最近四周的平均销售额基于R的条件

Question

我有几周内推广产品的数据集，我想计算过去4个非促销（由旗帜表示）周促销期间产品的平均销售额。如果交易是在非促销期间，我们应该进行销售，因为我们必须采取最近非促销周销售的平均销售，它们可能是不连续的。

请注意

structure(list(Product_group = structure(c(1L, 1L, 1L, 1L, 1L, 
1L), .Label = "A", class = "factor"), Promo = structure(c(1L, 
1L, 2L, 1L, 1L, 2L), .Label = c("0", "1"), class = "factor"), 
    Week = structure(c(1L, 2L, 2L, 3L, 4L, 5L), .Label = c("2017-01-01", 
    "2017-01-02", "2017-01-04", "2017-01-05", "2017-01-06", "2017-01-08", 
    "2017-01-09"), class = "factor"), Sales = c(50, 50, 60, 70, 
    50, 50)), .Names = c("Product_group", "Promo", "Week", "Sales"
), row.names = c(NA, 6L), class = "data.frame")

head(df)
  Product_group Promo       Week Sales
1             A     0 2017-01-01    50
2             A     0 2017-01-02    50
3             A     1 2017-01-02    60
4             A     0 2017-01-04    70
5             A     0 2017-01-05    50
6             A     1 2017-01-06    50

我正在寻找像这样的输出

          Product_group Promo       Week Sales Avg Pre Promo Sales
    1             A     0 2017-01-01    50      50 # Since it is non promo
    2             A     0 2017-01-02    50      50 
    3             A     1 2017-01-02    60      50 # 100/2
    4             A     0 2017-01-04    70      70 
    5             A     0 2017-01-05    50      50
    6             A     1 2017-01-06    50      55 # (50 +70 + 50 + 50 )/4

Answer 1

当Promo == 1，我看看Promo为零的指数。然后选择最后四周的最大值来获得平均值。

  df <- rbind(df, df) # get more rows to data

  df$AvgPrePromoSales <- 
   sapply(1 : nrow(df), function(x) if(df$Promo[x] == 1) {
        ind <- which(df[1:x,]$Promo == 0)
        mean(df$Sales[ind[max(1, length(ind) - 3) : length(ind)]])
      } else {
        df$Sales[x] 
      })

df[, c(2, 4, 5)]
#      Promo Sales AvgPrePromoSales
# 1      0    50               50
# 2      0    50               50
# 3      1    60               50
# 4      0    70               70
# 5      0    50               50
# 6      1    50               55
# 7      0    50               50
# 8      0    50               50
# 9      1    60               55
# 10     0    70               70
# 11     0    50               50
# 12     1    50               55