使用带时间戳的游程编码
使用带时间戳的游程编码
提问于 2017-12-26 14:32:31
我的目标是使用rle()函数找出自行车停靠站已经空了多长时间。使用下面的test数据，rle(test$bikes)将返回test$bikes中递归值的长度。
> rle(test$bikes)
Run Length Encoding
lengths: int [1:9] 3 2 3 1 5 1 7 1 1
values : num [1:9] 0 1 2 1 0 1 0 1 0
> test
station_id time bikes
1 1 2017-12-25 00:00:02 0
2 1 2017-12-25 00:01:02 0
3 1 2017-12-25 00:02:02 0
4 1 2017-12-25 00:03:02 1
5 2 2017-12-25 00:04:02 1
6 2 2017-12-25 00:05:02 2
7 2 2017-12-25 00:06:02 2
8 2 2017-12-25 00:07:02 2
9 2 2017-12-25 00:08:02 1
10 3 2017-12-25 00:09:02 0
11 3 2017-12-25 00:10:02 0
12 3 2017-12-25 00:11:02 0
13 3 2017-12-25 00:12:02 0
14 3 2017-12-25 00:13:02 0
15 4 2017-12-25 00:14:03 1
16 4 2017-12-25 00:15:02 0
17 4 2017-12-25 00:16:02 0
18 4 2017-12-25 00:17:02 0
19 4 2017-12-25 00:18:02 0
20 5 2017-12-25 00:19:02 0
21 5 2017-12-25 00:20:02 0
22 5 2017-12-25 00:21:02 0
23 5 2017-12-25 00:22:02 1
24 5 2017-12-25 00:23:02 0我的目标是更进一步，生成一个按staiton_id分组的输出，并且仅当test$bikes具有递归零时才返回时间差(以分钟为单位)。对于每个站点，这可能会发生多次(例如，对于test数据中的站点5)。最后，上述数据集将产生以下输出：
> output
station_id diff_time interval
1 1 2 00:00 - 00:02
2 3 4 00:09 - 00:13
3 4 3 00:15 - 00:18
4 5 2 00:19 -00:21
5 5 0 00:23 - 00:23任何关于如何使用dplyr和rle来做这件事的建议都将不胜感激！
测试数据如下：
> dput(test)
structure(list(station_id = c(1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3,
3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5), time = structure(c(1514178002.88487,
1514178062.99145, 1514178122.88463, 1514178182.63461, 1514178242.71401,
1514178302.20358, 1514178362.13263, 1514178422.88907, 1514178482.6502,
1514178542.59171, 1514178602.51222, 1514178662.23203, 1514178722.04015,
1514178782.87382, 1514178843.02124, 1514178902.71852, 1514178962.6987,
1514179022.42077, 1514179082.19535, 1514179142.97175, 1514179202.81556,
1514179262.85187, 1514179322.66264, 1514179382.50223), class = c("POSIXct",
"POSIXt"), tzone = ""), bikes = c(0, 0, 0, 1, 1, 2, 2, 2, 1,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0)), .Names = c("station_id",
"time", "bikes"), row.names = c(NA, 24L), class = "data.frame")回答 2
Stack Overflow用户
回答已采纳
发布于 2017-12-26 14:43:21
我们可以使用data.table中的rleid。基于'bikes‘的运行长度id创建一个分组变量( 'grp’)，然后按'station_id‘和’grp‘分组，指定'bikes’为0的时间，通过获取‘i’的最后一个和第一个观察值的difftime来总结输出，并对相应的‘paste’元素进行format
library(data.table)
setDT(test)[, grp:= rleid(bikes)][bikes==0,
.(diff_time = as.numeric(round(difftime(time[.N], time[1], unit = "min"))),
interval = paste(format(time[1], "%M:%S"), format(time[.N], "%M:%S"), sep=" - ")),
.(station_id, grp)]Stack Overflow用户
发布于 2017-12-26 16:26:17
在dplyr中没有类似的函数'rleid‘
因此这里创建了另一个额外的函数
rle_dplyr <- function(x){
x = rle(x)$lengths
rep(seq_along(x),times = x)
}
> head(test1)
station_id time bikes
1 1 2017-12-25 10:30:02 0
2 1 2017-12-25 10:31:02 0
3 1 2017-12-25 10:32:02 0
4 1 2017-12-25 10:33:02 1
5 2 2017-12-25 10:34:02 1
6 2 2017-12-25 10:35:02 2
library(tidyverse)
test1%>%
mutate(idrle = rle_dplyr(bikes))%>%
filter(bikes == 0)%>%
group_by(station_id,idrle)%>%
summarise(diff_time = last(minute(time)) - first(minute(time)),
interval = paste(format(first(time),"%M:%S"),format(last(time),"%M:%S"),sep = "-"))%>%
select(-idrle) 输出
# A tibble: 5 x 3
# Groups: station_id [4]
station_id diff_time interval
<dbl> <int> <chr>
1 1 2 30:02-32:02
2 3 4 39:02-43:02
3 4 3 45:02-48:02
4 5 2 49:02-51:02
5 5 0 53:02-53:02页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/47974759
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