如果满足多个条件，则带有重置选项的累积和答案

【问题标题】：Cumulative sum with reset option if multiple conditions are met如果满足多个条件，则带有重置选项的累积和
【发布时间】：2020-07-09 14:35:50
【问题描述】：

如果满足多个条件，我正在尝试使用重置选项进行累积总和。更具体地说，我想对由id 分组的变量amount 和count 进行累积求和，如果满足这两个条件，则再次从0 重置/开始：amount >= 10 和count >= 3。我还想创建一个新列，如果满足这些条件，则包含 1，否则为 0。

数据样本：

df <- data.frame(
    date = as.Date(c("2020-01-01", "2020-02-01", "2020-03-01", "2020-04-01", "2020-05-01", "2020-06-01", "2020-01-01", "2020-02-01", "2020-03-01", "2020-04-01", "2020-05-01", "2020-06-01", "2020-01-01", "2020-02-01", "2020-03-01", "2020-04-01", "2020-05-01", "2020-06-01")),
    id = c("A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C", "C"),
    amount = c(1, 9, 5, 5, 6, 2, 10, 4, 8, 10, 6, 5, 5, 1, 6, 5, 5, 5),
    count = c(0, 2, 5, 4, 5, 1, 0, 0, 0, 0, 2, 1, 1, 1, 1, 2, 1, 0)
)

期望的输出：

df <- data.frame(
    date = as.Date(c("2020-01-01", "2020-02-01", "2020-03-01", "2020-04-01", "2020-05-01", "2020-06-01", "2020-01-01", "2020-02-01", "2020-03-01", "2020-04-01", "2020-05-01", "2020-06-01", "2020-01-01", "2020-02-01", "2020-03-01", "2020-04-01", "2020-05-01", "2020-06-01")),
    id = c("A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "C", "C", "C", "C", "C", "C"),
    amount = c(1, 9, 5, 5, 6, 2, 10, 4, 8, 10, 6, 5, 5, 1, 6, 5, 5, 5),
    count = c(0, 2, 5, 4, 5, 1, 0, 0, 0, 0, 2, 1, 1, 1, 1, 2, 1, 0),
    amount_cumsum = c(1, 10, 15, 5, 11, 2, 10, 14, 22, 32, 38, 43, 5, 6, 12, 5, 10, 5),
    count_cumsum = c(0, 2, 7, 4, 9, 1, 0, 0, 0, 0, 2, 3, 1, 2, 3, 2, 3, 0),
    condition_met = c(0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0)
)

如果可能，我想要dplyr 解决方案，但也欢迎使用替代方案。谢谢！

更新：一个被作者删掉的答案几乎解决了问题：

df %>% group_by(id) %>%
    mutate(
        amount_cumsum = purrr::accumulate(.x = amount, .f = ~ if_else(condition = .x < 10, true = .x + .y, false = .y)),
        count_cumsum = purrr::accumulate(.x = count, .f = ~ if_else(condition = .x < 3, true = .x + .y, false = .y)),
        condition_met = as.integer(amount_cumsum >= 10 & count_cumsum >= 3)
 )

或者，或者：

df %>% group_by(id) %>%
    mutate(
        amount_cumsum = purrr::accumulate(.x = amount, .f = ~ case_when(.x < 10 ~ .x + .y, TRUE ~ .y)),
        count_cumsum = purrr::accumulate(.x = count, .f = ~ case_when(.x < 3 ~ .x + .y, TRUE ~ .y)),
        condition_met = as.integer(amount_cumsum >= 10 & count_cumsum >= 3)
    )

如果一个变量满足条件，上面的答案会重置累积总和，但不考虑是否满足另一个条件。

【问题讨论】：

标签： r dplyr

【解决方案1】：

贡献一个 base-R 解决方案：

df$amount_cumsum <- 0
df$count_cumsum <- 0    
df$condition_met <- 0  
reset = F
for (i in 1:nrow(df)) {
  if (i == 1 | reset) {
    df$amount_cumsum[i] = df$amount[i]
    df$count_cumsum[i] = df$count[i]
    reset = F
  } else if (df$id[i] != df$id[i-1]) {
    df$amount_cumsum[i] = df$amount[i]
    df$count_cumsum[i] = df$count[i]
    reset = F
  } else {
    df$amount_cumsum[i] = df$amount_cumsum[i-1] + df$amount[i]
    df$count_cumsum[i] = df$count_cumsum[i-1] + df$count[i]
  }
  
  if (df$amount_cumsum[i] >= 10 & df$count_cumsum[i] >= 3) {
    df$condition_met[i] = 1
    reset = T
  }
}

