R:计算自上次出现值以来的累计和和计数

Question

给定简化数据

set.seed(13)

user_id = rep(1:2, each = 10)
order_id = sample(1:20, replace = FALSE)
cost = round(runif(20, 1.5, 75),1)
category = sample( c("apples", "pears", "chicken"), 20, replace = TRUE)
pit = rep(c(0,0,0,0,1), 4)

df = data.frame(cbind(user_id, order_id, cost, category, pit))

user_id order_id cost category pit
      1       15 11.6    pears   0
      1        5 41.7   apples   0
      1        8 51.3  chicken   0
      1        2 40.3    pears   0
      1       16  7.9    pears   1
      1        1 47.1  chicken   0
      1        9  3.8   apples   0
      1       10 35.4   apples   0
      1       11 25.8  chicken   0
      1       20 48.1  chicken   1
      2        7 32.6    pears   0
      2       18 31.3    pears   0
      2       14   69   apples   0
      2        4 60.9  chicken   0
      2       13 41.2   apples   1
      2       17  9.4    pears   0
      2       19 34.9   apples   0
      2        6  5.3    pears   0
      2        3 57.3   apples   0
      2       12  7.7   apples   1

我想创建包含成本累计总和和不同类别计数的列自上次 pit == 1 以来。所以结果看起来像这样:

user_id order_id cost category pit cum_cost distinct_categories
      1       15 11.6    pears   0     11.6                   1
      1        5 41.7   apples   0     53.3                   2
      1        8 51.3  chicken   0    104.6                   3
      1        2 40.3    pears   0    144.9                   3
      1       16  7.9    pears   1    152.8                   3
      1        1 47.1  chicken   0     47.1                   1
      1        9  3.8   apples   0     50.9                   2
      1       10 35.4   apples   0     86.3                   2
      1       11 25.8  chicken   0    112.1                   3
      1       20 48.1  chicken   1    160.2                   3
      2        7 32.6    pears   0     32.6                   1
      2       18 31.3    pears   0     63.9                   1
      2       14   69   apples   0    132.9                   2
      2        4 60.9  chicken   0    193.8                   3
      2       13 41.2   apples   1    235.0                   3
      2       17  9.4    pears   0      9.4                   1
      2       19 34.9   apples   0     44.3                   2
      2        6  5.3    pears   0     49.6                   2
      2        3 57.3   apples   0    106.9                   2
      2       12  7.7   apples   1    114.6                   2

理想情况下，解决方案将在 dplyr 中，但我对其他包/方法持开放态度。非常感谢您的帮助!卡西娅

Answer 1

我们可以使用dplyr。按 'user_id' 和通过获取 'pit' 的累积和并获取其 lag 创建的分组变量进行分组，我们得到 'cost' 的 cumsum 作为 'cum_cost'和 match 索引的 cummax 在 'category' 和 unique 'category' 之间作为 'distinct_categories。

library(dplyr)
df %>%
    group_by(user_id, ind= lag(cumsum(pit), default=0)) %>% 
    mutate(cum_cost = cumsum(cost), 
           distinct_categories = cummax(match(category, unique(category))))
# user_id order_id  cost category   pit   ind cum_cost distinct_categories
#     <int>    <int> <dbl>    <chr> <int> <dbl>    <dbl>               <int>
#1        1        3  49.8   apples     0     0     49.8                   1
#2        1       13  14.8  chicken     0     0     64.6                   2
#3        1       18  11.4   apples     0     0     76.0                   2
#4        1       15  52.6  chicken     0     0    128.6                   2
#5        1       11  13.6  chicken     1     0    142.2                   2
#6        1       19  26.9  chicken     0     1     26.9                   1
#7        1        2  54.9  chicken     0     1     81.8                   1
#8        1        1  70.6  chicken     0     1    152.4                   1
#9        1       10  55.0  chicken     0     1    207.4                   1
#10       1       12  19.7  chicken     1     1    227.1                   1
#11       2        8  40.0    pears     0     2     40.0                   1
#12       2       16  37.4    pears     0     2     77.4                   1
#13       2       20  70.5    pears     0     2    147.9                   1
#14       2        5  63.8   apples     0     2    211.7                   2
#15       2       14  31.9   apples     1     2    243.6                   2
#16       2       17   9.1  chicken     0     3      9.1                   1
#17       2        4  21.9    pears     0     3     31.0                   2
#18       2        7  52.3   apples     0     3     83.3                   3
#19       2        9  43.3  chicken     0     3    126.6                   3
#20       2        6   9.9    pears     1     3    136.5                   3