单个tapply或聚合语句中的多个函数

是否有可能在单个tapply或聚合语句中包含两个函数？单个tapply或聚合语句中的多个函数

下面我使用两个tapply语句和两个汇总语句：一个用于平均值和一个用于标清。
我宁愿合并报表。

my.Data = read.table(text = " 
    animal age  sex weight 
     1 adult female  100 
     2 young male  75 
     3 adult male  90 
     4 adult female  95 
     5 young female  80 
", sep = "", header = TRUE) 

with(my.Data, tapply(weight, list(age, sex), function(x) {mean(x)})) 
with(my.Data, tapply(weight, list(age, sex), function(x) {sd(x) })) 

with(my.Data, aggregate(weight ~ age + sex, FUN = mean) 
with(my.Data, aggregate(weight ~ age + sex, FUN = sd) 

# this does not work: 

with(my.Data, tapply(weight, list(age, sex), function(x) {mean(x) ; sd(x)})) 

# I would also prefer that the output be formatted something similar to that 
# show below. `aggregate` formats the output perfectly. I just cannot figure 
# out how to implement two functions in one statement. 

    age sex mean  sd 
adult female 97.5 3.535534 
adult male  90  NA 
young female 80.0  NA 
young male  75  NA

我总是可以运行两个单独的语句并合并输出。我只是希望可能有一个更方便的解决方案。

我发现下面张贴在这里的答案：Apply multiple functions to column using tapply

f <- function(x) c(mean(x), sd(x)) 
do.call(rbind, with(my.Data, tapply(weight, list(age, sex), f)))

然而，无论是行或列的标签。

 [,1]  [,2] 
[1,] 97.5 3.535534 
[2,] 80.0  NA 
[3,] 90.0  NA 
[4,] 75.0  NA

我更喜欢base R中的解决方案。plyr软件包的解决方案发布在上面的链接中。如果我可以将正确的行和列标题添加到上面的输出中，那将是完美的。

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2013-03-05 Mark Miller

但这些应该有：

with(my.Data, aggregate(weight, list(age, sex), function(x) { c(MEAN=mean(x), SD=sd(x))})) 

with(my.Data, tapply(weight, list(age, sex), function(x) { c(mean(x) , sd(x))})) 
# Not a nice structure but the results are in there 

with(my.Data, aggregate(weight ~ age + sex, FUN = function(x) c(SD = sd(x), MN= mean(x)))) 
    age sex weight.SD weight.MN 
1 adult female 3.535534 97.500000 
2 young female  NA 80.000000 
3 adult male  NA 90.000000 
4 young male  NA 75.

原则被遵守是让你的函数返回“一件事”，这可以是一个向量或列表，但不能是两个函数的连续调用调用。

来源

2013-03-05 03:09:32

谢谢！这两个汇总语句有效。 tapply语句似乎不起作用，但我可以使用聚合方法。 – 2013-03-05 03:22:46

嗯，我认为它“有效”，只是不给你打印很好的东西。用（my.Data，tapply（weight，list）（age，sex），function（x）{c（mean（x），sd（x））}））[1,1]'试试''在列表矩阵内部查看索引。 – 2013-03-05 03:29:20

我明白了。谢谢。如果我将整个陈述放在colnames（）或rownames（）中，那么我就可以得到标签。 – 2013-03-05 03:35:55

如果您想使用data.table，它有with和by内置到它：

library(data.table) 
myDT <- data.table(my.Data, key="animal") 


myDT[, c("mean", "sd") := list(mean(weight), sd(weight)), by=list(age, sex)] 


myDT[, list(mean_Aggr=sum(mean(weight)), sd_Aggr=sum(sd(weight))), by=list(age, sex)] 
    age sex mean_Aggr sd_Aggr 
1: adult female  96.0 3.6055513 
2: young male  76.5 2.1213203 
3: adult male  91.0 1.4142136 
4: young female  84.5 0.7071068

我用一个稍微不同的数据集，从而不会有NA值SD

来源

2013-03-05 03:26:53

重塑让你传递2个函数; reshape2不会。

library(reshape) 
my.Data = read.table(text = " 
    animal age  sex weight 
     1 adult female  100 
     2 young male  75 
     3 adult male  90 
     4 adult female  95 
     5 young female  80 
", sep = "", header = TRUE) 
my.Data[,1]<- NULL 
(a1<- melt(my.Data, id=c("age", "sex"), measured=c("weight"))) 
(cast(a1, age + sex ~ variable, c(mean, sd), fill=NA)) 

#  age sex weight_mean weight_sd 
# 1 adult female  97.5 3.535534 
# 2 adult male  90.0  NA 
# 3 young female  80.0  NA 
# 4 young male  75.0  NA

我欠了@Ramnath，他在昨天注意到了this。

来源

2013-03-05 04:10:44

本着共享的精神，如果您熟悉SQL，您可能还会考虑“sqldf”包。（着重强调，因为你需要知道的，例如，该mean是为了得到你想要的结果avg）

sqldf("select age, sex, 
     avg(weight) `Wt.Mean`, 
     stdev(weight) `Wt.SD` 
     from `my.Data` 
     group by age, sex") 
    age sex Wt.Mean Wt.SD 
1 adult female 97.5 3.535534 
2 adult male 90.0 0.000000 
3 young female 80.0 0.000000 
4 young male 75.0 0.000000

来源

2013-03-05 18:26:29 A5C1D2H2I1M1N2O1R2T1

单个tapply或聚合语句中的多个函数

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