How to add some values between two points?

20 visualizzazioni (ultimi 30 giorni)
Dear everyone, Now I have some coordinates pairs and I want to add more points in the area.
For example, If I have the dataset
A = [1,2;2,2;1,3;2,3]
I want to get the matrix
B = [1, 2 ; 1.1, 2; 1, 2.1 ;1.1, 2.1; ..... ; 1.9, 2.9; 2.9, 3; 2, 2.9 ; 2, 3 ]
Thanks a lot

Risposta accettata

HONG CHENG
HONG CHENG il 20 Set 2017
In here, I didn't use any special function. but the result looks not bad
Xmin = 1;
Ymin = 2;
Xmax = 2;
Ymax = 3;
step = 0.1;
step_1 = 1/0.1;
K = (Xmax-Xmin)*(Ymax-Ymin)*step_1*step_1;
Co = zeros(K,2);
n_Co = 1;
for i = Xmin:step:Xmax
for j = Ymin:step:Ymax
Co(n_Co, 1) = i;
Co(n_Co, 2) = j;
n_Co = n_Co+1;
end
end
My result is in the following
  1 Commento
Image Analyst
Image Analyst il 20 Set 2017
Except that's not what you originally asked for. Originally you had the first column increasing every row.

Accedi per commentare.

Più risposte (2)

Walter Roberson
Walter Roberson il 20 Set 2017
Modificato: Walter Roberson il 20 Set 2017
r = linspace( min(A(:,1)), max(A(:,1)), 11 );
c = linspace( min(A(:,2)), max(A(:,2)), 11 );
11 is needed instead of 10 because you need to include the final value.
  1 Commento
HONG CHENG
HONG CHENG il 20 Set 2017
Thanks for your quick answer. And now we get the linspace for X and Y, but I need a matrix of them as the picture shows, I want to get the values as the blue block area shows~~~

Accedi per commentare.


Image Analyst
Image Analyst il 20 Set 2017
How about this:
n = 20 % Whatever...
A = [1,2;2,2;1,3;2,3]
% Make new linearly interpolated array.
A2 = imresize(A, [n, 2])
A2 = A2(3 : end-2,:)
  1 Commento
HONG CHENG
HONG CHENG il 20 Set 2017
Thanks for your answer.
And I have tried your method, but the result looks a little strange
0.876250000000000 2
0.886250000000000 2
0.906250000000000 2
0.936249999999999 2
0.976250000000000 2
1.02981250000000 1.99881250000000
1.11493750000000 1.99043750000000
1.22656250000000 1.97656250000000
they are not uniform
Now I write my own code in the following answer.maybe there is a better method
Thanks a lot

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by