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将儒略日转换为日/月/年

[英]Convert julian day to day/month/year

My post was apparently unclear, so I'm trying to fix it, don't hesitate to tell me if I'm still unclear!我的帖子显然不清楚,所以我正在尝试修复它,如果我仍然不清楚,请随时告诉我!

I got a dataframe of physical variables, with a data every minute.我得到一个物理变量的 dataframe,每分钟有一个数据。 I'd like to convert the 4 first columns into a single one: "%d/%m/%Y %H:%M" (GMT) in R.我想将前 4 列转换为一个:R 中的“%d/%m/%Y %H:%M”(格林威治标准时间)。

Year    Julian_day  Hour    Minute  Air_temp    Water_temp  Rel_hum   Wind_int  Wind_dir
2012    1   0   0   11.82   4.73    87.2    5.1 310
2012    1   0   1   11.92   4.743   87.2    5   310
2012    1   0   2   11.86   4.748   86.9    4.7 310
# This imports your data into a variable named "weather"
weather <- read.table(text = "Year    Julian_day  Hour    Minute  Air_temp    Water_temp  Rel_hum   Wind_int  Wind_dir
2012    1   0   0   11.82   4.73    87.2    5.1 310
2012    1   0   1   11.92   4.743   87.2    5   310
2012    1   0   2   11.86   4.748   86.9    4.7 310", header = TRUE)

# Combine the first four columns into a character vector
date_info <- with(weather, paste(Year, Julian_day, Hour, Minute))
# Parse that character vector
strptime(date_info, "%Y %j %H %M")

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