我已经扩展了您的数据集，并将此代码与 your solution 进行了基准测试。基准测试显示 Base-R 解决方案比 tidyverse 解决方案快 21 倍！

library(tidyverse)

dates = seq(as.Date("2019-01-01"), as.Date("2020-03-04"), by="days")

df <- data.frame(
  date = c(sample(dates, 300), sample(dates, 400), sample(dates, 350)),
  id = c(rep("A", 300), rep("B", 400), rep("C", 350)),
  amount = floor(runif(1050, 0, 15)),
  count = floor(runif(1050, 0, 5)),
  stringsAsFactors = F
)

rbenchmark::benchmark(
  "Tidy Solution" = {
    df_tidy <- df %>%
      group_by(id) %>%
      nest(data = c(amount, count)) %>%
      mutate(
        data_accumulate = purrr::accumulate(.x = data, .f = function(.x, .y) if (max(.x[1]) < 10 | max(.x[2]) < 3) .x + .y else .y)
      ) %>%
      unnest(cols = c(data_accumulate)) %>%
      rename(amount_cumsum = amount, count_cumsum = count) %>%
      unnest(cols = c(data)) %>%
      mutate(condition_met = case_when(
        amount_cumsum >= 10 & count_cumsum >= 3 ~ 1,
        TRUE ~ 0)
      )
  },
  "Base-R Solution" = {
    df_base <- df
    df_base$amount_cumsum <- 0
    df_base$count_cumsum <- 0    
    df_base$condition_met <- 0  
    reset = F  # to reset the counters
    for (i in 1:nrow(df_base)) {
      if (i == 1 | reset) {
        df_base$amount_cumsum[i] = df_base$amount[i]
        df_base$count_cumsum[i] = df_base$count[i]
        reset = F
      } else if (df_base$id[i] != df_base$id[i-1]) {
        df_base$amount_cumsum[i] = df_base$amount[i]
        df_base$count_cumsum[i] = df_base$count[i]
        reset = F
      } else {
        df_base$amount_cumsum[i] = df_base$amount_cumsum[i-1] + df_base$amount[i]
        df_base$count_cumsum[i] = df_base$count_cumsum[i-1] + df_base$count[i]
      }
      if (df_base$amount_cumsum[i] >= 10 & df_base$count_cumsum[i] >= 3) {
        df_base$condition_met[i] = 1
        reset = T
      }
    }
  },
  replications = 100)

gc()

           test replications elapsed relative user.self sys.self user.child sys.child
Base-R Solution          100    3.89    1.000      3.69      0.0         NA        NA
  Tidy Solution          100   84.00   21.594     78.65      0.2         NA        NA

【讨论】：

感谢您的回复 - 实际上您的 base 解决方案比 dplyr 解决方案快得多，但对于更大的数据集（+2 百万观察和 +700.000 个唯一组/ID）它不幸的是没有工作：dplyr 解决方案需要 13.55 分钟来计算，而base 解决方案即使在 1.81 小时后也没有完成计算。我已将您的答案标记为正确，因为我已经在较小的样本上对其进行了测试并且它有效。谢谢！

【解决方案2】：

我没有解决方案，但您可以先查看mess::cumsumbinning 函数，它或多或少是您正在寻找的。问题是mess::cumsumbinning只接受一个条件，我不知道如何将amount和count条件归纳为一个。

例如，如果您只寻找count>=3，您可以这样做：

df %>%
  group_by(id,group=cumsumbinning(count,3)) %>% 
  mutate(count_cumsum=cumsum(count))

# A tibble: 18 x 6
# Groups:   id, group [10]
   date       id    amount count group count_cumsum
   <date>     <fct>  <dbl> <dbl> <int>        <dbl>
 1 2020-01-01 A          1     1     1            1
 2 2020-02-01 A          9     3     2            3
 3 2020-03-01 A          5     1     3            1
 4 2020-04-01 A          5     1     3            2
 5 2020-05-01 A          6     4     4            4
 6 2020-06-01 A          2     1     5            1
 7 2020-01-01 B         10     0     5            0
 8 2020-02-01 B          4     0     5            0
 9 2020-03-01 B          8     0     5            0
10 2020-04-01 B         10     0     5            0
11 2020-05-01 B          6     2     5            2
12 2020-06-01 B          5     1     6            1
13 2020-01-01 C          5     1     6            1
14 2020-02-01 C          1     1     6            2
15 2020-03-01 C          6     1     7            1
16 2020-04-01 C          5     2     7            3
17 2020-05-01 C          5     1     8            1
18 2020-06-01 C          5     0     8            1

事实上，您的要求更加困难，因为您希望在达到限制后进行重置。

我知道这只是部分，但我希望它会帮助你！

【讨论】：

您好，谢谢您的回答！我刚刚用作者删除的答案更新了我的问题。该答案几乎解决了我的问题，并且与您的方法类似，但使用了purrr 包。
是的，这个答案和我的问题一样，因为purrr::accumulate 不能（或者我不知道如何）使用多个条件。

【解决方案3】：

我终于明白了。 This answer帮我解决了问题。

df <- df %>%
    group_by(id) %>%
    nest(data = c(amount, count)) %>%
    mutate(
        data_accumulate = purrr::accumulate(.x = data, .f = function(.x, .y) if (max(.x[1]) < 10 | max(.x[2]) < 3) .x + .y else .y)
    ) %>%
    unnest(cols = c(data_accumulate)) %>%
    rename(amount_cumsum = amount, count_cumsum = count) %>%
    unnest(cols = c(data)) %>%
    mutate(condition_met = case_when(
        amount_cumsum >= 10 & count_cumsum >= 3 ~ 1,
        TRUE ~ 0)
    )

【讨论】：