Converting Python to Matlab
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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 Commenti
Walter Roberson
il 24 Ott 2017
We do not have the data or NumPanels to test with.
Zach Dunagan
il 24 Ott 2017
Modificato: Zach Dunagan
il 25 Ott 2017
Risposta accettata
Più risposte (2)
Zach Dunagan
il 25 Ott 2017
Modificato: Walter Roberson
il 25 Ott 2017
15 Commenti
Walter Roberson
il 25 Ott 2017
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)
Zach Dunagan
il 26 Ott 2017
Walter Roberson
il 26 Ott 2017
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;
Zach Dunagan
il 26 Ott 2017
Walter Roberson
il 26 Ott 2017
M is the array whose diagonal is to be set. For example,
nSource(1:size(nSource,1)+1:end) = NewNumber;
if nSource is square.
Zach Dunagan
il 26 Ott 2017
Modificato: Walter Roberson
il 26 Ott 2017
Zach Dunagan
il 26 Ott 2017
Modificato: Walter Roberson
il 26 Ott 2017
Walter Roberson
il 26 Ott 2017
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);
Zach Dunagan
il 26 Ott 2017
Zach Dunagan
il 26 Ott 2017
Andrei Bobrov
il 26 Ott 2017
A(end, 1:end-1) = tSource(1, :) + tSource(end, :);
Zach Dunagan
il 27 Ott 2017
Zach Dunagan
il 28 Ott 2017
Modificato: Walter Roberson
il 28 Ott 2017
Zach Dunagan
il 28 Ott 2017
Modificato: Walter Roberson
il 28 Ott 2017
Zach Dunagan
il 28 Ott 2017
ASHOK KUMAR MEENA
il 18 Apr 2022
0 voti
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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