grouping rows and store is separate matrix

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Oday Shahadh
Oday Shahadh 2020 年 6 月 11 日
コメント済み: Oday Shahadh 2020 年 6 月 11 日
hi all,
in the script below I have p which is (m,3) vectors matrix, Itried to group this matrix according to z-value in L1 vector, so in case L1 is (0,0-2), rows of matrix p should be filtered according to z=-2, and each filtered rows have to be seperatly stored. I failed to do this pls help
clc
clear
close all
format long
x=[];
y=[];
z=[];
for L=-2:0.5:2
for r=0:.5:2
for theta=0:45:360;
x=[x;r*cos(theta*pi/180)];
y=[y;r*sin(theta*pi/180)];
z=[z,L];
L1=[x-x,x-x,z'];
end
end
end
lx=L1(:,1);ly=L1(:,2);lz=L1(:,3);
z=z';
p=[x,y,z];
a = cell(length(L1),1);
for ii = 1:length(L1);
a{ii} = p(p(:,3) == L1(ii,3),:);
end
celldisp(a)

採用された回答

Sai Gudlur
Sai Gudlur 2020 年 6 月 11 日
Hello Oday,
I couldn't understand you question completely but to to certain degree i could.
x=[x;r*cos(theta*pi/180)];
y=[y;r*sin(theta*pi/180)];
z=[z,L];
Do not think this is necessary. if i were u would do it this way.
x=[];
y=[];
z=[];
L = -2:0.5:2;
r=(0:.5:2)';
theta=0:45:360;
x = (r.*cos(theta*pi/180));
L1=[x-x,x-x,z'];
Again this part are you trying subtract elements in the 2nd column of matrix x from the 1st Column?
A little more detail might help.
  5 件のコメント
Sai Gudlur
Sai Gudlur 2020 年 6 月 11 日
I do not know how much would this help but try.
"ismember" or if i were you would try "intersect". This would return the values that have the same A&B "z" value. and then you could proceed to do that operation.
Another possibility is you could use "Unique" and find the ones that aren't the same and later use the common values of A & B.
Oday Shahadh
Oday Shahadh 2020 年 6 月 11 日
your first accepted answer :)

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