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**Added:** | ||
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* Functionalities to estimate sample mass density or capillary diameter from muD and related chemical info. | ||
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**Changed:** | ||
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* <news item> | ||
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**Deprecated:** | ||
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* <news item> | ||
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**Removed:** | ||
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* <news item> | ||
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**Fixed:** | ||
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* <news item> | ||
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**Security:** | ||
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* <news item> |
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import numpy as np | ||
from scipy.optimize import newton | ||
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from diffpy.utils.tools import compute_mu_using_xraydb | ||
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def estimate_mass_density(mud, diameter, sample_composition, energy): | ||
"""Estimate sample mass density (g/cm^3) from mu*D | ||
using capillary diameter, sample composition, and energy. | ||
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Parameters | ||
---------- | ||
mud : float | ||
The given product of attenuation coefficient mu | ||
in mm^{-1} and capillary diameter in mm. | ||
diameter : float | ||
The given diameter of the sample capillary in mm. | ||
sample_composition : str | ||
The chemical formula of the material. | ||
energy : float | ||
The energy of the incident x-rays in keV. | ||
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Returns | ||
------- | ||
estimated_density : float | ||
The estimated mass density of the packed powder/sample in g/cm^3. | ||
""" | ||
mu = mud / diameter | ||
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def residual_density(density): | ||
return np.abs( | ||
compute_mu_using_xraydb( | ||
sample_composition, energy, sample_mass_density=density | ||
) | ||
- mu | ||
) | ||
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estimated_density = newton(residual_density, x0=1.0, x1=10.0) | ||
return estimated_density | ||
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def estimate_diameter( | ||
mud, | ||
sample_composition, | ||
energy, | ||
sample_mass_density=None, | ||
packing_fraction=None, | ||
): | ||
"""Estimate capillary diameter (mm) from mu*D and mu. | ||
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Parameters | ||
---------- | ||
mud : float | ||
The given product of attenuation coefficient mu | ||
in mm^{-1} and capillary diameter in mm. | ||
sample_composition : str | ||
The chemical formula of the material. | ||
energy : float | ||
The energy of the incident x-rays in keV. | ||
sample_mass_density : float | ||
The mass density of the packed powder/sample in g/cm^3. | ||
packing_fraction : float, optional, Default is None | ||
The fraction of sample in the capillary (between 0 and 1). | ||
Specify either sample_mass_density or packing_fraction but not both. | ||
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Returns | ||
------- | ||
diameter : float | ||
The given diameter of the sample capillary in mm. | ||
""" | ||
mu = compute_mu_using_xraydb( | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. comment out and replace with pass. Let's just work on tests to begin with, until we learn what we want the function to do. |
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sample_composition, energy, sample_mass_density, packing_fraction | ||
) | ||
return mud / mu | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. isn't mud/ mu = d? |
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import pytest | ||
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from diffpy.labpdfproc.getmud import estimate_diameter, estimate_mass_density | ||
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@pytest.mark.parametrize( | ||
"inputs, expected_mass_density", | ||
[ | ||
( | ||
{ | ||
"mud": 2.0, | ||
"diameter": 1.5, | ||
"sample_composition": "ZrO2", | ||
"energy": 20, | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. it is probably a better test if we use reasonable energies. These would be Cu Kalpha, Mo Kalpha, Ag Kalpha. They don't have to be exact but ballpark, |
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}, | ||
0.25, | ||
), | ||
], | ||
) | ||
def test_estimate_mass_density(inputs, expected_mass_density): | ||
actual_mass_density = estimate_mass_density( | ||
inputs["mud"], | ||
inputs["diameter"], | ||
inputs["sample_composition"], | ||
inputs["energy"], | ||
) | ||
assert actual_mass_density == pytest.approx( | ||
expected_mass_density, rel=0.01, abs=0.1 | ||
) | ||
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@pytest.mark.parametrize( | ||
"inputs, expected_diameter", | ||
[ | ||
( # C1: user specifies a sample mass density | ||
{ | ||
"mud": 2.0, | ||
"sample_composition": "ZrO2", | ||
"energy": 20, | ||
"sample_mass_density": 0.25, | ||
}, | ||
1.5, | ||
), | ||
# ( # C2: user specifies a packing fraction | ||
# { | ||
# "mud": 2.0, | ||
# "sample_composition": "ZrO2", | ||
# "energy": 20, | ||
# "packing_fraction": 0.3 | ||
# }, | ||
# 1.5 | ||
# ), | ||
], | ||
) | ||
def test_estimate_diameter(inputs, expected_diameter): | ||
actual_diameter = estimate_diameter(**inputs) | ||
assert actual_diameter == pytest.approx( | ||
expected_diameter, rel=0.01, abs=0.1 | ||
) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
These are not needed presumably. If I know mud and I know mu then nothing else matters.