Generate timeseries vector (spatial average) from NETCDF R

Question

I have a NETCDF file with attributes being lon,lat,time,precipitation. The data covers a certain spatial domain. It is daily data from 1960 to 2100.

1) I will like to subset the data spatially (e.g lat[45,50] and lon[-78,-85])from the main domain

2) From the subset, I will like to average over all grids and produce a single column daily time series then write it to a .csv file.

NB: my data contains missing values

Robert Hijmans · Accepted Answer · 2015-11-12T19:24:59.307

3

Something along these lines ought to work

library(raster)
b <- brick('file.nc')
be <- crop(b, extent(-85, -78, 45, 50))
a <- aggregate(be, dim(be)[2:1], na.rm=TRUE)
v <- values(a)
write.csv(v, 'precip.csv', row.names=FALSE)

Normally, to get the date as well:

date <- getZ(be)

Or

date <- names(a)
date <- gsub('X', '', date)

And then

v <- data.frame(date=date, prec=v)
write.csv(v, 'precip.csv', row.names=FALSE)

Whether or not this date is human readable depends on how it is stored in the ncdf file (i.e. whether certain conventions are followed or not).

edited Nov 12 '15 at 19:24

answered Nov 11 '15 at 20:14

Robert Hijmans

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Great! Thanks very much @RobertH. Exactly what I was looking for. How can I make the date column to appear human readable? Many thanks. – code123 Nov 12 '15 at 02:11
I have edited the answer to address this additional question – Robert Hijmans Nov 12 '15 at 05:04

ClimateUnboxed · Answer 2 · 2018-09-26T07:18:16.377

3

This would do the cut and the averaging but writes the answer to another netcdf. If you really need CSV, then you would need to use that part of the solution above.

cdo fldmean -sellonlatbox,-85,-78,45,50 in.nc out.nc

edited Sep 26 '18 at 07:18

answered Mar 29 '18 at 11:57

ClimateUnboxed

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thanks for this solution. It's great. I now use cdo, ncl or nco for most of these operations. – code123 Mar 31 '18 at 01:08

score 1 · Answer 3 · answered Dec 03 '20 at 16:08

1

library (ncdf4)

nc <- nc_open("netcdf.nc")

lon <- ncvar_get(nc,"lon")
lat <- ncvar_get(nc,"lat")
time <- ncvar_get(nc,"time")

lon_lim <- c(45,50)
lat_lim <- c(-78,-85)

lon_ind <- which(lon >= lon_lim[1] & lon <= lon_lim[2])
lat_ind <- which(lat >= lat_lim[1] & lat <= lat_lim[2])

precip <- ncvar_get(nc,"precip",start = c(lon_ind[1],lat_ind[1],time),count = c(length(lon_ind),length(lat_ind),length(time)))

ts <- apply(precip,3,mean,na.rm=TRUE)

answered Dec 03 '20 at 16:08

Shanto Khandoker

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please share some explanation about how this code solve the problem – OMR Dec 03 '20 at 18:34
warning: this solution will give the INCORRECT answer as it does not weight for the cells getting smaller at higher latitudes... See this answer for example: https://stackoverflow.com/questions/60571445/get-mean-of-netcdf-file-using-xarray/65353505#65353505 – ClimateUnboxed Dec 18 '20 at 08:59

Generate timeseries vector (spatial average) from NETCDF R

3 Answers3