i want to find distances between coordinate points as i explained below?? any help??

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M.Prasanna kumar 2019-8-7

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https://ww2.mathworks.cn/matlabcentral/answers/475179-i-want-to-find-distances-between-coordinate-points-as-i-explained-below-any-help

评论： darova 2019-8-10

采纳的回答： darova

i have coordinates as

A = (0,0); B = (1,0); C = (2,0)

D = (0,1); E = (1,1); F = (2,1)

G = (0,2); H = (1,2); I = (2,2)

i want to distance between A to D plus D to B and so on......

matrix 1 (2*3)

[(AtoD+DtoB) (AtoE+EtoB) (AtoF+FtoB)

(AtoG+GtoB) (AtoH+HtoB) (AtoI+ItoB)

matrix 2 (2*3)

[(AtoD+DtoC) (AtoE+EtoC) (AtoF+FtoC)

(AtoG+GtoC) (AtoH+HtoC) (AtoI+ItoC)

matrix 3 (2*3)

[(BtoD+DtoA) (BtoE+EtoA) (BtoF+FtoA)

(BtoG+GtoA) (BtoH+HtoA) (BtoI+ItoA) ]

matrix 4 (2*3)

[(BtoD+DtoC) (BtoE+EtoC) (BtoF+FtoC)

(BtoG+GtoC) (BtoH+HtoC) (BtoI+ItoC) ]

matrix 5 (2*3)

[(CtoD+DtoA) (CtoE+EtoA) (CtoF+FtoA)

(CtoG+GtoA) (CtoH+HtoA) (CtoI+ItoA) ]

matrix 6 (2*3)

[(CtoD+DtoB) (CtoE+EtoB) (CtoF+FtoB)

(CtoG+GtoB) (CtoH+HtoB) (CtoI+ItoB) ]

finally i want six matrices as i explained above

thank you

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darova 2019-8-8

matrix 1 == matrix 3

matrix 2 == matrix 5

matrix 4 == matrix 6

What do you think? Does it make your task easier?

M.Prasanna kumar 2019-8-8

no they are not equal..they are different sir

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采纳的回答

darova 2019-8-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/475179-i-want-to-find-distances-between-coordinate-points-as-i-explained-below-any-help#answer_386807

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With use of pdist2()

A = [0 0];    B = [1 0];    C = [2 0];
[x,y] = meshgrid(0:2,1:2);        % D G E H F I
D_I = [x(:) y(:)];
AtoXY = reshape(pdist2(A, D_I),size(x));    % matrix distance from A to (D G E H F I)
BtoXY = reshape(pdist2(B, D_I),size(x));    % distance from B
CtoXY = reshape(pdist2(C, D_I),size(x));    % distance from C
matrix1 = AtoXY + BtoXY;
matrix2 = AtoXY + CtoXY;
matrix4 = BtoXY + CtoXY;

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M.Prasanna kumar 2019-8-9

i will explain with your code

A = [0 0]; B = [1 0]; C = [2 0];

[x,y] = meshgrid(0:2,1:2); % D G E H F I

D_I = [x(:) y(:)];

AtoXY = reshape(pdist2(A, D_I),size(x)); % matrix distance from A to (D G E H F I)

BtoXY = reshape(pdist2(B, D_I),size(x)); % distance from B

CtoXY = reshape(pdist2(C, D_I),size(x)); % distance from C

matrix1 = AtoXY + BtoXY;

matrix2 = AtoXY + CtoXY;

matrix4 = BtoXY + CtoXY;

lets fix A (A to B via (x y))

from distance from A to every point in (x y) plus from every point in (x y) to B = matrix 1

2. fix A (A to C via (x y))

from distance from A to every point in (x y) plus from every point in (x y) to C = matrix 2

3. fix B (B to A via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 3

3. fix B and C(B to C via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 4

3. fix C (C to A via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 5

3. fix C (C to B via (x y))

from distance from B to every point in (x y) plus from every point in (x y) to A = matrix 6

as u said

matrix 1 == matrix 3

matrix 2 == matrix 5

matrix 4 == matrix 6

when we fix A remaing points are B and C (SO WE GET TWO MATRICES)

when we fix B remaing points are A and C (SO WE GET TWO MATRICES)

when we fix C remaing points are A and B (SO WE GET TWO MATRICES)

M.Prasanna kumar 2019-8-9

编辑：M.Prasanna kumar 2019-8-9

sir,

thank your very much for your kind reply. you were replying very patiently

code is working perfectly

darova 2019-8-10

You are welcome

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