Question

我试图获取数据帧中组（“a”和“b”）的变量（v）的累积和。如何将结果在底部 - 其行的编号正确 - 得到我的数据帧的列cs？

> library(nlme)
> g <- factor(c("a","b","a","b","a","b","a","b","a","b","a","b"))
> v <- c(1,4,1,4,1,4,2,8,2,8,2,8)
> cs <- rep(0,12)
> d <- data.frame(g,v,cs)

> d
   g v cs
1  a 1 0
2  b 4 0
3  a 1 0
4  b 4 0
5  a 1 0
6  b 4 0
7  a 2 0
8  b 8 0
9  a 2 0
10 b 8 0
11 a 2 0
12 b 8 0

> r=gapply(d,FUN="cumsum",form=~g, which="v")
>r

$a     
   v   
1  1   
3  2   
5  3  
7  5  
9  7  
11 9  

$b    
    v 
2   4 
4   8 
6  12 
8  20 
10 28 
12 36 

> str(r)
List of 2
 $ a:'data.frame':  6 obs. of  1 variable:
  ..$ v: num [1:6] 1 2 3 5 7 9
 $ b:'data.frame':  6 obs. of  1 variable:
  ..$ v: num [1:6] 4 8 12 20 28 36

我想我可以找出一些费力的方法将这些数据帧中的数据转换为d $ cs，但是我必须进行一些简单的调整。

Answer 1

split<-是一个非常奇怪的野兽

split(d$cs, d$g) <- lapply(split(d$v, d$g), cumsum)

导致

Answer 2

我会使用ave。如果你看一下ave的来源，你会发现它基本上包裹了马丁摩根的solution。

R> g <- factor(c("a","b","a","b","a","b","a","b","a","b","a","b"))
R> v <- c(1,4,1,4,1,4,2,8,2,8,2,8)
R> d <- data.frame(g,v)
R> d$cs <- ave(v, g, FUN=cumsum)
R> d
   g v cs
1  a 1  1
2  b 4  4
3  a 1  2
4  b 4  8
5  a 1  3
6  b 4 12
7  a 2  5
8  b 8 20
9  a 2  7
10 b 8 28
11 a 2  9
12 b 8 36

Answer 3

我选择的工具是 plyr 包：

require(plyr)
> ddply(d,.(g),transform,cs = cumsum(v))
   g v cs
1  a 1  1
2  a 1  2
3  a 1  3
4  a 2  5
5  a 2  7
6  a 2  9
7  b 4  4
8  b 4  8
9  b 4 12
10 b 8 20
11 b 8 28
12 b 8 36

Answer 4

> library(nlme)
> g <- factor(c("a","b","a","b","a","b","a","b","a","b","a","b"))
> v <- c(1,4,1,4,1,4,2,8,2,8,2,8)
> cs <- rep(0,12)
> d <- data.frame(g,v,cs)
> d <- d[order(d$g),]
> temp <- by(d$v,d$g,cumsum)
> d$cs <- do.call("c",temp)
> d
   g v cs
1  a 1  1
3  a 1  2
5  a 1  3
7  a 2  5
9  a 2  7
11 a 2  9
2  b 4  4
4  b 4  8
6  b 4 12
8  b 8 20
10 b 8 28
12 b 8 36

使用by函数的另一种解决方案，但我必须先订购数据

将函数应用于R中data.frame中的组

4 个答案: