r - 将 netcdf 时间变量转换为 R 日期对象

Question

我有一个带有时间序列的 netcdf 文件，时间变量具有以下典型元数据:

    double time(time) ;
            time:standard_name = "time" ;
            time:bounds = "time_bnds" ;
            time:units = "days since 1979-1-1 00:00:00" ;
            time:calendar = "standard" ;
            time:axis = "T" ;

在 R 中，我想将时间转换为 R 日期对象。我目前通过读取单位属性并拆分字符串并使用第三个条目作为我的原点(因此假设间距为“天”，时间为 00:00 等)以硬连线方式实现了这一点:

require("ncdf4")
f1<-nc_open("file.nc")
time<-ncvar_get(f1,"time")
tunits<-ncatt_get(f1,"time",attname="units")
tustr<-strsplit(tunits$value, " ")
dates<-as.Date(time,origin=unlist(tustr)[3])

这个硬接线解决方案适用于我的特定示例，但我希望 R 中可能有一个包可以很好地处理时间单位的 UNIDATA netcdf 数据约定并将它们安全地转换为 R 日期对象？

Answer 1

没有，据我所知。我使用 lubridate 有这个方便的功能，这与您的基本相同。

getNcTime <- function(nc) {
    require(lubridate)
    ncdims <- names(nc$dim) #get netcdf dimensions
    timevar <- ncdims[which(ncdims %in% c("time", "Time", "datetime", "Datetime", "date", "Date"))[1]] #find time variable
    times <- ncvar_get(nc, timevar)
    if (length(timevar)==0) stop("ERROR! Could not identify the correct time variable")
    timeatt <- ncatt_get(nc, timevar) #get attributes
    timedef <- strsplit(timeatt$units, " ")[[1]]
    timeunit <- timedef[1]
    tz <- timedef[5]
    timestart <- strsplit(timedef[4], ":")[[1]]
    if (length(timestart) != 3 || timestart[1] > 24 || timestart[2] > 60 || timestart[3] > 60 || any(timestart < 0)) {
        cat("Warning:", timestart, "not a valid start time. Assuming 00:00:00\n")
        warning(paste("Warning:", timestart, "not a valid start time. Assuming 00:00:00\n"))
        timedef[4] <- "00:00:00"
    }
    if (! tz %in% OlsonNames()) {
        cat("Warning:", tz, "not a valid timezone. Assuming UTC\n")
        warning(paste("Warning:", timestart, "not a valid start time. Assuming 00:00:00\n"))
        tz <- "UTC"
    }
    timestart <- ymd_hms(paste(timedef[3], timedef[4]), tz=tz)
    f <- switch(tolower(timeunit), #Find the correct lubridate time function based on the unit
        seconds=seconds, second=seconds, sec=seconds,
        minutes=minutes, minute=minutes, min=minutes,
        hours=hours,     hour=hours,     h=hours,
        days=days,       day=days,       d=days,
        months=months,   month=months,   m=months,
        years=years,     year=years,     yr=years,
        NA
    )
    suppressWarnings(if (is.na(f)) stop("Could not understand the time unit format"))
    timestart + f(times)
}

编辑:人们可能还想看看 ncdf4.helpers::nc.get.time.series
EDIT2:请注意，新提议且目前正在开发中的令人敬畏的 stars包会自动处理日期，见 the first blog post举个例子。

EDIT3:另一种方法是使用 units直接打包，这是什么 stars使用。可以做这样的事情:(仍然没有正确处理日历，我不确定 units 可以)

getNcTime <- function(nc) { ##NEW VERSION, with the units package
    require(units)
    require(ncdf4)
    options(warn=1) #show warnings by default
    if (is.character(nc)) nc <- nc_open(nc)
    ncdims <- names(nc$dim) #get netcdf dimensions
    timevar <- ncdims[which(ncdims %in% c("time", "Time", "datetime", "Datetime", "date", "Date"))] #find (first) time variable
    if (length(timevar) > 1) {
        warning(paste("Found more than one time var. Using the first:", timevar[1]))
        timevar <- timevar[1]
    }
    if (length(timevar)!=1) stop("ERROR! Could not identify the correct time variable")
    times <- ncvar_get(nc, timevar) #get time data
    timeatt <- ncatt_get(nc, timevar) #get attributes
    timeunit <- timeatt$units
    units(times) <- make_unit(timeunit)
    as.POSIXct(time)
}