如果我的分组变量是一个因素，我如何生成按组汇总的统计数据? [英] How can I generate by-group summary statistics if my grouping variable is a factor?
问题描述
假设我想获得数据集 mtcars(基本 R 版本 2.12.1 的一部分)的一些汇总统计信息.下面，我根据汽车的发动机气缸数对汽车进行分组，并对 mtcars 中剩余的变量取每组平均值.
Suppose I wanted to get some summary statistics on the dataset mtcars (part of base R version 2.12.1).
Below, I group the cars according to the number of engine cylinders they have and take the per-group means of the remaining variables in mtcars.
> 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 ...
> ddply(mtcars, .(cyl), mean)
mpg cyl disp hp drat wt qsec vs am gear
1 26.66364 4 105.1364 82.63636 4.070909 2.285727 19.13727 0.9090909 0.7272727 4.090909
2 19.74286 6 183.3143 122.28571 3.585714 3.117143 17.97714 0.5714286 0.4285714 3.857143
3 15.10000 8 353.1000 209.21429 3.229286 3.999214 16.77214 0.0000000 0.1428571 3.285714
carb
1 1.545455
2 3.428571
3 3.500000
但是，如果我的分组变量恰好是一个因素，事情就会变得棘手.ddply() 对因子的每个级别都发出警告，因为不能取一个因子的 mean().
But, if my grouping variable happens to be a factor things get trickier. ddply() throws a warning for each level of the factor,
since one can't take the mean() of a factor.
> mtcars$cyl <- as.factor(mtcars$cyl)
> 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 : Factor w/ 3 levels "4","6","8": 2 2 1 2 3 2 3 1 1 2 ...
$ 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 ...
> ddply(mtcars, .(cyl), mean)
mpg cyl disp hp drat wt qsec vs am gear
1 26.66364 NA 105.1364 82.63636 4.070909 2.285727 19.13727 0.9090909 0.7272727 4.090909
2 19.74286 NA 183.3143 122.28571 3.585714 3.117143 17.97714 0.5714286 0.4285714 3.857143
3 15.10000 NA 353.1000 209.21429 3.229286 3.999214 16.77214 0.0000000 0.1428571 3.285714
carb
1 1.545455
2 3.428571
3 3.500000
Warning messages:
1: In mean.default(X[[2L]], ...) :
argument is not numeric or logical: returning NA
2: In mean.default(X[[2L]], ...) :
argument is not numeric or logical: returning NA
3: In mean.default(X[[2L]], ...) :
argument is not numeric or logical: returning NA
>
所以，我想知道我是否只是以错误的方式生成汇总统计数据.
So, I'm wondering if I'm just going about generating summary statistics the wrong way.
通常如何生成按因子或按组汇总统计的数据结构(如均值、标准差等)?我应该使用 ddply() 以外的其他东西吗?如果我可以使用 ddply()，我可以做些什么来避免在尝试取分组因子的平均值时导致的错误?
How does one usually generate data structures of by-factor or by-group summary statistics (like means, standard deviations, etc.)? Should I be using something other than ddply()? If I can use ddply(), what can I do to avoid the errors that result when trying to take the mean of my grouping factor?
推荐答案
Use numcolwise(mean):numcolwise 函数将其参数(函数)转换为函数仅对数字列进行操作(并忽略分类/因子列).
Use numcolwise(mean): the numcolwise function converts its argument (a function) into a function that operates only on numerical columns (and ignores the categorical/factor columns).
> ddply(mtcars, .(cyl), numcolwise(mean))
cyl mpg disp hp drat wt qsec vs
1 4 26.66364 105.1364 82.63636 4.070909 2.285727 19.13727 0.9090909
2 6 19.74286 183.3143 122.28571 3.585714 3.117143 17.97714 0.5714286
3 8 15.10000 353.1000 209.21429 3.229286 3.999214 16.77214 0.0000000
am gear carb
1 0.7272727 4.090909 1.545455
2 0.4285714 3.857143 3.428571
3 0.1428571 3.285714 3.500000
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