# How can I repeat this line plot?

조회 수: 2 (최근 30일)
Thomas 2016년 6월 30일
댓글: Thomas 2016년 7월 4일
Hi everyone,
The code I am using is
clear all;
a0 = 1;
b0 = a0./4 .* [1 1 1];
a1 = a0./2 .* [0 1 1];
a2 = a0./2 .* [1 0 1];
a3 = a0./2 .* [1 1 0];
N1=2;
N2=2;
N3=2;
In = zeros(N1,N2,N3,3);
P = zeros(N1,N2,N3,3);
for n = 0:N1-1
for m = 0:N2-1
for l = 0:N3-1
In(n+1,m+1,l+1,1:3) = n.*a1+m.*a2+l.*a3;
P(n+1,m+1,l+1,1:3) = n.*a1+m.*a2+l.*a3 + b0;
end
end
end
In_plt = reshape(In,[N1*N2*N3,3]);
hold off
scatter3(In_plt(:,1),In_plt(:,2),In_plt(:,3), 'b','filled');
P_plt = reshape(P,[N1*N2*N3,3]);
hold on
scatter3(P_plt(:,1),P_plt(:,2),P_plt(:,3), 'r','filled');
line([P(1,1,1,1) In(1,1,1,1)],[P(1,1,1,2) In(1,1,1,2)],[P(1,1,1,3) In(1,1,1,3)]);
line([P(1,1,1,1) In(1+1,1,1,1)],[P(1,1,1,2) In(1+1,1,1,2)],[P(1,1,1,3) In(1+1,1,1,3)]);
line([P(1,1,1,1) In(1,1+1,1,1)],[P(1,1,1,2) In(1,1,1,2)],[P(1,1,l+1,3) In(1,1+1,1,3)]);
line([P(1,1,1,1) In(1,1,1,1)],[P(1,1,1,2) In(1,1,1+1,2)],[P(1,1,1,3) In(1,1,1+1,3)]);
hold off
view(3), axis vis3d
axis off
camproj perspective, rotate3d on
title('Diamond structure of Indium Phosphide')
And this produces
What I want to happen is to connect all of these points using line. The section of code which produces the line connections is
line([P(1,1,1,1) In(1,1,1,1)],[P(1,1,1,2) In(1,1,1,2)],[P(1,1,1,3) In(1,1,1,3)]);
line([P(1,1,1,1) In(1+1,1,1,1)],[P(1,1,1,2) In(1+1,1,1,2)],[P(1,1,1,3) In(1+1,1,1,3)]);
line([P(1,1,1,1) In(1,1+1,1,1)],[P(1,1,1,2) In(1,1,1,2)],[P(1,1,l+1,3) In(1,1+1,1,3)]);
line([P(1,1,1,1) In(1,1,1,1)],[P(1,1,1,2) In(1,1,1+1,2)],[P(1,1,1,3) In(1,1,1+1,3)]);
What can I do to reproduce this section of code for every connection?

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### 채택된 답변

Muhammad Usman Saleem 2016년 6월 30일
why you do not copy paste it into code, i think due to bluck in coding
try to access these lines through for loop.
Check it
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Muhammad Usman Saleem 2016년 7월 1일
Your code is bucky can you point out where are the lines you wish to repeat?
Thomas 2016년 7월 4일
Thanks for the response Muhammad, I figured the problem out and my code is now fully functional.

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