Converting Python to Matlab

Spent a few hours on trying to figure out why the outputs are different, no luck. Python and Matlab are in a txt file along with their outputs.
Suggestions on what I should be looking at to resolve the issue?

2 Comments

We do not have the data or NumPanels to test with.
Zach Dunagan
Zach Dunagan on 24 Oct 2017
Edited: Zach Dunagan on 25 Oct 2017
I figured it out.

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 Accepted Answer

Andrei Bobrov
Andrei Bobrov on 24 Oct 2017
Edited: Andrei Bobrov on 24 Oct 2017
Small typo in MATLAB code in nu2
nu2 = atan2(-(xc(k)-xp(n))*sin(theta(k))+(yc(k)-yp(n))*cos(theta(n))+(xp(n+1)-xp(n))*sin(theta(n))-(yp(n+1)-yp(n))*cos(theta(n)),(xc(k)-xp(n))*cos(theta(n))+(yc(k)-yp(n))*sin(theta(n))-(xp(n+1)-xp(n))*...
cos(theta(n))-(yp(n+1)-yp(n))*sin(theta(n))); % fixed!

3 Comments

I figured it out. I don't need the multiplication, '\' in Python means '...' in Matlab. The nu2 = atan2(x1, x2) similar to this line of code in my script. atan2(yp(k+1)-yp(k), xp(k+1)-xp(k)).
Stop Zach!
How?
cos(theta(n)-(yp(n+1)-yp(n))*sin(theta(n)))
or
cos(theta(n))-(yp(n+1)-yp(n))*sin(theta(n))
Oh... Thanks.

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More Answers (2)

Zach Dunagan
Zach Dunagan on 25 Oct 2017
Edited: Walter Roberson on 25 Oct 2017
How would I insert zeros in the diagonal line of these 128 x 128 matrices?
nSource = -(sin(theta-transpose(theta)))*ups + (cos(theta-transpose(theta)))*wps;
ftSource = cos(theta-transpose(theta))*ups + sin(theta-transpose(theta))*wps;
nVortex = sin(theta-transpose(theta))*upv + cos(theta-transpose(theta))*wpv;
tVortex = cos(theta-transpose(theta))*upv + sin(theta-transpose(theta))*wpv;

15 Comments

For the special case of square matrices, often the easiest way is
M - diag(diag(M))
For other matrices,
tril(M,1) + triu(M,1)
What if I wanted to insert a number, other than zero?
Just in case you have unsigned integers or the case where the existing value is so different from the new value that you have to worry about loss of precision:
Square matrix:
M(1:size(M,1)+1:end) = NewNumber;
Non-square matrix:
shorter = min(size(M,1),size(M,2));
lastidx = size(M,1)*(shorter-1)+shorter;
M(1:size(M,1)+1:lastidx) = NewNumber;
M = M = nSource = 128 x 128 ?
Is the M inside of the size() the vector that lays in the diag of nSource?
M is the array whose diagonal is to be set. For example,
nSource(1:size(nSource,1)+1:end) = NewNumber;
if nSource is square.
Actually this seem to have worked for my situation.
ups(logical(eye(size(ups)))) = 0;
wps(logical(eye(size(wps)))) = 0.5;
Can you please help me with the Matlab equivalent?
A=np.zeros((numPanels+1,numPanels+1))
A[:numPanels,:numPanels]=nSource
I just need the equivalent to this code.
A[:numPanels,-1]=np.sum(nVortex,axis=1)
A[:numPanels,:numPanels]=nSource
would be
A(1:end-1, 1:end-1) = nSource;
I am not certain about
A[:numPanels,-1]=np.sum(nVortex,axis=1)
but I suspect
A(1:end-1, end) = sum(nVortex, 2);
It worked! Thank you.
A[-1,:numPanels]=tSource[0,:]+tSource[-1,:]
Here is what I have.
A(end, 1:end) = tSource(1, :) + tSource(end, :);
Error:
Subscripted assignment dimension mismatch.
Error in ssPanelMethod (line 112) A(end, 1:end) = tSource(1, :) + tSource(end, :);
A(end, 1:end-1) = tSource(1, :) + tSource(end, :);
Thank you.
I really want to understand these little operations, rather than constantly asking for help here...
Python code
normVelFoil=np.dot(nSource,x[:-1])+np.dot(nVortex,x[-1]*np.ones((num
Panels,1)))+normU
tangVelFoil=np.dot(tSource,x[:-1])+np.dot(tVortex,x[-1]*np.ones((numPanels,1)))+tangU
Matlab code
normVelFoil = dot(nSource, x(1:end, 1:end-1)) + dot(nVortex, x(1:end, 1:end-1) * ones(numPanels, 1)) + normU;
tangVelFoil = dot(tSource, x(1:end, 1:end-1)) + dot(tVortex, x(1:emd, 1:end-1) * ones(numPanels, 1)) + tangU;
tSource, tVortex are 128 x 128, while x is a 129 x 1.
I changed dot to mtimes() or I could use .*
I compare the first dot to second dot of python and everything looks to be lining up. When I compare the normVelFoil outputs they don't line up.
Here is what I changed..
normVelFoil = mtimes(nSource, x(1:end-1, 1)) + mtimes(nVortex, x(end)*ones(numPanels, 1)) + normU;
I have come to a conclusion the outputs are the same. They are both extremely small values, practically zero.

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ASHOK KUMAR MEENA
ASHOK KUMAR MEENA on 18 Apr 2022
def Lagrange(x, y, n, xx):
sum = 0
for i in range(0, n + 1):
product = y[i]
for j in range(0, n + 1):
if (i != j):
product = product * (xx - x[j]) / (x[i] - x[j])
sum += product
return sum
def Trapezoidal(h, n, f):
sum = f[0]
for i in range (1, n):
sum = sum + 2 * f[i]
sum = sum + f[n]
ans = h * sum / 2
return ans

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