R:日期之间的天数顺序
R: sequence of days between dates
我有以下数据框:
AllDays
2012-01-01
2012-01-02
2012-01-03
...
2015-08-18
Leases
StartDate EndDate
2012-01-01 2013-01-01
2012-05-07 2013-05-06
2013-09-05 2013-12-01
我想要做的是,对于 allDays 数据框中的每个日期,计算有效的租约数量。例如如果有 4 个开始日期 <= 2015-01-01 和结束日期 >= 2015-01-01 的租约,那么我想在该数据框中放置一个 4。
我有以下代码
for (i in 1:nrow(leases))
{
occupied = seq(leases$StartDate[i],leases$EndDate[i],by="days")
occupied = occupied[occupied < dateOfInt]
matching = match(occupied,allDays$Date)
allDays$Occupancy[matching] = allDays$Occupancy[matching] + 1
}
有效,但由于我有大约 5000 个租约,因此大约需要 1.1 秒。有没有人有更有效的方法需要更少的计算时间?
利息日期只是当前日期,仅用于确保它不计算未来的租赁日期。
使用 seq
几乎肯定是低效的——假设您的数据租期长达 10000 年。 seq
将永远花费 return 10000*365-1 天,这对我们来说无关紧要。然后我们必须使用 %in%
这也会进行相同数量的不必要的比较。
我不确定以下是最好的方法(我相信有一个完全矢量化的解决方案)但它更接近问题的核心。
数据
set.seed(102349)
days<-data.frame(AllDays=seq(as.Date("2012-01-01"),
as.Date("2015-08-18"),"day"))
leases<-data.frame(StartDate=sample(days$AllDays,5000L,T))
leases$EndDate<-leases$StartDate+round(rnorm(5000,mean=365,sd=100))
方法
使用data.table
和sapply
:
library(data.table)
setDT(leases); setDT(days)
days[,lease_count:=
sapply(AllDays,function(x)
leases[StartDate<=x&EndDate>=x,.N])][]
AllDays lease_count
1: 2012-01-01 5
2: 2012-01-02 8
3: 2012-01-03 11
4: 2012-01-04 16
5: 2012-01-05 18
---
1322: 2015-08-14 1358
1323: 2015-08-15 1358
1324: 2015-08-16 1360
1325: 2015-08-17 1363
1326: 2015-08-18 1359
另一种方法,但我不确定它是否更快。
library(lubridate)
library(dplyr)
AllDays = data.frame(dates = c("2012-02-01","2012-03-02","2012-04-03"))
Lease = data.frame(start = c("2012-01-03","2012-03-01","2012-04-02"),
end = c("2012-02-05","2012-04-15","2012-07-11"))
# transform to dates
AllDays$dates = ymd(AllDays$dates)
Lease$start = ymd(Lease$start)
Lease$end = ymd(Lease$end)
# create the range id
Lease$id = 1:nrow(Lease)
AllDays
# dates
# 1 2012-02-01
# 2 2012-03-02
# 3 2012-04-03
Lease
# start end id
# 1 2012-01-03 2012-02-05 1
# 2 2012-03-01 2012-04-15 2
# 3 2012-04-02 2012-07-11 3
data.frame(expand.grid(AllDays$dates,Lease$id)) %>% # create combinations of dates and ranges
select(dates=Var1, id=Var2) %>%
inner_join(Lease, by="id") %>% # join information
rowwise %>%
do(data.frame(dates=.$dates,
flag = ifelse(.$dates %in% seq(.$start,.$end,by="1 day"),1,0))) %>% # create ranges and check if the date is in there
ungroup %>%
group_by(dates) %>%
summarise(N=sum(flag))
# dates N
# 1 2012-02-01 1
# 2 2012-03-02 1
# 3 2012-04-03 2
试试 lubridate 包。为每个租约创建一个间隔。然后统计每个日期所在的租约间隔。
# make some data
AllDays <- data.frame("Days" = seq.Date(as.Date("2012-01-01"), as.Date("2012-02-01"), by = 1))
