# How to concatenate horizontally a cell into another cell based on a vector

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endystrike on 13 Apr 2021 at 14:20
Commented: endystrike on 13 Apr 2021 at 18:15
Hi,
I want to concatenate horizzontally with a criteria based on a vector this cell into another cell. All the double matrices inside the "wl_tmp" cell have the same number of rows, but not the same number of colums!
I have this vector and in my idea, the elements with the same value should be grouped together (concatenating them horizontally from left to right).
For example: the final cell is "wl_final":
wl_final{1} = [wl_tmp{1}, wl_tmp{2}, wl_tmp{3}, wl_tmp{4}, wl_tmp{5}, wl_tmp{6}, wl_tmp{7}, wl_tmp{8}, wl_tmp{9}, wl_tmp{10}, wl_tmp{11}];
wl_final{2} = [wl_tmp{12}, wl_tmp{13}];
%and so on...
I found a solution with a for loop but because of the high amount of data, it's terribly slow and I wanted to figure out how I could use "accumarray" instead, but I cannot understand how that function works exactly and how to use it for my purposes.
wl_tot = length(wl_idx);
wl_final = cell(1,length(pX_i_ref_wl)); %"pX_i_ref_wl" contains just the name of the groups: in this case is a 1x11 string
clear tmp;
for k=1:wl_tot
if wl_idx(k)~=wl_idx(max(k-1,1)) || k==1
tmp = wl_tmp{k};
else
tmp = [tmp, wl_tmp{k}]; %#ok<AGROW>
end
if k==wl_tot || wl_idx(k+1)~=wl_idx(k)
wl_final{wl_idx(k)} = tmp;
clear tmp;
end
end
P.S. I cannot group them before, because the "wl_tmp" comes from a parfor loop with sliced variables and I'd prefer to group them later!

Stephen Cobeldick on 13 Apr 2021 at 14:42
Edited: Stephen Cobeldick on 13 Apr 2021 at 14:54
X = [1,1,2,2];
C = {rand(3,2),rand(3,1),rand(3,2),rand(3,1)};
C{:}
ans = 3×2
0.3878 0.8399 0.9283 0.7467 0.7731 0.5606
ans = 3×1
0.5279 0.0653 0.8680
ans = 3×2
0.5664 0.3253 0.0695 0.4880 0.0237 0.4730
ans = 3×1
0.5007 0.0762 0.1294
Method one: arrayfun
F = @(x)[C{x==X}];
D = arrayfun(F,1:max(X),'uni',0);
D{:}
ans = 3×3
0.3878 0.8399 0.5279 0.9283 0.7467 0.0653 0.7731 0.5606 0.8680
ans = 3×3
0.5664 0.3253 0.5007 0.0695 0.4880 0.0762 0.0237 0.4730 0.1294
Method two: accumarray
Y = 1:numel(X);
F = @(y){[C{y}]};
D = accumarray(X(:),Y(:),[],F);
D{:}
ans = 3×3
0.3878 0.8399 0.5279 0.9283 0.7467 0.0653 0.7731 0.5606 0.8680
ans = 3×3
0.5664 0.3253 0.5007 0.0695 0.4880 0.0762 0.0237 0.4730 0.1294
endystrike on 13 Apr 2021 at 18:15
Thanks a lot, your solution works perfectly! :)

R2020a

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