Question

我想写一个简单的函数来计算Var1的平均值：

data<-structure(list(time = structure(c(1358832600, 1358832600), class = c("POSIXct", 
"POSIXt"), tzone = ""), Var1 = c(0.4, 0.2)), .Names = c("time", 
"Var1"), row.names = c(NA, -2L), class = "data.frame")

data<- data.table(data)

time                Var1
2013-01-22 09:30:00 0.4
2013-01-22 09:30:00 0.2

Aggregated.Data<- function(data, col) {
 aggregated <- ddply(data, time, summarise, col= mean(eval(col)))
 return(aggregated)
}

aggregated.data <- Aggregated.Data(data, quote(Var1))
Everything works, output:

time           Col
2013-01-22 09:30:00 0.3

问题：

这是正确的方法吗？我的意思是使用quote和eval？
为什么输出中的第二列名称为col，如何将其更改为Var1？

编辑：使用data.table

Aggregated.Data<- function(data, col) {
 aggregated <- data(data, list(col=mean(eval(substitute(col)))), by=list(time=time))
 return(aggregated)
}

Answer 1

一种选择是使用eval(as.name)并获取mean。我们稍后可以使用setnames更改列名称。

f1 <- function(dat, col){
 DT <- dat[, mean(eval(as.name(col))), time]
 v1 <- setdiff(colnames(DT), colnames(dat))
 setnames(DT, v1, col)
 DT 
}


f1(data, 'Var1')
#                  time Var1
#1: 2013-01-22 00:30:00  0.3

Answer 2

library(dplyr)
library(lazyeval)

summarize_column = function(data, variable_name)
  data %>% 
    group_by(time) %>% 
    select_(lazy(variable_name)) %>% 
    summarize_each(funs(mean))

data %>% summarize_column(Var1)

将data.table的列（名称和值）传递给函数

2 个答案: