如果满足多个条件，则具有重置选项的累积和 [英] Cumulative sum with reset option if multiple conditions are met

问题描述

如果满足多个条件，我将尝试使用reset选项进行累加。更具体地说，我想对按 id分组的变量金额和 count 进行累加和并在满足以下两个条件的情况下再次从0重新开始：金额> = 10和 count > =3。如果满足这些条件，我还想创建一个包含1的新列，否则满足0。

I am trying to make a cumulative sum with reset option if multiple conditions are met. More specifically, I want to cumulative sum the variables amount and count grouped by id and reset/start from 0 again if these two conditions are met: amount >= 10 and count >= 3. I also would like to create a new column that contains 1 if these conditions are met and 0 otherwise.

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)
)

所需的输出：

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)
)

如果可能的话，我想使用 dplyr 解决方案，但也欢迎使用其他方法。谢谢！

If possible, I would like a dplyr solution, but alternatives are also welcome. Thanks!

更新：作者删除的答案几乎解决了问题：

Update: an answer that was deleted by the author almost got the problem solved:

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)
    )

答案a如果一个变量满足条件，bove会重置累积总和，但是如果另一个条件满足，bove将重置总和。

The answer above resets the cumulative sum if the condition is met for one single variable, but not taking into account if the other condition was met.

推荐答案

使用base-R解决方案进行贡献：

Contributing with a base-R solution:

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，并根据您的解决方案。基准测试显示，Base-R解决方案比tidyverse解决方案快21倍！

I've expanded your dataset and benchmarked this code against your solution. Benchmark shows the Base-R solution 21 times faster than the tidyverse one!

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

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