I have as input
Y =[[[0 0 0 ..., 0 0 0]
[1 0 0 ..., 0 0 0]
[0 0 0 ..., 0 0 0]
...,
[0 1 0 ..., 0 0 0]
[0 0 1 ..., 0 0 0]
[0 0 0 ..., 0 0 0]]]
Meaning the shape is (1,103,10)
What I want is an array of shape (103,10)
So my array should look like:
Y =[[0 0 0 ..., 0 0 0]
[1 0 0 ..., 0 0 0]
[0 0 0 ..., 0 0 0]
...,
[0 1 0 ..., 0 0 0]
[0 0 1 ..., 0 0 0]
[0 0 0 ..., 0 0 0]]
How can we do this?
Related
I try to solve system of lineare equation using the Gauss Jordan methode. I try my code with a smaller matrice and I got the rigth answer. Now I try it with a bigger matrice and I got NAN(not a number) as answer. I don't what I am doing wrong.
a = [-1 0 0 0 0 0 1 0 0 0 1 0 0 0;
0 -1 0 0 0 0 0 1 0 0 0 1 0 0 ;
1 0 -1 0 -1 0 0 0 0 0 0 0 0 0;
0 1 0 -1 0 -1 0 0 0 0 0 0 0 0;
0 0 0 0 1 0 0 0 0 0 -1 0 -1 0;
0 0 0 0 0 1 0 0 0 0 0 -1 0 -1;
0 0 1 0 0 0 -1 0 -1 0 0 0 0 0;
0 0 0 1 0 0 0 -1 0 -1 0 0 0 0;
-0.025 0 1 0 0 0 0 0 0 0 0 0 0 0;
0 -0.8 0 1 0 0 0 0 0 0 0 0 0 0;
0 0 0 0 -0.04 0 0 0 0 0 1 0 0 0;
0 0 0 0 0 -0.96 0 0 0 0 0 1 0 0;
0 0 -0.04 0 0 0 0 0 1 0 0 0 0 0;
0 0 0 -0.96 0 0 0 0 0 1 0 0 0 0];
b = [-50;-50;0;0;0;0;0;0;0;0;0;0;0;0];
Gaussjordan(a,b);
% Function Gauss Jordan
function x=Gaussjordan(a,b)
ab=[a b];
[m,n]=size(ab); %size of matrix
for i=1:m
ab(i,:)=ab(i,:)/ab(i,i);
for j=1:m
if j==i; continue;
end
ab(j,:)=ab(j,:)-ab(j,i)*ab(i,:);
end;
end;
x=ab(:,length(b)+1);
end
with the samll matrix a = [1 1 1; 2 -3 4; 3 4 5]
b = [9; 13; 40], I got the answer ans = [1 3 5].
But with the bigger one upon in code I got ans = [NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN, NaN]
Did someone has an idea?
Ps: I solve this systen of equation with fsolve to be sure that it exist an answer on the system equation
Essentially I am trying to create a volume by rotating a closed curve. I have attempted to use cylinder to achieve this, but I am getting undesirable results. I also have tried directly using surf. Here are those two attempts (newEl defined below).
elPatternPlot = plot3(newEl(:,1),newEl(:,2),newEl(:,3),'gr-','LineWidth',0.5);
xlabel('x')
ylabel('y')
zlabel('z')
grid on
Next, I rotate the 2D closed curve to the correct angle.
rotate(elPatternPlot, [1 0 0], 90);
To generate data points in the desired volume, I use rotate:
% Now rotate around z and store in the data
axObjs = azElFig.Children;
dataObjs = axObjs.Children;
xElData = dataObjs(1).XData;
yElData = dataObjs(1).YData;
zElData = dataObjs(1).ZData;
for a1=1:5:180
rotate(elPatternPlot, [0 0 1], a1);
axObjs = azElFig.Children;
dataObjs = axObjs.Children;
xElData = [xElData dataObjs(1).XData];
yElData = [yElData dataObjs(1).YData];
zElData = [zElData dataObjs(1).ZData];
end
plot3(xElData,yElData,zElData,'r-','LineWidth',0.5)
And the result:
This this yields the correct outside geometry, but I am seeking to create a volume. From here I use surf:
xv = linspace(min(xElData), max(xElData), 150);
yv = linspace(min(yElData), max(yElData), 150);
[X, Y] = meshgrid(xv, yv);
Z = griddata(xElData,yElData,zElData,X,Y);
figure; surf(X,Y,Z);
The output is:
Using cylinder,
[x,y,z] = cylinder(Z);
figure; surf(x,y,z)
I obtain:
The desired surface should look more like (from the bottom):
And from the side, a filled in version of this:
newEl =
-906.5000 -36.5000 0
-909.5000 -65.5000 0
-903.5000 -91.5000 0
-891.5000 -139.5000 0
-877.5000 -203.5000 0
-863.5000 -238.5000 0
-855.5000 -254.5000 0
-847.5000 -269.5000 0
-839.5000 -284.5000 0
-784.5000 -369.5000 0
-771.5000 -384.5000 0
-746.5000 -406.5000 0
-733.5000 -421.5000 0
