Save variable number in matrix for later iteration
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Ok, this is the code I have:
clc, clearvars, home,
lim = input("Limit value: ");
V = zeros(1,lim);
V(1) = 5;
for i = 2:lim
if mod(i,2) == 0
V(i) = V(i-1) + 2;
else
V(i) = V(i-1) + 4;
end
end
I = zeros(1,lim);
for i = 1:lim
I(i) = (V(i)^2-1)/3;
end
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;
for i = 2:lim
if mod(i,2) == 0
J1(i) = J1(i-1) + 2;
J2(i) = J2(i-1) + 2;
else
J1(i) = J1(i-1) + 2;
J2(i) = J2(i-1) + 6;
end
end
J12 = [J1; J2];
J = sum(J12,1);
C = J-I;
K(1) = @(x) (x+C(1))/J(1);
for i = 2:lim
K(i) = @(x) K{i-1} + (x+C(i))/J(i);
end
Val = zeros(1,max(I));
K = 0;
count = 0;
for i = 1:max(I)
if ismember(i,I) == 1
count = count + 1;
K = K + K1(count)*i + K2(count);
Val(i) = i/4 - (K);
else
Val(i) = i/4 - (K);
end
end
It basicaly calculates three matrixes with values I, J and C with length "lim" and now I want to make a for loop where the value is for each i
Val(i) = i/4 - K
where I want K to be
K = i-C/J % count times
The thing is I want K to grow each time i appears in the matrix I, so for 2 occurences its desired value is
K = (i-C(1))/J(2) + (i-C(2))/J(2)
for 4 occurences is
K = (i-C(1))/J(2) + (i-C(2))/J(2) + (i-C(3))/J(3) + (i-C(4))/J(4)
and so.
The problem I'm facing is how to call a variable that contains a value that can change with the current index in the loop. I tried with functions, with the symbolic toolbox... And I'm currently stucked as all I make contains some kind of error.
If it's not clear what I want to make respond so I will clear it out. Thanks in advance for the help.
Réponse acceptée
Maybe like this ?
I'm not sure whether you want to take the elements of C and J according to the order of the elements in I or sort(I). In the case below, I and sort(I) are identical since I is strictly increasing.
lim = 4;
V = zeros(1,lim);
V(1) = 5;
for i = 2:lim
if mod(i,2) == 0
V(i) = V(i-1) + 2;
else
V(i) = V(i-1) + 4;
end
end
I = zeros(1,lim);
for i = 1:lim
I(i) = (V(i)^2-1)/3;
end
I
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;
for i = 2:lim
if mod(i,2) == 0
J1(i) = J1(i-1) + 2;
J2(i) = J2(i-1) + 2;
else
J1(i) = J1(i-1) + 2;
J2(i) = J2(i-1) + 6;
end
end
J12 = [J1; J2];
J = sum(J12,1)
C = J-I
Val = (1:max(I))/lim;
count = 0;
for k = I
count = count + 1;
for j = k:max(I)
Val(j) = Val(j) - (j + C(count))/J(count);
end
end
Val(8)
8/4 - ((8+2)/10)
Val(9)
9/4 - ((9+2)/10)
Val(18)
18/4 - ((18+2)/10 + (18-2)/14)
1 commentaire
Joan
le 19 Jan 2026
Plus de réponses (1)
"The problem I'm facing is how to call a variable that contains a value that can change with the current index in the loop. I tried with functions, with the symbolic toolbox... "
All you are doing is adding numbers, so you do not need complicated things like function handles or the symbolic toolbox. Here is a much simpler approach using SUM:
lim = 4;
% Vectorized V calculation using alternating pattern
increments = 2*ones(1,lim-1);
increments(2:2:end) = 4; % [2, 4, 2, 4, ...]
V = 5*ones(1,lim-1);
V(2:lim) = 5+cumsum(increments)
% Vectorized I calculation
I = (V.^2 - 1) / 3;
I = round(I)
% Vectorized J1 and J2 calculation
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;
increments_J1 = 2*ones(1,lim-1);
J1(2:lim) = 3 + cumsum(increments_J1)
increments_J2 = 2*ones(1,lim-1);
increments_J2(2:2:end) = 6; % [2, 6, 2, 6, ...]
J2(2:lim) = 7 + cumsum(increments_J2)
J = J1 + J2
C = J - I
% Vectorized Val calculation with cumulative K
Val = zeros(1,max(I));
K = 0;
for i = 1:max(I)
% Find all indices where I equals the current i
indices = find(I==i);
% Vectorized accumulation for all matching indices
if ~isempty(indices)
K = K + sum((i - C(indices)) ./ J(indices));
end
Val(i) = i/4 - K;
end
K
Val
3 commentaires
Compared to your original (non-vectorized) code:
lim = 4;
V = zeros(1,lim);
V(1) = 5;
for i = 2:lim
if mod(i,2) == 0
V(i) = V(i-1) + 2;
else
V(i) = V(i-1) + 4;
end
end
V
I = zeros(1,lim);
for i = 1:lim
I(i) = (V(i)^2-1)/3;
end
I
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;
for i = 2:lim
if mod(i,2) == 0
J1(i) = J1(i-1) + 2;
J2(i) = J2(i-1) + 2;
else
J1(i) = J1(i-1) + 2;
J2(i) = J2(i-1) + 6;
end
end
J1
J2
J12 = [J1; J2];
J = sum(J12,1)
C = J-I
lim = 4;
% Vectorized V calculation using alternating pattern
increments = 2*ones(1,lim-1);
increments(2:2:end) = 4; % [2, 4, 2, 4, ...]
V = 5*ones(1,lim);
V(2:lim) = 5 + cumsum(increments)
% Vectorized I calculation
I = round((V.^2 - 1) / 3)
% Vectorized J1 and J2 calculation
J1 = zeros(1,lim);
J2 = zeros(1,lim);
J1(1) = 3;
J2(1) = 7;
increments_J1 = 2*ones(1,lim-1); % [2, 2, 2, ...]
J1(2:lim) = 3 + cumsum(increments_J1)
increments_J2 = 2*ones(1,lim-1);
increments_J2(2:2:end) = 6; % [2, 6, 2, 6, ...]
J2(2:lim) = 7 + cumsum(increments_J2)
J = J1 + J2
C = J - I
% Initialize Val
Val = (1:max(I)) / 4;
% Apply corrections for each element in I
for count = 1:lim
k = I(count);
% Subtract (j + C(count))/J(count) from all Val(j) where j >= k
Val(k:end) = Val(k:end) - ((k:max(I)) + C(count)) ./ J(count);
end
% Display results
Val(8)
8/4 - ((8+2)/10)
Val(9)
9/4 - ((9+2)/10)
23/20
Val(18)
18/4 - ((18+2)/10 + (18-2)/14)
19/14
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