i want to code such that the loop is constraint for two different values?

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kitty varghese
kitty varghese 2017-8-28
编辑： KL 2017-8-28
I want to code this equation
s(i)=x(i)+(l1-sum(x(i))/N
here the loop must be i=1:108300
and the value for l1 and x(i) must be same for first 300
here I something I tried.
for i2=1:108300
for i3=1:300
s(i2)=W(i2)+(l1-sum(W(:,i3)))/N;
end
end
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KL
KL 2017-8-28
编辑：KL 2017-8-28
Your question is still not clear.
What do you mean by the value for l1 and x(i) must be same for first 300? Is l1 a constant or a vector (like x)? If so, does it change only every 300 rows?
and what do you mean by sum(x(i).. Do you wanna make a sum of every 300 rows and keep it constant?
kitty varghese
kitty varghese 2017-8-28
l1 is a constant which is same for batches of 300. sum(x(i)) is the sum of rows which i need to keep it constant for batches of 300.
therefore for easy, i have already calculated sum(x(i)) and kept in array k.

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采纳的回答

KL
KL 2017-8-28
编辑：KL 2017-8-28
N = 108300;
x = rand(N,1);
sum_rows = 300;
l1 = x(1:sum_rows:N); %take every 300th value
l1_r = reshape(repmat(l1,1,sum_rows)',N,1); %repeat it
sum_x = sum(reshape(x,N/sum_rows,sum_rows),2); %create a sum of every 300 rows
sum_x_r = reshape(repmat(sum_x,1,sum_rows)',N,1); %repeat 361 times to create a vector of N length
s=x+(l1_r-sum_x_r)./N;

更多回答（1 个）

Walter Roberson
Walter Roberson 2017-8-28
N = 108300;
x = rand(1, N);
x(1:300) = l1(1:300);
i = 1 : N;
s(i) = x(i)+(l1-sum(x(i))/N;
  1 个评论
kitty varghese
kitty varghese 2017-8-28
@Walter Roberson I want this process for the batch of every 300. ie the first 300 will take one value and second batch of 300 will take one value and so on ...for 361 times.

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