Removing Points Along a Diagonal Line

I have a slope that i have discretized into points. However along the diagonal line I would like to remove all the points to the right of this line. Is this possible? All I have been able to do is remove the points as shown in the second figure but my slope actually looks like the first figure. The points to be reomved are all those lying within the shaded region. Is this possible?

Answer 1

[infinity]

Hello,

I would suggest you a solution,

Since you have known two points of slope, which are (40,35) and (74.4, 15), you can easily get a segment passing to these points. Let say, y = ax + b.

Then, you can define a function, for example phi(x,y) = ax + b - y. As it can seen that phi(x,y) > 0 if (x,y) is in the left of the slope and phi(x,y) < 0 if (x,y) is in the right of the slope and phi(x,y) = 0 if (x,y) is in the slope. This technique is called "levelset method".

Now, what you can do is to use the phi funtion to check a point is in the left or the right of the slope. If it is in the right, you just eleminate it.

[racqs]

Thanx for your feedback. But now I have two arrays of different dimensions. One that contains all the x,y coordinates and one containing only those which should be removed. For example

A = [ 14.25 1.25

14.75 1.25

15.25 2.25

15.75 2.25

.........

90.25 14.25

90.75 14.75]

B = [17.25 0.25

17.75 0.25

18.25 0.75

18.75 0.75]

where the x coordinates are listed on the left and the y coordinates on the right. The values of B to be removed exist in matrix A and I want to remove them so I am left with a new matrix without these values listed in B.

[infinity]

Hello,

What is your point in this comment?

You have A and B, A contains all nodes and B contains removed nodes. Then, what do you dealing with?

[racqs]

B contains the nodes I would like to remove from the full matrix A. I would like to get the remaining nodes. ie A without the nodes B

[infinity]

Hello,

So, you could use for loop (or even don't need for loop) to check wheather a nodes in A is in B or not. If it is not, you store it into a new variable. For example,

clear
A = [1 2; 3 4; 5 6];
B = [3 4; 5 6];
C = [];
for i = 1:size(A,1)
    if ~ismember(A(i,1),B(:,1)) || ~ismember(A(i,2),B(:,2))
    C = [C; A(i,:)];
    end
end

[darova]

Read about inpolygon() also