在 R 中按组乘以中值

Question

我有数据集

df=structure(list(SKU = c(11202L, 11202L, 11202L, 11202L, 11202L, 
11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L
), stuff = c(8.85947691, 9.450108704, 10.0407405, 10.0407405, 
10.63137229, 11.22200409, 11.22200409, 11.81263588, 12.40326767, 
12.40326767, 12.40326767, 12.99389947, 13.58453126, 14.17516306, 
14.76579485, 15.94705844, 17.12832203, 17.71895382, 21.26274458, 
25.98779894, 63.19760196), action = c(0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L), 
    acnumber = c(137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 
    137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 
    137L, 137L, 137L), year = c(2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L)), .Names = c("SKU", 
"stuff", "action", "acnumber", "year"), class = "data.frame", row.names = c(NA, 
-21L))

action 列只有两个值 0 和 1。正如我们所见，1 类事物有 3 个观测值，0 类事物有 18 个观测值。

我需要 -仅针对类别 1（等于 25.98779894）计算不带零的填充变量的中值。正如我们所看到的，1 和 1 之间有零，它们需要被删除，负值（如果存在）也是如此。即，好像数据集是这样的：

structure(list(SKU = c(11202L, 11202L, 11202L, 11202L, 11202L, 
11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 
11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L, 11202L
), stuff = c(8.85947691, 9.450108704, 10.0407405, 10.0407405, 
10.63137229, 11.22200409, 11.22200409, 11.81263588, 12.40326767, 
12.40326767, 12.40326767, 12.99389947, 13.58453126, 14.17516306, 
14.76579485, 15.94705844, 17.12832203, 17.71895382, 21.26274458, 
25.98779894, 63.19760196), action = c(0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 1L, NA, NA, NA, NA, NA, NA, NA, NA, 1L, 1L), 
    acnumber = c(137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 
    137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 137L, 
    137L, 137L, 137L), year = c(2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 
    2018L, 2018L, 2018L, 2018L, 2018L, 2018L, 2018L)), .Names = c("SKU", 
"stuff", "action", "acnumber", "year"), class = "data.frame", row.names = c(NA, 
-21L))

我还需要根据类别 0 的填充变量计算最后三个观察值的中值，它在第一个观察值之前，在我们的例子中是 12,40326767

然后从类别 1 的中位数中减去类别 0 的中位数，然后乘以一的数量，在本例中为 3。

(25,98779894-12,40326767)*3=40,75359381

这个解决方案

df %>%
  group_by(SKU,acnumber,year) %>%
  summarize(value = 3*(median(stuff[action==1]) - median(stuff[match(1,action)-3:1])),
            stuff=first(stuff),
            action = sum(action)) %>%
  select(SKU,stuff,action,acnumber,year,value)

Moody_Mudskipper 帮助了我

但是！在这个例子中，action的个数是3，所以我们乘以3，但个数可以大于 3 也可以小于 3。如何乘以实数？例如，如果我们有 2 个 by action for stuff，那么

summarize(value = 2*(median(stuff[action==1]) - median(stuff[match(1,action)-3:1])),

免得每次都手动输入。

解决方法 sum(df$action == 1)不适合

summarize(value = sum(df$action == 1)*(median(stuff[action==1]) - median(stuff[match(1,action)-3:1])),

因为它把dataset的所有的加起来，然后乘法不正确。总个数=692，这个数字乘以

 summarize(value = 692*(median(stuff[action==1]) - median(stuff[match(1,action)-3:1])),

错了 1 的乘法必须针对每个特定组 SKU、acnumber、year

111-23-2018 is first group has 3 ones
112-24-2018 is second group has 2 ones

等等

如何做正确？

Answer 1

df%>%
   group_by(SKU,acnumber,year)%>%
   summarise(s=sum(action),k=which(action==1)[1],
            l=s*(median(stuff[action==1])-median(stuff[(k-s+1):k])))%>%
   data.frame()
    SKU acnumber year s  k        l
1 11202      137 2018 3 11 40.75359

在 R 中按组乘以中值

multiply values on median by group in R

r

plyr

lapply

dplyr