R语言 aggregate函数
前言
这个函数的功能比较强大，它首先将数据进行分组（按行），然后对每一组数据进行函数统计，最后把结果组合成一个比较nice的表格返回。根据数据对象不同它有三种用法，分别应用于数据框（data.frame）、公式（formula）和时间序列（ts）：
aggregate(x, by, FUN, ..., simplify = TRUE) aggregate(formula, data, FUN, ..., subset, na.action = na.omit) aggregate(x, nfrequency = 1, FUN = sum, ndeltat = 1, ts.eps = getOption("ts.eps"), ...)
语法
aggregate(x, ...)
## S3 method for class 'default':
aggregate((x, ...))
## S3 method for class 'data.frame':
aggregate((x, by, FUN, ..., simplify = TRUE))
## S3 method for class 'formula':
aggregate((formula, data, FUN, ...,
subset, na.action = na.omit))
## S3 method for class 'ts':
aggregate((x, nfrequency = 1, FUN = sum, ndeltat = 1,
ts.eps = getOption("ts.eps"), ...))
###细节查看 ?aggregate
Example1
我们通过 mtcars 数据集的操作对这个函数进行简单了解。mtcars 是不同类型汽车道路测试的数据框类型数据：
> str(mtcars)
'data.frame': 32 obs. of 11 variables:
$ mpg : num 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
$ cyl : num 6 6 4 6 8 6 8 4 4 6 ...
$ disp: num 160 160 108 258 360 ...
$ hp : num 110 110 93 110 175 105 245 62 95 123 ...
$ drat: num 3.9 3.9 3.85 3.08 3.15 2.76 3.21 3.69 3.92 3.92 ...
$ wt : num 2.62 2.88 2.32 3.21 3.44 ...
$ qsec: num 16.5 17 18.6 19.4 17 ...
$ vs : num 0 0 1 1 0 1 0 1 1 1 ...
$ am : num 1 1 1 0 0 0 0 0 0 0 ...
$ gear: num 4 4 4 3 3 3 3 4 4 4 ...
$ carb: num 4 4 1 1 2 1 4 2 2 4 ...
先用attach函数把mtcars的列变量名称加入到变量搜索范围内，然后使用aggregate函数按cyl（汽缸数）进行分类计算平均值：
> attach(mtcars)
> aggregate(mtcars, by=list(cyl), FUN=mean)
Group.1 mpg cyl disp hp drat wt qsec vs am gear carb
1 4 26.66364 4 105.1364 82.63636 4.070909 2.285727 19.13727 0.9090909 0.7272727 4.090909 1.545455
2 6 19.74286 6 183.3143 122.28571 3.585714 3.117143 17.97714 0.5714286 0.4285714 3.857143 3.428571
3 8 15.10000 8 353.1000 209.21429 3.229286 3.999214 16.77214 0.0000000 0.1428571 3.285714 3.500000
by参数也可以包含多个类型的因子，得到的就是每个不同因子组合的统计结果：
> aggregate(mtcars, by=list(cyl, gear), FUN=mean)
Group.1 Group.2 mpg cyl disp hp drat wt qsec vs am gear carb
1 4 3 21.500 4 120.1000 97.0000 3.700000 2.465000 20.0100 1.0 0.00 3 1.000000
2 6 3 19.750 6 241.5000 107.5000 2.920000 3.337500 19.8300 1.0 0.00 3 1.000000
3 8 3 15.050 8 357.6167 194.1667 3.120833 4.104083 17.1425 0.0 0.00 3 3.083333
4 4 4 26.925 4 102.6250 76.0000 4.110000 2.378125 19.6125 1.0 0.75 4 1.500000
5 6 4 19.750 6 163.8000 116.5000 3.910000 3.093750 17.6700 0.5 0.50 4 4.000000
6 4 5 28.200 4 107.7000 102.0000 4.100000 1.826500 16.8000 0.5 1.00 5 2.000000
7 6 5 19.700 6 145.0000 175.0000 3.620000 2.770000 15.5000 0.0 1.00 5 6.000000
8 8 5 15.400 8 326.0000 299.5000 3.880000 3.370000 14.5500 0.0 1.00 5 6.000000
公式（formula）是一种特殊的R数据对象，在aggregate函数中使用公式参数可以对数据框的部分指标进行统计：
> aggregate(cbind(mpg,hp) ~ cyl+gear, FUN=mean)
cyl gear mpg hp
1 4 3 21.500 97.0000
2 6 3 19.750 107.5000
3 8 3 15.050 194.1667
4 4 4 26.925 76.0000
5 6 4 19.750 116.5000
6 4 5 28.200 102.0000
7 6 5 19.700 175.0000
8 8 5 15.400 299.5000
上面的公式 cbind(mpg,hp) ~ cyl+gear 表示使用 cyl 和 gear 的因子组合对 cbind(mpg,hp) 数据进行操作。aggregate在时间序列数据上的应用请参考R的函数说明文档。
