使用CDO将netcdf文件中的累积变量转换为时间步长值

问题描述

我在一个带有一个变量的网格上有一个大约100个时间步长的netcdf文件，该文件在时间步长上累积.我现在有兴趣计算每个时间步长对变量值的贡献(即连续时间步长之差).

I have a netcdf-file with about 100 timesteps on a grid with one variable, which is accumulated over the timesteps. I am now interested in calculating the contribution of each timestep to the variable's value (i.e. the difference of consecutive timesteps).

目前，我使用以下顺序:

Currently I use the following sequence:

1. 要将每个时间步提取到一个新文件中，我使用cdo seltimestep,$i ...，
2. 使用cdo sub $i ${i-1} ...
将每个差异计算到一个新文件中
3. ，最后将这些新文件与cdo mergetime ...合并为一个结果文件.

1. To extract every single timestep into a new file I use cdo seltimestep,$i ...,
2. calculate each difference into a new file with cdo sub $i ${i-1} ...
3. and merge those new files in the end with cdo mergetime ... into one single result file.

在我看来，这很麻烦，而且对性能而言并不理想.由于时间步长很大，因此我无法使用cdo管道，因此需要同时创建许多文件.

That seems to me to be very cumbersome and not ideal regarding to performance. Because of the amount of timesteps I cannot use a cdo pipeline and need to create many files in the meantime therefore.

是否有更好的解决方案，可以使用cdo(或其他类似nco/ncl的东西)将累加变量转换为时间步长值?

Is there one better solution to convert an accumulated variable to timestep values with cdo (or something else like nco/ncl?)

推荐答案

numpy的差异计算连续条目的差.

我怀疑您的文件中包含一个多维变量，因此这是一个通用示例:

I suspect you have a multi-dimension variable in your file, so here is a generic example of how to do it:

import netCDF4
import numpy as np

ncfile = netCDF4.Dataset('./myfile.nc', 'r')
var = ncfile.variables['variable'][:,:,:] # [time x lat x lon]

# Differences with a step of 1 along the 'time' axis (0) 
var_diff = np.diff(var, n=1, axis=0) 
ncfile.close()

# Write out the new variable to a new file     
ntim, nlat, nlon = np.shape(var_diff)

ncfile_out = netCDF4.Dataset('./outfile.nc', 'w')
ncfile_out.createDimension('time', ntim)
ncfile_out.createDimension('lat', nlat)
ncfile_out.createDimension('lon', nlon)
var_out = ncfile_out.createVariable('variable', 'f4', ('time', 'lat', 'lon',))
var_out[:,:,:] = var_diff[:,:,:]
ncfile_out.close()

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