如何计算一天是否在两个日期之间？答案

【问题标题】：How to calculate if a day is between two dates?如何计算一天是否在两个日期之间？
【发布时间】：2021-11-26 09:23:39
【问题描述】：

我有每个人的进入和退出日期的数据。我的数据如下所示：

  ID      entry       exit
1  1 01.01.2010 31/01/2016
2  2 01/02/2016 01/01/2021
3  3 01/10/2010 30/09/2019
4  4 01.09.2013 31/12/2015
5  5 01/01/2016 30/09/2020

我想创建额外的列，我想在其中确定个人是否在给定日期（进入和退出日期之间）处于活动状态。

数据应如下所示：

 ID      entry    exit           active_31_12_2011    active_31_12_2016
1  1 01/01/2010 31/01/2016               yes                no
2  2 01/02/2016 01/01/2021                no               yes
3  3 01/10/2010 30/09/2019               yes               yes
4  4 01/09/2013 31/12/2015                no                no
5  5 01/01/2016 30/09/2020                no               yes

有人知道我该怎么做吗？

【问题讨论】：

标签： r dataframe date

【解决方案1】：

使用data.table::between

df %>%
  mutate(entry = as.Date(entry, format = "%d/%m/%Y"),
         exit = as.Date(exit, format = "%d/%m/%Y")) %>%
  rowwise %>%
  mutate(active_31_12_2011 = between(as.Date("2011/12/31"),lower = entry, upper = exit),
         active_31_12_2016 = between(as.Date("2016/12/31"),lower = entry, upper = exit))

     ID entry      exit       active_31_12_2011 active_31_12_2016
  <int> <date>     <date>     <lgl>             <lgl>            
1     1 2010-01-01 2016-01-31 TRUE              FALSE            
2     2 2016-02-01 2021-01-01 FALSE             TRUE             
3     3 2010-10-01 2019-09-30 TRUE              TRUE             
4     4 2013-09-01 2015-12-31 FALSE             FALSE            
5     5 2016-01-01 2020-09-30 FALSE             TRUE

要获得是/否输出，

df %>%
  mutate(entry = as.Date(entry, format = "%d/%m/%Y"),
         exit = as.Date(exit, format = "%d/%m/%Y")) %>%
  rowwise %>%
  mutate(active_31_12_2011 = ifelse(between(as.Date("2011/12/31"),lower = entry, upper = exit), "Yes", "No"),
         active_31_12_2016 = ifelse(between(as.Date("2016/12/31"),lower = entry, upper = exit), "Yes", "No"))

     ID entry      exit       active_31_12_2011 active_31_12_2016
  <int> <date>     <date>     <chr>             <chr>            
1     1 2010-01-01 2016-01-31 Yes               No               
2     2 2016-02-01 2021-01-01 No                Yes              
3     3 2010-10-01 2019-09-30 Yes               Yes              
4     4 2013-09-01 2015-12-31 No                No               
5     5 2016-01-01 2020-09-30 No                Yes

【讨论】：

谢谢，成功了！

【解决方案2】：

这是使用lubridate的方法：

library(tidyverse)
library(lubridate)
#> 
#> Attaching package: 'lubridate'
#> The following objects are masked from 'package:base':
#> 
#>     date, intersect, setdiff, union

data <- tribble(
  ~ID, ~entry, ~exit,
  1, "01/01/2010", "31/01/2016",
  2, "01/02/2016", "01/01/2021",
  3, "01/10/2010", "30/09/2019",
  4, "01/09/2013", "31/12/2015",
  5, "01/01/2016", "30/09/2020"
)

data %>%
  mutate_at(c("entry", "exit"), ~ .x %>% parse_date(format = "%d/%m/%Y")) %>%
  mutate(
    interval = interval(entry, exit),
    active_31_12_2011 = ymd("2011-12-31") %within% interval,
    active_31_12_2016 = ymd("2016-12-31") %within% interval
  )
#> # A tibble: 5 x 6
#>      ID entry      exit       interval                       active_31_12_2011
#>   <dbl> <date>     <date>     <Interval>                     <lgl>            
#> 1     1 2010-01-01 2016-01-31 2010-01-01 UTC--2016-01-31 UTC TRUE             
#> 2     2 2016-02-01 2021-01-01 2016-02-01 UTC--2021-01-01 UTC FALSE            
#> 3     3 2010-10-01 2019-09-30 2010-10-01 UTC--2019-09-30 UTC TRUE             
#> 4     4 2013-09-01 2015-12-31 2013-09-01 UTC--2015-12-31 UTC FALSE            
#> 5     5 2016-01-01 2020-09-30 2016-01-01 UTC--2020-09-30 UTC FALSE            
#> # … with 1 more variable: active_31_12_2016 <lgl>

^{由reprex package (v2.0.1) 于 2021-10-06 创建}

【讨论】：

【解决方案3】：

当日期为 dates 或格式为 2010-01-01 时，您可以使用 ==、>、<、>= 或 <=。

x$active_31_12_2011 <- c("no", "yes")[1 + (x$entry <= "2011-12-31" & "2011-12-31" <= x$exit)]
x$active_31_12_2016 <- c("no", "yes")[1 + (x$entry <= "2016-12-31" & "2016-12-31" <= x$exit)]

x
#  ID      entry       exit active_31_12_2011 active_31_12_2016
#1  1 2010-01-01 2016-01-31               yes                no
#2  2 2016-02-01 2021-01-01                no               yes
#3  3 2010-10-01 2019-09-30               yes               yes
#4  4 2013-09-01 2015-12-31                no                no
#5  5 2016-01-01 2020-09-30                no               yes

数据：

x <- read.table(header=TRUE, text="ID      entry       exit
1  1 01/01/2010 31/01/2016
2  2 01/02/2016 01/01/2021
3  3 01/10/2010 30/09/2019
4  4 01/09/2013 31/12/2015
5  5 01/01/2016 30/09/2020")
x$entry <- as.Date(x$entry, "%d/%m/%Y")
x$exit <- as.Date(x$exit, "%d/%m/%Y")

【讨论】：

【解决方案4】：

您可以自愿将日期强制转换为整数并使用data.table::between。

dat_n <- as.data.frame(lapply(dat, as.integer))

library(data.table)
act <- t(apply(dat_n[-1], 1, \(z) between(as.numeric(x), z[1], z[2]))) |>
  `colnames<-`(paste0('active_', x))
res <- cbind(dat, act)
res
#   ID      entry       exit active_2011-12-31 active_2016-12-31
# 1  1 2010-01-01 2016-01-31              TRUE             FALSE
# 2  2 2016-02-01 2021-01-01             FALSE              TRUE
# 3  3 2010-10-01 2019-09-30              TRUE              TRUE
# 4  4 2013-09-01 2015-12-31             FALSE             FALSE
# 5  5 2016-01-01 2020-09-30             FALSE              TRUE

数据

dat <- structure(list(ID = 1:5, entry = structure(c(14610, 16832, 14883, 
15949, 16801), class = "Date"), exit = structure(c(16831, 18628, 
18169, 16800, 18535), class = "Date")), row.names = c("1", "2", 
"3", "4", "5"), class = "data.frame")

x <- c(as.Date('2011-12-31'), as.Date('2016-12-31'))

【讨论】：