使用带有时间戳的 运行 长度编码
Using run-length encoding with timestamps
我的 objective 是使用 rle() 函数来找出自行车站空了多长时间。使用下面的 test 数据，rle(test$bikes) 将 return test$bikes.
中重复值的长度
> rle(test$bikes)
Run Length Encoding
lengths: int [1:9] 3 2 3 1 5 1 7 1 1
values : num [1:9] 0 1 2 1 0 1 0 1 0
> test
station_id time bikes
1 1 2017-12-25 00:00:02 0
2 1 2017-12-25 00:01:02 0
3 1 2017-12-25 00:02:02 0
4 1 2017-12-25 00:03:02 1
5 2 2017-12-25 00:04:02 1
6 2 2017-12-25 00:05:02 2
7 2 2017-12-25 00:06:02 2
8 2 2017-12-25 00:07:02 2
9 2 2017-12-25 00:08:02 1
10 3 2017-12-25 00:09:02 0
11 3 2017-12-25 00:10:02 0
12 3 2017-12-25 00:11:02 0
13 3 2017-12-25 00:12:02 0
14 3 2017-12-25 00:13:02 0
15 4 2017-12-25 00:14:03 1
16 4 2017-12-25 00:15:02 0
17 4 2017-12-25 00:16:02 0
18 4 2017-12-25 00:17:02 0
19 4 2017-12-25 00:18:02 0
20 5 2017-12-25 00:19:02 0
21 5 2017-12-25 00:20:02 0
22 5 2017-12-25 00:21:02 0
23 5 2017-12-25 00:22:02 1
24 5 2017-12-25 00:23:02 0
我的目标是更进一步，并产生一个输出，该输出按 staiton_id 分组，并且 returns 仅当 test$bikes 具有时的时间差（以分钟为单位）重复的零。对于每个站点，这可能会发生多次（例如，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
任何有关如何使用 dplyr 和 rle 执行此操作的建议，我们将不胜感激！
这里是测试数据：
> 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")
我们可以使用 data.table 中的 rleid。根据'bikes'的运行-length-id创建分组变量（'grp'），然后按'station_id'和'grp'分组，指定i 其中 'bikes' 为 0，通过获取 'time' 的最后一次和第一次观察的 difftime 以及 paste format 对应的 paste 总结输出=24=] 个元素
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)]
dplyr中没有类似的函数'rleid'
所以这里创建了另一个附加功能
rle_dplyr <- function(x){
x = rle(x)$lengths
rep(seq_along(x),times = x)
}
> head(test1)
station_id time bikes
1 1 2017-12-25 10:30:02 0
2 1 2017-12-25 10:31:02 0
3 1 2017-12-25 10:32:02 0
4 1 2017-12-25 10:33:02 1
5 2 2017-12-25 10:34:02 1
6 2 2017-12-25 10:35:02 2
library(tidyverse)
test1%>%
mutate(idrle = rle_dplyr(bikes))%>%
filter(bikes == 0)%>%
group_by(station_id,idrle)%>%
summarise(diff_time = last(minute(time)) - first(minute(time)),
interval = paste(format(first(time),"%M:%S"),format(last(time),"%M:%S"),sep = "-"))%>%
select(-idrle)
输出
# A tibble: 5 x 3
# Groups: station_id [4]
station_id diff_time interval
<dbl> <int> <chr>
1 1 2 30:02-32:02
2 3 4 39:02-43:02
3 4 3 45:02-48:02
4 5 2 49:02-51:02
5 5 0 53:02-53:02
我的 objective 是使用 rle() 函数来找出自行车站空了多长时间。使用下面的 test 数据，rle(test$bikes) 将 return test$bikes.
> rle(test$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，并产生一个输出，该输出按 staiton_id 分组，并且 returns 仅当 test$bikes 具有时的时间差（以分钟为单位）重复的零。对于每个站点，这可能会发生多次（例如，test 数据中的站点 5）。最后，上述数据集将产生以下输出：
> output
station_id diff_time interval
1 1 2 00:00 - 00:02
2 3 4 00:09 - 00:13
3 4 3 00:15 - 00:18
4 5 2 00:19 -00:21
5 5 0 00:23 - 00:23
任何有关如何使用 dplyr 和 rle 执行此操作的建议，我们将不胜感激！
这里是测试数据：
> dput(test)
structure(list(station_id = c(1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3,
3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5), time = structure(c(1514178002.88487,
1514178062.99145, 1514178122.88463, 1514178182.63461, 1514178242.71401,
1514178302.20358, 1514178362.13263, 1514178422.88907, 1514178482.6502,
1514178542.59171, 1514178602.51222, 1514178662.23203, 1514178722.04015,
1514178782.87382, 1514178843.02124, 1514178902.71852, 1514178962.6987,
1514179022.42077, 1514179082.19535, 1514179142.97175, 1514179202.81556,
1514179262.85187, 1514179322.66264, 1514179382.50223), class = c("POSIXct",
"POSIXt"), tzone = ""), bikes = c(0, 0, 0, 1, 1, 2, 2, 2, 1,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0)), .Names = c("station_id",
"time", "bikes"), row.names = c(NA, 24L), class = "data.frame")
我们可以使用 data.table 中的 rleid。根据'bikes'的运行-length-id创建分组变量（'grp'），然后按'station_id'和'grp'分组，指定i 其中 'bikes' 为 0，通过获取 'time' 的最后一次和第一次观察的 difftime 以及 paste format 对应的 paste 总结输出=24=] 个元素
library(data.table)
setDT(test)[, grp:= rleid(bikes)][bikes==0,
.(diff_time = as.numeric(round(difftime(time[.N], time[1], unit = "min"))),
interval = paste(format(time[1], "%M:%S"), format(time[.N], "%M:%S"), sep=" - ")),
.(station_id, grp)]
dplyr中没有类似的函数'rleid'
所以这里创建了另一个附加功能
rle_dplyr <- function(x){
x = rle(x)$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