R/lubridate:计算两个期间之间的重叠天数 [英] R / lubridate: Calculate number of overlapping days between two periods

问题描述

我正在尝试计算两个时间段之间的重叠天数.一个期间固定在开始日期和结束日期，另一个期间记录在数据框中作为开始日期和结束日期.

I am trying to calculate the number of overlapping days between two time periods. One period is fixed in a start and end date, the other is recorded as start and end dates in a data frame.

我正在处理具有发布日期(df $ start)和未发布日期(df $ end)的广告.我要找出的是他们在特定月份的上网天数(my.start = 2018-01-01，my.end = 2018-08-31).

I'm dealing with ads that have publish date (df$start) and an unpublish date (df$end). What I'm trying to find out is how many days they have been online in a specific month (my.start = 2018-01-01, my.end = 2018-08-31).

library(dplyr)
library(lubridate)

my.start <- ymd("2018-08-01")
my.end <- ymd("2018-08-31")

df <- data.frame(start = c("2018-07-15", "2018-07-20", "2018-08-15", "2018-08-20", "2018-09-01"), 
                 end   = c("2018-07-20", "2018-08-05", "2018-08-19", "2018-09-15", "2018-09-15"))

# strings to dates
df <- mutate(df, start = ymd(start), end = ymd(end))

# does not work - calculate overlap in days
df <- mutate(df, overlap = intersect(interval(my.start, my.end), interval(start, end)))

结果应为0、5、4、12、0天:

Results should be 0, 5, 4, 12, 0 days:

   my.start |-------------------------------| my.end

|-----| (0)
        |---------| (5)
                            |----| (4)
                                   |------------------| (12)
                                             |---------------| (0)

在Excel中，我将使用

In Excel, I would use

=MAX(MIN(my.end, end) - MAX(my.start, start) + 1, 0)

但这也不起作用:

# does not work - calculate via min/max
df <- mutate(df, overlap = max(min(my.end, end) - max(my.start, start) + 1, 0))

在尝试使用日期上使用as.numeric()的Excel方法之前，我想知道是否有更聪明的方法来做到这一点.

Before I try to use the Excel approach using as.numeric() on the dates, I wondered if there is a cleverer way to do this.

实际上，Excel数值方法似乎也不起作用(所有结果均为零):

Actually, the Excel numeric approach doesn't seem two work either (all results are zero):

# does not work - calculate via numeric

ms.num <- as.numeric(my.start)
me.num <- as.numeric(my.end)

df <- df %>% 
  mutate(s.num = as.numeric(start),
         e.num = as.numeric(end),

         overlap = max(min(e.num, me.num) - max(s.num, ms.num) + 1, 0))

@akrun的方法似乎适用于ymd日期.但是，它似乎在ymd_hms时间不起作用:

The approach by @akrun seems to work for ymd dates. However, it doesn't seem to work for ymd_hms times:

library(dplyr)
library(lubridate)
library(purrr)

my.start <- ymd("2018-08-01")
my.end <- ymd("2018-08-31")

df <- data.frame(start = c("2018-07-15 10:00:00", "2018-07-20 10:00:00", "2018-08-15 10:00:00", "2018-08-20 10:00:00", "2018-09-01 10:00:00"), 
                 end   = c("2018-07-20 10:00:00", "2018-08-05 10:00:00", "2018-08-19 10:00:00", "2018-09-15 10:00:00", "2018-09-15 10:00:00"))

# strings to dates
df <- mutate(df, start = ymd_hms(start), end = ymd_hms(end))

# leads to 0 results
df %>% mutate(overlap = map2(start, end, ~ sum(seq(.x, .y, by = '1 day') %in% seq(my.start, my.end, by = '1 day'))))

推荐答案

我认为您可能会遇到max和min与pmax和pmin的问题:

I think you may be running into issues with max and min vs pmax and pmin:

library(dplyr)

df %>%
  mutate(overlap = pmax(pmin(my.end, end) - pmax(my.start, start) + 1,0))

       start        end overlap
1 2018-07-15 2018-07-20  0 days
2 2018-07-20 2018-08-05  5 days
3 2018-08-15 2018-08-19  5 days
4 2018-08-20 2018-09-15 12 days
5 2018-09-01 2018-09-15  0 days

这篇关于R/lubridate:计算两个期间之间的重叠天数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！