Question

我正在使用不平衡，不规则间隔的横截面时间序列。我的目标是获得“数量”向量的滞后移动平均向量，按“主题”分段。

换句话说，对于Subject_1，已经观察到以下Quanatities： [1,2,3,4,5]。我首先需要将其滞后1，产生[NA，1,2,3,4]。

然后我需要采用3阶移动平均线，产生[NA，NA，NA，（3 + 2 + 1）/ 3，（4 + 3 + 2）/ 3]

需要对所有受试者进行上述操作。

# Construct example balanced panel DF
panel <- data.frame(
  as.factor(sort(rep(1:6,5))),
  rep(1:5,6),
  rnorm(30)                
)
colnames(panel)<- c("Subject","Day","Quantity")

#Make panel DF unbalanced
panelUNB <- subset(panel,as.numeric(Subject)!= Day)
panelUNB <- panelUNB[-c(15,16),]

如果面板是平衡的，我将首先使用包plm和函数lag滞后“数量”变量。然后，我将使用包rollmean中的函数zoo来获取滞后“Quanatity”的移动平均值：

panel$QuantityMA <- ave(panel$Quantity, panel$Subject, FUN = function(x) rollmean(
                     x,3,align="right",fill=NA,na.rm=TRUE))

当应用于平衡的“面板”DF时，这将产生正确的结果。

问题在于plm和lag依赖于均匀间隔的系列来生成索引变量，而rollapply要求所有主体的观察数（windowsize）相等。

StackExchange上有解决方案，data.table提示我的问题的解决方案：Producing a rolling average of an unbalanced panel data set

也许可以修改此解决方案以产生固定长度的移动平均线而不是“滚动累积平均值”。

Answer 1

这会给你带来预期的效果吗？

library(reshape2)
library(zoo)

# create time series where each subject have an observation at each time step
d1 <- data.frame(subject = rep(letters[1:4], each = 5),
                 day = rep(1:5, 4),
                 quantity = sample(x = 1:4, size = 20, replace = TRUE))
d1

# select some random observations
d2 <- d1[sample(x = seq_len(nrow(d1)), size = 15), ]
d2

# reshape to wide format with dcast
# -> 'automatic' extension from irregular to regular series for each subject,
# _given_ that all time steps are represented.
# Alternative method below more explicit

# fill for structural missings defaults to NA
d3 <- dcast(d2, day ~ subject, value.var = "quantity")
d3

# convert to zoo time series 
z1 <- zoo(x = d3[ , -1], order.by = d3$day)

################################
# alternative method to extend time series
# time steps to include are given explicitly

# create a zero-dimensional zoo series
z0 <- zoo(, min(d1$day):max(d1$day))

# extend z1 to contain the same time indices as z0 
z1 <- merge(z1, z0) 
################################

# lag, defaults to one unit 
z2 <- lag(x = z1)
z2

# calculate rolling mean with window width 3
rollmeanr(x = z2, k = 3)

# Handling of NAs:
# from ?rollmean:
# "The default method of rollmean does not handle inputs that contain NAs.
# In such cases, use rollapply instead.": 
rollapplyr(data = z2, width = 3, FUN = mean, na.rm = TRUE)

Answer 2

所以，回答我自己的问题，一种方法是通过split-lapply（rollingaverage）-unlist：

Temp <-with(panelUNB, split(Quantity, Subject))
Temp <- lapply(Temp, FUN=function (x) rollapplyr(
   x,2,align="right",fill=NA,na.rm=TRUE, FUN=mean))
QuantityMA <-unlist(Temp)

然后必须将“QuantityMA”向量添加回主“panelUNB”帧。似乎工作。滞后可以在ddply的不平衡面板上完成。

如果有人有另一个，也许更优雅的解决方案，欢迎您分享。

R中不平衡面板上的简单移动平均值

2 个答案: