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ENH use log1p and expm1 in Yeo-Johnson transformation and its inverse #27868

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ENH use log1p and expm1 in Yeo-Johnson transformation and its inverse #27868

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e531ebd
ENH use log1p and expm1 in Yeo-Johnson transformation and its inverse
xuefeng-xu Nov 29, 2023
92bba12
skip computation for lambda=1
xuefeng-xu Nov 29, 2023
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17 changes: 11 additions & 6 deletions sklearn/preprocessing/_data.py
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Original file line number Diff line number Diff line change
Expand Up @@ -3304,27 +3304,32 @@ def _yeo_johnson_inverse_transform(self, x, lmbda):
"""Return inverse-transformed input x following Yeo-Johnson inverse
transform with parameter lambda.
"""
if lmbda == 1:
return x

x_inv = np.zeros_like(x)
pos = x >= 0

# when x >= 0
if abs(lmbda) < np.spacing(1.0):
x_inv[pos] = np.exp(x[pos]) - 1
x_inv[pos] = np.expm1(x[pos])
else: # lmbda != 0
x_inv[pos] = np.power(x[pos] * lmbda + 1, 1 / lmbda) - 1
x_inv[pos] = np.expm1(np.log1p(x[pos] * lmbda) / lmbda)

# when x < 0
if abs(lmbda - 2) > np.spacing(1.0):
x_inv[~pos] = 1 - np.power(-(2 - lmbda) * x[~pos] + 1, 1 / (2 - lmbda))
x_inv[~pos] = -np.expm1(np.log1p((lmbda - 2) * x[~pos]) / (2 - lmbda))
else: # lmbda == 2
x_inv[~pos] = 1 - np.exp(-x[~pos])
x_inv[~pos] = -np.expm1(-x[~pos])

return x_inv

def _yeo_johnson_transform(self, x, lmbda):
"""Return transformed input x following Yeo-Johnson transform with
parameter lambda.
"""
if lmbda == 1:
return x

out = np.zeros_like(x)
pos = x >= 0 # binary mask
Expand All @@ -3333,11 +3338,11 @@ def _yeo_johnson_transform(self, x, lmbda):
if abs(lmbda) < np.spacing(1.0):
out[pos] = np.log1p(x[pos])
else: # lmbda != 0
out[pos] = (np.power(x[pos] + 1, lmbda) - 1) / lmbda
out[pos] = np.expm1(lmbda * np.log1p(x[pos])) / lmbda

# when x < 0
if abs(lmbda - 2) > np.spacing(1.0):
out[~pos] = -(np.power(-x[~pos] + 1, 2 - lmbda) - 1) / (2 - lmbda)
out[~pos] = -np.expm1((2 - lmbda) * np.log1p(-x[~pos])) / (2 - lmbda)
else: # lmbda == 2
out[~pos] = -np.log1p(-x[~pos])

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