I have a data set that has column containing year, and a column containing the ordinal (yday) day for each year. I would like to convert this information into a calendar date, so I can better filter the data set. The data looks like this, but is from 1980-2016 and has an entry for each day of the year:
year yday temp
1980 1 0.5
1980 2 -5.0
1980 3 -3.5
1980 4 1.0
1980 5 -1.0
temps<-structure(list(year = c(1980L, 1980L, 1980L, 1980L, 1980L), yday = 1:5,
temp = c(0.5, -5, -3.5, 1, -1)), row.names = c(NA, -5L), class = c("tbl_df",
"tbl", "data.frame"), .Names = c("year", "yday", "temp"))
I tried the following code, but could not get the correct calendar date: Convert day of year to date
library(lubridate)
library(dplyr)
temps <- tibble::tribble(
~year, ~yday, ~temp,
1980L, 1L, 0.5,
1980L, 2L, -5,
1980L, 3L, -3.5,
1980L, 4L, 1,
1980L, 5L, -1,
1980L, 99L, -1,
1980L, 50L, -1
)
temps %>%
mutate(date = make_date(year) + yday - 1)
#> # A tibble: 7 x 4
#> year yday temp date
#> <int> <int> <dbl> <date>
#> 1 1980 1 0.5 1980-01-01
#> 2 1980 2 -5.0 1980-01-02
#> 3 1980 3 -3.5 1980-01-03
#> 4 1980 4 1.0 1980-01-04
#> 5 1980 5 -1.0 1980-01-05
#> 6 1980 99 -1.0 1980-04-08
#> 7 1980 50 -1.0 1980-02-19
You could start with January 1st, then calculate it by adding your yday
values.
with(temps, as.Date(paste0(year, "-01-01")) + (yday - 1))
# [1] "1980-01-01" "1980-01-02" "1980-01-03" "1980-01-04" "1980-01-05"
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