如何計算R中開始日期結束日期間隔的記錄？

Question

我有一個包含項目的數據框，並且每個項目都有一個開始日期和結束日期。 我想知道在一段時間內每天有多少項是活躍的。

示例數據集：

ItemId <- c(1,2,3)
StartDate <- c(ymd("2014-01-01"),ymd("2014-02-01"),ymd("2014-03-01"))
EndDate <- c(ymd("2014-02-15"),ymd("2014-02-07"),ymd("2014-03-03"))
data.frame(ItemId,StartDate,EndDate)
  ItemId           StartDate             EndDate
1      1 2014-01-01 01:00:00 2014-02-15 01:00:00
2      2 2014-02-01 01:00:00 2014-02-07 01:00:00
3      3 2014-03-01 01:00:00 2014-03-03 01:00:00

結果看起來應該是這樣的（每天一個條目）：

Date        ActiveCount
2014-01-01  1
2014-01-02  1
...
2014-02-01  2
...

我有一個使用sqldf的解決方案，但不知道如何在R中執行此操作

select d.date
,      ( select count(ItemID)
         from   items
         where  startdate <= d.date
         and    enddate >= d.date
       ) activecount
from   (select distinct startdate from items
        union
        select distinct enddate from items
       ) d
order by 1

（我的每天包含多個條目，因此對於R中的sqlite，這是有效的。在postgresql上，我可以生成一系列更好的日期。）

謝謝。

Answer 1

調用你的數據df ：

dates = seq(min(df$StartDate), max(df$EndDate), by = "day")

counts = data.frame(date = dates,
                    count = sapply(dates, function(x) sum(x <= df$EndDate & x >= df$StartDate)))

Answer 2

每當R任務類似於SQL任務時，可能需要將dplyr帶出櫥櫃：

library(dplyr) 
ItemId <- c(1,2,3)
StartDate <- c(ymd("2014-01-01"),ymd("2014-02-01"),ymd("2014-03-01"))
EndDate <- c(ymd("2014-02-15"),ymd("2014-02-07"),ymd("2014-03-03"))

jim <- data.frame(ItemId,StartDate,EndDate)

# One technique that's often useful especially in R, is to take your 
# iterator, and define it as a variable.  You can use that to implement
# a vectorised version of whatever you were thinking of doing.*/

ed <- data.frame(rng = seq(min(jim$StartDate), max(jim$EndDate), by = 'day'))
merge(jim, ed, all=TRUE) %>% 
     filter(rng >= StartDate, rng <= EndDate) %>%
     group_by(rng) %>% 
     summarise(n())

這會給你：

    rng         n()
1   2014-01-01  1 
2   2014-01-02  1
3   2014-01-03  1
...

Answer 3

我已經多次回到這個問題，我一直在尋找最有效的方法。

我之前使用過map-reduce方法，但發現它不能很好地擴展到具有寬日期間隔的大型數據幀。 我只是嘗試使用lubridate包中的interval類，並發現它是目前為止最快的實現。

這是最終的代碼：

library(tidyverse)
library(lubridate)

# Initialize a dataframe with start and end "active" dates per object
N = 1000
id_dates = tibble(id = 1 : N) %>%
  mutate(
    start = sample(seq(as.Date('2018-1-1'), as.Date('2019-1-1'), by = "day"), size = N, replace = TRUE),
    end   = start + sample(7 : 100, size = N, replace = TRUE),
    interval = interval(start, end))

# Use the %within% command to calculate the number of active items per date
queue_history = tibble(Date = seq(min(id_dates$start), max(id_dates$end), by = "1 day")) %>% 
  rowwise() %>% 
  mutate(numInWIP = sum(Date %within% id_dates$interval)) %>%
  ungroup()

以下是一些基准測試，表明潤滑劑解決方案比目前的答案和地圖減少方法快得多

library(tidyverse)
library(lubridate)

# Initialize a dataframe with start and end "active" dates per object
N = 1000
id_dates = tibble(id = 1 : N) %>%
  mutate(
    start = sample(seq(as.Date('2018-1-1'), as.Date('2019-1-1'), by = "day"), size = N, replace = TRUE),
    end   = start + sample(7 : 100, size = N, replace = TRUE),
    interval = interval(start, end))

# a map-reduce solution
method_mapreduce = function() {
  queue_history = as.tibble(table(reduce(map2(id_dates$start, id_dates$end, seq, by = 1), c)))
  queue_history = queue_history %>%
    rename(Date = Var1, numInWIP = Freq) %>%
    mutate(Date = as_date(Date))

  return (queue_history)
}

# a lubridate interval solution
method_intervals = function() {
  date_df = tibble(Date = seq(min(id_dates$start), max(id_dates$end), by = "1 day"))
  queue_history = date_df %>% 
    rowwise() %>% 
    mutate(numInWIP = sum(Date %within% id_dates$interval))

  return (queue_history)
}

# current best answer
method_currentsolution = function() {
  date_df = tibble(Date = seq(min(id_dates$start), max(id_dates$end), by = "1 day"))
  queue_history = merge(id_dates, date_df, all=TRUE) %>% 
    filter(Date >= start, Date <= end) %>%
    group_by(Date) %>% 
    summarise(n())

}

# Compare with benchmarks
tst = microbenchmark::microbenchmark(
  method_mapreduce(),
  method_intervals(),
  method_currentsolution(),
  times = 5)

microbenchmark::autoplot.microbenchmark(tst) +
  scale_y_log10(
    name   = sprintf("Time [%s]", attr(summary(tst), "unit")),
    breaks = scales::trans_breaks("log10", function(x) round(10^x)))

Answer 4

您首先要獲得至少有一個活動項目的所有日期，然后您想要計算每天活動項目的數量。 如果我們將數據存儲在itemDates那么這應該照顧它：

dates <- min(itemDates$StartDate) + days(0:as.numeric(max(itemDates$EndDate) - min(itemDates$StartDate)))
dateCounts <- data.frame(
    row.names=dates,
    counts=sapply(dates, function(date)
        sum(date >= itemDates$StartDate & date <= itemDates$EndDate)))