Leases <- data.frame("StartDate" = as.Date(c("2012-01-01", "2012-01-08")),
"EndDate" = as.Date(c("2012-01-10", "2012-01-21")))
library(lubridate)
x <- new_interval(Leases$StartDate, Leases$EndDate, tzone = "UTC")
AllDays$NumberInEffect <- sapply(AllDays$Days, function(a){sum(a %within% x)})
输出
head(AllDays)
Days NumberInEffect
1 2012-01-01 1
2 2012-01-02 1
3 2012-01-03 1
4 2012-01-04 1
5 2012-01-05 1
6 2012-01-06 1
没有你的数据,我无法测试这是否更快,但它可以用更少的代码完成工作:
for (i in 1:nrow(AllDays)) AllDays$tally[i] = sum(AllDays$AllDays[i] >= Leases$Start.Date & AllDays$AllDays[i] <= Leases$End.Date)
我用下面的来测试它;请注意,两个数据框中的相关列都被格式化为日期:
AllDays = data.frame(AllDays = seq(from=as.Date("2012-01-01"), to=as.Date("2015-08-18"), by=1))
Leases = data.frame(Start.Date = as.Date(c("2013-01-01", "2012-08-20", "2014-06-01")), End.Date = as.Date(c("2013-12-31", "2014-12-31", "2015-05-31")))
这正是 foverlaps 的亮点所在:根据另一个 data.frame 对 data.frame 进行子集化(foverlaps
似乎是为此目的量身定制的)。
基于@MichaelChirico 的数据。
setkey(days[, AllDays1:=AllDays,], AllDays, AllDays1)
setkey(leases, StartDate, EndDate)
foverlaps(leases, days)[, .(lease_count=.N), AllDays]
# user system elapsed
# 0.114 0.018 0.136
# @MichaelChirico's approach
# user system elapsed
# 0.909 0.000 0.907
Here 是@Arun 对它如何工作的简要解释,这让我开始使用 data.table
。
我有以下数据框:
AllDays
2012-01-01
2012-01-02
2012-01-03
...
2015-08-18
Leases
StartDate EndDate
2012-01-01 2013-01-01
2012-05-07 2013-05-06
2013-09-05 2013-12-01
我想要做的是,对于 allDays 数据框中的每个日期,计算有效的租约数量。例如如果有 4 个开始日期 <= 2015-01-01 和结束日期 >= 2015-01-01 的租约,那么我想在该数据框中放置一个 4。
我有以下代码
for (i in 1:nrow(leases))
{
occupied = seq(leases$StartDate[i],leases$EndDate[i],by="days")
occupied = occupied[occupied < dateOfInt]
matching = match(occupied,allDays$Date)
allDays$Occupancy[matching] = allDays$Occupancy[matching] + 1
}
有效,但由于我有大约 5000 个租约,因此大约需要 1.1 秒。有没有人有更有效的方法需要更少的计算时间? 利息日期只是当前日期,仅用于确保它不计算未来的租赁日期。
使用 seq
几乎肯定是低效的——假设您的数据租期长达 10000 年。 seq
将永远花费 return 10000*365-1 天,这对我们来说无关紧要。然后我们必须使用 %in%
这也会进行相同数量的不必要的比较。
我不确定以下是最好的方法(我相信有一个完全矢量化的解决方案)但它更接近问题的核心。
数据
set.seed(102349)
days<-data.frame(AllDays=seq(as.Date("2012-01-01"),
as.Date("2015-08-18"),"day"))
leases<-data.frame(StartDate=sample(days$AllDays,5000L,T))
leases$EndDate<-leases$StartDate+round(rnorm(5000,mean=365,sd=100))
方法
使用data.table
和sapply
:
library(data.table)
setDT(leases); setDT(days)
days[,lease_count:=
sapply(AllDays,function(x)
leases[StartDate<=x&EndDate>=x,.N])][]
AllDays lease_count
1: 2012-01-01 5
2: 2012-01-02 8
3: 2012-01-03 11
4: 2012-01-04 16
5: 2012-01-05 18
---