Example2
## Compute the averages for the variables in 'state.x77', grouped
## according to the region (Northeast, South, North Central, West) that
## each state belongs to.
aggregate(state.x77, list(Region = state.region), mean)
## Compute the averages according to region and the occurrence of more
## than 130 days of frost.
aggregate(state.x77,
list(Region = state.region,
Cold = state.x77[,"Frost"] > 130),
mean)
## (Note that no state in 'South' is THAT cold.)
## example with character variables and NAs
testDF <- data.frame(v1 = c(1,3,5,7,8,3,5,NA,4,5,7,9),
v2 = c(11,33,55,77,88,33,55,NA,44,55,77,99) )
by1 <- c("red", "blue", 1, 2, NA, "big", 1, 2, "red", 1, NA, 12)
by2 <- c("wet", "dry", 99, 95, NA, "damp", 95, 99, "red", 99, NA, NA)
aggregate(x = testDF, by = list(by1, by2), FUN = "mean")
# and if you want to treat NAs as a group
fby1 <- factor(by1, exclude = "")
fby2 <- factor(by2, exclude = "")
aggregate(x = testDF, by = list(fby1, fby2), FUN = "mean")
## Formulas, one ~ one, one ~ many, many ~ one, and many ~ many:
aggregate(weight ~ feed, data = chickwts, mean)
aggregate(breaks ~ wool + tension, data = warpbreaks, mean)
aggregate(cbind(Ozone, Temp) ~ Month, data = airquality, mean)
aggregate(cbind(ncases, ncontrols) ~ alcgp + tobgp, data = esoph, sum)
## Dot notation:
aggregate(. ~ Species, data = iris, mean)
aggregate(len ~ ., data = ToothGrowth, mean)
## Often followed by xtabs():
ag <- aggregate(len ~ ., data = ToothGrowth, mean)
xtabs(len ~ ., data = ag)
## Compute the average annual approval ratings for American presidents.
aggregate(presidents, nfrequency = 1, FUN = mean)
## Give the summer less weight.
aggregate(presidents, nfrequency = 1,
FUN = weighted.mean, w = c(1, 1, 0.5, 1))
Example3
------------------------------------------------------
#load data
data <- ChickWeight
head(data)
weight Time Chick Diet
1 42 0 1 1
2 51 2 1 1
3 59 4 1 1
4 64 6 1 1
5 76 8 1 1
6 93 10 1 1
#dimension of the data
dim(data)
[1] 578 4
#how many chickens
unique(data$Chick)
[1] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
[31] 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
50 Levels: 18 < 16 < 15 < 13 < 9 < 20 < 10 < 8 < 17 < 19 < 4 < 6 < 11 < 3 < 1 < 12 < ... < 48
#how many diets
unique(data$Diet)
[1] 1 2 3 4
Levels: 1 2 3 4
#how many time points
unique(data$Time)
[1] 0 2 4 6 8 10 12 14 16 18 20 21
library(ggplot2)
ggplot(data=data, aes(x=Time, y=weight, group=Chick, colour=Chick)) +
geom_line() +
geom_point()
------------------------------------------------------
## S3 method for class 'data.frame'
## aggregate(x, by, FUN, ..., simplify = TRUE)
#find the mean weight depending on diet
aggregate(data$weight, list(diet = data$Diet), mean)
diet x
1 1 102.6455
2 2 122.6167
3 3 142.9500
4 4 135.2627
#aggregate on time
aggregate(data$weight, list(time=data$Time), mean)
time x
1 0 41.06000
2 2 49.22000
3 4 59.95918
4 6 74.30612
5 8 91.24490
6 10 107.83673
7 12 129.24490
8 14 143.81250
9 16 168.08511
10 18 190.19149
11 20 209.71739
12 21 218.68889
#use a different function
aggregate(data$weight, list(time=data$Time), sd)
time x
1 0 1.132272
2 2 3.688316
3 4 4.495179
4