nested cell array into single cell array

5 次查看（过去 30 天）
Mausmi Verma
Mausmi Verma 2021-12-26
评论： Voss 2022-1-5
A={<1x1 cell> <1x3 cell> 4 <1x4 cell> }
<1x1>= A{1, 1}{1, 1}{1, 1} <1x3 double> =[2 3 4]
<1x3>= A{1, 2}{1, 1} <0x0 double> = [ ]
A{1, 2}{1, 2} <1x2 double> = [ 3 4]
A{1, 2}{1, 3}{1, 1} <0x0 double> =[ ] A{1, 2}{1, 3}{1, 2} <1x3 double> =[3 8 13]
4= A{1, 3} <1x1 double> = [4]
<1x4>= A{1, 4}{1, 1} <1x2 double> = [9 4]
A{1, 7}{1, 2}{1, 1} <0x0 double> =[ ]
A{1, 7}{1, 2}{1, 2} <1x3 double> =[ 9 8 13 ]
A{1, 7}{1, 3} <0x0 cell> = [ ]
A{1, 7}{1, 4}{1, 1} <1x3 double> = [ 9 14 13 ] A{1, 7}{1, 4}{1, 2} <1x3 double> =[ 9 14 19 ]
and based on the output given above, i want my final answer to look like this:
B={ [2 3 4] [3 4; 3 8 13] [4] [9 4 ; 9 8 13 ; 9 14 13 ; 9 14 19] }
I hope my question is understandable
Thanks in advance
  4 个评论
显示 2更早的评论
Mausmi Verma
Mausmi Verma 2021-12-27
Thank you Benjamin for reply, and i got your point.
can i still get the desired output by the help of zero padding for matching the dimension
Mausmi Verma
Mausmi Verma 2021-12-27
@Image Analyst thank you for pointing out that but its exactly not the same because the pattern of the desired output is a bit different

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Voss
Voss 2021-12-27
编辑：Voss 2021-12-30
Ran in:
As far as I can tell, the variable A has this structure (I'm going to assume those 7's in the question should actually be 4's):
A = { ...
{{[2 3 4]}} ...
{[] [3 4] {[] [3 8 13]}} ...
4 ...
{[9 4] {[] [9 8 13]} [] {[9 14 13] [9 14 19]}} ...
};
To verify each entry (and leaving in redundant row index "1," throughout, for consistency with the notation in the question):
format('compact');
A{1,1}
ans = 1×1 cell array {1×1 cell}
A{1,1}{1,1}
ans = 1×1 cell array {[2 3 4]}
A{1,1}{1,1}{1,1}
ans = 1×3
2 3 4
A{1,2}
ans = 1×3 cell array {0×0 double} {[3 4]} {1×2 cell}
A{1,2}{1,1}
ans = []
A{1,2}{1,2}
ans = 1×2
3 4
A{1,2}{1,3}
ans = 1×2 cell array {0×0 double} {[3 8 13]}
A{1,2}{1,3}{1,1}
ans = []
A{1,2}{1,3}{1,2}
ans = 1×3
3 8 13
A{1,3}
ans = 4
A{1,4}
ans = 1×4 cell array {[9 4]} {1×2 cell} {0×0 double} {1×2 cell}
A{1,4}{1,1}
ans = 1×2
9 4
A{1,4}{1,2}
ans = 1×2 cell array {0×0 double} {[9 8 13]}
A{1,4}{1,2}{1,1}
ans = []
A{1,4}{1,2}{1,2}
ans = 1×3
9 8 13
A{1,4}{1,3}
ans = []
A{1,4}{1,4}
ans = 1×2 cell array {[9 14 13]} {[9 14 19]}
A{1,4}{1,4}{1,1}
ans = 1×3
9 14 13
A{1,4}{1,4}{1,2}
ans = 1×3
9 14 19
Below is a function you can use to do what you want. You can call it like this to get the B you want from the A you have:
B = cell(size(A));
for i = 1:numel(A)
B{i} = build_matrix(A{i});
end
display(B)
B = 1×4 cell array {[2 3 4]} {2×3 double} {[4]} {4×3 double}
B{:}
ans = 1×3
2 3 4
ans = 2×3
3 4 0 3 8 13
ans = 4
ans = 4×3
9 4 0 9 8 13 9 14 13 9 14 19
function M = build_matrix(C,M)
if nargin < 2
M = [];
end
if iscell(C)
for i = 1:numel(C)
M = build_matrix(C{i},M);
end
elseif ~isempty(C)
if ~isempty(M)
sC = size(C,2);
sM = size(M,2);
if sM < sC
M(:,end+1:sC) = 0;
elseif sM > sC
C(:,end+1:sM) = 0;
end
end
M = [M; C];
end
end
% function M = build_matrix(C,M)
% if nargin < 2
% M = [];
% end
% if iscell(C)
% for i = 1:numel(C)
% M = build_matrix(C{i},M);
% end
% else
% if ~isempty(M) && size(M,2) ~= size(C,2)
% M(:,end+1:size(C,2)) = 0;
% end
% M = [M; C];
% end
% end
  4 个评论
显示 2更早的评论
Mausmi Verma
Mausmi Verma 2022-1-5
Thanks benjamin for ur support,
after applying the way mentioned above in my work i am getting my answer as:
[1,2] 2 [3,2,0;3,8,13] [4,3,2] [5,10,15] [6,1,2;6,7,2] [7,2,0;7,8,13;7,12,13] [8,3,2;8,7,2;8,13,0] <4x3 double>
[10,15] [11,12,13] [12,7,2;12,13,0;12,17,22] 13 [14,13;14,15] 15 [16,17,22;16,21,22] [17,12,13;17,18,13;17,22,0] [18,13,0;18,17,22;18,23,22] <4x3 double>
[20,15] [21,22] 22 [23,18,13;23,22,0] [24,23,22] [25,20,15]
where, <4x3 double> =
9 8 13
9 10 15
9 14 13
9 14 15
now here i have stuck with my further work as i want my answer to be like,
in place of <4x3> i want answer to be seen as [9,8,13;9,10,15;9,14,13;9,14,15]
and same goes for other <4x3 double> as [19,14,13; 19,14,15;19,18,13;19,20,15]
so finally i want my answer to look like this,
[1,2] 2 [3,2,0;3,8,13] [4,3,2] [5,10,15] [6,1,2;6,7,2] [7,2,0;7,8,13;7,12,13] [8,3,2;8,7,2;8,13,0] [9,8,13;9,10,15;9,14,13;9,14,15] [10,15] [11,12,13] [12,7,2;12,13,0;12,17,22] 13 [14,13;14,15] 15 [16,17,22;16,21,22] [17,12,13;17,18,13;17,22,0] [18,13,0;18,17,22;18,23,22] [19,14,13; 19,14,15;19,18,13;19,20,15] [20,15] [21,22] 22 [23,18,13;23,22,0] [24,23,22] [25,20,15]
Voss
Voss 2022-1-5
Looks like that's just how MATLAB prints it to the command line, not a difference between the answer you get and the answer you want.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 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