Question

我想用R季度推断季度数据。我找到deseasonalize包，但它说它可以仅将月度数据延长。

这是我可以重现的例子：

## How to deseasonalize quarterly data? ##
df <- as.data.frame(rnorm(40, 0, 1))
library(lubridate); library(zoo)
row.names(df) <- as.yearqtr(seq(ymd('2000-01-01'), by = '1 quarter', length.out=(40)))
df
seriestobedeseasonalized <- ts(df[,1])

Answer 1

肯定有更好的方法。我正在使用基函数decompose和ts

names(df) <- "value"
decom <- decompose(ts(df$value, frequency=4, start=c(2000, 2009)))$figure
df$deseasonalize <- df$value - rep(decom, dim(df)[1]/4 + 4)[3:(dim(df)[1]+2)]

Answer 2

我不知道使用了哪个过滤器dimitris_ps，但他的解决方案中的平滑度并不好。最终我设计了以下具有非常好的平滑度的解决方案。我使用了＃34;居中的移动平均线＆＃34;作为过滤器。

#######deseasonalizeQ: deseasonalize quarterly data #######
# Inspired by excellent work of Jason Delaney on Quarterly Deseasonalize: https://www.youtube.com/watch?v=Jr_2nj6M7L8
# Mutatis mutandis replica of Jason's logic in R 

sales <- ts(c(6,15,10,4,10,18,15,7,14,26,23,12,19,28,25,18,22,34,28,21,24,36,30,20,28,40,35,27))

deseasonalizeQ <- function (x){
x <- ts(x)
#Step1: Centered moving averages: create cma time series having the same length with the original time series x
# cma has 2 NAs on both ends.
cma <- filter(x, filter = c(1/8, 1/4, 1/4, 1/4, 1/8), sides=2)

#Step2: Ratios = Original time series / centered moving averages
ratio <- x/cma

#Step3: Unadjusted 4 seasonal indexes
unadj4si <- ts(1:4)
# floor((length(x)-4)/4)  #"-4" is 4 NA at both ends; below "-1" is due to starting "0:" in multiplication

unadj4si[1] <- mean(ratio[3+4*(0:(floor((length(x)-4)/4) - 1))])
unadj4si[2] <- mean(ratio[4+4*(0:(floor((length(x)-4)/4) - 1))])
unadj4si[3] <- mean(ratio[5+4*(0:(floor((length(x)-4)/4) - 1))])
unadj4si[4] <- mean(ratio[6+4*(0:(floor((length(x)-4)/4) - 1))])

#Step4: Adjusted 4 seasonal indexes
adj4si <- ts(1:4)
adj4si[1] <- unadj4si[1]/mean(c(unadj4si[1],unadj4si[2],unadj4si[3],unadj4si[4]))
adj4si[2] <- unadj4si[2]/mean(c(unadj4si[1],unadj4si[2],unadj4si[3],unadj4si[4]))
adj4si[3] <- unadj4si[3]/mean(c(unadj4si[1],unadj4si[2],unadj4si[3],unadj4si[4]))
adj4si[4] <- unadj4si[4]/mean(c(unadj4si[1],unadj4si[2],unadj4si[3],unadj4si[4]))

#Step5: Propogated adjusted seasonal indexes
propadjsi <- ts(1:length(x))

propadjsi[3+4*(0:(floor((length(x)-4)/4) - 1))] <- adj4si[1]
propadjsi[4+4*(0:(floor((length(x)-4)/4) - 1))] <- adj4si[2]
propadjsi[5+4*(0:(floor((length(x)-4)/4) - 1))] <- adj4si[3]
propadjsi[6+4*(0:(floor((length(x)-4)/4) - 1))] <- adj4si[4]

propadjsi[1] <- adj4si[3]
propadjsi[2] <- adj4si[4]
propadjsi[length(x)-1] <- adj4si[1]
propadjsi[length(x)] <- adj4si[2]

#Step6: Deseasonalized values
out <- x/propadjsi  # deseasonalized = x/propadjsi
out
}

deseasonalizeQ(sales)

#Time Series:Start = 1, End = 28, Frequency = 1  
[1]  6.673117 11.015814  8.941810  6.442787 11.121862 13.218976 13.412714 
[8] 11.274878 15.570607 19.094077 20.566162 19.328362 21.131538 20.562852
[15] 22.354524 28.992543 24.468097 24.969177 25.037067 33.824633 26.692469
[22] 26.437953 26.825429 32.213936 31.141214 29.375503 31.296333 43.488814

###### Plots ########
salesSA <- deseasonalizeQ(sales)
salesSAsalesORJ <- cbind(salesSA, sales)

plot(salesSAsalesORJ, plot.type="single", main="Compare", ylab="values", col=c("blue", "red"), lty=1:2)
legend(10, 40, legend=c("salesSA","sales"), col=c("blue", "red"), lty=1:2)

我通过销售（28obs）和增长（56obs）时间序列检查了这个deseasonalizeQ。两者都可以。无论如何，这个解决方案很长，但它的平滑性比上面的答案要好得多。

如何在R中延长季度数据的时间？

2 个答案: