How may I convert Matlab code into Python?

Question

I tried to convert Matlab code to Python but in this xx = diff(xx>setpoint); situation I did not get the same result in python.

fs = 50;
t = [0:1/fs:1];
xx = sin(2*pi*300*t)+2*sin(2*pi*600*t);
xx = xx(:)'/max(abs(xx)); %-- normalize xx
Lx = length(xx);
Lz = round(0.01*fs);
setpoint = 0.02; 
%xx = filter( ones(1,Lz)/Lz, 1, abs(xx) );
xx = diff(xx>setpoint);

Actually I don't understand what the statement xx = diff(xx>setpoint) do.

• Daniel: You are stating that your python code behaves different, please add your python code to your question.

Answer 1

TL;DR Python is different, you can make Python like Matlab, in this case I prefer the approach of Python.

---

In Python xx>0.002 is an array of Booleans, False and True and np.diff treats the Boolean values as such, in Matlab xx>0.002 is a matrix of logical values as well but diff converts them to 0 and 1 before taking the differences and this imply that we have more possibilities in Matlab

In [15]: for a, b in ((0,0), (0,1), (1,0), (1,1)): print(np.diff((a,b))) 
[0]
[1]
[-1]
[0]
In [16]: f, t = False, True 
 ...: for a, b in ((f,f), (f,t), (t,f), (t,t)): print(np.diff((a,b))) 
[False]
[ True]
[ True]
[False]

When I plot xx and diff(xx>0.02) in Matlab (oh well, in Octave...) I have

enter image description here

When I plot xx and np.diff(xx>0.02) in Python+Numpy+Matplotlib I get

enter image description here

To have exactly the results of Matlab we can convert the boolean array to an array of floats, just multiplying by 1.0 is OK — so this is the plot of xx and np.diff( 1.0*(xx>0.02) )

enter image description here

If the aim of the OP is to show where the signal is larger than 0.02 I dare say that the native Python (no conversion to floats) is better at that...

• Daniel: Your explanation about MATLAB is slightly off, xx>0.002 is of data type logical, but the result of false-true is a double of value -1.
• gboffi: @Daniel I'll correct the answer, thank you for the correction.