我有一个带有时间序列的netcdf文件,并且时间变量具有以下典型的元数据:
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" ;
RI内部希望将时间转换为R日期对象。目前,我通过读取units属性并拆分字符串并将第三个条目用作我的来源(因此假设间距为“天”,时间为00:00,以此类推)来实现此目的:
require("ncdf4")
f1<-nc_open("file.nc")
time<-ncvar_get(f1,"time")
tunits<-ncatt_get(f1,"time",attname="units")
tustr<-strsplit(tunits$value, " ")
dates<-as.Date(time,origin=unlist(tustr)[3])
这种硬连线的解决方案适用于我的特定示例,但是我希望R中可能有一个程序包,可以很好地处理时间单位的UNIDATA netcdf数据约定并将它们安全地转换为R date对象?
我所没有的。我使用有了这个方便的功能lubridate
,该功能基本上与您的功能相同。
getNcTime <- function(nc) {
require(lubridate)
ncdims <- names(nc$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$units, " ")[[1]]
timeunit <- timedef[1]
tz <- timedef[5]
timestart <- strsplit(timedef[4], ":")[[1]]
if (length(timestart) != 3 || timestart[1] > 24 || timestart[2] > 60 || timestart[3] > 60 || any(timestart < 0)) {
cat("Warning:", timestart, "not a valid start time. Assuming 00:00:00\n")
warning(paste("Warning:", timestart, "not a valid start time. Assuming 00:00:00\n"))
timedef[4] <- "00:00:00"
}
if (! tz %in% OlsonNames()) {
cat("Warning:", tz, "not a valid timezone. Assuming UTC\n")
warning(paste("Warning:", timestart, "not a valid start time. Assuming 00:00:00\n"))
tz <- "UTC"
}
timestart <- ymd_hms(paste(timedef[3], timedef[4]), tz=tz)
f <- switch(tolower(timeunit), #Find the correct lubridate time function based on the unit
seconds=seconds, second=seconds, sec=seconds,
minutes=minutes, minute=minutes, min=minutes,
hours=hours, hour=hours, h=hours,
days=days, day=days, d=days,
months=months, month=months, m=months,
years=years, year=years, yr=years,
NA
)
suppressWarnings(if (is.na(f)) stop("Could not understand the time unit format"))
timestart + f(times)
}
编辑:也许还想看看 ncdf4.helpers::nc.get.time.series
EDIT2:请注意,新提议且当前正在开发的超棒stars
软件包将自动处理日期,有关示例,请参阅第一篇博客文章。
EDIT3:另一种方法是units
直接使用包,这就是stars
使用的方法。一个人可以做这样的事情:(仍然无法正确处理日历,我不确定是否units
可以)
getNcTime <- function(nc) { ##NEW VERSION, with the units package
require(units)
require(ncdf4)
options(warn=1) #show warnings by default
if (is.character(nc)) nc <- nc_open(nc)
ncdims <- names(nc$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)
}
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