Plotting Two Consecutive Columns of Matrices

Hi,

I have 4 x 202 matrix, and I want to plot the columns in 2's e.g columns 1 and 2, next columns 3 and 4 as they represent X and Y coordinates. How can I do this?

Thanks

Answer 1

[Star Strider]

First a (4 x 202) matrix defines a matrix of 4 rows and 202 columns. Assuming that it’s actually a (202 x 4) matrix, perhaps something like this —

M = randn(15, 4)                                           % Prototype

M = 15×4

-1.4757 0.4730 -1.7963 0.5677 -1.4236 0.6301 -0.9039 -0.0171 -1.1303 1.3193 0.1080 0.2304 -1.2160 -0.9149 -0.3458 -1.0942 0.7360 0.9042 -0.1741 1.7954 0.1131 0.5694 1.0606 -0.2865 0.1741 -0.7245 -0.7010 1.2694 -1.8673 -0.5442 -0.6397 -2.0804 0.4837 0.0083 -0.6583 0.5971 -1.0150 -2.1755 0.5630 -0.2446

Mr = reshape(M, [], 2)

Mr = 30×2

-1.4757 -1.7963 -1.4236 -0.9039 -1.1303 0.1080 -1.2160 -0.3458 0.7360 -0.1741 0.1131 1.0606 0.1741 -0.7010 -1.8673 -0.6397 0.4837 -0.6583 -1.0150 0.5630

figure
plot(Mr(:,1), Mr(:,2), '.')

If the matrix actually is (4 x 202), transpose it first, then use the approach in this code example.

.

[Femi O. Jinadu]

Thank you very much, it's not given the expected response yet. I probably didn't explain well enough.

The matrix is a 4 by 202 matrix. The first 2 columns and 4 rows are the location of an object at time 0, the next 2 columns (3 and 4) are the location at time 1, so it's like the same 4 points are converging towards something. I tried using a for loops can't seem to get it yet. See the image. The lines are meant to converge to the scatter plots which they are, but the way i constructed the for and while loops are creating some extra lines which aren't supposed to be there.

t

[Femi O. Jinadu]

l= size(Xf) % Xf is the 4x202 matrix

for n=1:4

i=1

while i<=l(2)

plot(Xf(n,i:2:end-1),Xf(n,i+1:2:end))

i=i+1

end

end

[Star Strider]

I am not certain what the matrix is (not supplied), nor am I certain what the desired result is.

See if this produces the desired result —

M = randn(4, 16)                                           % Prototype Matrix (Random)

M = 4×16

0.0752 -0.2438 -1.1305 0.5674 -0.8264 -0.8058 -0.0557 0.5916 0.5528 -0.7535 0.6855 -0.0317 -0.5626 -1.2305 -1.7698 -0.1324 -0.7432 0.4237 -1.8701 -1.1656 -1.3737 0.9483 1.0826 0.8271 1.4154 0.2680 1.7482 -0.3266 -0.7582 2.0894 -0.4560 0.2552 1.5404 0.1558 -0.9183 -0.8252 0.9294 -1.2118 -0.2726 1.3016 -0.1283 0.1639 -0.1338 0.1248 -0.1141 1.1451 -0.5234 -0.6048 -0.6458 0.2220 0.9233 0.0739 0.8112 -0.6546 -0.4364 0.6713 0.8739 -0.2792 -0.1227 1.6188 1.1815 -0.2486 -0.1096 0.5278

figure
hold on
for k = 1:2:fix(size(M,2))/2
    colpair = [1 2]+(k-1)                                   % Check Indices
    xvct = M(:,colpair(:,1))                                % Check X-Vector
    yvct = M(:,colpair(:,2))                                % Check Y-Vector
    plot(M(:,colpair(:,1)), M(:,colpair(:,2)), '.-')
end

colpair = 1×2

1 2

xvct = 4×1

0.0752 -0.7432 1.5404 -0.6458

yvct = 4×1

-0.2438 0.4237 0.1558 0.2220

colpair = 1×2

3 4

xvct = 4×1

-1.1305 -1.8701 -0.9183 0.9233

yvct = 4×1

0.5674 -1.1656 -0.8252 0.0739

colpair = 1×2

5 6

xvct = 4×1

-0.8264 -1.3737 0.9294 0.8112

yvct = 4×1

-0.8058 0.9483 -1.2118 -0.6546

colpair = 1×2

7 8

xvct = 4×1

-0.0557 1.0826 -0.2726 -0.4364

yvct = 4×1

0.5916 0.8271 1.3016 0.6713

hold off

This reports the results of each loop iteration as well as plots them, so check to be certain it references the appropriate matrix columns and produces the desired result (since I still have no idea what that is). First, use a subset of the maatrix, for example the first 16 columns as I did here (there must be an even number of columns). If that works, see if the entire matrix plots correctly.

.

[Femi O. Jinadu]

Still no luck, here's the data though. Thanks a lot

[Star Strider]

No luck, indeed!

I have no idea which of those 155 variables is the one to work with. My guess would be ‘Xf’ or‘Xf1’ so I’ll wait for confirmation before I proceed.

Meanwhile, what does ‘no luck’ mean? What should the code do? (I still have absolutely no idea.)

.

[Femi O. Jinadu]

I'm sorry, Xf.

We have 4 rows, and 202 columns. Each row represents a node and the node location (coordinates) are changing - like column 1 and column 2 are the first cordinates of the 4 nodes, the next 2 columns are the next location, etc.

[Femi O. Jinadu]

Hi, thanks

I've figured it out, thanks so much.

[Star Strider]

My pleasure!

If my Answer helped you solve your problem, please Accept it!

.

Answer 2

[Chunru]

x = rand(202, 4);  % 202x4 (not 4x202)
plot(x(:,1), x(:, 2), 'ro', x(:,3), x(:, 4), 'b*')