R 如果满足多个条件，则具有重置选项的累积和

如果满足多个条件，我将尝试使用重置选项进行累计和。更具体地说，我想将变量

amount

和

count

累加起来，按

id

分组，如果满足这两个条件，则重新设置/从0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，否则包含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)
)

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)
    )

df%>%group\u by（id）%>%
变异(
金额_cumsum=purrr:：累计（.x=amount，.f=~case_当（.x<10~.x+.y，TRUE~.y）），
count_cumsum=purrr:：累计（.x=count，.f=~case_当（.x<3~.x+.y，TRUE~.y）），
条件满足=整数（金额总和>=10，计数总和>=3）
)

---

如果一个变量满足条件，上面的答案将重置累积和，但不考虑是否满足其他条件。

我没有解决方案，但您可以从查看

mess:：cumsumbining

函数开始，该函数或多或少就是您要找的。问题是只接受一个条件，我不知道如何将金额和计数条件汇总成一个条件

例如，如果您只查找

count>=3

，则可以执行以下操作：

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

df%>%
分组依据（id，group=Cumsumbining（计数，3））%>%
变异（count_cumsum=cumsum（count））
#A tibble:18x6
#组：id，组[10]
日期id金额计数组计数总和
1202-01-01A1
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
52020-05-01A4
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 80 5 0
102020-04-01B1050
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 C1 1 6 2
15 2020-03-01 C 6 1 7 1
16 2020-04-01 C 5 2 7 3
172020-05-01C5181
182020-06-01C5081

事实上，你所要求的更为困难，因为你希望在达到极限后进行重置

我知道这只是局部的，但我希望它能帮助你

我终于明白了。帮我解决了这个问题

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)
    )

df%
分组依据（id）%>%
嵌套（数据=c（数量、计数））%>%
变异(
如果（max（.x[1]）<10 | max（.x[2]）<3.x+.y else.y），则data_acculate=purrr:：acculate（.x=data，.f=function（.x，.y）
) %>%
unnest（cols=c（数据_累计））%>%
重命名（金额总和=金额，计数总和=计数）%>%
unnest（cols=c（数据））%>%
变异（条件满足=情况满足时）(
金额>=10，计数>=3~1，
对（0）
)

为base-R解决方案做出贡献：

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
  }
}

---

嗨，谢谢你的回答！我刚刚用作者删除的答案更新了我的问题。这个答案几乎解决了我的问题，与您的方法类似，但使用了

purr

包。是的，这个答案与我的问题相同，因为

purr:：accumulate

不能（或者我不知道如何）使用多种条件。感谢您的回复-事实上，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（+200万个观察值和+700.000个唯一组/id），不幸的是它不起作用：

dplyr

解决方案花了13,55分钟进行计算，而

base

溶液即使在1,81小时后也没有完成计算。我已将你的答案标记为正确，因为我已在较小的样本上进行了测试，结果有效。谢谢

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)
    )

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
  }
}

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