Question

我期待制作一个功能，它接受年份的输入并给出星期日的所有日期。

library(lubridate)
yearSunday <- function(year) {
      ymd() + weeks(0:51)

}

我需要帮助才能保持闰年的选择。

Answer 1

稍微快一点的方法是

生成一年中前七天的序列
确定哪一天是星期日
生成从第一个星期日到年末的序列，递增7。

下面，我将Ronak的yearSunday与其他yearSunday2进行了比较。 yearSunday2运行速度快4倍。

yearSunday <- function(year) {
 dates <- seq(as.Date(paste(year, "-01-01", sep = "")), as.Date(paste(year, "-12-31", sep = "")), by="+1 day")
 days <- weekdays(dates) == "Sunday"
 dates[days]
}

yearSunday2 <- function(year){
  start <- seq(as.Date(paste0(year, "-01-01")), 
               as.Date(paste0(year, "-01-07")),
               by = 1)
  seq(start[which(weekdays(start) == "Sunday")],
      as.Date(paste0(year, "-12-31")), 
      by = 7)
}


library(microbenchmark)
microbenchmark(
  yearSunday(2015),
  yearSunday2(2015)
)

Unit: microseconds
              expr      min       lq      mean    median       uq      max neval cld
  yearSunday(2015) 1499.074 1513.149 1543.4064 1527.8120 1543.207 2616.634   100   b
 yearSunday2(2015)  357.173  372.569  386.5511  383.7125  398.375  444.854   100  a

Answer 2

试试这个，

yearSunday <- function(year) {
 dates <- seq(as.Date(paste(year, "-01-01", sep = "")), as.Date(paste(year, "-12-31", sep = "")), by="+1 day")
 days <- weekdays(dates) == "Sunday"
 dates[days]
}

yearSunday(2015)
#[1] "2015-01-04" "2015-01-11" "2015-01-18" "2015-01-25" "2015-02-01" "2015-02-08" "2015-02-15"
#[8] "2015-02-22" "2015-03-01" "2015-03-08" "2015-03-15" "2015-03-22" "2015-03-29" "2015-04-05"
#[15] "2015-04-12" "2015-04-19" "2015-04-26" "2015-05-03" "2015-05-10" "2015-05-17" "2015-05-24"
#[22] "2015-05-31" "2015-06-07" "2015-06-14" "2015-06-21" "2015-06-28" "2015-07-05" "2015-07-12"
#[29] "2015-07-19" "2015-07-26" "2015-08-02" "2015-08-09" "2015-08-16" "2015-08-23" "2015-08-30" 
#[36] "2015-09-06" "2015-09-13" "2015-09-20" "2015-09-27" "2015-10-04" "2015-10-11" "2015-10-18"
#[43] "2015-10-25" "2015-11-01" "2015-11-08" "2015-11-15" "2015-11-22" "2015-11-29" "2015-12-06"
#[50] "2015-12-13" "2015-12-20" "2015-12-27"

这将自动照顾闰年。

yearSunday(2016)
#[1] "2016-01-03" "2016-01-10" "2016-01-17" "2016-01-24" "2016-01-31" "2016-02-07" "2016-02-14"
#[8] "2016-02-21" "2016-02-28" "2016-03-06" "2016-03-13" "2016-03-20" "2016-03-27" "2016-04-03"
#[15] "2016-04-10" "2016-04-17" "2016-04-24" "2016-05-01" "2016-05-08" "2016-05-15" "2016-05-22"
#[22] "2016-05-29" "2016-06-05" "2016-06-12" "2016-06-19" "2016-06-26" "2016-07-03" "2016-07-10"
#[29] "2016-07-17" "2016-07-24" "2016-07-31" "2016-08-07" "2016-08-14" "2016-08-21" "2016-08-28"
#[36] "2016-09-04" "2016-09-11" "2016-09-18" "2016-09-25" "2016-10-02" "2016-10-09" "2016-10-16"
#[43] "2016-10-23" "2016-10-30" "2016-11-06" "2016-11-13" "2016-11-20" "2016-11-27" "2016-12-04"
#[50] "2016-12-11" "2016-12-18" "2016-12-25"

修改

因此，如果我们需要更快的方法，那么我们可以获得该年的第一个星期日，然后将其增加7天，

yearSunday <- function(year) { dates <- as.Date(paste(year, "-01-01", sep = "")) + 0:6 days <- weekdays(dates) == "Sunday" seq(dates[days], as.Date(paste(year, "-12-31", sep = "")), by = "+7 day") }

比本杰明的回答略快一点

library(microbenchmark) > microbenchmark( + yearSunday(2015), + yearSunday2(2015) + ) # Unit: microseconds # expr min lq mean median uq max neval # yearSunday(2015) 270.277 279.343 308.9434 294.054 312.529 623.005 100 # yearSunday2(2015) 357.176 379.927 416.4793 397.717 411.402 1295.960 100

根据输入（年）获取一年中每个星期日的日期

2 个答案: