# Modify a cell array referring to another cell array

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luca on 24 Sep 2019
Commented: luca on 24 Sep 2019
Given the following code
G= {[1 2 1 2 1 2 3 4 5 4 5 4 6 3 9 3 9; 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 85],[1 1 1 1 2 3 4 4 4 5 4 6 3 6 3 9; 5 10 15 20 25 30 35 40 45 50 55 60 65 70 75 80 ]};
SP= [1 2 3 4 5 6 9];
t1 = 20; t2 = 20 ; t3 = 20 ; t4 = 20 ; t5 = 20 ; t6 = 20 ; t7 = 20; t8=20 ; t9=20;
for i = 1:size (G,2)
result = cell(size(G));
for gidx = 1:numel(G)
[uval, loc1, ids] = unique(G{gidx}(1, :));
count = accumarray(ids, 1)';
result{gidx} = arrayfun(@(v, s, n) [repelem(v, n); s, zeros(1, n-1)], uval, G{gidx}(2, loc1), count, 'UniformOutput', false);
end
AB = G{i}(1,:);
end
With the value contained in the fist raw of each cell G, I want to modify the arrays contained in cell "result"
I for example we take the first raw of the first cell of G, reported as AB in each cycle:
1 2 1 2 1 2 3 4 5 4 5 4 6 3 9 3 9
at each iteration everytime I meet a specified element, I want to modify its array in result.
In particular, if I consider result{1, 1}{1, 1}" that was
1 1 1
5 0 0
Should now become
1 1 1
5+t1 5+t1+t1 5+t1+t1+t1
so
1 1 1
25 45 65
That is: every time I meet 1 in AB, I want to modify the values inside "result" that refer to that element and add everytime the value t1 to the previous one.
For element 2, at the beginning the result was
2 2 2
10 0 0
Then has to become
2 2 2
10+t2 10+t2+t2 10+t2+t2+t2
...
For element 9, at the beginning the result was
9 9
75 0
Then has to become
9 9
75+t9 75+t9+t9
I hope that it's clear what I would like to obtain
May someone help me with this task?
luca on 24 Sep 2019
yes !

Adam Danz on 24 Sep 2019
Edited: Adam Danz on 24 Sep 2019
Replace your t1,t2, t3... with this
tt = [20 20 20 20 20 20 20 20 20]; %Much easier to work with
Then replace the result{} line in the for(gidx...) loop with this
result{gidx} = arrayfun(@(v, s, n, t) [repelem(v, n); s + t.*(1:n)], uval, G{gidx}(2, loc1), count, tt(uval), 'UniformOutput', false);
% Changes > > > > > > > > > > > 1) ^ > > > > > > > 2) ^^^^^^^^^^ > > > > > > > > > > > > > > > 3) ^^^^^^^^
luca on 24 Sep 2019
Thanks a lot

R2019b

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