从两个整数(1, 5)
可以通过以下方式创建范围
1:5
[1] 1 2 3 4 5
如果您提供两个日期("2014-09-04 JST", "2014-09-11 JST")
输出必须是
[1](“2014-09-04 JST”,“2014-09-05 JST”,“2014-09-06 JST”,“2014-09-07 JST”,“2014-09-08 JST “)
答案 0 :(得分:20)
这有帮助吗?
seq(as.Date("2014/09/04"), by = "day", length.out = 5)
# [1] "2014-09-04" "2014-09-05" "2014-09-06" "2014-09-07" "2014-09-08"
编辑:添加关于时区的内容
这适用于我当前的时区
seq(c(ISOdate(2014,4,9)), by = "DSTday", length.out = 5)
#[1] "2014-04-09 08:00:00 EDT" "2014-04-10 08:00:00 EDT" "2014-04-11 08:00:00 EDT" "2014-04-12 08:00:00 EDT"
#[5] "2014-04-13 08:00:00 EDT"
<强> EDIT2:强>
OlsonNames() # I used this to find out what to write for the JST tz - it's "Japan"
x <- as.POSIXct("2014-09-04 23:59:59", tz="Japan")
format(seq(x, by="day", length.out=5), "%Y-%m-%d %Z")
# [1] "2014-09-04 JST" "2014-09-05 JST" "2014-09-06 JST" "2014-09-07 JST" "2014-09-08 JST"
答案 1 :(得分:12)
要仅使用开始日期和结束日期来获取一系列日期(天,周,......),您可以使用:
seq(as.Date("2014/1/1"), as.Date("2014/1/10"), "days”)
[1] "2014-01-01" "2014-01-02" "2014-01-03" "2014-01-04" "2014-01-05" "2014-01-06" "2014-01-07"
[8] "2014-01-08" "2014-01-09" "2014-01-10”
答案 2 :(得分:0)
这里的答案肯定比@ jalapic更糟糕,它不使用seq而是使用for循环:
date1 <- "2014-09-04"
date2 <- "2014-09-11"
dif <- as.numeric(abs(as.Date(date1) - as.Date(date2)))
dates <- vector()
for (i in 1:dif) {
date <- (as.Date(date1) + i)
dates <- append(dates, date)
}
# [1] "2014-09-05" "2014-09-06" "2014-09-07" "2014-09-08" "2014-09-09" "2014-09-10" "2014-09-11
答案 3 :(得分:0)
虽然我的系统无法识别时区JST,但这是一个镜头
d1<-ISOdate(year=2014,month=9,day=4,tz="GMT")
seq(from=d1,by="day",length.out=5)
[1] "2014-09-04 12:00:00 GMT" "2014-09-05 12:00:00 GMT" "2014-09-06 12:00:00 GMT" "2014-09-07 12:00:00 GMT" "2014-09-08 12:00:00 GMT"