如何在 R 中按时间间隔分组

Question

我下面有一组数据

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安排我的时间

然后将时间分组为每小时间隔

然后汇总得到下面的列

我想知道一小时内处理了多少笔交易

然后一小时内的值

然后有一列说明有多少人成功，然后另一列说明一小时内有多少人失败

在下方查看所需汇总 table 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)

我不太确定您原来的 Date 列的格式。这里我假设它是一个字符串。因为 Interval 列的确切格式对您很重要，所以使用您想要显示的字符串制作和连接一个单独的小标题似乎更容易。

重要的是用零替换缺失的 NA 值，否则 Count = Successful + Failed 将在只有其中一个存在时秘密失败。

如何在 R 中按时间间隔分组

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