Zhaitan - 3 months ago 17x

Python Question

So I've been trying to fit to an exponentially modified gaussian function (if interested, https://en.wikipedia.org/wiki/Exponentially_modified_Gaussian_distribution)

`import numpy as np`

import scipy.optimize as sio

import scipy.special as sps

def exp_gaussian(x, h, u, c, t):

z = 1.0/sqrt(2.0) * (c/t - (x-u)/c) #not important

k1 = k2 = h * c / t * sqrt(pi / 2) #not important

n1 = 1/2 * (c / t)**2 - (x-u)/t #not important

n2 = -1 / 2 * ((x - u) / c)**2 #not important

y = np.zeros(len(x))

y += (k1 * np.exp(n1) * sps.erfc(z)) * (z < 0)

y += (k2 * np.exp(n2) * sps.erfcx(z)) * (z >= 0)

return y

In order to prevent overflow problems, one of two equivilent functions must be used depending on whether z is positive or negative (see Alternative forms for computation from previous wikipedia page).

`y += (k2 * np.exp(n2) * sps.erfcx(z)) * (z >= 0)`

`sps.erfcx(-30)`

`inf`

`x = np.linspace(400, 1000, 1001)`

y = exp_gaussian(x, 100, 400, 10, 5)

y

array([ 84.27384586, 86.04516723, 87.57518493, ..., nan,

nan, nan])

I tried the replacing the line in question with the following:

`y += numpy.nan_to_num((k2 * np.exp(n2) * sps.erfcx(z)) * (z >= 0))`

But doing this ran into serious runtime issues. Is there a way to only evaluate

`(k2 * np.exp(n2) * sps.erfcx(z))`

`(z >= 0)`

Thanks!

EDIT: After Rishi's advice, the following code seems to work much better:

`def exp_gaussian(x, h, u, c, t):`

z = 1.0/sqrt(2.0) * (c/t - (x-u)/c)

k1 = k2 = h * c / t * sqrt(pi / 2)

n1 = 1/2 * (c / t)**2 - (x-u)/t

n2 = -1 / 2 * ((x - u) / c)**2

return = np.where(z >= 0, k2 * np.exp(n2) * sps.erfcx(z), k1 * np.exp(n1) * sps.erfc(z))

Answer

How about using numpy.where with something like: `np.where(z >= 0, sps.erfcx(z), sps.erfc(z))`

. I'm no numpy expert, so don't know if it's efficient. Looks elegant at least!

Source (Stackoverflow)

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