[英]R: Calculating the number of occurrences within a specific time period in the past for each unique individual in a dataset in R
我正在嘗試計算過去特定時間段內給定個人發生事件的次數。 在這種特殊情況下,我需要知道,對於每個新觀察(反映單個調度請求),個人在前 60 天內安排了多少次旅行(trip_scheduled)。 最終,我需要計算該人在前 60 天的預定行程的同一天取消的次數。 但我只是從“移動”60 天期間的計數開始。
我在這篇文章中找到了一些類似但略有不同的問題的優雅答案: R:計算特定時間未來特定事件的發生次數
我的情況在幾個方面有所不同:第一,我正在嘗試查看以前的時間段,我不知道這是否會改變我的方法,第二,我需要對 40,000 多人進行分析,我一直試圖通過我在另一個答案中找到的代碼的混合來完成,一個 for 循環(我知道這是不贊成的)和 dplyr 分組。 它根本不起作用。
有人能幫我指出正確的方向嗎? 我很想堅持使用 dplyr 和 base。 我只是不太了解data.table。
這是我一直在嘗試處理的代碼和測試數據:
test_set2 <- structure(list(tripID = c("20180112-100037-674-101", "20180112-100037-674-201",
"20180112-100037-674-301", "20180113-100037-676-101", "20180113-100037-676-201",
"20180115-100037-675-101", "20180115-100037-675-201", "20180116-100037-677-101",
"20180116-100037-677-201", "20180131-100037-678-101", "20180101-100146-707-101",
"20180101-100146-707-201", "20180102-100146-708-101", "20180102-100146-708-201",
"20180103-100146-709-101", "20180103-100146-709-201", "20180104-100146-710-101",
"20180104-100146-710-201", "20180105-100146-711-101", "20180105-100146-711-201",
"20180403-100532-223-101", "20180403-100532-223-201", "20180620-100532-224-101",
"20180620-100532-224-201", "20180704-100532-225-101", "20180704-100532-225-201",
"20180926-100532-228-101", "20180926-100532-228-201", "20180927-100532-226-101",
"20180927-100532-226-201"), CUSTOMER_ID = c(100037L, 100037L,
100037L, 100037L, 100037L, 100037L, 100037L, 100037L, 100037L,
100037L, 100146L, 100146L, 100146L, 100146L, 100146L, 100146L,
100146L, 100146L, 100146L, 100146L, 100532L, 100532L, 100532L,
100532L, 100532L, 100532L, 100532L, 100532L, 100532L, 100532L
), trip_date = structure(c(17543, 17543, 17543, 17544, 17544,
17546, 17546, 17547, 17547, 17562, 17532, 17532, 17533, 17533,
17534, 17534, 17535, 17535, 17536, 17536, 17624, 17624, 17702,
17702, 17716, 17716, 17800, 17800, 17801, 17801), class = "Date"),
trip_scheduled = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1), same_day_cancel = c(1,
1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0)), row.names = c(NA, -30L), groups = structure(list(
CUSTOMER_ID = c(100037L, 100146L, 100532L), .rows = list(
1:10, 11:20, 21:30)), row.names = c(NA, -3L), class = c("tbl_df",
"tbl", "data.frame"), .drop = TRUE), class = c("grouped_df",
"tbl_df", "tbl", "data.frame"))
running_frame <- test_set2[1,]
unique_customers <- unique(test_set2$CUSTOMER_ID)
for (cust in unique_customers){
temp_events <- test_set2 %>% filter(CUSTOMER_ID == i)
cs = cumsum(temp_events$trip_scheduled) # cumulative number of trips of individual
output_temp <- data.frame(temp_events,
trips_minus_60 = cs[findInterval(temp_events$trip_date - 60, temp_events$trip_date, left.open = TRUE)] - cs)
new_table <- rbind(new_table,output_temp)
}
這是我最近產生的錯誤:
data.frame(temp_events, trips_minus_60 = cs[findInterval(temp_events$trip_date - : 參數意味着不同的行數:10, 0
我不確定這是否滿足您的需求,但這是基於您鏈接到的tidyverse
的tidyverse
解決方案。 在group_by
您的CUSTOMER_ID
您可以將所有行與trip_scheduled
為 1,並且日期介於當前日期和 60 天之前。 我希望你也可以為same_day_cancel
做類似的same_day_cancel
。
library(tidyverse)
test_set2 %>%
group_by(CUSTOMER_ID) %>%
mutate(schedule_60 = unlist(map(trip_date, ~sum(trip_scheduled == 1 & between(trip_date, . - 60, .))))) %>%
print(n=30)
# A tibble: 30 x 6
# Groups: CUSTOMER_ID [3]
tripID CUSTOMER_ID trip_date trip_scheduled same_day_cancel schedule_60
<chr> <int> <date> <dbl> <dbl> <int>
1 20180112-100037-674-101 100037 2018-01-12 1 1 3
2 20180112-100037-674-201 100037 2018-01-12 1 1 3
3 20180112-100037-674-301 100037 2018-01-12 1 1 3
4 20180113-100037-676-101 100037 2018-01-13 1 0 5
5 20180113-100037-676-201 100037 2018-01-13 1 0 5
6 20180115-100037-675-101 100037 2018-01-15 1 1 7
7 20180115-100037-675-201 100037 2018-01-15 1 1 7
8 20180116-100037-677-101 100037 2018-01-16 1 0 9
9 20180116-100037-677-201 100037 2018-01-16 1 0 9
10 20180131-100037-678-101 100037 2018-01-31 1 0 10
11 20180101-100146-707-101 100146 2018-01-01 1 1 2
12 20180101-100146-707-201 100146 2018-01-01 1 1 2
13 20180102-100146-708-101 100146 2018-01-02 1 1 4
14 20180102-100146-708-201 100146 2018-01-02 1 1 4
15 20180103-100146-709-101 100146 2018-01-03 1 1 6
16 20180103-100146-709-201 100146 2018-01-03 1 1 6
17 20180104-100146-710-101 100146 2018-01-04 1 1 8
18 20180104-100146-710-201 100146 2018-01-04 1 1 8
19 20180105-100146-711-101 100146 2018-01-05 1 1 10
20 20180105-100146-711-201 100146 2018-01-05 1 1 10
21 20180403-100532-223-101 100532 2018-04-03 1 0 2
22 20180403-100532-223-201 100532 2018-04-03 1 0 2
23 20180620-100532-224-101 100532 2018-06-20 1 0 2
24 20180620-100532-224-201 100532 2018-06-20 1 0 2
25 20180704-100532-225-101 100532 2018-07-04 1 0 4
26 20180704-100532-225-201 100532 2018-07-04 1 0 4
27 20180926-100532-228-101 100532 2018-09-26 1 0 2
28 20180926-100532-228-201 100532 2018-09-26 1 0 2
29 20180927-100532-226-101 100532 2018-09-27 1 0 4
30 20180927-100532-226-201 100532 2018-09-27 1 0 4
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