我正在处理宽格式的数据集,如
> data=read.csv("http://www.kuleuven.be/bio/ento/temp/data.csv")
> data
factor1 factor2 count_1 count_2 count_3
1 a a 1 2 0
2 a b 3 0 0
3 b a 1 2 3
4 b b 2 2 0
5 c a 3 4 0
6 c b 1 1 0
其中因子1和因子2是我想要采用的不同因素(实际上我有2个以上,但这不重要),而count_1到count_3是顺序量表上的积极交互计数(3> 2→1)。我现在想将此数据集转换为长格式,以获得类似
的内容 factor1 factor2 aggression
1 a a 1
2 a a 2
3 a a 2
4 a b 1
5 a b 1
6 a b 1
7 b a 1
8 b a 2
9 b a 2
10 b a 3
11 b a 3
12 b a 3
13 b b 1
14 b b 1
15 b b 2
16 b b 2
17 c a 1
18 c a 1
19 c a 1
20 c a 2
21 c a 2
22 c a 2
23 c a 2
24 c b 1
25 c b 2
如果没有使用for ... to循环,有人会碰巧知道如何做到这一点,例如使用包reshape2
? (我意识到它应该可以使用melt
,但我还没有找到正确的语法)
编辑:对于那些也需要这种功能的人来说,下面是Ananda的回答,包含在一个小功能中:
widetolong.ordinal<-function(data,factors,responses,responsename) {
library(reshape2)
data$ID=1:nrow(data) # add an ID to preserve row order
dL=melt(data, id.vars=c("ID", factors)) # `melt` the data
dL=dL[order(dL$ID), ] # sort the molten data
dL[,responsename]=match(dL$variable,responses) # convert reponses to ordinal scores
dL[,responsename]=factor(dL[,responsename],ordered=T)
dL=dL[dL$value != 0, ] # drop rows where `value == 0`
out=dL[rep(rownames(dL), dL$value), c(factors, responsename)] # use `rep` to "expand" `data.frame` & drop unwanted columns
rownames(out) <- NULL
return(out)
}
# example
data <- read.csv("http://www.kuleuven.be/bio/ento/temp/data.csv")
widetolong.ordinal(data,c("factor1","factor2"),c("count_1","count_2","count_3"),"aggression")
答案 0 :(得分:2)
melt
只会让你解决这个问题的一部分。要完成剩下的工作,您只需要使用基础R中的rep
:
data <- read.csv("http://www.kuleuven.be/bio/ento/temp/data.csv")
library(reshape2)
## Add an ID if the row order is importantt o you
data$ID <- 1:nrow(data)
## `melt` the data
dL <- melt(data, id.vars=c("ID", "factor1", "factor2"))
## Sort the molten data, if necessary
dL <- dL[order(dL$ID), ]
## Extract the numeric portion of the "variable" variable
dL$aggression <- gsub("count_", "", dL$variable)
## Drop rows where `value == 0`
dL <- dL[dL$value != 0, ]
## Use `rep` to "expand" your `data.frame`.
## Drop any unwanted columns at this point.
out <- dL[rep(rownames(dL), dL$value), c("factor1", "factor2", "aggression")]
这就是输出最终的样子。如果要删除有趣的行名称,只需使用rownames(out) <- NULL
。
out
# factor1 factor2 aggression
# 1 a a 1
# 7 a a 2
# 7.1 a a 2
# 2 a b 1
# 2.1 a b 1
# 2.2 a b 1
# 3 b a 1
# 9 b a 2
# 9.1 b a 2
# 15 b a 3
# 15.1 b a 3
# 15.2 b a 3
# 4 b b 1
# 4.1 b b 1
# 10 b b 2
# 10.1 b b 2
# 5 c a 1
# 5.1 c a 1
# 5.2 c a 1
# 11 c a 2
# 11.1 c a 2
# 11.2 c a 2
# 11.3 c a 2
# 6 c b 1
# 12 c b 2