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，我想对由amount分组的变量count和id求和，并在满足以下两个条件的情况下再次从0重置/开始：amount> = 10和count> =3。如果满足这些条件，我还想创建一个新列，其中包含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)
)
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，我想要一个dplyr解决方案，但也欢迎使用其他方法。谢谢！
更新：作者删除的答案几乎可以解决问题：
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，以上答案将重置累计和，但不满足其他条件。
答案 0 :(得分：1)
我没有解决方案，但是您可以先查看mess::cumsumbinning函数，此函数或多或少都在寻找。问题是mess::cumsumbinning仅接受一个条件，而我不知道如何将amount和count条件概括为一个条件。
例如，如果您只寻找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
实际上，您要问的问题甚至更加困难，因为您希望在达到限制后 进行重置。
我知道这只是部分内容，但希望对您有所帮助！
答案 1 :(得分：1)
我终于明白了。 This answer帮助我解决了这个问题。
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)
)
答案 2 :(得分：1)
提供基本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，并针对your solution对该代码进行了基准测试。基准测试显示，Base-R解决方案比整洁解决方案快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