如何计算Python中水汽通量辐合的垂直积分？ |
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如何计算Python中水汽通量辐合的垂直积分？
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我是刚接触python的人，我用metpy.calc函数HMC_LE = (np.array(mpcalc.divergence(uq, vq, dx=dx, dy=dy)))计算了印度在压力水平850 hPa时的水汽通量散度。我遵循了这里描述的步骤：How to calculate moisture flux divergence in python。我的密码- import numpy as np import xarray as xr import matplotlib.pyplot as plt import cartopy.crs as ccrs from cartopy.mpl.ticker import LongitudeFormatter, LatitudeFormatter from cartopy.util import add_cyclic_point import pandas as pd import cartopy.feature as cf from netCDF4 import Dataset import metpy.calc as mpcalc ds1 = xr.open_dataset('Downloads/uwnd.mon.mean.nc') ds2 = xr.open_dataset('Downloads/vwnd.mon.mean.nc') ds3 = xr.open_dataset('Downloads/shum.mon.mean.nc') lons = ds1['lon'][:] lats = ds1['lat'][:] levs = ds1['level'][2] # --- select all JJAS months uJJAS_ = ds1['uwnd'].sel(time=np.in1d(ds1['time.month'], [6, 7, 8,9])) uJJAS_ uJJAS=uJJAS_[:,2,:,:] uJJAS # Select JJAS 850 hPa from 1981-2010 u_1=uJJAS[132:252,:,:] u_1 # Prepare the climatology u_1_mean= np.mean(u_1,axis=0) u_1_mean # --- For v vJJAS_ = ds2['vwnd'].sel(time=np.in1d(ds2['time.month'], [6, 7, 8,9])) vJJAS_ vJJAS=vJJAS_[:,2,:,:] vJJAS v_1=vJJAS[132:252,:,:] v_1 v_1_mean= np.mean(v_1,axis=0) v_1_mean # --- For specific humidity shJJAS_ = ds3['shum'].sel(time=np.in1d(ds3['time.month'], [6, 7, 8,9])) shJJAS_ shJJAS=shJJAS_[:,2,:,:] shJJAS sh_1=shJJAS[132:252,:,:] sh_1 sh_1_mean= np.mean(sh_1,axis=0) sh_1_mean uq=(u_1_mean)*(sh_1_mean) vq=(v_1_mean)*(sh_1_mean) # Compute dx and dy spacing for use in divergence calculation dx, dy = mpcalc.lat_lon_grid_deltas(lons, lats) # Calculate moisture flux divergence HMC_LE = (np.array(mpcalc.divergence(uq, vq, dx=dx, dy=dy))) lat=lats.to_numpy() lon=lons.to_numpy() u_=uq.to_numpy() v_=vq.to_numpy() ax2 = plt.axes(projection=ccrs.PlateCarree(central_longitude=180)) clevs = [-0.06,-0.05,-0.04,-0.03,-0.02,-0.01, 0,0.01,0.02,0.03,0.04,0.05,0.06] vp_fill = ax2.contourf(lons,lats,HMC_LE*1000,clevs,transform=ccrs.PlateCarree(),cmap=plt.cm.RdBu_r, extend='both') plt.colorbar(vp_fill, orientation='vertical') q2=ax2.quiver(lon, lat, u_, v_,width=0.003,headwidth=5, scale_units='xy',scale=25,transform=ccrs.PlateCarree()) ax2.coastlines(alpha=0.8) ax2.set_xticks([60,70,80,90,100], crs=ccrs.PlateCarree()) ax2.set_yticks([-10,0,10,20,30,40], crs=ccrs.PlateCarree()) lon_formatter = LongitudeFormatter(zero_direction_label=True,number_format='.0f') lat_formatter = LatitudeFormatter() ax2.xaxis.set_major_formatter(lon_formatter) ax2.yaxis.set_major_formatter(lat_formatter) ax2.set_extent([57,101,-10,45]) ax2.add_feature(cf.BORDERS,edgecolor='black',alpha=0.8) plt.title('Moisture flux divergence at 850 hPa ', fontsize=16) plt.show()由此产生的情节如下所示-  我有1000 hPa，925 hPa，850 hPa，700 hPa，600 hPa，500 hPa，400 hPa，300 hPa的值。现在，我想把MFD值从1000 hPa垂直积分到300 hPa压力级。请告诉我如何进行这种整合？(({d(Qu) hPa }+{d(Qv)/dy}) dP.) 谢谢! |
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