在多个变量中使用tapply

时间:2014-12-10 17:56:42

标签: r tapply

我有一组数据,其中包含有关客户及其花费的信息,每位客户只出现一次:

customer<-c("Andy","Bobby","Oscar","Oliver","Jane","Cathy","Emma","Chris")
age<-c(25,34,20,35,23,35,34,22)
gender<-c("male","male","male","male","female","female","female","female")
moneyspent<-c(100,100,200,200,400,400,500,200)

data<-data.frame(customer=customer,age=age,gender=gender,moneyspent=moneyspent)

如果我想计算男性和女性客户平均花费的金额,我可以使用tapply:

tapply(moneyspent,gender,mean)

给出:

female   male 
  375    150

但是,我现在想要找到性别和年龄组所花费的平均金额,而我的目标是:

 Male Age 20-30      Female Age 20-30      Male Age 30-40      Female Age 30-40
    150                     300                 150                   450

我如何修改tapply代码以便它提供这些结果?

谢谢你

2 个答案:

答案 0 :(得分:2)

您可能需要使用cut

mat <- tapply(moneyspent, list(gender, age=cut(age, breaks=c(20,30,40), 
                include.lowest=TRUE)), mean)

nm1 <- outer(rownames(mat), colnames(mat), FUN=paste)
setNames(c(mat), nm1)
#female [20,30]   male [20,30] female (30,40]   male (30,40] 
#       300            150            450            150 

其他选项包括

library(dplyr)
data %>% 
     group_by(gender, age=cut(age, breaks=c(20,30,40), 
              include.lowest=TRUE)) %>% 
     summarise(moneyspent=mean(moneyspent))

或者

 library(data.table)
 setDT(data)[, list(moneyspent=mean(moneyspent)),
     by=list(gender, age=cut(age, breaks= c(20,30,40), include.lowest=TRUE))]

答案 1 :(得分:0)

使用plyr软件包

library(plyr)

ddply(data,.(gender, age=cut(age, breaks=c(20,30,40), 
                  include.lowest=TRUE)), summarize, moneyspent=mean(moneyspent))

也会给出相同的结果。

注意:Summari z e和Summari s e执行相同的功能。

警告:加载plyr会掩盖dplyr的摘要!您需要detach plyr,然后再使用Summarize之类的功能。