如何在数据表中插入新日期和旧日期之间的所有缺失周来计算每周库存R

Question

所以我有一个列ID为DATE和STOCK 的数据表df

在该表中，同一ID具有多个日期和库存值：

ID        DATE        STOCK
a1     2017-05-04       1
a1     2017-06-04       4
a1     2017-06-05       1
a1     2018-05-04       1
a1     2018-06-04       3
a1     2018-06-05       1
a2     2016-11-26       2
a2      ...             ..

使用lubridate我可以得到哪一周的日期如下：

dfWeeks=df[,"WEEK" := floor_date(df$`Date`, "week")]
ID        DATE        STOCK        WEEK
a1     2017-05-04       1       2017-04-30
a1     2017-06-04       4       2017-06-04
a1     2017-06-05       1       2017-06-04
a1     2018-05-04       1       2018-04-29
a1     2018-06-04       3       2018-06-03
a1     2018-06-05       1       2018-06-03
a2     2016-11-26       2       2016-11-20
a2      ...             ..

因此，从DATE列中，我知道我的旧日期是2017-05-04，最新日期为2018-06-05，大约有56.71429周：

dates <- c( "2017-05-04","2018-06-05")
dif <- diff(as.numeric(strptime(dates, format = "%Y-%m-%d")))/(60 * 60 * 24 * 7) 

我的表只有4个唯一的周，所以我们的想法是对每周的库存进行汇总，并插入库存中缺失的(57-4=53周(0值的库存。

然后我可以像一样计算所有周的平均值

meanStock<- dfWeeks[, .(mean=sum(Stock, na.rm = T)/dif <- diff(as.numeric(strptime(c(min(Date), max(Date)), format = "%Y-%m-%d")))/(60 * 60 * 24 * 7) ), by = .(ID)]

但我不知道它是否有效，希望我已经明确表示，欢迎任何建议或方法。

更新：

这就是我获取最大和最小日期的方式

max = aggregate(df$`Date`,by=list(df$ID),max)
colnames(max) = c("ID", "MAX")
min = aggregate(df$`Date`,by=list(df$ID),min)
colnames(min) = c("ID", "MIN")
test <- merge(max, min, by="ID", all=T)

Answer 1

类似于：

library(data.table)
setDT(df)[, DATE := as.Date(DATE)][, `:=` (st = min(DATE), end = max(DATE) + 7), by = ID][
, .(ID = ID, DATE = DATE, STOCK = STOCK, Expanded = seq(st, end, by = "week")), by = 1:nrow(df)][
, `:=` (WEEK = floor_date(Expanded, "week"), WEEK2 = floor_date(DATE, "week"))][
WEEK != WEEK2, STOCK := 0][
, .(SUM_STOCK = sum(STOCK)), by = .(WEEK, ID)]

输出(2017-04-02至2017-06-11和ID至a1周的行(：

WEEK ID SUM_STOCK
1: 2017-04-02 a1         0
2: 2017-04-09 a1         0
3: 2017-04-16 a1         0
4: 2017-04-23 a1         0
5: 2017-04-30 a1         1
6: 2017-05-07 a1         0
7: 2017-05-14 a1         0
8: 2017-05-21 a1         0
9: 2017-05-28 a1         0
10: 2017-06-04 a1         5
11: 2017-06-11 a1         0