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平均一个月的 NetCDF 数据到一天 24 小时

[英]Average a Months' worth of NetCDF data to one 24hr-day

I hope you're well --我希望你很好 -

I have multiple files all having one month of hourly data.我有多个文件都有一个月的每小时数据。

Wind = Jun1.variables['wind'][:,:,:]
Wind = Jun2.variables['winds'][:,:,:]
etc

Below shows the dimensions下图显示尺寸

I want to average each individual files in a way to calculate the hourly average to get the diurnal cycle for each month with array shape (time, lon, lat) of:我想以一种计算每小时平均值的方式对每个单独的文件进行平均,以获得每个月的昼夜周期,数组形状(时间、经度、纬度)为:

If I understand your question correctly, you want to compute the mean diurnal cycle from a month of hourly data.如果我正确理解您的问题,您想根据一个月的每小时数据计算平均昼夜周期。 NCO's ncra does this with a feature called "interleaving": NCO 的ncra使用称为“交错”的功能来做到这一点:

ncra -d time,,,,720,24 in.nc out.nc

If Charlie's interpretation is correct (I think it is from your desired array shape) then you can work out the diurnal cycle over a multi day period with CDO using如果查理的解释是正确的(我认为它来自您想要的阵列形状),那么您可以使用 CDO 计算出多天的昼夜周期

cdo dhourmean in.nc out.nc 

I think you need to make sure you have a recent version of cdo for this.我认为您需要确保为此拥有最新版本的 cdo。 v1.9.5 should be okay. v1.9.5 应该没问题。

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