Question

我用以下代码创建了cols和行

    library(lubridate)    

date1<-ymd_hms("2000-01-01 05:30:00",tz = "US/Eastern")  
date2<- ymd_hms("2000-01-01 00:00:00",tz = "US/Eastern")

cols <- lapply(0:14, function(x){
  mapply(function(y,z){
    interval((date1+days(x)+minutes(y)), (date1+days(x)+minutes(y+z)))
  }, y = c(0,150,420,600,720,840,1080), z = c(600,570,600,600,600,600,600), SIMPLIFY = FALSE)
})

rows <- lapply(0:14, function(x){
  mapply(function(y,z){
    interval((date2+days(x)+minutes(y)), (date2+days(x)+minutes(y+z)))
  }, y = seq(0,1380,60), rep(c(60),24), SIMPLIFY = FALSE)
})

要求是创建一个具有cols和rows以及1和0的矩阵。如果行间隔在列间隔中则为1，否则为0

Answer 1

间隔由start：开始日期，.Data：间隔持续时间的秒数和tzone：时区组成。

假设tzone是稳定的，则只需重构要比较的两个间隔的开始日期，并检查一个间隔在第二个间隔之后开始并在第二个间隔之前结束。

示例：

    isIncluded <- function(a,b){
      (a[[1]]@start>=b[[1]]@start & (a[[1]]@start+seconds(a[[1]]@.Data))<=(b[[1]]@start+seconds(b[[1]]@.Data)))
    }

 library(lubridate)    

date1<-ymd_hms("2000-01-01 05:30:00",tz = "US/Eastern")  
date2<- ymd_hms("2000-01-01 00:00:00",tz = "US/Eastern")

cols <- lapply(0:14, function(x){
  mapply(function(y,z){
    interval((date1+days(x)+minutes(y)), (date1+days(x)+minutes(y+z)))
  }, y = c(0,150,420,600,720,840,1080), z = c(600,570,600,600,600,600,600), SIMPLIFY = FALSE)
})

rows <- lapply(0:14, function(x){
  mapply(function(y,z){
    interval((date2+days(x)+minutes(y)), (date2+days(x)+minutes(y+z)))
  }, y = seq(0,1380,60), rep(c(60),24), SIMPLIFY = FALSE)
})  
    x <- cols[[1]][1]
    y <- cols[[1]][2]
    isIncluded(x,y) ## Is included x in y? False
    isIncluded(x,x) ## Is included x in x? True

The rest is just populating the matrix!
Best!

如何检查日期间隔是否在另一个间隔中

1 个答案: