如何按时间间隔和列值对R数据帧进行分组？

Question

我有一个数据集,它是特定设施发生事件的时间列表:

> head(facility_events);
facility_id          event_time
1   20248 2018-01-01 00:00:01
2   12445 2018-01-01 00:00:04
3   20248 2018-01-01 00:00:08
4   17567 2018-01-01 00:00:47
5   17567 2018-01-01 00:03:50
6   10459 2018-01-01 00:04:01

我想通过按设施对数据进行分组并将事件分组为3分钟的间隔来生成具有总和的数据帧.输出看起来像这样:

count facility interval
2      20248   0 
1      12445   0
1      17567   0
1      17567   1
1      10459   1

你如何在R中实现这一目标？

Answer 1

您可以使用tidyverse与lubridate此:

df <- data.frame(facility_id = c(20248, 12445, 20248, 17567, 17567, 10459),
                 event_time = as.POSIXct(c("2018-01-01 00:00:01", "2018-01-01 00:00:04", "2018-01-01 00:00:08", "2018-01-01 00:00:47", "2018-01-01 00:03:50", "2018-01-01 00:04:01")))

library(tidyverse)

df %>%
    mutate(interval = lubridate::minute(event_time) %/% 3) %>%
    group_by(facility_id, interval) %>%
    summarise(count = n())

# A tibble: 5 x 3
# Groups: facility_id [?]
  facility_id interval count
        <dbl>    <int> <int>
1       10459        1     1
2       12445        0     1
3       17567        0     1
4       17567        1     1
5       20248        0     2

Answer 2

这是一个解决方案data.table。同样的逻辑：

• 创建区间组
• 按此组汇总（计数）

这是一个具有data.table简洁语法的单行。

df <- data.frame(facility_id = c(20248, 12445, 20248, 17567, 17567, 10459),
                 event_time = as.POSIXct(c("2018-01-01 00:00:01", "2018-01-01 00:00:04", "2018-01-01 00:00:08", "2018-01-01 00:00:47", "2018-01-01 00:03:50", "2018-01-01 00:04:01")))

library(data.table)

setDT(df)
df[, .(count = .N), by = .(facility_id, interval= minute(event_time) %/% 3)]
#>    facility_id interval count
#> 1:       20248        0     2
#> 2:       12445        0     1
#> 3:       17567        0     1
#> 4:       17567        1     1
#> 5:       10459        1     1

由reprex 包(v0.1.1.9000)于2018年 1 月 14 日创建。