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How to concatenate table variables with underscores in their name

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I have variables like B1_1_3(53x1) B2_1_3(53x1) until B56_1_3(53x1) stored in my table A. I want to concatenate them to receive a new vector(53x56). I tried:
for k in 1:56
a = [A.B{k}_1_3, A.B{k}_1_3];
end
However this does not work. Any suggestions? Thank you in advance!
  1 Comment
infinity
infinity on 23 Jul 2019
What do you mean by "concatenate them to receive a new vector (53x56) "?
For example,
A = [1 2 3]
B = [1 2 3 4]
what is your output of a new matrix?

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Accepted Answer

Adam Danz
Adam Danz on 23 Jul 2019
Edited: Adam Danz on 23 Jul 2019
No loop needed. Just list all of the table headers, pull out the ones that take the form "B#_1_3", and then create a matrix based on values from those columns.
% Create demo table
A = array2table(rand(6,4),'variableNames',{'B1_1_3','B2_1_3','B3_1_3','total'});
% List all headers
headers = A.Properties.VariableNames;
% Extract headers that have the form B#_1_3
goodHeaders = regexp(headers,'B\d+_1_3','Match');
goodHeaders = [goodHeaders{:}]';
% Create matrix of values from good header columns *see Guillaume's comment below
m = cell2mat(cellfun(@(x)A.(x)',goodHeaders,'UniformOutput',false))';
  2 Comments
Adam Danz
Adam Danz on 23 Jul 2019
Another approach would be to remove the columns you don't want and keep the data in a table which has its advantages. That solution would look like this.
% Create demo table
A = array2table(rand(6,4),'variableNames',{'B1_1_3','B2_1_3','B3_1_3','total'});
% List all headers
headers = A.Properties.VariableNames;
% Extract headers that have the form B#_1_3
goodHeaderIdx = regexp(headers,'B\d+_1_3');
badHeaderIdx = cellfun(@isempty,goodHeaderIdx);
% Create matrix of values from good header columns
A2 = A;
A2(:,badHeaderIdx) = [];

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More Answers (1)

llueg
llueg on 23 Jul 2019
Edited: llueg on 24 Jul 2019
You could use eval for this. Assuming your data in the table is numeric and you want a matrix as result:
a = zeros(53,56);
for i=1:56
eval(['a(:,i) = A.B' num2string(i) '_1_3'])
end
Edit: It has been pointed out that using eval is unnecessary here and bad in general. See the comments below for better solutions.
I just considered this to be closest to what OP was trying to do, but there are certainly better methods.
  3 Comments
Adam Danz
Adam Danz on 23 Jul 2019
Edited: Adam Danz on 24 Jul 2019
There's no need for a loop and certainly no need to use eval(). This is bad advice (please look at the link provided by Guillaume above).
[update]
@llueg, thanks for clarifying your answer with your [edit].
Guillaume
Guillaume on 23 Jul 2019
Gah! Why did you accept this answer. Again, eval is bad! extremely bad!. See the link in my comment above for all the ways that it is bad.
Do not use eval and certainly not for this. I've demonstrated a much more reliable method (and simpler, and faster, and etc... read the link).
Adam's method is even more reliable in that it doesn't assume that variables 1 to 56 do exist but check which ones exist.

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