Why won't cells populate via if-else statements?

1 次查看（过去 30 天）
Kyle Lyman
Kyle Lyman 2021-5-21
评论： J. Alex Lee 2021-5-22
I am taking the transpose of the last three rows of matrix "con". What I want to do is search each newly transposed column for specific numbers using find() (i.e. 1,2,3,4,5,6,7,8,9). I want each of these specific numbers to correspond to a 2x1 matrix, each of which is to be inserted into a cell structure. Then I want to use cell2array to form a matrix. For example [1;2;3;4;5;6] maps to [1;2;3;4;5;6;7;8;9;10;11;12]. My issue is that when I run the code T is displaying as an empty matrix (T=[ ]). I have attached the code down below. Thank you in advance!
clc
clear all
close all
con=[1 2 3 4 5 6; %Connectivity matrix
1 2 3 4 5 6;
3 5 4 8 7 9;
4 5 6 2 5 7];
Con_Transpose=con(2:end,:).' %Transpose of last three rows of connectivity matrix
width=1:1:size(Con_Transpose,2);
for i=1:length(width)
a={};
if Con_Transpose(:,width(i))==1;
a{find(Con_Transpose==1,1),1}=[1;2]
end
if Con_Transpose(:,width(i))==2;
a{find(Con_Transpose==2,1),1}=[3;4]
end
if Con_Transpose(:,width(i))==3;
a{find(Con_Transpose==3,1),1}=[5;6]
end
if Con_Transpose(:,width(i))==4;
a{find(Con_Transpose==4,1),1}=[7;8]
end
if Con_Transpose(:,width(i))==5;
a{find(Con_Transpose==5,1),1}=[9;10]
end
if Con_Transpose(:,width(i))==6;
a{find(Con_Transpose==6,1),1}=[11;12]
end
if Con_Transpose(:,width(i))==7;
a{find(Con_Transpose==7,1),1}=[13;14]
end
if Con_Transpose(:,width(i))==8;
a{find(Con_Transpose==8,1),1}=[15;16]
end
if Con_Transpose(:,width(i))==9;
a{find(Con_Transpose==9,1),1}=[17;18]
end
T=cell2mat(a)
end

请先登录，再回答此问题。

回答（1 个）

J. Alex Lee
J. Alex Lee 2021-5-21
编辑：J. Alex Lee 2021-5-21
Ran in:
The if statements are probably your problem...you are comparing a column vector with scalars, which off the top of my head I don't know the behavior of.
I won't attempt to fix your code, but consider this strategy that's based on indexing, doesn't need to loop, and doesn't touch cell arrays...does it do what you want:
% generate your "map" from 1->[1;2], 2->[3;4], 4->[7;8], etc.
rmap = reshape(1:18,2,[])
rmap = 2×9
1 3 5 7 9 11 13 15 17 2 4 6 8 10 12 14 16 18
% test the mechanism
x = [1 4 6]
x = 1×3
1 4 6
a = reshape(rmap(:,x),[],1)
a = 6×1
1 2 7 8 11 12
% now solve your problem
con=[1 2 3 4 5 6; %Connectivity matrix
1 2 3 4 5 6;
3 5 4 8 7 9;
4 5 6 2 5 7];
Con_Transpose=con(2:end,:).'
Con_Transpose = 6×3
1 3 4 2 5 5 3 4 6 4 8 2 5 7 5 6 9 7
sz = [2,1].*size(Con_Transpose)
sz = 1×2
12 3
T = reshape(rmap(:,Con_Transpose),sz)
T = 12×3
1 5 7 2 6 8 3 9 9 4 10 10 5 7 11 6 8 12 7 15 3 8 16 4 9 13 9 10 14 10
  2 个评论
Kyle Lyman
Kyle Lyman 2021-5-22
Hey J. Alex Lee! Thank you for the prompt response. The method that you presented certainly works for matrices that have an even number of elements (i.e. size(con,2)*(size(con,1)-1)=18 in the example you replied to). However, when I change the matrix "con" to have an odd number of elements, I receive the error:
"Error using reshape
Product of known dimensions, 2, not divisible into total number of elements, 9.
Error in DegreeOfFreedom (line 8)
rmap = reshape(1:range,2,[])"
The code that reproduces this error is listed below:
con=[1 2 3; %The first line is local degrees of freedom, thus we focus on the second row through the end row, which sums to 9 total elements.
6 7 9;
4 1 6;
9 7 2]
Con_Transpose=con(2:end,:).'
range=numel(con)-size(con,2);
% generate your "map" from 1->[1;2], 2->[3;4], 4->[7;8], etc.
rmap = reshape(1:range,2,[])
x=[con(2:end,:)];
a = reshape(rmap(:,x),[],1);
sz = [2,1].*size(Con_Transpose);
T = reshape(rmap(:,Con_Transpose),sz)
J. Alex Lee
J. Alex Lee 2021-5-22
So what does your conversion map look like...
1->1,2
2->3,4
3->5,6
4->7,8
5->9
6->?
7->?
8->?
9->?

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by