Stumped again... Converting Python to Matlab

2017 10 22
2 个回答
回答已采纳
更新时间：2017 10 24
7 次查看（30 天）

0 个投票

Here is the bit of Python code I am trying to convert to Matlab. code:
# thetas for each panel in global frame
theta=np.zeros((numPanels,1))
for i in range(numPanels):
theta[i]=np.arctan2(yp[i+1]-yp[i],xp[i+1]-xp[i])
output:
-2.63356
-2.92712
-2.98475
-3.01668
-3.03837
-3.05471
-3.06785
-3.07891
-3.08853
-3.09713
-3.10498
-3.11228
-3.11917
-3.12577
-3.13217
-3.13846
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.14159
3.13846
3.13217
3.12577
3.11917
3.11228
3.10498
3.09713
3.08853
3.07891
3.06785
3.05471
3.03837
3.01668
2.98475
2.92712
2.63356
0.508031
0.21447
0.156842
0.124909
0.103226
0.0868853
0.0737421
0.0626828
0.053059
0.0444609
0.0366123
0.0293165
0.0224263
0.0158261
0.00942088
0.00312805
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
-0.00312805
-0.00942088
-0.0158261
-0.0224263
-0.0293165
-0.0366123
-0.0444609
-0.053059
-0.0626828
-0.0737421
-0.0868853
-0.103226
-0.124909
-0.156842
-0.21447
-0.508031
These are my Matlab arrays, where xpAdd = xp[i+1] and xp = xp[i]
xpAdd =
0.0600
0.0581
0.0562
0.0544
0.0525
0.0506
0.0488
0.0469
0.0450
0.0431
0.0412
0.0394
0.0375
0.0356
0.0319
0.0300
0.0281
0.0262
0.0244
0.0225
0.0206
0.0188
0.0169
0.0150
0.0131
0.0112
0.0094
0.0075
0.0056
0.0038
0.0019
0
-0.0019
-0.0038
-0.0056
-0.0075
-0.0094
-0.0112
-0.0131
-0.0150
-0.0169
-0.0188
-0.0206
-0.0225
-0.0244
-0.0262
-0.0281
-0.0300
-0.0319
-0.0338
-0.0356
-0.0375
-0.0394
-0.0412
-0.0431
-0.0450
-0.0469
-0.0488
-0.0506
-0.0525
-0.0544
-0.0562
-0.0581
-0.0600
-0.0600
-0.0581
-0.0562
-0.0544
-0.0525
-0.0506
-0.0488
-0.0469
-0.0450
-0.0431
-0.0412
-0.0394
-0.0375
-0.0356
-0.0319
-0.0300
-0.0281
-0.0262
-0.0244
-0.0225
-0.0206
-0.0188
-0.0169
-0.0150
-0.0131
-0.0112
-0.0094
-0.0075
-0.0056
-0.0038
-0.0019
0
0.0019
0.0038
0.0056
0.0075
0.0094
0.0112
0.0131
0.0150
0.0169
0.0188
0.0206
0.0225
0.0244
0.0262
0.0281
0.0300
0.0319
0.0338
0.0356
0.0375
0.0394
0.0412
0.0431
0.0450
0.0469
0.0488
0.0506
0.0525
0.0544
0.0562
0.0581
0.0600
xp' =
0.0600
0.0581
0.0562
0.0544
0.0525
0.0506
0.0488
0.0469
0.0450
0.0431
0.0412
0.0394
0.0375
0.0356
0.0319
0.0300
0.0281
0.0262
0.0244
0.0225
0.0206
0.0188
0.0169
0.0150
0.0131
0.0112
0.0094
0.0075
0.0056
0.0038
0.0019
0
-0.0019
-0.0038
-0.0056
-0.0075
-0.0094
-0.0112
-0.0131
-0.0150
-0.0169
-0.0188
-0.0206
-0.0225
-0.0244
-0.0262
-0.0281
-0.0300
-0.0319
-0.0338
-0.0356
-0.0375
-0.0394
-0.0412
-0.0431
-0.0450
-0.0469
-0.0488
-0.0506
-0.0525
-0.0544
-0.0562
-0.0581
-0.0600
Does anyone have an idea what this is doing theta[i]=np.arctan2(yp[i+1]-yp[i],xp[i+1]-xp[i]) ? I believe it's finding the angle and x1 and x2 points.

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 采纳的回答

Walter Roberson
Walter Roberson 2017-10-22

0 个投票

It is finding the angle from (xp(i),yp(i)) to (xp(i+1),yp(i+1))

3 个评论
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Zach Dunagan
Zach Dunagan 2017-10-22
Suggestions? Am I needing to find angles at each xp(i) and yp(i)? Take the difference and find the angles? Take the difference between xp(i) and xp(i+1) ?
Walter Roberson
Walter Roberson 2017-10-22
编辑：Walter Roberson 2017-10-22
Find the angles between what and each xp(i) and yp(i) ?
The code that is there now is answering the question: "if you are at (xp(i), yp(i)) and you want to go to (xp(i+1),yp(i+1)) then what angle do you need to travel?"
If that is not what you need to calculate, then please explain in words what you do need to calculate ?
If you are just looking for the MATLAB equivalent,
theta[i]=np.arctan2(yp[i+1]-yp[i],xp[i+1]-xp[i])
translates as
theta(i) = atan2(yp(i+1)-yp(i), xp(i+1)-xp(i));
and the code can be written without a loop as:
theta = atan2( diff(yp), diff(xp) );
When I glance at the python code I think it might possibly try to go past the end of the vector, but I am not certain as I do not use python.
Walter Roberson
Walter Roberson 2017-10-22
编辑：Walter Roberson 2017-10-24
The MATLAB equivalent code is:
% thetas for each panel in global frame
theta = zeros(numPanels, 1);
for i = 1 : numPanels
theta(i) = atan2( yp(i+1)-yp(i), xp(i+1)-xp(i) );
end
If numPanels is not at least one less than length(xp) then this code will crash trying to go past the end of xp or yp. (You said you wanted the equivalent MATLAB code, not that you wanted correct code.)

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更多回答（1 个）

Zach Dunagan
Zach Dunagan 2017-10-24
编辑：Walter Roberson 2017-10-24

0 个投票

The only value that appears is the last one in the array.
xp(66:end) = flipud(xp(1:numPanels/2));
yp(66:end) = -flipud(yp(1:numPanels/2));
% collocation points
xc = zeros(numPanels, 1);
yc = zeros(numPanels, 1);
for k = numPanels
xc(k) = (xp(k) + xp(k+1))/2;
yc(k) = (yp(k) + yp(k+1))/2;
end
I have some code that will output the values, but I am trying to learn how to use these loops.

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