我正在尝试计算扩展窗口上的均值，但是数据结构使之成为可能，因此先前的答案至少缺少了所需要的一点（最接近的位置：link）。
我的数据如下：
Company TimePeriod IndividualID Date.Indiv.Acted Value
1 1 2015 A 2015-01-01 400
2 1 2015 B 2015-02-01 200
3 1 2015 A 2015-06-15 400
4 1 2015 C 2015-07-12 300
5 1 2016 A 2016-07-15 400
6 1 2016 B 2016-08-09 100
7 1 2016 C 2016-09-10 400
8 1 2016 A 2016-10-11 100
9 2 2004 A 2004-07-12 200
10 2 2004 B 2004-08-12 300
我需要为每个Date.Indiv.Acted取按Company-TimePeriod的值的累积平均值。但是，我需要在保留最新副本的同时删除重复副本。因此，对于前两个平均值，没有问题-它们将包括第1行，第1行和第2行。但是，第1、2和3行应删除第1行，因为IndividualID是重复的。本质上，我具有预测信息，并且希望在每次均值计算中仅使用个人的最新预测。
所以我的最终数据看起来像（添加行以便于解释-数据中不需要）
Company TimePeriod IndividualID Date.Indiv.Acted Value CumMean
1 1 2015 A 2015-01-01 400 400
2 1 2015 B 2015-02-01 200 300 (row 1 and 2)
3 1 2015 A 2015-06-15 400 300 (row 2 and 3)
4 1 2015 C 2015-07-12 300 300 (2,3,4)
5 1 2016 A 2016-07-15 400 400 (5)
6 1 2016 B 2016-08-09 100 250 (5,6)
7 1 2016 C 2016-09-10 400 300 (5,6,7)
8 1 2016 A 2016-10-11 100 200 (6,7,8)
9 2 2004 A 2004-07-12 200 200 (9)
10 2 2004 B 2004-08-12 300 250 (9,10)
一个data.table解决方案将是理想的，但是我并不挑剔，只要它可以在相当大的数据（大约2000万行）上运行并且直到宇宙热死为止才需要。
任何帮助，我们将不胜感激。
答案 0 :(得分：3)
setDT(dt)
dt[, occ := 1:.N, by = .(Company, TimePeriod, IndividualID)]
dt[, n := cumsum(!duplicated(IndividualID)), by = .(Company, TimePeriod)]
dt[, Value1 := Value,]
dt[, x := c(0, diff(Value)), by = .(Company, TimePeriod, IndividualID)]
dt[occ>1, Value1 := x,]
dt[, Cummean := cumsum(Value1)/n, by = .(Company, TimePeriod)]
dt[, c("occ", "n", "Value1", "x") := NULL][]
# Company TimePeriod IndividualID Date.Indiv.Acted Value Cummean
# 1: 1 2015 A 2015-01-01 400 400
# 2: 1 2015 B 2015-02-01 200 300
# 3: 1 2015 A 2015-06-15 400 300
# 4: 1 2015 C 2015-07-12 300 300
# 5: 1 2016 A 2016-07-15 400 400
# 6: 1 2016 B 2016-08-09 100 250
# 7: 1 2016 C 2016-09-10 400 300
# 8: 1 2016 A 2016-10-11 100 200
# 9: 2 2004 A 2004-07-12 200 200
#10: 2 2004 B 2004-08-12 300 250
dt <- structure(list(Company = c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2), TimePeriod = c(2015,
2015, 2015, 2015, 2016, 2016, 2016, 2016, 2004, 2004), IndividualID = c("A",
"B", "A", "C", "A", "B", "C", "A", "A", "B"), Date.Indiv.Acted = c("2015-01-01",
"2015-02-01", "2015-06-15", "2015-07-12", "2016-07-15", "2016-08-09",
"2016-09-10", "2016-10-11", "2004-07-12", "2004-08-12"), Value = c(400,
200, 400, 300, 400, 100, 400, 100, 200, 300)), row.names = c(NA,
-10L), class = "data.frame")
答案 1 :(得分：1)
我特别不喜欢循环，但是我认为这个循环足够简单，可以逐步理解。可以很容易地将其更改为运行任何其他指标，而不是平均值（例如累积方差）
# function that drops duplicates and calculates cumulative mean
fun.attempt <- function(dat, dup, value){
#dat: data set
#dup: string column to look for duplicates
#value: string column to calculate the mean
x <- dat[!duplicated(get(dup), fromLast = T), .(get(value))]
y <- cumsum(x) / 1:nrow(x)
y <- y[nrow(y)]
return(y)
}
foo[, grp := .GRP, by = .(Company, TimePeriod)] # to create a more efficient loop
hl <- list() # as storage
for(k in unique(foo$grp)){
got <- foo[grp == k] # running the cumulative mean for each grouping
for(y in 1:nrow(got)){
# applying customized function
got[y, cummean2:= fun.attempt(got[1:y], 'IndividualID', 'Value')]
}
hl[[k]] <- got # storing the subsetted data.tables
}
现在只需编译data.tables的列表。 CumMean列是您的原始计算，cummean2是我的。
rbindlist(hl)
Company TimePeriod IndividualID Date.Indiv.Acted Value CumMean grp cummean2
1: 1 2015 A 2015-01-01 400 400 1 400
2: 1 2015 B 2015-02-01 200 300 1 300
3: 1 2015 A 2015-06-15 400 300 1 300
4: 1 2015 C 2015-07-12 300 300 1 300
5: 1 2016 A 2016-07-15 400 400 2 400
6: 1 2016 B 2016-08-09 100 250 2 250
7: 1 2016 C 2016-09-10 400 300 2 300
8: 1 2016 A 2016-10-11 100 200 2 200
9: 2 2004 A 2004-07-12 200 200 3 200
10: 2 2004 B 2004-08-12 300 250 3 250