從寬格式到長格式重塑時間序列數據

Question

給定四個客戶端的時間序列數據，客戶端 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"

請注意，每個客戶數據之后是同一時期的季節性、趨勢和剩余部分。

我想以一種看起來像的方式重塑為長格式

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

[客戶端編號：1,2,3,4; 類型：客戶、季節性、趨勢、剩余]

（額外幫助理解：如果數據（寬格式）有 30 行/實例，那么我們必須轉換成長格式（30*4*4= 480 行/實例）

Answer 1

看起來您的數據是一種笨拙的重復寬格式。 正如ssdecontrol所說，使用reshape2中的melt，我會結合像lapply這樣的東西來分成幾個部分。 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)

我發現這篇文章對寬格式和長格式非常清楚： http : //seananderson.ca/2013/10/19/reshape.html