r - 如何在数据表中插入新旧日期之间所有缺失的周数以计算每周库存 R

Question

所以我有一个包含 ID DATE 和 STOCK 列的数据表 df

在此表中，同一个 ID 具有多个值及其日期和库存:

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 我可以获得日期是哪一周，如下所示:

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      ...             ..

所以从 DATE 列我知道我的旧日期是 2017-05-04和最新日期 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 个独特的周，所以我的想法是对每周的库存求和，并插入缺失的(57-4=53 周)库存值为 0 的周。

然后我可以像这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)]

但我不知道它是否会起作用，希望我说清楚了，欢迎任何建议或方法。

更新:

这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 到 2017-06-11 和 ID 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