# How to reconstruct a cell with different sizes from a matrix

2 ビュー (過去 30 日間)
NHJ 2022 年 9 月 27 日

Hi all,
I have a cell array of complex double and I want to convert them to matrix for the next process. Then, I want to reconstruct back the matrix to cell array. I do like the following codes, where I try with a simple input cells. I know this is not a good method to do, because when I try to change the input into a bigger size, it hard to manually code the position of the cells. Is there any efficient way to do this? Thank you in advance
clc
clear all
% Input cells
F = {{[0.04,0.2,0.56;0.31,0.67,0.22]},{...
[6+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[5+6j,8+5j;6+8j,7-6j;5+6j,3-9j],...
[6+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[5+6j,8+5j;6+8j,7-6j;5+6j,3-9j]},{...
[16+6j,7+3j,8-6j;6+8j,7-6j,3-3j],...
[16+6j,7+3j,8-6j;6+8j,7-6j,3-3j]},{...
[0+0j,0+0j;0+0j,0+0j;0+0j,0+0j]}}
% Extract the cells
G={}
dim = 2
for k=1:length(F)
G = [G,F{k}]
end
% Convert the cell to matrix
% Delete NaN in the matrix
V = B(~isnan(B))
% Another process apply to the matrix
%%%%%%%%%%%%%%%
% Inverse Reconstruction:
% Convert the matrix to cell
H =num2cell(V)
% Reshape the cell
M1 = {}
M = H'
for s=1:6
M1 = [M1,M{s}]
end
M1s = [M1(1) M1(3) M1(5); M1(2) M1(4) M1(6)]
M1s = cell2mat(M1s)
M2 = {}
for s=7:18
M2 = [M2,M{s}]
end
M2s{1,1}{1,1} = [M2{1,1} M2{1,3} M2{1,5};M2{1,2} M2{1,4} M2{1,6}]
M2s{1,1}{1,2} = [M2{1,7} M2{1,10}; M2{1,8} M{1,11}; M{1,9} M2{1,12}]
M3 = {}
for s=19:24
M3 = [M3,M{s}]
end
M3s = [M3(1) M3(3) M3(5); M3(2) M3(4) M3(6)]
M3ss = cell2mat(M3s)
M4 = {}
for s=25:30
M4 = [M4,M{s}]
end
M4s = [M4(1) M4(3); M4(5) M4(2); M4(4) M4(6)]
M4ss = cell2mat(M4s)
% Combine all cells to become only one cell aray
% Create 1x4 cells
C = {M1s {M2s{1,1}{1,1} M2s{1,1}{1,2}} M3ss M4ss}
##### 11 件のコメント9 件の古いコメントを表示9 件の古いコメントを非表示
NHJ 2022 年 9 月 27 日
Missing another part in the code:
% Delete symmetry cells
for kk = 1:numel(F)
for ii = numel(F{kk}):-1:((numel(F{kk}))/2)+1
for oo = 1:ii-1
F{kk}(oo) = [];
break
end
end
end
Thats why the previous output C is doesn't match. At initial I want to ignore this code. By the way, is there any efficient way/method to do my previous question? Thanks for your response
Jan 2022 年 9 月 27 日

Just a hint: Replace
G={};
dim = 2;
for k=1:length(F)
G = [G,F{k}];
end
by
G = [F{:}];
and
M2 = {}
for s=7:18
M2 = [M2,M{s}]
end
by
M2 = [M{7:18}];
But the main problem is the switching from cells of cells to cells to numerical arrays and the other way around. What is the purpose of this confusing procedure?

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