• 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     6  9.012038
    	5     8 16.239780
    	6    10 23.987277
    	7    12 34.119600
    	8    14 38.300412
    	9    16 46.904079
    	10   18 57.394757
    	11   20 66.511708
    	12   21 71.510273
    	 
    	#we could also aggregate on time and diet
    	head(aggregate(data$weight,
    	               list(time = data$Time, diet = data$Diet),
    	               mean
    	              )
    	    )
    	  time diet        x
    	1    0    1 41.40000
    	2    2    1 47.25000
    	3    4    1 56.47368
    	4    6    1 66.78947
    	5    8    1 79.68421
    	6   10    1 93.05263
    	tail(aggregate(data$weight,
    	               list(time = data$Time, diet = data$Diet),
    	               mean
    	              )
    	    )
    	   time diet        x
    	43   12    4 151.4000
    	44   14    4 161.8000
    	45   16    4 182.0000
    	46   18    4 202.9000
    	47   20    4 233.8889
    	48   21    4 238.5556
    	 
    	#to see the weights over time across different diets
    	ggplot(data) + geom_line(aes(x=Time, y=weight, colour=Chick)) +
    	             facet_wrap(~Diet) +
    	             guides(col=guide_legend(ncol=3))
    
    
    	------------------------------------------------------
    

    Example4

      The aggregate function is more difficult to use, but it is included in the base R installation and does not require the installation of another package.

    	# Get a count of number of subjects in each category (sex*condition)
    	cdata <- aggregate(data["subject"], by=data[c("sex","condition")], FUN=length)
    	cdata
    	#>   sex condition subject
    	#> 1   F   aspirin       5
    	#> 2   M   aspirin       9
    	#> 3   F   placebo      12
    	#> 4   M   placebo       4
    	
    	# Rename "subject" column to "N"
    	names(cdata)[names(cdata)=="subject"] <- "N"
    	cdata
    	#>   sex condition  N
    	#> 1   F   aspirin  5
    	#> 2   M   aspirin  9
    	#> 3   F   placebo 12
    	#> 4   M   placebo  4
    	
    	# Sort by sex first
    	cdata <- cdata[order(cdata$sex),]
    	cdata
    	#>   sex condition  N
    	#> 1   F   aspirin  5
    	#> 3   F   placebo 12
    	#> 2   M   aspirin  9
    	#> 4   M   placebo  4
    	
    	# We also keep the __before__ and __after__ columns:
    	# Get the average effect size by sex and condition
    	cdata.means <- aggregate(data[c("before","after","change")], 
    	                         by = data[c("sex","condition")], FUN=mean)
    	cdata.means
    	#>   sex condition   before     after    change
    	#> 1   F   aspirin 11.06000  7.640000 -3.420000
    	#> 2   M   aspirin 11.26667  5.855556 -5.411111
    	#> 3   F   placebo 10.13333  8.075000 -2.058333
    	#> 4   M   placebo 11.47500 10.500000 -0.975000
    	
    	# Merge the data frames
    	cdata <- merge(cdata, cdata.means)
    	cdata
    	#>   sex condition  N   before     after    change
    	#> 1   F   aspirin  5 11.06000  7.640000 -3.420000
    	#> 2   F   placebo 12 10.13333  8.075000 -2.058333
    	#> 3   M   aspirin  9 11.26667  5.855556 -5.411111
    	#> 4   M   placebo  4 11.47500 10.500000 -0.975000
    	
    	# Get the sample (n-1) standard deviation for "change"
    	cdata.sd <- aggregate(data["change"],
    	                      by = data[c("sex","condition")], FUN=sd)
    	# Rename the column to change.sd
    	names(cdata.sd)[names(cdata.sd)=="change"] <- "change.sd"
    	cdata.sd
    	#>   sex condition change.sd
    	#> 1   F   aspirin 0.8642916
    	#> 2   M   aspirin 1.1307569
    	#> 3   F   placebo 0.5247655
    	#> 4   M   placebo 0.7804913
    	
    	# Merge
    	cdata <- merge(cdata, cdata.sd)
    	cdata
    	#>   sex condition  N   before     after    change change.sd
    	#> 1   F   aspirin  5 11.06000  7.640000 -3.420000 0.8642916
    	#> 2   F   placebo 12 10.13333  8.075000 -2.058333 0.5247655
    	#> 3   M   aspirin  9 11.26667  5.855556 -5.411111 1.1307569
    	#> 4   M   placebo  4 11.47500 10.500000 -0.975000 0.7804913
    	
    	# Calculate standard error of the mean
    	cdata$change.se <- cdata$change.sd / sqrt(cdata$N)
    	cdata
    	#>   sex condition  N   before     after    change change.sd change.se
    	#> 1   F   aspirin  5 11.06000  7.640000 -3.420000 0.8642916 0.3865230
    	#> 2   F   placebo 12 10.13333  8.075000 -2.058333 0.5247655 0.1514867
    	#> 3   M   aspirin  9 11.26667  5.855556 -5.411111 1.1307569 0.3769190
    	#> 4   M   placebo  4 11.47500 10.500000 -0.975000 0.7804913 0.3902456
    

      If you have NA’s in your data and wish to skip them, use na.rm=TRUE:

    	cdata.means <- aggregate(data[c("before","after","change")], 
    	                         by = data[c("sex","condition")],
    	                         FUN=mean, na.rm=TRUE)
    	cdata.means
    	#>   sex condition   before     after    change
    	#> 1   F   aspirin 11.06000  7.640000 -3.420000
    	#> 2   M   aspirin 11.26667  5.855556 -5.411111
    	#> 3   F   placebo 10.13333  8.075000 -2.058333
    	#> 4   M   placebo 11.47500 10.500000 -0.975000
    人前一杯酒,各自饮完;人后一片海,独自上岸
  • 相关阅读:
    每日日报63
    每日日报62
    每日日报61
    每日日报60
    每日日报59
    每日日报58
    el-table表格拖动排序
    vue/eslint
    $attrs $listeners
    table封装成全局组件
  • 原文地址:https://www.cnblogs.com/kisen/p/12579946.html
Copyright © 2020-2023  润新知