-695.5000 -457.5000 0
-668.5000 -479.5000 0
-641.5000 -496.5000 0
-614.5000 -513.5000 0
-600.5000 -525.5000 0
-573.5000 -544.5000 0
-561.5000 -549.5000 0
-549.5000 -553.5000 0
-488.5000 -579.5000 0
-474.5000 -585.5000 0
-460.5000 -592.5000 0
-446.5000 -599.5000 0
-417.5000 -603.5000 0
-388.5000 -607.5000 0
-293.5000 -611.5000 0
-277.5000 -608.5000 0
-246.5000 -597.5000 0
-231.5000 -593.5000 0
-187.5000 -578.5000 0
-160.5000 -553.5000 0
-133.5000 -528.5000 0
-120.5000 -512.5000 0
-85.5000 -478.5000 0
-61.5000 -428.5000 0
-56.5000 -411.5000 0
-47.5000 -359.5000 0
-42.5000 -342.5000 0
-33.5000 -289.5000 0
-28.5000 -271.5000 0
-19.5000 -220.5000 0
-14.5000 -202.5000 0
-5.5000 -149.5000 0
-0.5000 -132.5000 0
4.5000 -128.5000 0
9.5000 -161.5000 0
18.5000 -194.5000 0
23.5000 -226.5000 0
32.5000 -261.5000 0
37.5000 -293.5000 0
42.5000 -324.5000 0
51.5000 -359.5000 0
56.5000 -390.5000 0
65.5000 -426.5000 0
70.5000 -455.5000 0
125.5000 -517.5000 0
138.5000 -528.5000 0
165.5000 -556.5000 0
178.5000 -567.5000 0
192.5000 -578.5000 0
207.5000 -585.5000 0
251.5000 -596.5000 0
267.5000 -604.5000 0
298.5000 -611.5000 0
393.5000 -607.5000 0
422.5000 -602.5000 0
451.5000 -597.5000 0
465.5000 -592.5000 0
479.5000 -586.5000 0
493.5000 -580.5000 0
530.5000 -565.5000 0
542.5000 -559.5000 0
554.5000 -552.5000 0
592.5000 -534.5000 0
619.5000 -515.5000 0
646.5000 -496.5000 0
673.5000 -477.5000 0
725.5000 -430.5000 0
738.5000 -419.5000 0
776.5000 -379.5000 0
789.5000 -368.5000 0
801.5000 -353.5000 0
811.5000 -336.5000 0
836.5000 -292.5000 0
844.5000 -279.5000 0
852.5000 -268.5000 0
860.5000 -253.5000 0
881.5000 -200.5000 0
886.5000 -184.5000 0
891.5000 -167.5000 0
895.5000 -146.5000 0
902.5000 -104.5000 0
907.5000 -90.5000 0
909.5000 -46.5000 0
904.5000 6.5000 0
899.5000 58.5000 0
897.5000 60.5000 0
875.5000 126.5000 0
867.5000 142.5000 0
828.5000 208.5000 0
820.5000 224.5000 0
763.5000 276.5000 0
750.5000 289.5000 0
712.5000 319.5000 0
699.5000 332.5000 0
686.5000 341.5000 0
659.5000 360.5000 0
632.5000 378.5000 0
605.5000 396.5000 0
578.5000 416.5000 0
566.5000 431.5000 0
517.5000 496.5000 0
505.5000 509.5000 0
436.5000 567.5000 0
407.5000 587.5000 0
377.5000 603.5000 0
361.5000 604.5000 0
345.5000 606.5000 0
329.5000 607.5000 0
313.5000 608.5000 0
282.5000 610.5000 0
236.5000 606.5000 0
221.5000 601.5000 0
151.5000 567.5000 0
111.5000 547.5000 0
100.5000 533.5000 0
89.5000 520.5000 0
79.5000 500.5000 0
60.5000 477.5000 0
46.5000 454.5000 0
27.5000 427.5000 0
13.5000 403.5000 0
-10.5000 403.5000 0
-24.5000 423.5000 0
-38.5000 444.5000 0
-52.5000 465.5000 0
-66.5000 486.5000 0
-75.5000 502.5000 0
-96.5000 531.5000 0
-107.5000 548.5000 0
-147.5000 566.5000 0
-174.5000 581.5000 0
-202.5000 591.5000 0
-217.5000 600.5000 0
-262.5000 611.5000 0
-309.5000 607.5000 0
-325.5000 607.5000 0
-341.5000 606.5000 0
-357.5000 604.5000 0
-373.5000 603.5000 0
-403.5000 587.5000 0
-432.5000 566.5000 0
-501.5000 511.5000 0
-513.5000 494.5000 0
-525.5000 478.5000 0
-537.5000 463.5000 0
-587.5000 407.5000 0
-628.5000 380.5000 0
-655.5000 361.5000 0
-682.5000 341.5000 0
-708.5000 320.5000 0
-721.5000 311.5000 0
-759.5000 277.5000 0
-797.5000 244.5000 0
-807.5000 237.5000 0
-816.5000 224.5000 0
-824.5000 213.5000 0
-832.5000 199.5000 0
-871.5000 124.5000 0
-882.5000 104.5000 0
-887.5000 98.5000 0
-894.5000 72.5000 0
-896.5000 37.5000 0
-899.5000 34.5000 0
-901.5000 1.5000 0
-906.5000 -36.5000 0
You almost had it right. The problem is within how you rotate inside the for loop.
