正如文档中所述，by函数可以作为tapply的“包装器”。当我们想要计算tapply无法处理的任务时，by的幂函数就出现了。例如下面的代码:

ct <- tapply(iris$Sepal.Width , iris$Species , summary )
cb <- by(iris$Sepal.Width , iris$Species , summary )

 cb
iris$Species: setosa
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.300   3.200   3.400   3.428   3.675   4.400 
-------------------------------------------------------------- 
iris$Species: versicolor
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.000   2.525   2.800   2.770   3.000   3.400 
-------------------------------------------------------------- 
iris$Species: virginica
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.200   2.800   3.000   2.974   3.175   3.800 


ct
$setosa
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.300   3.200   3.400   3.428   3.675   4.400 

$versicolor
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.000   2.525   2.800   2.770   3.000   3.400 

$virginica
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  2.200   2.800   3.000   2.974   3.175   3.800 

如果我们打印这两个对象，ct和cb，我们“本质上”有相同的结果，唯一的区别是它们的显示方式和不同的类属性，分别是by for cb和array for ct。

正如我所说，当我们不能使用tapply时，by的力量就出现了;下面的代码是一个例子:

 tapply(iris, iris$Species, summary )
Error in tapply(iris, iris$Species, summary) : 
  arguments must have same length

R说参数必须具有相同的长度，比如“我们想要计算虹膜中所有变量沿因子Species的总和”:但是R不能这样做，因为它不知道如何处理。

使用by函数R为数据帧类分派一个特定的方法，然后让summary函数工作，即使第一个参数的长度(以及类型)不同。

bywork <- by(iris, iris$Species, summary )

bywork
iris$Species: setosa
  Sepal.Length    Sepal.Width     Petal.Length    Petal.Width          Species  
 Min.   :4.300   Min.   :2.300   Min.   :1.000   Min.   :0.100   setosa    :50  
 1st Qu.:4.800   1st Qu.:3.200   1st Qu.:1.400   1st Qu.:0.200   versicolor: 0  
 Median :5.000   Median :3.400   Median :1.500   Median :0.200   virginica : 0  
 Mean   :5.006   Mean   :3.428   Mean   :1.462   Mean   :0.246                  
 3rd Qu.:5.200   3rd Qu.:3.675   3rd Qu.:1.575   3rd Qu.:0.300                  
 Max.   :5.800   Max.   :4.400   Max.   :1.900   Max.   :0.600                  
-------------------------------------------------------------- 
iris$Species: versicolor
  Sepal.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: virginica
  Sepal.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     

它确实起作用了，结果非常令人惊讶。它是一个类的对象，即沿着Species(例如，对于它们中的每一个)计算每个变量的摘要。

注意，如果第一个参数是一个数据帧，分派的函数必须有针对该类对象的方法。例如，我们将这段代码与均值函数一起使用我们将得到这段完全没有意义的代码:

 by(iris, iris$Species, mean)
iris$Species: setosa
[1] NA
------------------------------------------- 
iris$Species: versicolor
[1] NA
------------------------------------------- 
iris$Species: virginica
[1] NA
Warning messages:
1: In mean.default(data[x, , drop = FALSE], ...) :
  argument is not numeric or logical: returning NA
2: In mean.default(data[x, , drop = FALSE], ...) :
  argument is not numeric or logical: returning NA
3: In mean.default(data[x, , drop = FALSE], ...) :
  argument is not numeric or logical: returning NA

总

如果我们以这种方式使用tapply，那么Aggregate可以被视为tapply的另一种不同的使用方式。

at <- tapply(iris$Sepal.Length , iris$Species , mean)
ag <- aggregate(iris$Sepal.Length , list(iris$Species), mean)

 at
    setosa versicolor  virginica 
     5.006      5.936      6.588 
 ag
     Group.1     x
1     setosa 5.006
2 versicolor 5.936
3  virginica 6.588

两个直接的区别是，aggregate的第二个参数必须是一个列表，而tapply可以(非强制)是一个列表，aggregate的输出是一个数据帧，而tapply的输出是一个数组。

aggregate的强大之处在于它可以轻松地处理带有子集参数的数据子集，并且它还有ts对象和公式的方法。

这些元素使聚合体在某些情况下更容易与tapply一起工作。 以下是一些例子(可在文档中找到):

ag <- aggregate(len ~ ., data = ToothGrowth, mean)

 ag
  supp dose   len
1   OJ  0.5 13.23
2   VC  0.5  7.98
3   OJ  1.0 22.70
4   VC  1.0 16.77
5   OJ  2.0 26.06
6   VC  2.0 26.14

我们可以用tapply实现同样的效果，但语法略难，输出(在某些情况下)可读性较差:

att <- tapply(ToothGrowth$len, list(ToothGrowth$dose, ToothGrowth$supp), mean)

 att
       OJ    VC
0.5 13.23  7.98
1   22.70 16.77
2   26.06 26.14

还有一些时候，我们不能使用by或tapply，而必须使用aggregate。

 ag1 <- aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, mean)

 ag1
  Month    Ozone     Temp
1     5 23.61538 66.73077
2     6 29.44444 78.22222
3     7 59.11538 83.88462
4     8 59.96154 83.96154
5     9 31.44828 76.89655

我们不能在一次调用中使用tapply获得之前的结果，但我们必须计算每个元素的平均值沿着Month，然后将它们组合起来(还要注意，我们必须调用na。rm = TRUE，因为聚合函数的公式方法默认为na。Action = na.省略):

ta1 <- tapply(airquality$Ozone, airquality$Month, mean, na.rm = TRUE)
ta2 <- tapply(airquality$Temp, airquality$Month, mean, na.rm = TRUE)

 cbind(ta1, ta2)
       ta1      ta2
5 23.61538 65.54839
6 29.44444 79.10000
7 59.11538 83.90323
8 59.96154 83.96774
9 31.44828 76.90000

虽然使用by我们无法实现这一点，但实际上下面的函数调用会返回一个错误(但很可能与所提供的函数有关，mean):

by(airquality[c("Ozone", "Temp")], airquality$Month, mean, na.rm = TRUE)

其他时候，结果是相同的，差异只是在类(然后它是如何显示/打印的，而不仅仅是——例如，如何子集)对象:

byagg <- by(airquality[c("Ozone", "Temp")], airquality$Month, summary)
aggagg <- aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, summary)

前面的代码实现了相同的目标和结果，在某些情况下使用什么工具只是个人口味和需求的问题;前面两个对象在子集方面有非常不同的需求。