Compute Autocorrelation (lag correlation) in cdo

Added by Lyndon Mark Olaguera about 8 years ago

Hi all,

I would like to perform autocorrelation (lag 1) of a netcdf file.
Is this possible in cdo?
I tried to use the fldcor and timcor functions, but I dont know how to implement this.
Would it be correct if I just delete a time step in the netcdf file and correlate it with the original netcdf file?

Ill appreciate any help

Replies (5)

RE: Compute Autocorrelation (lag correlation) in cdo - Added by Sebastian Bathiany almost 8 years ago

Hi Lyndon,

I just had the same problem.
timcor seems to do what you want if you copy your time series and shift it by the required lag, then use timcor on the original and shifted time series.

There is one caveat I struggled with:
The cdo command does not seem to recognize the date. So the "t" in the cdo documentation does actually not stand for absolute time, but for the number of the record in the file. So it is not sufficient to just relabel the dates, you have to remove a record entry.
I did it the following way:

noftimesteps=`cdo showtime dataset.nc |wc -w`        #counts time steps (this and the next command depend on the shell you are using)
        (( noftimesteps2 = ${noftimesteps} - 1 ))            #number of time steps minus 1 (because my lag is one month and the data are monthly)
        cdo shifttime,1month dataset.nc dataset_shift.nc
        cdo seltimestep,2/${noftimesteps} dataset_shift.nc dataset_shift1.nc
        cdo seltimestep,1/${noftimesteps2} dataset.nc dataset_shift2.nc

cdo timcor dataset_shift1.nc dataset_shift2.nc dataset_AC.nc

I hope that this may help anyone.

Sebastian

RE: Compute Autocorrelation (lag correlation) in cdo - Added by Lyndon Mark Olaguera almost 8 years ago

Dear Sebastian,

Deleting the timestep before using timcor also worked for my case.
Your solution is more elegant than my method.
Many many thanks for the help.

RE: Compute Autocorrelation (lag correlation) in cdo - Added by Ralf Mueller almost 8 years ago

Instead of

noftimesteps=`cdo showtime dataset.nc |wc -w`

you could use the ntime operator. see here for more

RE: Compute Autocorrelation (lag correlation) in cdo - Added by Ralf Mueller almost 8 years ago

Instead of selecting cdo infon: Started cdo(2) for: Set default -1 : Date 1 : 0001-01-01 00:00:00 2 : 0001-01-02 00:00:00 3 : 0001-01-03 00:00:00 4 : 0001-01-04 00:00:00 5 : 0001-01-05 00:00:00 6 : 0001-01-06 00:00:00 7 : 0001-01-07 00:00:00 8 : 0001-01-08 00:00:00 9 : 0001-01-09 00:00:00 10 : 0001-01-10 00:00:00 cdo(2) for: ( 0.00s ) cdo infon: Processed ram@luthien:~ cdo -infov cdo infon: Started cdo(2) delete: Started cdo(3) for: Set default -1 : Date 1 : 0001-01-02 00:00:00 2 : 0001-01-03 00:00:00 3 : 0001-01-04 00:00:00 4 : 0001-01-05 00:00:00 5 : 0001-01-06 00:00:00 6 : 0001-01-07 00:00:00 7 : 0001-01-08 00:00:00 8 : 0001-01-09 00:00:00 9 : 0001-01-10 00:00:00 cdo(3) for: ( 0.00s ) cdo(2) delete: Processed cdo infon: Processed ram@luthien:~ cdo -infov cdo infon: Started cdo(2) delete: Started cdo(3) for: Set default -1 : Date 1 : 0001-01-04 00:00:00 2 : 0001-01-05 00:00:00 3 : 0001-01-06 00:00:00 4 : 0001-01-07 00:00:00 5 : 0001-01-08 00:00:00 6 : 0001-01-09 00:00:00 7 : 0001-01-10 00:00:00 cdo(3) for: ( 0.00s ) cdo(2) delete: Processed cdo infon: Processed the timesteps you want, you could delete the ones, you don't want with the delete operator:

ram@luthien:~ cdo -infov -for,1,10 [10:30:10|16-12-02] child process "for,1,10 (pipe1.1)". filetype to GRIB Time Level Gridsize Miss : Minimum Mean Maximum : Parameter name 0 1 0 : 1.0000 : for 0 1 0 : 2.0000 : for 0 1 0 : 3.0000 : for 0 1 0 : 4.0000 : for 0 1 0 : 5.0000 : for 0 1 0 : 6.0000 : for 0 1 0 : 7.0000 : for 0 1 0 : 8.0000 : for 0 1 0 : 9.0000 : for 0 1 0 : 10.000 : for 10 values from 1 variable over 10 timesteps ( 0.00s ) -delete,timestep=1 -for,1,10 [10:30:39|16-12-02] child process "delete,timestep=1 -for,1,10 (pipe1.1)". child process "for,1,10 (pipe2.1)". filetype to GRIB Time Level Gridsize Miss : Minimum Mean Maximum : Parameter name 0 1 0 : 2.0000 : for 0 1 0 : 3.0000 : for 0 1 0 : 4.0000 : for 0 1 0 : 5.0000 : for 0 1 0 : 6.0000 : for 0 1 0 : 7.0000 : for 0 1 0 : 8.0000 : for 0 1 0 : 9.0000 : for 0 1 0 : 10.000 : for 9 values from 1 variable over 10 timesteps ( 0.00s ) 9 values from 1 variable over 9 timesteps ( 0.00s ) -delete,timestep=1/3 -for,1,10 [10:30:48|16-12-02] child process "delete,timestep=1/3 -for,1,10 (pipe1.1)". child process "for,1,10 (pipe2.1)". filetype to GRIB Time Level Gridsize Miss : Minimum Mean Maximum : Parameter name 0 1 0 : 4.0000 : for 0 1 0 : 5.0000 : for 0 1 0 : 6.0000 : for 0 1 0 : 7.0000 : for 0 1 0 : 8.0000 : for 0 1 0 : 9.0000 : for 0 1 0 : 10.000 : for 7 values from 1 variable over 10 timesteps ( 0.00s ) 7 values from 1 variable over 7 timesteps ( 0.00s )

hth
ralf

RE: Compute Autocorrelation (lag correlation) in cdo - Added by Noor Shazwani almost 7 years ago

noftimesteps=`cdo showtime dataset.nc |wc -w` #counts time steps (this and the next command depend on the shell you are using)
(( noftimesteps2 = ${noftimesteps} - 1 )) #number of time steps minus 1 (because my lag is one month and the data are monthly)
cdo shifttime,1month dataset.nc dataset_shift.nc
cdo seltimestep,2/${noftimesteps} dataset_shift.nc dataset_shift1.nc
cdo seltimestep,1/${noftimesteps2} dataset.nc dataset_shift2.nc

The codes above for monthly data. How about daily data? Do I need to minus one? I already try it but it shows invalid arithmetic operator. Maybe because of my data is in daily data. Do anyone know how to solve it? Thanks in advance.

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