how to find sum different data for two different variables using loop in matlab

Question

M.Rameswari Sudha 2022-4-28

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1707140-how-to-find-sum-different-data-for-two-different-variables-using-loop-in-matlab

评论： Andy 2022-4-28

I have different datas for two different variables (lambdai and b1). I need to find the separate M1 values and need to find sum M1 also. I couldn't find the answer. let me know if you will get the answer.

clc
clear all 
lambdai=[0 5.6 22.4 57.4];
b1 = [100 80 120 90 70];
k=1
j=1;
l=1;
m=1;
for i=1:length(lambdai)
    for s=1:length(b1)
        M1=(b1(s).*lambdai(i))./lambda;
        M(j,k)=M1;
        j=j+1;
        k=k+1;
    end
end
sumM=sum(M)

4 个评论
显示 2更早的评论隐藏 2更早的评论

M.Rameswari Sudha 2022-4-28

yes. lamda = 100

Andy 2022-4-28

Once I set a value for lambda it works, All the values for M1 are stored in the array M(. , . ) . sum(M) is calculated as expected but perhaps you want to sum all numbers in the array in which case it should be

sumM = sum(M , 'all');

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Jan 2022-4-28

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1707140-how-to-find-sum-different-data-for-two-different-variables-using-loop-in-matlab#answer_952910

在 MATLAB Online 中打开

Maybe you want to do: "Multiply each element of vector a with all elements of vector b, devide by 100 and get the sum of all results."

If you formulate the procedure clearly in naturla language, this is a good structure for the Matlab code also.

lambdai = [0 5.6 22.4 57.4];
b1      = [100 80 120 90 70];
lambda  = 100;
M       = zeros(length(lambdai), length(b1));
for i = 1:length(lambdai)
    for s = 1:length(b1)
        M(i, s) = b1(s) * lambdai(i) / lambda;
    end
end
sumM = sum(M, 'all')
sumM = 392.8400

Of course you do not have to devide each element by lambda, because it is sufficient to divide the result only.

Matlab offers a way to do this without loops:

lambdai = [0 5.6 22.4 57.4];
b1      = [100 80 120 90 70];
lambda  = 100;
M       = lambdai.' * b1;   % dyadic product: [4x1]*[1x5] = [4x5]
% Or:
% M     = lambdai .* b1.';  % elementwise product: [1x4]*[5x1] = [5x4]
sumM    = sum(M, 'all') / lambda
sumM = 392.8400