Your code:
for a1=1:5:180
rotate(elPatternPlot, [0 0 1], a1);
...
end
a1 is incremented by 5 as it iterates. Therefore, you are not rotating it by a step of 5.
To fix this, simply change
rotate(elPatternPlot, [0 0 1], a1);
to
rotate(elPatternPlot, [0 0 1], 5);
To make it easier to plot your figures, you can keep your data in matrix form:
for a1=1:5:180
rotate(elPatternPlot, [0 0 1], 5);
xElData = [xElData; elPatternPlot.XData];
yElData = [yElData; elPatternPlot.YData];
zElData = [zElData; elPatternPlot.ZData];
end
To plot:
figure
plot3(xElData, yElData, zElData)
grid on; axis equal
figure
surf(xElData, yElData, zElData)
grid on; axis equal
figure
mesh(xElData, yElData, zElData)
grid on; axis equal
So I need to find the minimum value of corresponding elements in multiple matrices. I did find this page but I would like to extend my question a little further. My matrices has positive and negative values (and zeros) and I want to find the minimum values excluding zeros.
Example:
Z(:,:,1) = [-5 0 5
0 0 0
1 0 3];
Z(:,:,2) = [1 0 2
0 0 0
0 0 0];
Z(:,:,3) = [0 0 0
-9 0 4
0 0 0];
Z(:,:,4) = [0 0 0
-2 0 0
0 0 0];
Here's what I'm using as of now:
Zmin = min(Z,[],3);
But that gives me:
[-5 0 0
-9 0 0
0 0 0]
But I want my result to be:
[-5 0 2
-9 0 4
1 0 3]
Any ideas? When I use nonzeros, it messes up everything.
Here's a workaround:
Replace all the 0s in Z by NaN, calculate the min, then switch back to 0:
clear all
clc
close all
Z(:,:,1) = [-5 0 5
0 0 0
1 0 3];
Z(:,:,2) = [1 0 2
0 0 0
0 0 0];
Z(:,:,3) = [0 0 0
-9 0 4
0 0 0];
Z(:,:,4) = [0 0 0
-2 0 0
0 0 0];
%// Assign NaN to 0 elements
Z(Z ==0) = NaN;
Zmin = min(Z,[],3);
%// Switch back with 0
Zmin(isnan(Zmin)) = 0;
%// Same for Z;
Z(isnan(Z)) =0;
The output looks like this:
Zmin
Z
Zmin =
-5 0 2
-9 0 4
1 0 3
Z(:,:,1) =
-5 0 5
0 0 0
1 0 3
Z(:,:,2) =
1 0 2
0 0 0
0 0 0
Z(:,:,3) =
0 0 0
-9 0 4
0 0 0
Z(:,:,4) =
0 0 0
-2 0 0
0 0 0
One option is to set all zero elements to NaN that way:
Z(Z==0)=NaN;
And then use nanmean
Zmin = nanmin(Z,[],3);
Say I have a matrix
A = zeros(5, 5);
Instead of looping with a for loop, I wish to batch-modify some of the elements. For example, I wish to change elements marked by pts_to_modify to 1, where
pts_to_modify=[[2 3]; [3 2]];
So I wish A to become
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 0 0 0 0
However, when I do
A(pts_to_modify(:, 1), pts_to_modify(:, 2)) = 1,
I get
A =
0 0 0 0 0
0 1 1 0 0
0 1 1 0 0
0 0 0 0 0
0 0 0 0 0
How can I do it correctly?
You can use sub2ind:
>> ind = sub2ind(size(A), pts_to_modify(1,:), pts_to_modify(2,:))
ind =
12 8
>> A(ind) = 1
A =
0 0 0 0 0
0 0 1 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 0
sub2ind
linear indexing
the matlab variables browser in the IDE GUI per default only shows 10 digits. longer variables have their dimension shown.
[0 0 0 0 0 0 0 0 0 1]
[0 0 0 0 0 1 1 0 0]
<1x11 double>
[1 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]
<1x11 double>
[0 0 1 0 0 0 0 0 0 0]
[0 0 0 1 1 1 0 0 0]
[1 0 0 1 0 0 0 1 1 0]
is there a way to modify this behavior in matlab 8.1 ?
As you may already have guessed from the comments:
No, there is no setting that can be changed via regular means to modify this behavior.
Of course everything is possible, but the 'easiest' way would probably require a java hack, like the one for changing the workspace preferences.