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HOW CAN I SOLVE THIS PROBLEM???​???!!!!!!!​!!!!!!!!!!

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ali hadjer
ali hadjer 2015 年 11 月 20 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
HELLO I WANT TO FULLFIL MY MATRIX bet with just the value of matrix net for which degree==1 or the max(degree)>1 and distance<r (after each iteration i want to eliminate the points that have distance <r by putting the value 100 in it) but it does not work. It gives a result of bet equal to net.
clc;
clear all;
n=10;
x=100;
y=100;
r=25;
net = [1:n;rand([1,n])*x;rand([1,n])*y];
for i = 1:n
for j = 1:n
X1 = net(2,i);
Y1 = net(3,i);
X2 = net(2,j);
Y2 = net(3,j);
xSide = abs(X2-X1);
ySide = abs(Y2-Y1);
d(i,j) = sqrt(xSide^2+ySide^2);% distance euclidienne ENTRE NOEUD
end
end
degree = zeros(n,1);
for i = 1:n
%for j = i+1:n
for j = 1:n
if (d(i,j)<=r )
degree(i)= degree(i)+1;
end
end
end
bet=zeros(2,n)
for i=1:n
for j=1:n
while d~=100
if degree(i)== 1
bet(2,i)=net(2,i);
bet(3,i)=net(3,i);
else
if max(degree(i))>1 && d(i,j)<r
bet(2,i)=net(2,i);
bet(3,i)=net(3,i);
end
end
d(i,:)=100;
d(:,j)=100;
end
end
end
  2 件のコメント
Image Analyst
Image Analyst 2015 年 11 月 20 日
I'm not sure I understand. degree is a 1-D vector while bet and net are 2-D matrices. Some comments or example input and output matrices would help.
ali hadjer
ali hadjer 2015 年 11 月 21 日
YES YOU ARE RIGHT. to understand my probleme i will explaine to you: I want to create a nw matrix (bet) from the old matrix (net) but i have a conditions to do that ; so i calculat a distance between all the point of matrix net (d) and a degree of each point(nombre of point whos distance are inferieur to r(r is a range) ; my condition to creat the matrix bet is if the degree is equal to 1 or i take the maximum value of degree max(degree) and put his corespondant point in matrix *bet*and to eliminat all the point that have a distance <r whit this point i change all his distance to 100

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