从宽格式到长格式重塑时间序列数据

Question

给定四个客户端的时间序列数据，客户端 1、2、3、4。

names(data)
"Date" "X1.CLIENT"   "X1seasonal"  "X1trend"  "X1remainder" "X2.CLIENT" 
  "X2seasonal"  "X2trend"     "X2remainder" "X3.CLIENT"   "X3seasonal"  
"X3trend"     "X3remainder" "X4.CLIENT"   "X4seasonal"  "X4trend"     
"X4remainder"

请注意，每个客户数据之后是同一时期的季节性、趋势和剩余部分。

我想以一种看起来像的方式重塑为长格式

"Date" "CLIENT_Number" "Type"  "Value"

[客户端编号：1,2,3,4; 类型：客户、季节性、趋势、剩余]

（额外帮助理解：如果数据（宽格式）有 30 行/实例，那么我们必须转换成长格式（30*4*4= 480 行/实例）

Answer 1

看起来您的数据是一种笨拙的重复宽格式。 正如ssdecontrol所说，使用reshape2中的melt，我会结合像lapply这样的东西来分成几个部分。 例如：

# Set up
install.packages("reshape2")
library(reshape2)

# Dummy data
d = data.frame(date=seq.Date(as.Date("2015-01-01"), length.out=100, by=1),
Client.1="A", trend.1=rnorm(100), remainder.1=rpois(10, 3), Client.2="B",
trend.2=rnorm(100), remainder.2=rpois(10, 3), Client.3="C", trend.3=rnorm(100),
remainder.3=rpois(10, 3))

# Split data by columns with client numbers
d = lapply(c(2,5,8), function(i){
    # Take columns relating to client i
    x = d[,c(1, seq(i, length.out=3, by=1))]
    # Rename so you have consistent factor labels
    names(x)[2:4] = c("client", "trend", "remainder")
    # Conververt from wide to long
    x = melt(x, id.vars=c("date", "client"))
})

# Make list into a data frame
d = do.call("rbind", d)

我发现这篇文章对宽格式和长格式非常清楚： http : //seananderson.ca/2013/10/19/reshape.html