Moisture flux divergence using numpy and metpy differ

Question

As described in this question I want to calculate moisture flux divergence at 850 hPa. For this, I have used the code described here , which makes use of the np.gradient function from the numpy package. The code I am using is this:

from matplotlib import pyplot
from matplotlib.cm import get_cmap
from __future__ import print_function
from netCDF4 import Dataset,num2date,date2num
from matplotlib.colors import from_levels_and_colors
from cartopy import crs
from cartopy.feature import NaturalEarthFeature, COLORS
#
import metpy.calc as mpcalc
import xarray as xr
import cartopy.crs as ccrs
import matplotlib
import cartopy.feature as cfe
import numpy as np
import matplotlib.pyplot as plt
import datetime
#
##################################################################################################################
########################################## Calculate Mosture Divergence ########################################## 
##################################################################################################################
#
root_dir = '/users/pr007/mkaryp/'
nc = Dataset(root_dir+'ERA5.nc')
#
v = np.array(nc.variables['v'][0,:,:])                    
u = np.array(nc.variables['u'][0,:,:])
q = np.array(nc.variables['q'][0,:,:])        
#
lon=nc.variables['longitude'][:]
lat=nc.variables['latitude'][:]
#
qu = q*u
qv = q*v  
#
[dqu_dx, dqu_dy] = np.gradient(qu)
[dqv_dx, dqv_dy] = np.gradient(qv)
#
uq = [dqu_dx, dqu_dy]
vq = [dqv_dx, dqv_dy]
#
divg = np.sum([uq, vq], axis=0)
HMD = divg
# 
##################################################################################################################
#################################################### Plot Map #################################################### 
##################################################################################################################
#
min_val = HMD[0].min()
max_val = HMD[0].max()
diff = max_val-min_val
step = diff/14
#
# Set the figure size, projection, and extent
fig = plt.figure(figsize=(4.5,4))
ax = plt.axes(projection=ccrs.Robinson())
#ax.set_global()
ax.coastlines(resolution="110m",linewidth=1)
ax.gridlines(linestyle='--',color='black')
#
# Set contour levels, then draw the plot and a colorbar
clevs = np.arange(min_val,max_val,step)
plt.contourf(lon, lat, HMD[0], clevs, transform=ccrs.PlateCarree(), cmap=get_cmap("seismic"), extend="both")
plt.title('HMD from ERA5 using numpy 0', size=14)
cb = plt.colorbar(ax=ax, orientation="vertical", pad=0.02, aspect=16, shrink=0.8)
cb.set_label('kg kg-1 ms-1',size=12,rotation=270,labelpad=15)
cb.ax.tick_params(labelsize=10)
#
# Save the plot as a PNG image
#
fig.savefig('HMD_ERA5_numpy0.png', format='png', dpi=300)

The plot produced is shown here .

The HMD array produced has the following dimensions:

np.shape(HMD)
(2, 125, 145)

125 and 145 refer to lats and lons and 2 is the number of variables in HMD as produced by the np.gradient and np.sum functions. The plot displayed above uses HMD[0]. When instead HMD(1) is plotted, the output map is this .

When instead the metpy.divergence () function is used, the code looks like that:

#
root_dir = '/users/pr007/mkaryp/'
nc = Dataset(root_dir+'ERA5.nc')
#
v = np.array(nc.variables['v'][0,:,:])                    
u = np.array(nc.variables['u'][0,:,:])
q = np.array(nc.variables['q'][0,:,:])        
#
lon=nc.variables['longitude'][:]
lat=nc.variables['latitude'][:]
#
qu = q*u
qv = q*v  
#
# Compute dx and dy spacing for use in divergence calculation
dx, dy = mpcalc.lat_lon_grid_deltas(lon, lat)
# 
HMD = (np.array(mpcalc.divergence(qu, qv, dx=dx, dy=dy)))
#

When the metpy.divergence() function is used the output map looks like this .

So my question is why the 2 different ways (using numpy and metpy) differ so much, or am I terribly wrong in the way I am trying to do the calculations?

The file I use is this .

EDIT: After a suggestion from the comments section the code was changed to:

#
qu = q*u
qv = q*v  
#
dqu_dx = np.gradient(qu)[0]  # take d/dx of qu 
dqv_dy = np.gradient(qv)[1]  # take d/dy of qv
#
divg = np.sum([dqu_dx, dqv_dy], axis=0)
HMD = divg
# 

The resulting map is this .

Answer 1

If I understood your code correctly I think the error in the former is that you are including the terms d(qu)/dy and d(qv)/dx in your calculation of flux divergence, which I think is incorrect. Basically the moisture flux divergence is only d(qu)/dx + d(qv)/dy, so I think you need to change the lines in the code to pick out the x derivative of qu gradient command and the y derivative of qv. In fact an alternative could be to use np.diff instead of gradient, but try the following?

dqu_dx = np.gradient(qu)[0]  # take d/dx of qu 
dqv_dy = np.gradient(qv)[1]  # take d/dy of qv

divg = np.add(dqu_dx,dqv_dy)

Hopefully that should reproduce the metpy result...