我的目标是使用data.table计算离开电台的自行车数量，然后按station_id，hour和date进行汇总。< / p>
如果之前的记录 - bikes_available的当前记录是正数，那么这就是自行车丢失的数量。如果前一记录 - 当前记录为负或零，则表示自行车数量保持不变或增加，因此不应计算这些情况。
> head(dat, n = 10)
station_id bikes_available time date hour
1: 3 2 2018-01-15 01:58:02 2018-01-15 1
2: 3 1 2018-01-15 01:59:01 2018-01-15 1
3: 3 1 2018-01-15 02:00:03 2018-01-15 2
4: 3 4 2018-01-15 02:01:02 2018-01-15 2
5: 3 4 2018-01-15 02:02:02 2018-01-15 2
6: 3 1 2018-01-15 02:03:02 2018-01-15 2
7: 3 1 2018-01-15 02:04:02 2018-01-15 2
8: 3 1 2018-01-15 02:05:02 2018-01-15 2
9: 3 7 2018-01-15 02:06:02 2018-01-15 2
10: 3 3 2018-01-15 02:07:02 2018-01-15 2
lead函数可用于查找先前记录和当前记录之间的差异，然后仅使用以下过滤掉的正值：
dat[,ba_lead:=shift(bikes_available, 1, type='lead')]
dat$diff <- dat$bikes_available - dat$ba_lead
但是如何使用station_id将3个变量 - time date和data.table分组？
例如，可以从提供的数据中获得以下输出
> output
station_id bikes_taken hour date
1 3 1 1 2018-01-15
2 3 7 2 2018-01-15
3 4 4 1 2018-01-15
4 4 1 2 2018-01-15
5 5 0 1 2018-01-15
6 5 2 2 2018-01-15
（下面的完整数据集）
> dput(dat)
structure(list(station_id = c(3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L,
3L, 3L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L,
5L, 5L, 5L, 5L, 5L, 5L), bikes_available = c(2, 1, 1, 4, 4, 1,
1, 1, 7, 3, 4, 0, 0, 0, 0, 0, 1, 1, 1, 0, 5, 5, 5, 5, 4, 4, 4,
4, 3, 3), time = structure(c(1516010282, 1516010341, 1516010403,
1516010462, 1516010522, 1516010582, 1516010642, 1516010702, 1516010762,
1516010822, 1516010282, 1516010341, 1516010403, 1516010462, 1516010522,
1516010582, 1516010642, 1516010702, 1516010762, 1516010822, 1516010282,
1516010341, 1516010403, 1516010462, 1516010522, 1516010582, 1516010642,
1516010702, 1516010762, 1516010822), class = c("POSIXct", "POSIXt"
), tzone = ""), date = structure(c(17546, 17546, 17546, 17546,
17546, 17546, 17546, 17546, 17546, 17546, 17546, 17546, 17546,
17546, 17546, 17546, 17546, 17546, 17546, 17546, 17546, 17546,
17546, 17546, 17546, 17546, 17546, 17546, 17546, 17546), class = "Date"),
hour = c(1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L)), .Names = c("station_id", "bikes_available",
"time", "date", "hour"), row.names = c(NA, -30L), class = c("data.table",
"data.frame"), .internal.selfref = <pointer: 0x102800778>)
答案 0 :(得分：1)
使用tidyverse函数，您可以尝试：
df %>%
group_by(station_id, date, hour) %>%
mutate( b_taken = bikes_available - lead(bikes_available)) %>%
filter(b_taken >= 0) %>%
mutate(b_taken = sum(b_taken)) %>%
select(b_taken) %>%
unique()
给出：
station_id date hour b_taken
<int> <date> <int> <dbl>
1 3 2018-01-15 1 1
2 3 2018-01-15 2 7
3 4 2018-01-15 1 4
4 4 2018-01-15 2 1
5 5 2018-01-15 1 0
6 5 2018-01-15 2 2
答案 1 :(得分：1)
library("data.table")
setDT(dat)
dat[,
j = .(bikes_taken = bikes_available - shift( x = bikes_available, n = 1, type = 'lead')),
by = .(station_id, date, hour)][ i = bikes_taken >= 0,
j = .(bikes_taken = sum(bikes_taken)),
by = .(station_id, date, hour)]
# station_id date hour bikes_taken
# 1: 3 2018-01-15 1 1
# 2: 3 2018-01-15 2 7
# 3: 4 2018-01-15 1 4
# 4: 4 2018-01-15 2 1
# 5: 5 2018-01-15 1 0
# 6: 5 2018-01-15 2 2
答案 2 :(得分：1)
data.table的另一种看法：
dat[, .(bikes_taken = diff(bikes_available)), by = .(station_id, date, hour)
][bikes_taken <= 0, .(bikes_taken = sum(bikes_taken*-1)), by = .(station_id, date, hour)]
给出：
station_id date hour bikes_taken 1: 3 2018-01-15 1 1 2: 3 2018-01-15 2 7 3: 4 2018-01-15 1 4 4: 4 2018-01-15 2 1 5: 5 2018-01-15 1 0 6: 5 2018-01-15 2 2