It has been almost 2 weeks since I started getting familiar with netCDF files and operators (NCO) and climate data operators (cdo) and I am not an experienced programmer.

I have a folder which contains (monthly) nc files from January 1980 to December 2015 (in total 432 nc files). Each of these nc files contains hourly data of different compounds (such as PM2.5, O3, NO2, NO, etc) for one month in UTC and I need to have netCDF files which will contain daily averages in local time.

A colleague of mine gave me the following instructions: first I need to identify the zone (let's name it z0) with [long1,long2], then identify the time (UTC--> local time, where UTC= 0h and LT=+1 and so on) and simultaneously derive the daily averages. He also told me that the final result e.g for one month should be an nc file with daily averages where 1 day should contain the averaged data of the different compounds where z0= UTC[0:23], z1= UTC[1:24]....etc....(This doesn't make any sense to me).

As I am quite new at this, it is very difficult for me to understand what and how I need to do it.

Since the time is in this format:

int time(time) ;
                time:units = "seconds since 1979-12-02 00:00:00 UTC" ;
                time:long_name = "time" ;
                time:axis = "T" ;
                time:calendar = "standard" ;
                time:standard_name = "time" ;

I thought the change from UTC to local time could be done with the command ncatted, and then I could move on with the code in order to get the daily averages. Any hints?

  • Do your netcdf files contain only data for 1 spatial location (i.e. one pair of lat-long), or is your data spatial? In other words, which dimensions are defined in the netcdf files, only "time" or also something like "longitude" and "latitude"? Commented Jul 25, 2019 at 14:02
  • The data are from different parts in the world. The dimensions are time, longitude and latitude and my variables are for example: time(time), lon(lon),lat(lat), PM2.5 (time, lon, lat), PM10 (time, lon, lat), O3 (time, lon, lat) and so on..
    – Nina
    Commented Jul 28, 2019 at 11:05


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.