1322: 2015-08-14 1358
1323: 2015-08-15 1358
1324: 2015-08-16 1360
1325: 2015-08-17 1363
1326: 2015-08-18 1359
另一种方法,但我不确定它是否更快。
library(lubridate)
library(dplyr)
AllDays = data.frame(dates = c("2012-02-01","2012-03-02","2012-04-03"))
Lease = data.frame(start = c("2012-01-03","2012-03-01","2012-04-02"),
end = c("2012-02-05","2012-04-15","2012-07-11"))
# transform to dates
AllDays$dates = ymd(AllDays$dates)
Lease$start = ymd(Lease$start)
Lease$end = ymd(Lease$end)
# create the range id
Lease$id = 1:nrow(Lease)
AllDays
# dates
# 1 2012-02-01
# 2 2012-03-02
# 3 2012-04-03
Lease
# start end id
# 1 2012-01-03 2012-02-05 1
# 2 2012-03-01 2012-04-15 2
# 3 2012-04-02 2012-07-11 3
data.frame(expand.grid(AllDays$dates,Lease$id)) %>% # create combinations of dates and ranges
select(dates=Var1, id=Var2) %>%
inner_join(Lease, by="id") %>% # join information
rowwise %>%
do(data.frame(dates=.$dates,
flag = ifelse(.$dates %in% seq(.$start,.$end,by="1 day"),1,0))) %>% # create ranges and check if the date is in there
ungroup %>%
group_by(dates) %>%
summarise(N=sum(flag))
# dates N
# 1 2012-02-01 1
# 2 2012-03-02 1
# 3 2012-04-03 2
试试 lubridate 包。为每个租约创建一个间隔。然后统计每个日期所在的租约间隔。
# make some data
AllDays <- data.frame("Days" = seq.Date(as.Date("2012-01-01"), as.Date("2012-02-01"), by = 1))
Leases <- data.frame("StartDate" = as.Date(c("2012-01-01", "2012-01-08")),
"EndDate" = as.Date(c("2012-01-10", "2012-01-21")))
library(lubridate)
x <- new_interval(Leases$StartDate, Leases$EndDate, tzone = "UTC")
AllDays$NumberInEffect <- sapply(AllDays$Days, function(a){sum(a %within% x)})
输出
head(AllDays)
Days NumberInEffect
1 2012-01-01 1
2 2012-01-02 1
3 2012-01-03 1
4 2012-01-04 1
5 2012-01-05 1
6 2012-01-06 1
没有你的数据,我无法测试这是否更快,但它可以用更少的代码完成工作:
for (i in 1:nrow(AllDays)) AllDays$tally[i] = sum(AllDays$AllDays[i] >= Leases$Start.Date & AllDays$AllDays[i] <= Leases$End.Date)
我用下面的来测试它;请注意,两个数据框中的相关列都被格式化为日期:
AllDays = data.frame(AllDays = seq(from=as.Date("2012-01-01"), to=as.Date("2015-08-18"), by=1))
Leases = data.frame(Start.Date = as.Date(c("2013-01-01", "2012-08-20", "2014-06-01")), End.Date = as.Date(c("2013-12-31", "2014-12-31", "2015-05-31")))
这正是 foverlaps 的亮点所在:根据另一个 data.frame 对 data.frame 进行子集化(foverlaps
似乎是为此目的量身定制的)。
基于@MichaelChirico 的数据。
setkey(days[, AllDays1:=AllDays,], AllDays, AllDays1)
setkey(leases, StartDate, EndDate)
foverlaps(leases, days)[, .(lease_count=.N), AllDays]
# user system elapsed
# 0.114 0.018 0.136
# @MichaelChirico's approach
# user system elapsed
# 0.909 0.000 0.907
Here 是@Arun 对它如何工作的简要解释,这让我开始使用 data.table
。