dat <- structure(list(crop_name = c("a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "a", "a",
"a", "a", "a", "a", "a", "a", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b", "b",
"b", "b"),
year = c(2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L, 2010L), doy = c("Sep_24", "Oct_1", "Oct_8", "Oct_15", "Oct_22", "Oct_29", "Nov_5", "Nov_12", "Nov_19", "Nov_26", "Dec_3", "Sep_30", "Oct_7", "Oct_14", "Oct_21", "Oct_28", "Nov_4", "Nov_11", "Nov_18", "Nov_25", "Dec_2", "Jan_22", "Jan_29", "Feb_5",
"Feb_12", "Feb_19", "Feb_25", "March_5", "March_10", "March_19",
"Jan_14","Jan_21", "Jan_28", "Feb_4", "Feb_11", "Feb_18",
"Feb_25", "March_4","March_11", "March_18"),
per_f = c(1, 4, 13, 18, 30, 42, 68, 83, 93, 100, 100, 0, 0, 2, 8, 20, 35, 65, 78, 90, 100, 0, 2, 18, 26, 28, 70, 76, 84, 100, 1, 2, 10, 25, 50, 80, 92, 98, 100, 100)),
class = "data.frame", row.names = c(NA, -40L))
在上述数据中,我需要将doy
列转换为儒略日,假设全年有365天。到目前为止,这是我设法做的事情:
doy.col <- as.data.frame(str_split_fixed(dat$doy, "_", 2))
dat$month <- doy.col$V1
dat$doy <- doy.col$V2
dat$date <- paste0(dat$doy,"/",dat$month,"/",dat$year)
我现在如何将字符dat$date
隐藏到每年的朱利安日?
答案 0 :(得分:1)
使用您当前拥有的内容,只需将其转换为日期,然后使用strftime
指定儒略格式。
dat$date2 <- as.Date(dat$date, format = "%d/%b/%Y")
dat$doy <- strftime(dat$date2, format = "%j")
head(dat)
crop_name year doy per_f month date date2
1 a 2009 267 1 Sep 24/Sep/2009 2009-09-24
2 a 2009 274 4 Oct 1/Oct/2009 2009-10-01
3 a 2009 281 13 Oct 8/Oct/2009 2009-10-08
4 a 2009 288 18 Oct 15/Oct/2009 2009-10-15
5 a 2009 295 30 Oct 22/Oct/2009 2009-10-22
6 a 2009 302 42 Oct 29/Oct/2009 2009-10-29