How would I write the following MatLab code in Python?

Sorry for the extremely newb question. I am currently using the Matlab trial to rewrite some Matlab code into Python. I've gotten started with one document and have made good progress. However this line doesn't make sense to me.

f_spec(:,n) = Xs(parameter.coefficientRange(1):parameter.coefficientRange(end)))

This is my best guess in Python:

calc = Xs[self.coefficientRange[0]:int(self.coefficientRange[-1])]
self.f_spec = self.f_spec[:, calc]

But I don't know if that is right. It certainly isn't working anyway!

Please let me know if my understanding is wrong. It seems to me that:

f_spec(:,n)

This is trying to index the vector named 'f_spec' and it's trying to index from the start to the value of 'n'. And the value of 'n' is being resolved from everything after the '=' sign:

Xs(parameter.coefficientRange(1):parameter.coefficientRange(end)))

Do I have that right?

I would really appreciate the help!

[Balakrishnan Rajan]

I dont know what this code is a part of, but, why did you typecast the variable 'self.coefficientRange[-1]' to int? If the MATLAB line works it means that the MATLAB counterpart is already a positive integer. If not, then probably some other part of the code is not working as it is meant to be.

[tri stan]

Firstly, if you are to rewrite Matlab code to python you may find this page helpful

The line

parameter.coefficientRange(1):parameter.coefficientRange(end))

generates vector of consecutive numbers from parameter.coefficientRange(1) to parameter.coefficientRange(end). I assume both these values are integer.

Then

Xs(parameter.coefficientRange(1):parameter.coefficientRange(end)))

takes from array Xs consecutive elements from index parameter.coefficientRange(1) to index

parameter.coefficientRange(end)), so you may think of this line as a vector of length the same as the length of vector

parameter.coefficientRange(1):parameter.coefficientRange(end) 

Then, I assume that

f_spec(:,n) 

refers to n-th column of two-dimensional array f_spec. You need to make sure that

number of rows of f_spec is the same as length of

Xs(parameter.coefficientRange(1):parameter.coefficientRange(end)))

Anyway, here's the Python code that more or less does what you need (at least I hope so):

import numpy as np


beginingIndex = 3
endIndex      = 7 

parameter_coefficientRange = np.array([ beginingIndex, endIndex])
# given array
Xs = np.array( [4, 5, 21,  13, 0 , 4,  -1, 9 , 13, ] )

n = 2   # number of column you need to fill
temp = Xs[parameter_coefficientRange[0]:parameter_coefficientRange[-1] ]
rowsNo = len(temp)
colsNo = 10 # or whatever number you need

f_spec = np.zeros([rowsNo, colsNo])

f_spec[:,n] = temp

Of course, the code above can be easily shortened, but I'm leaving this part for you.

Helpful tip for those who work simultaneously in Matlab and Python: Always remember that arrays in Matlab are indexed from 1, while in Python - from 0 ;)

[Mootrax]

Thanks guys, this is actually what I ended up with:

fftRange = Xs[int(self.coefficientRange[0]):int(self.coefficientRange[-1])]
self.f_spec[:, i] = fftRange[:, 0]

It was simpler than I thought. My 'i' loop iterator is what 'n' was in the original code.

I solved this and got pretty much the entire class ported over now, and the outputs match - well.... as much as complex floating point numbers can match across the two different programs! Seems like there is a precision issue between MATLAB and Python; or by default Numpy offers more precision.

I need to fix that int-typecasting in the original coefficientRange variable, so that I don't have to do it every single time I use it. Simple really.

Thanks for the links!