# A compact way to horizontally concatenate rows of many cell arrays ?

1 Ansicht (letzte 30 Tage)
Sim am 9 Feb. 2023
Bearbeitet: Sim am 9 Feb. 2023
A compact way to horizontally concatenate rows of many cell arrays
a{1},a{2},a{3},..,a{N} % especially for N>>1
as in the following example ?
% Input
a{1} = {[0 0 1 4 1]
[9 9 0 0 0]
[2 3 5 1 2]};
a{2} = {[1 0 7 4 0]
[2 8 8 2 6]
[1 9 8 1 1]};
a{3} = {[7 7 6 0 6]
[9 0 9 1 2]
[3 9 2 4 6]};
a{4} = {[3 3 3 2 3]
[4 1 5 6 3]
[9 0 0 0 7]};
% A non-compact way to get my desired output, and that I would like to avoid or improve:
for i = 1 : 3
a2{i} = {[a{1}{i,:} a{2}{i,:} a{3}{i,:} a{4}{i,:}]};
end
c = vertcat(a2{:});
% Desired Output
c
c = 3×1 cell array
{[0 0 1 4 1 1 0 7 4 0 7 7 6 0 6 3 3 3 2 3]} {[9 9 0 0 0 2 8 8 2 6 9 0 9 1 2 4 1 5 6 3]} {[2 3 5 1 2 1 9 8 1 1 3 9 2 4 6 9 0 0 0 7]}
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Sim am 9 Feb. 2023
Bearbeitet: Sim am 9 Feb. 2023
The reason of my question is the following.
If I use the abovementioned way to build up the concatenation of cell arrays, when I have many cell arrays, the following piece of code
{[a{1}{i,:} a{2}{i,:} a{3}{i,:} .. a{N}{i,:}]} % for N>>1
becomes very long, and I guess I should write down all the cell arrays
a{j}{i,:} % where j = 1, 2, 3, ..., N
manually, right ?
Therefore, is there any way to build up that concatenation in a "more automatic" way when there are many (i.e. N>>1) cell arrays?

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### Akzeptierte Antwort

Stephen23 am 9 Feb. 2023
a = {{[0,0,1,4,1];[9,9,0,0,0];[2,3,5,1,2]};{[1 0 7 4,0];[2,8,8,2,6];[1,9,8,1,1]};{[7,7,6,0,6];[9,0,9,1,2];[3,9,2,4,6]};{[3,3,3,2,3];[4,1,5,6,3];[9,0,0,0,7]}}
a = 4×1 cell array
{3×1 cell} {3×1 cell} {3×1 cell} {3×1 cell}
tmp = [a{:}];
c = tmp(:,1);
for k = 1:numel(c)
c{k} = [tmp{k,:}];
end
display(c)
c = 3×1 cell array
{[0 0 1 4 1 1 0 7 4 0 7 7 6 0 6 3 3 3 2 3]} {[9 9 0 0 0 2 8 8 2 6 9 0 9 1 2 4 1 5 6 3]} {[2 3 5 1 2 1 9 8 1 1 3 9 2 4 6 9 0 0 0 7]}
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### Weitere Antworten (1)

Rik am 9 Feb. 2023
Two calls to horzcat and a loop do the trick, although this can probably be improved a lot.
% Input
a{1} = {[0 0 1 4 1]
[9 9 0 0 0]
[2 3 5 1 2]};
a{2} = {[1 0 7 4 0]
[2 8 8 2 6]
[1 9 8 1 1]};
a{3} = {[7 7 6 0 6]
[9 0 9 1 2]
[3 9 2 4 6]};
a{4} = {[3 3 3 2 3]
[4 1 5 6 3]
[9 0 0 0 7]};
% A non-compact way to get my desired output, and that I would like to avoid or improve:
for i = 1 : 3
a2{i} = {[a{1}{i,:} a{2}{i,:} a{3}{i,:} a{4}{i,:}]};
end
c = vertcat(a2{:});
% Almost what you want
b = horzcat(a{:})
for row=1:size(b,1)
b{row,1} = horzcat(b{row,:});
end
b(:,2:end)=[]; % remove excess columns
%test equality
isequal(b,c)
ans = logical
1
##### 1 KommentarKeine anzeigenKeine ausblenden
Sim am 9 Feb. 2023
Bearbeitet: Sim am 9 Feb. 2023
Dear @Stephen23, Dear @Rik, thanks a lot for your contributions!
Yes, the trick was made by
[a{:}] % or horzcat(a{:})
and
[tmp{k,:}] % or horzcat(b{row,:})
.......and I did not think about it! :-)
Please, consider that I would have accepted both answers, since they are very similar!
However, for this time, I will accept the @Stephen23's answer since it is slightly more compact than the @Rik's one. Indeed, it has not the @Rik's command to remove the excess columns:
b(:,2:end)=[]; % remove excess columns
Again, many many thanks to both of you for your time and contributions!!

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