How do I convert this code from matlab to python?

Question

I am trying to convert a calculation in matlab to python. This is code in matlab:

% Solving for the bandpass correction

T = 8050; % temperature in kelvin
c = 2.99e10; % speed of light in cm/s
h = 6.626e-27; % planck constant in ergs
k = 1.38e-16; % boltzmann constant in erg/K
x1 = 3e-4; % lower wavelength in cm
x2 = 13e-4; % upper wavelength in cm

fun = @(x) ((2 * h * c^2) ./ x.^5) ./ (exp((h * c) ./ (x * k * T)) - 1); % planck function
f_deriv = @(x) ((2 * h^2 * c^3) ./ (x.^6 * k)) .* (exp((h .* c) ./ (x* k * T)) ./ (T * (exp((h * c) ./ (x * k * T)) - 1).^2));
numerator = integral(f_deriv,x1,x2);
denominator = (4 * integral(fun,x1,x2));
beta = numerator / denominator;
fprintf('The numerator is %f \n',numerator)
fprintf('The denominator is %f \n',denominator)
fprintf('The bandpass value is %f \n',beta)

Here is my attempt to code it in python:

# Solving for the bandpass correction:

from scipy.integrate import quad
import numpy as np

T = 8050  # temperature in kelvin
c = 2.99e10  # speed of light in cm/s
h = 6.626e-27  # planck constant in ergs
k = 1.38e-16  # boltzmann constant in erg/K
x1 = 3e-4  # lower wavelength in cm
x2 = 13e-4  # upper wavelength in cm


def p_function(x):
    return ((2 * h * c ** 2) / x ** 5) / (np.exp((h * c)/(x * k * T)) - 1)  # The Planck function


def p_deriv(x):
    return ((2 * h ** 2 * c ** 3) / (x ** 6 * k)) * (np.exp((h * c)/(x * k * T))/(T * (np.exp((h * c) / (x * k * T)) - 1) ** 2))


numerator = quad(p_deriv, x1, x2)
denominator = 4 * quad(p_function, x1, x2)
beta = numerator[0] / denominator[0]
print("The numerator is", numerator[0])
print("The denominator is", denominator[0])
print("The bandpass value is", beta)

The beta value is not consistent between the two and I cannot see what is causing the difference. I would appreciate any help in reconciling this. Thanks!

I don't know matlab, but could those dots . before some of the operators (like the ./, the .*, and the .^) have anything to do with it? What do those dots mean? — Sylvester is on codidact.com
– Sylvester is on codidact.com, Commented Mar 7, 2022 at 23:59
This should not be the cause of your problem but... Your denominator should probably be set to 4 * quad(p_function, x1, x2)[0]. Because multiplying a tuple by 4 results in a tuple of length 4 (in python 3 at least). — dekuShrub
– dekuShrub, Commented Mar 8, 2022 at 0:04
@SylvesterKruin those are element-wise operations if I remember correctly. — dekuShrub
– dekuShrub, Commented Mar 8, 2022 at 0:05
@dekuShrub I made the change you suggested and it worked. The values are consistent now. Thank you for your help! — AppleBetaPi
– AppleBetaPi, Commented Mar 8, 2022 at 0:16

hpaulj · Accepted Answer · 2022-03-08 02:03:02Z

Often when translating MATLAB it's important to get shapes/sizes correct. But when I run your code in Octave I see all variables are (1,1), "scalar". So dimensions shouldn't be an issue.

Let's check function values:

>> fun(1)
ans =  0.066426
>> f_deriv(1)
ans =  0.066432

Your numpy values look the same:

In [3]: p_function(1)
Out[3]: 0.06642589646577152
In [4]: p_deriv(1)
Out[4]: 0.06643181982384715

And integration results:

>> numerator
numerator =  795765635.47589

In [7]: numerator = quad(p_deriv, x1, x2)
In [8]: numerator
Out[8]: (795765635.4758892, 0.030880182071769013)

READING THE DOCS we see that quad returns (as a default) 2 values, the integral and an error estimate. The first of the tuple matches the MATLAB. You do use numerator[0].

Same for the denominator - taking care to extract the value from the tuple before multiplication:

In [10]: denominator = 4 * quad(p_function, x1, x2)[0]
In [11]: denominator
Out[11]: 2568704689.659281
In [12]: numerator[0] / denominator
Out[12]: 0.309792573151506

>> beta
beta =  0.30979

Without the correction, the result is 4x too large

In [13]: denominator = 4 * quad(p_function, x1, x2)
In [14]: denominator
Out[14]: 
(642176172.4148202,
 0.006319781838840299,
 642176172.4148202,
 0.006319781838840299,
 642176172.4148202,
 0.006319781838840299,
 642176172.4148202,
 0.006319781838840299)
In [15]: numerator[0] / denominator[0]
Out[15]: 1.239170292606024

Sometimes (or even often) we need to test values step by step. Even when developing numpy code from scratch, I test all steps.

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