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more consistency in naming, better docstring

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Matt Terry
2013-08-17 11:06:00 -07:00
parent 0182f7c98a
commit 9fe1e26583
1 changed files with 27 additions and 20 deletions
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skimage/color/delta_e.py
+27 -20
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@@ -108,7 +108,7 @@ def deltaE_ciede94(lab1, lab2, kH=1, kC=1, kL=1, k1=0.045, k2=0.015):
dL = L1 - L2
dC = C1 - C2
dH = get_dH(lab1, lab2)
dH2 = get_dH2(lab1, lab2)
SL = 1
SC = 1 + k1 * C1
@@ -116,7 +116,7 @@ def deltaE_ciede94(lab1, lab2, kH=1, kC=1, kL=1, k1=0.045, k2=0.015):
dE2 = (dL / (kL * SL)) ** 2
dE2 += (dC / (kC * SC)) ** 2
dE2 += (dH / (kH * SH)) ** 2
dE2 += dH2 / (kH * SH) ** 2
return np.sqrt(dE2)
@@ -289,7 +289,7 @@ def deltaE_cmc(lab1, lab2, kL=1, kC=1):
dC = C1 - C2
dL = L1 - L2
dH = get_dH(lab1, lab2)
dH2 = get_dH2(lab1, lab2)
T = np.where(np.logical_and(np.rad2deg(h1) >= 164, np.rad2deg(h1) <= 345),
0.56 + 0.2 * np.abs(np.cos(h1 + np.deg2rad(168))),
@@ -304,29 +304,36 @@ def deltaE_cmc(lab1, lab2, kL=1, kC=1):
dE2 = (dL / (kL * SL)) ** 2
dE2 += (dC / (kC * SC)) ** 2
dE2 += (dH / SH) ** 2
dE2 += dH2 / (SH ** 2)
return np.sqrt(dE2)
def get_dH(lab1, lab2):
"""numerically well behaved calculation of dH
def get_dH2(lab1, lab2):
"""squared hue difference term occurring in deltaE_cmc and deltaE_ciede94
Given a1, b1, a2, b2
c1 = sqrt(a1**2 + b1**2)
c2 = sqrt(a2**2 + b2**2)
term = (a1-a2)**2 + (b1-b2)**2 - (c1-c2)**2
dH = sqrt(term)
Despite its name "dH" is not a simple difference of hue values. We avoid
working directly with the hue value directly since differencing angles is
troublesome. The hue term is usually written as:
c1 = sqrt(a1**2 + b1**2)
c2 = sqrt(a2**2 + b2**2)
term = (a1-a2)**2 + (b1-b2)**2 - (c1-c2)**2
dH = sqrt(term)
However, this has poor roundoff properties when a or b is dominant.
Instead, r is a vector with elements a and b. The same dH term can be
re-written as:
|r1-r2|**2 - (|r1| - |r2|)**2
and then simplified to:
2*|r1|*|r2| - 2*dot(r1, r2)
"""
lab1 = np.asarray(lab1)
lab2 = np.asarray(lab2)
a1, b1 = np.rollaxis(lab1, -1)[1:3]
a2, b2 = np.rollaxis(lab2, -1)[1:3]
r1 = np.rollaxis(lab1, -1)[1:3]
r2 = np.rollaxis(lab2, -1)[1:3]
# magnitude of (a, b) is the chroma
C1 = np.hypot(a1, b1)
C2 = np.hypot(a2, b2)
C1 = np.hypot(r1[0], r1[1])
C2 = np.hypot(r2[0], r2[1])
term = C1 * C2
term -= (r1 * r2).sum(0)
term *= 2
return np.sqrt(term)
term = (C1 * C2) - (a1 * a2 + b1 * b2)
return 2*term