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Prevent frontogenesis from returning nans #3696

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Nov 21, 2024
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Prevent frontogenesis from returning nans #3696

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12f4285
Prevent frontogenesis from returning nans
sgdecker Nov 19, 2024
7a820ef
Avoid nans more accurately
sgdecker Nov 21, 2024
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15 changes: 12 additions & 3 deletions src/metpy/calc/kinematics.py
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Original file line number Diff line number Diff line change
Expand Up @@ -567,9 +567,18 @@ def frontogenesis(potential_temperature, u, v, dx=None, dy=None, x_dim=-1, y_dim

# Compute the angle (beta) between the wind field and the gradient of potential temperature
psi = 0.5 * np.arctan2(shrd, strd)
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If you're interested in eliminating all trig:

We have $\tan(2 \psi) =\frac{shrd}{strd}$, which implies $\cos(2\psi)=\frac{strd}{\sqrt{shrd^2+strd^2}}$. We use psi next line to find $\cos(\psi)$ and $\sin(\psi)$.

We show later that $\cos(2\psi) = 1 - 2 \sin^2(\psi) = 2 \cos^2(\psi) - 1$. Rearranging those identities gives
$\cos(\psi) = \sqrt{\frac{1 + \cos(2 \psi)}{2} }$
$\sin(\psi) = \sqrt{\frac{1 - \cos(2 \psi)}{2} } = \sqrt{1-\cos^2(\psi)}$
This lets you replace three trig functions and a multiplication with three or four multiplications, three additions, three square roots, and a division, and four trips through the data with eight. I have no idea whether the eliminating the trig calls makes up for the extra trips through the data.

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I ran through that too and wasn't wild about all the square roots. I'm not passionate about trying it but if someone wants to see about performance/accuracy, I'd entertain it.

beta = np.arcsin((-ddx_theta * np.cos(psi) - ddy_theta * np.sin(psi)) / mag_theta)

return 0.5 * mag_theta * (tdef * np.cos(2 * beta) - div)
# We need to be careful to avoid division by zero. When mag_theta
# is zero, the frontogenesis will also be zero. The minus signs
# are omitted from the numerator since this expression is squared
# to compute the frontogenesis.
sin_beta = np.divide(ddx_theta * np.cos(psi) + ddy_theta * np.sin(psi), mag_theta,
out=np.zeros_like(mag_theta), where=mag_theta != 0)

# The textbook definition of frontogenesis includes the term
# cos(2*beta). However, using trig identities, one can show that
# cos(2*beta) = 1 - 2 * sin(beta)**2, and the expression involving
# sin(beta) is more numerically stable.
return 0.5 * mag_theta * (tdef * (1 - 2 * sin_beta**2) - div)


@exporter.export
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53 changes: 53 additions & 0 deletions tests/calc/test_kinematics.py
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Expand Up @@ -307,6 +307,59 @@ def test_frontogenesis_asym():
assert_almost_equal(fronto, true_fronto)


def test_frontogenesis_nan():
"""Test that frontogenesis calculation does not result in nan."""
x = np.array([-4142340.8, -4061069.8, -3979798.8, -3898527.8, -3817256.8],
dtype=np.float32)
y = np.array([-832207.56, -750936.56, -669665.56, -588394.5, -507123.56],
dtype=np.float32)
lat = np.array([[12.38805122, 12.58268367, 12.77387305, 12.96159447, 13.14582354],
[13.07545967, 13.27159197, 13.46425116, 13.65341235, 13.83905111],
[13.76423766, 13.96186003, 14.15597929, 14.34657051, 14.53360928],
[14.45429168, 14.65339375, 14.84896275, 15.04097373, 15.22940228],
[15.14552468, 15.3460955, 15.54310332, 15.73652321, 15.92633074]])
lon = np.array([[-132.75696788, -132.05286812, -131.34671228, -130.63852983,
-129.92835084],
[-132.9590417, -132.25156385, -131.54199837, -130.83037505,
-130.11672431],
[-133.16323241, -132.45234731, -131.73934239, -131.02424779,
-130.30709426],
[-133.36957268, -132.65525085, -131.93877637, -131.22017972,
-130.49949199],
[-133.57809517, -132.86030681, -132.14033233, -131.41820252,
-130.69394884]])
uvals = np.array([[0.165024, 0.055023, -0.454977, -1.534977, -2.744976],
[0.155024, -0.434977, -2.114977, -3.474977, -4.034977],
[-1.554977, -2.714977, -2.084976, -5.274977, -3.334976],
[-3.424976, -7.644977, -7.654977, -5.384976, -3.224977],
[-9.564977, -9.934977, -7.454977, -6.004977, -4.144977]]) * units('m/s')
vvals = (np.array([[2.6594982, 1.9194984, 2.979498, 2.149498, 2.6394978],
[3.4994984, 4.0794983, 4.8494987, 5.2594986, 3.1694984],
[5.159498, 6.4594975, 6.559498, 5.9694977, 3.189499],
[6.5994987, 9.799498, 7.4594975, 4.2894993, 3.729498],
[11.3394985, 6.779499, 4.0994987, 4.819498, 4.9994984]])
* units('m/s'))
tvals = np.array([[290.2, 290.1, 290.2, 290.30002, 290.30002],
[290.5, 290.30002, 290.30002, 290.30002, 290.2],
[290.80002, 290.40002, 290.2, 290.40002, 289.90002],
[290.7, 290.90002, 290.7, 290.1, 289.7],
[290.90002, 290.40002, 289.7, 289.7, 289.30002]]) * units('degK')

x = xr.DataArray(data=x, coords={'x': x}, dims='x', attrs={'units': 'meters'})
y = xr.DataArray(data=y, coords={'y': y}, dims='y', attrs={'units': 'meters'})

dims = ['y', 'x']
coords = {'x': x, 'y': y, 'latitude': (dims, lat), 'longitude': (dims, lon)}

u = xr.DataArray(data=uvals, coords=coords, dims=dims)
v = xr.DataArray(data=vvals, coords=coords, dims=dims)
t = xr.DataArray(data=tvals, coords=coords, dims=dims)

th = potential_temperature(850 * units('hPa'), t)
f = frontogenesis(th, u, v)
assert not np.any(np.isnan(f))


def test_advection_uniform():
"""Test advection calculation for a uniform 1D field."""
u = np.ones((3,)) * units('m/s')
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