我需要知道如何根据其他2个变量计算变量的mean, max，sd。例如，这是下面的数据集：我想得到牛奶的区域，渠道方向，牛奶最大区域方向通道等的平均值
Rg CHn Milk Grc
1 1 7209 4897
1 1 2154 6824
2 1 2280 2112
2 2 11487 9490
3 1 685 2216
3 2 891 5226
答案 0 :(得分：0)
强烈建议您使用dplyr软件包以获得更清晰的工作流程。以下是iris的示例：
data(iris)
iris %>%
select(-Species) %>% # remove "Species" variable from iris for following function
summarise_each(funs(mean, max, sd))
Sepal.Length_mean Sepal.Width_mean Petal.Length_mean Petal.Width_mean Sepal.Length_max Sepal.Width_max Petal.Length_max Petal.Width_max Sepal.Length_sd
1 5.843333 3.057333 3.758 1.199333 7.9 4.4 6.9 2.5 0.8280661
Sepal.Width_sd Petal.Length_sd Petal.Width_sd
1 0.4358663 1.765298 0.7622377
按物种获得mean，max和sd：
iris %>%
group_by(Species) %>%
summarise_each(funs(mean, max, sd))
Source: local data frame [3 x 13]
Species Sepal.Length_mean Sepal.Width_mean Petal.Length_mean Petal.Width_mean Sepal.Length_max Sepal.Width_max Petal.Length_max Petal.Width_max
(fctr) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
1 setosa 5.006 3.428 1.462 0.246 5.8 4.4 1.9 0.6
2 versicolor 5.936 2.770 4.260 1.326 7.0 3.4 5.1 1.8
3 virginica 6.588 2.974 5.552 2.026 7.9 3.8 6.9 2.5
Variables not shown: Sepal.Length_sd (dbl), Sepal.Width_sd (dbl), Petal.Length_sd (dbl), Petal.Width_sd (dbl)
从2个变量获得平均值max和sd的另一个例子：
data(mtcars)
mtcars %>%
group_by(gear, carb) %>% # grouping by two variables
summarise_each(funs(mean, max, sd))
Source: local data frame [11 x 29]
Groups: gear [?]
gear carb mpg_mean cyl_mean disp_mean hp_mean drat_mean wt_mean qsec_mean vs_mean am_mean mpg_max cyl_max disp_max hp_max drat_max wt_max qsec_max vs_max
(dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl) (dbl)
1 3 1 20.33333 5.333333 201.0333 104.0 3.1800 3.046667 19.89000 1.0 0.0 21.5 6 258.0 110 3.70 3.460 20.22 1
2 3 2 17.15000 8.000000 345.5000 162.5 3.0350 3.560000 17.06000 0.0 0.0 19.2 8 400.0 175 3.15 3.845 17.30 0
3 3 3 16.30000 8.000000 275.8000 180.0 3.0700 3.860000 17.66667 0.0 0.0 17.3 8 275.8 180 3.07 4.070 18.00 0
4 3 4 12.62000 8.000000 416.4000 228.0 3.2200 4.685800 16.89400 0.0 0.0 14.7 8 472.0 245 3.73 5.424 17.98 0
5 4 1 29.10000 4.000000 84.2000 72.5 4.0575 2.072500 19.22000 1.0 1.0 33.9 4 108.0 93 4.22 2.320 19.90 1
6 4 2 24.75000 4.000000 121.0500 79.5 4.1625 2.683750 20.00500 1.0 0.5 30.4 4 146.7 109 4.93 3.190 22.90 1
7 4 4 19.75000 6.000000 163.8000 116.5 3.9100 3.093750 17.67000 0.5 0.5 21.0 6 167.6 123 3.92 3.440 18.90 1
8 5 2 28.20000 4.000000 107.7000 102.0 4.1000 1.826500 16.80000 0.5 1.0 30.4 4 120.3 113 4.43 2.140 16.90 1
9 5 4 15.80000 8.000000 351.0000 264.0 4.2200 3.170000 14.50000 0.0 1.0 15.8 8 351.0 264 4.22 3.170 14.50 0
10 5 6 19.70000 6.000000 145.0000 175.0 3.6200 2.770000 15.50000 0.0 1.0 19.7 6 145.0 175 3.62 2.770 15.50 0
11 5 8 15.00000 8.000000 301.0000 335.0 3.5400 3.570000 14.60000 0.0 1.0 15.0 8 301.0 335 3.54 3.570 14.60 0
Variables not shown: am_max (dbl), mpg_sd (dbl), cyl_sd (dbl), disp_sd (dbl), hp_sd (dbl), drat_sd (dbl), wt_sd (dbl), qsec_sd (dbl), vs_sd (dbl), am_sd (dbl)
请参阅https://www.rstudio.com/wp-content/uploads/2015/02/data-wrangling-cheatsheet.pdf了解可以节省大量数据处理时间的有用技巧。
答案 1 :(得分：0)
我会使用dplyr包。如果您的数据位于名为df的data.frame中，则会为您提供带有区域摘要的data.frame：
library(dplyr)
df %>% group_by(Rg) %>% summarize(mean=mean(Milk), sd=sd(Milk), max=max(Milk))
# Source: local data frame [3 x 4]
#
# Rg mean sd max
# (int) (dbl) (dbl) (int)
# 1 1 4681.5 3574.425 7209
# 2 2 6883.5 6510.332 11487
# 3 3 788.0 145.664 891
编辑：如果您需要同时执行这两项操作：
df %>% group_by(Rg, CHn) %>% summarize(mean=mean(Milk), sd=sd(Milk), max=max(Milk))
# Source: local data frame [5 x 5]
# Groups: Rg [?]
#
# Rg CHn mean sd max
# (int) (int) (dbl) (dbl) (int)
# 1 1 1 4681.5 3574.425 7209
# 2 2 1 2280.0 NaN 2280
# 3 2 2 11487.0 NaN 11487
# 4 3 1 685.0 NaN 685
# 5 3 2 891.0 NaN 891