如何在R中按时间间隔分组

Question

我在下面有一个数据集

Date	Status	Value
05/12/2021 23:59	Failed	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Failed	1500
05/12/2021 23:59	Failed	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Failed	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Failed	500
05/12/2021 23:59	Failed	1500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Failed	1500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Failed	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Failed	500
05/12/2021 23:59	Failed	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	500
05/12/2021 23:59	Successful	1500
05/12/2021 23:59	Successful	500

我希望能够拆分日期时间列以获得我的时间

然后将每小时间隔的时间分组

然后总结得到下面的列

我想知道一个小时内处理了多少笔交易

那么一小时内的值

然后有一列说明有多少成功，然后另一列说明一小时内有多少失败

请参阅下面所需汇总表的输出

Interval	Value	Count	Successful	Failed
00:00 am - 00:59 am	32,000	54	40	15
00:59 am - 01:00 am	42,000	55	41	14
01:00 am - 02:59 am	21,400	56	42	14
03:00 am - 03:59 am	4,00	57	43	14
04:00 am - 04:59 am	543,000	58	2	56
05:00 am - 05:59 am	411,000	59	6	53

Answer 1

这个怎么样：

library(tidyverse)
library(lubridate)
library(glue)

df <- tribble(~Date,    ~Status,    ~Value,
              "05/12/2021 23:59",   "Failed",   500,
              "05/12/2021 23:59",   "Successful",   1500,
              "05/12/2021 23:59",   "Successful",   500,
              "05/12/2021 23:59",   "Successful",   1500,
              "05/12/2021 23:59",   "Successful",   1500,
              "05/12/2021 23:59",   "Failed",   1500)

df2 <- df %>% 
    mutate(Datetime = dmy_hms(Date), # convert to datetime format
           Date = as.Date(Datetime), # extract date, if you need it later
           Hour = hour(Datetime))    # extract hour

hourly_value <- df2 %>% 
    group_by(Hour) %>% 
    summarize(Value = sum(Value),
              .groups = "drop")

hourly_count <- df2 %>% 
    count(Hour, Status) %>% 
    pivot_wider(names_from = "Status", values_from = "n")

interval_helper <- tibble(Hour = 0:23,
                          display_hour = str_pad(Hour %% 12, 2, pad = '0'),
                          ampm = if_else(Hour < 12, "am", "pm"),
                          Interval = glue("{display_hour}:00 {ampm} - {display_hour}:59 {ampm}"))

full_join(hourly_value, hourly_count, by = "Hour") %>% 
    replace_na(list(Successful = 0L, Failed = 0L, Value = 0)) %>% 
    left_join(interval_helper, by = "Hour") %>% 
    mutate(Count = Successful + Failed) %>% 
    select(Interval, Value, Count, Successful, Failed)

我不确定您原来的 Date 列的格式。在这里，我假设它是一个字符串。因为Interval 列的确切格式对您来说很重要，所以使用您想要显示的字符串创建和连接一个单独的小标题似乎更容易。

用零替换缺少的NA 值很重要，否则Count = Successful + Failed 在只有其中一个存在时会秘密失败。