concatanate arrays upon condition

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Dear MatLAb Community,
I would like to concatanate selected coloumns of arrays, some of which are empty arrays, and this is varying dataset by dataset.
How can I exclude the empty arrays from concatanation?
Let's say now , distr3 and distr6 are empty but this my vary.
distr = cat(1,distr1(:,ch1),distr2(:,ch2),distr3(:,ch3),distr4(:,ch4),distr5(:,ch5),distr6(:,ch6),distr7(:,ch7),distr8(:,ch8));
Thans for you suggestions
BEst
lg

Accepted Answer

Matt J
Matt J on 20 Jun 2025
Edited: Matt J on 20 Jun 2025
The way your data is held, as separate enumerated variables, doesn't leave many options. The first thing would be to undo that:
distrCell = arrayfun(@(i)eval("distr"+i),1:8,'uni',0); %should be unnecessary
chCell = arrayfun(@(i)eval("ch"+i),1:8,'uni',0); %should be unnecessary
Then, you can do,
keep=~cellfun('isempty',distrCell);
distrCell=distrCell(keep);
chCell=chCell(keep);
distr=cell2mat( cellfun(@(d,c) d(:,c), distrCell,chCell ,'uni',0 ) )
  2 Comments
Levente Gellért
Levente Gellért on 20 Jun 2025
Moved: Torsten on 20 Jun 2025
Dear Matt J, first I started to understand and then run your first line. Although the distr1 variable is in the workspace, the following error message comes:
Error using eval
Undefined function or variable 'distr1'.
Error in @(i)eval("distr"+i)
matlab.internal.language.introspective.errorDocCallback('@(i)eval(
Error: Character vector is not terminated properly.
Can you help further?
best
lg
Matt J
Matt J on 20 Jun 2025
Edited: Matt J on 20 Jun 2025
Modify along the following lines:
distr1=17; distr2=5; distr3=68;
distrCell = arrayfun(@(i)evalin('caller',"distr"+i),1:3,'uni',0) %should be unnecessary
distrCell = 1×3 cell array
{[17]} {[5]} {[68]}
Or, even better, recreate the data from scratch, but instead of distr1, distr2,... store the arrays in a distrCell variable directly.

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More Answers (1)

Walter Roberson
Walter Roberson on 20 Jun 2025
At least in R2025a, there is no problem in using "," to concatenate empty datasets, as long as they are empty along either the first or second dimension. (There can be problems if they are empty along the third dimension and the sizes of the first and second dimension do not match the existing data.)
A = zeros(5,3);
B = ones(5,0);
C = zeros(5,2);
D = []; size(D)
ans = 1×2
0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
E = ones(3,0);
F = ones(0,3);
G = ones(10,9,0);
[A,B,C]
ans = 5×5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[A,D,C]
ans = 5×5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[A,E,C]
ans = 5×5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[A,F,C]
ans = 5×5
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
[A,G,C]
Error using horzcat
Dimensions of arrays being concatenated are not consistent.
  4 Comments
Walter Roberson
Walter Roberson on 23 Jun 2025
Edited: Walter Roberson on 23 Jun 2025
Is the error you recieved due to the fact that you are subscripting an empty array at columns that do not exist?
T1 = randi(9,5,3)
T1 = 5×3
2 2 1 9 8 3 9 1 5 6 5 1 5 4 5
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
T2 = randi(9,7,4)
T2 = 7×4
5 7 6 3 6 9 9 3 2 3 1 3 1 6 7 5 9 4 2 3 8 3 8 7 1 1 2 4
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
T3 = randi(9,4,1)
T3 = 4×1
2 2 4 4
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
T4 = randi(9,6,3)
T4 = 6×3
3 2 8 2 9 7 2 4 7 6 1 5 3 7 7 1 8 1
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
S = @(A,c) A(:,c:c.*(c<=size(A,2)));
%use the auxillary function S to select columns
[S(T1,2); S(T2,2); S(T3,2); S(T4,2)]
ans = 18×1
2 8 1 5 4 7 9 3 6 4 3 1 2 9 4
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
%whereas directly indexing might fail
[T1(:,2); T2(:,2); T3(:,2); T4(:,2)]
Index in position 2 exceeds array bounds. Index must not exceed 1.
Walter Roberson
Walter Roberson on 23 Jun 2025
Huh, there is a difference between cat(1,A,B) and [A;B] ...
A = zeros(5,1);
B = ones(3,0);
[A;B]
ans = 5×1
0 0 0 0 0
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
cat(1, A, B)
Error using cat
Dimensions of arrays being concatenated are not consistent.

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