简体   繁体   English

如何每 15 分钟计算一个“批次”

[英]how to count a “batch” every 15 minutes

1 2021-01-01 12:59:38
2 2021-01-01 14:08:59
3 2021-01-01 14:09:08
4 2021-01-01 14:11:30
5 2021-01-01 14:22:19
6 2021-01-01 14:41:07

I want to be able to count the number of entries every 15 minutes but on a rolling basis.我希望能够每 15 分钟计算一次条目数,但要滚动计算。 Eg 12:59 would be 1 within 15 mins, 14:08 => 14:22 would all be within 15 minutes so this would return 4 in this batch and finally 14:41 would be by itself in another 15 minute batch.例如 12:59 将在 15 分钟内为 1, 14:08 => 14:22 将在 15 分钟内,因此这将在这批中返回 4,最后 14:41 将在另一个 15 分钟内单独出现。

I hope this makes sense and thanks in advance.我希望这是有道理的,并提前感谢。

Apologies for not including this抱歉没有包括这个

> dput(df)
structure(list(ClickedDate = structure(c(1609460198.707, 1609462979.593, 
1609465088.437, 1609476270.88, 1609478479.177, 1609479667.373, 
1609493081.887, 1609499187.29, 1609507506.37, 1609510989.533, 
1609511522.023, 1609511894.067, 1609512194.773, 1609512377.227, 
1609514474.153), tzone = "UTC", class = c("POSIXct", "POSIXt"
)), batch_no = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 
12L, 12L, 12L, 13L), batch_size = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 3L, 3L, 3L, 1L)), row.names = c(NA, -15L), class = c("tbl_df", 
"tbl", "data.frame"))

NEW EDIT - Thank you for working on this.新编辑 - 感谢您为此工作。 I am getting an error我收到一个错误

Error in UseMethod("mutate") : 
  no applicable method for 'mutate' applied to an object of class "c('integer', 'numeric')"

This seems odd, my variable is in class这看起来很奇怪,我的变量在 class

> class(df$ClickedDate)
[1] "POSIXct" "POSIXt" 

Does this work with mutate, or do I need to convert this?这是否适用于变异,还是我需要转换它?

> dput(df)
structure(list(ClickedDate = structure(c(1609460198.707, 1609462979.593, 
1609465088.437, 1609476270.88, 1609478479.177, 1609479667.373, 
1609493081.887, 1609499187.29, 1609507506.37, 1609510989.533, 
1609511522.023, 1609511894.067, 1609512194.773, 1609512377.227, 
1609514474.153), tzone = "UTC", class = c("POSIXct", "POSIXt"
)), batch_no = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 
12L, 12L, 12L, 13L), batch_size = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 3L, 3L, 3L, 1L)), row.names = c(NA, -15L), class = c("tbl_df", 
"tbl", "data.frame"))

Thanks in advance提前致谢

Usage of runner package will help in this scenario.在这种情况下,使用runner package 会有所帮助。 Use the following strategy使用以下策略

library(tidyverse)
library(runner)

df %>% mutate(b_len = runner::runner(x = ClickedDate,
                             idx = ClickedDate,
                             k = "15 mins",
                             lag = "-14 mins",
                             f = length),
              b_no = purrr::accumulate(seq_len(length(b_len)-1), .init = b_len[1], ~ifelse(.x > .y, .x, .x + b_len[.x +1])),
              b_no = as.integer(as.factor(b_no))) %>%
  group_by(b_no) %>%
  mutate(b_len = n())

# A tibble: 15 x 3
# Groups:   b_no [12]
   ClickedDate         b_len  b_no
   <dttm>              <int> <int>
 1 2021-01-01 00:16:38     1     1
 2 2021-01-01 01:02:59     1     2
 3 2021-01-01 01:38:08     1     3
 4 2021-01-01 04:44:30     1     4
 5 2021-01-01 05:21:19     1     5
 6 2021-01-01 05:41:07     1     6
 7 2021-01-01 09:24:41     1     7
 8 2021-01-01 11:06:27     1     8
 9 2021-01-01 13:25:06     1     9
10 2021-01-01 14:23:09     2    10
11 2021-01-01 14:32:02     2    10
12 2021-01-01 14:38:14     3    11
13 2021-01-01 14:43:14     3    11
14 2021-01-01 14:46:17     3    11
15 2021-01-01 15:21:14     1    12

Notes -备注 -

  • lag argument in runner function allows a backward time window (rolling) so I am using negative lag to use forward time window. runner function 中的lag参数允许向后时间 window(滚动)所以我使用负滞后来使用向前时间 window。
  • k argument in runner is for given length of rolling window runner中的k参数是给定的滚动长度 window
  • b_no column initially identifies the sliding/rolling window upto the earliest window is exhausted and thereafter takes new window. b_no列最初标识滑动/滚动 window 直到最早的 window 用完,然后取新的 window。
  • dense_rank can also be used (see alternative below)也可以使用dense_rank (参见下面的替代方案)

Alternatively或者

df %>% mutate(b_len = runner::runner(x = ClickedDate,
                             idx = ClickedDate,
                             k = "15 mins",
                             lag = "-14 mins",
                             f = length),
              b_no = purrr::accumulate(seq_len(length(b_len)-1), .init = b_len[1], ~ifelse(.x > .y, .x, .x + b_len[.x +1])),
              b_no = dense_rank(b_no)) %>%
  group_by(b_no) %>%
  mutate(b_len = n()) %>%
  ungroup()
# A tibble: 15 x 3
   ClickedDate         b_len  b_no
   <dttm>              <int> <int>
 1 2021-01-01 00:16:38     1     1
 2 2021-01-01 01:02:59     1     2
 3 2021-01-01 01:38:08     1     3
 4 2021-01-01 04:44:30     1     4
 5 2021-01-01 05:21:19     1     5
 6 2021-01-01 05:41:07     1     6
 7 2021-01-01 09:24:41     1     7
 8 2021-01-01 11:06:27     1     8
 9 2021-01-01 13:25:06     1     9
10 2021-01-01 14:23:09     2    10
11 2021-01-01 14:32:02     2    10
12 2021-01-01 14:38:14     3    11
13 2021-01-01 14:43:14     3    11
14 2021-01-01 14:46:17     3    11
15 2021-01-01 15:21:14     1    12

data used使用的数据

df
> df
# A tibble: 15 x 1
   ClickedDate        
   <dttm>             
 1 2021-01-01 00:16:38
 2 2021-01-01 01:02:59
 3 2021-01-01 01:38:08
 4 2021-01-01 04:44:30
 5 2021-01-01 05:21:19
 6 2021-01-01 05:41:07
 7 2021-01-01 09:24:41
 8 2021-01-01 11:06:27
 9 2021-01-01 13:25:06
10 2021-01-01 14:23:09
11 2021-01-01 14:32:02
12 2021-01-01 14:38:14
13 2021-01-01 14:43:14
14 2021-01-01 14:46:17
15 2021-01-01 15:21:14

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM