Do you see an error?

Do any of you see a problem with the Matlab code? The output from Python and Matlab are the same except 'wps'.

The outputs for wps is attached as an image. When I was testing the primary script, I thought maybe the function files are outputting different numbers which is seen in the image. I made up several inputs in terms of arrays for both Python and Matlab.

xc = 1:5;
yc = 1:5;
xp = 1:5;
yp = 1:5;
theta = 1:5;
ups = zeros(length(xc), length(xp) - 1);
wps = zeros(length(xc), length(xp) - 1);

for i = 1:length(xc)
    for j = 1:(length(xp) - 1)
        r1 = ((xc(i) - xp(j)).^2 + (yc(i) - yp(j)).^2).^(1/2);
        r2 = ((xc(i) - xp(j+1)).^2 + (yc(i) - yp(j+1)).^2).^(1/2);
        nu2 = atan2(-(xc(i) - xp(j)) * sin(theta(j)) + (yc(i) - yp(j)) * cos(theta(j))+(xp(j+1) - xp(j)) * sin(theta(j))-(yp(j+1) - yp(j)) * cos(theta(j)),...
                    (xc(i) - xp(j)) * cos(theta(j)) + (yc(i) - yp(j)) * sin(theta(j)) - (xp(j + 1) - xp(j)) * cos(theta(j)) - (yp(j + 1) - yp(j)) * sin(theta(j)));
        nu1 = atan2(-(xc(i) - xp(j)) * sin(theta(j)) + (yc(i) - yp(j)) * cos(theta(j)), (xc(i) - xp(j)) * cos(theta(j)) + (yc(i) - yp(j)) * sin(theta(j)));
        %
        ups(i, j) = 1 / (4 * pi) * log(r1.^2 / r2.^2);
        wps(i, j) = 1 / (2 * pi) * (nu2 - nu1);
    end
end

Here is the Python

    import numpy as np

    xc = np.array([1,2,3,4,5])
    yc = np.array([1,2,3,4,5])
    xp = np.array([1,2,3,4,5])
    yp = np.array([1,2,3,4,5])
    theta = np.array([1,2,3,4,5])

    ups=np.zeros((len(xc),len(xp)-1))
    wps=np.zeros((len(xc),len(xp)-1)) 
    for i in range(len(xc)):
        for j in range(len(xp)-1):
            r1=((xc[i]-xp[j])**2+(yc[i]-yp[j])**2)**(1/2)
            r2=((xc[i]-xp[j+1])**2+(yc[i]-yp[j+1])**2)**(1/2)
            nu2=np.arctan2(-(xc[i]-xp[j])*np.sin(theta[j])+(yc[i]-yp[j])*np.cos(theta[j])+(xp[j+1]-xp[j])*np.sin(theta[j])-(yp[j+1]-yp[j])*np.cos(theta[j]),\
                                 (xc[i]-xp[j])*np.cos(theta[j])+(yc[i]-yp[j])*np.sin(theta[j])-(xp[j+1]-xp[j])*np.cos(theta[j])-(yp[j+1]-yp[j])*np.sin(theta[j]))
            nu1=np.arctan2(-(xc[i]-xp[j])*np.sin(theta[j])+(yc[i]-yp[j])*np.cos(theta[j]),(xc[i]-xp[j])*np.cos(theta[j])+(yc[i]-yp[j])*np.sin(theta[j]))
            ups[i,j]=1/(4*np.pi)*np.log(r1**2/r2**2)
            wps[i,j]=1/(2*np.pi)*(nu2-nu1)

[per isakson]

"The output from Python and Matlab are the same except 'wps'." which variables are output?

It's only the value of the Matlab wps(4,4), which differs significantly from the Python value.

Proposal: Set i=4; and j=4; and remove the for-loops. Compare the values of all variables

[Zach Dunagan]

ups and wps are 5x4 matrices. I call these outputs because xc, yc, xp, yp, etc. are the inputs. If my terminology isn't correct, please correct me. I will try setting i = 4 and j = 4.

EDIT: So this is the output for ups and wps when letting i = 4 and j = 4.

[per isakson]

Split the large expressions into smaller and introduce intermediate variables. Compare the values of the intermediate variables. Do this until you tracked down the source of the difference. I don't know of any better way.

[Zach Dunagan]

Okay, thank you Per isakson for your assistance.

[David Goodmanson]

Hi Zach, the latest jpg for python wps(4,4) appears to conflict with the original jpg for python wps(4,4) which was .48838.

[Stephen23]

@Zach Dunagan: Follow per isakson's advice, because that is exactly what we would do for you anyway. There is no "magic tool" that experienced users have to investigate these kind of problems, it simply just comes down to really looking at what the code is doing: break the code down and check each operation and each intermediate value. It is tedious and boring, but you will get there.

[David Goodmanson]

Hi Zach, if the original value of .48838 is correct for python wps(4,4), then for that matrix element the difference between matlab and python = (1/2), to five decimal places. That means that for the quantity (nu2 - nu1), the difference between matlab and python = pi. Since atans are involved to get the nu's, this seems more than coincidental.

[Zach Dunagan]

@Stephen Cobeldick. I was braking the code down as Per isakson suggested. Maybe my reply didn't look like I was doing this. Thanks.

[Christoph F.]

What are the commas doing in the nu2 and nu1 expressions? I would assume that Python and Matlab have slightly different ideas of the function of a comma operator.