How to apply the same operation to an ever increasing number of columns?

Question

Zulfiqar Alibhai 2023-11-9

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2044897-how-to-apply-the-same-operation-to-an-ever-increasing-number-of-columns

评论： William Rose 2023-11-9

采纳的回答： Dyuman Joshi

lets say you have an array called A = (1:7) and a matrix called B = (1:20,1:7)

if I want to multiply each row from B with a value from A and than sum each row, it's: sum(A.*B')

If I want to do a specific row, it's: (A(2).*B(1:20,2))

if I want to do the first 5 rows, it's: sum(A(1:5).*B(1:20,1:5))

But what I want to do is to create a new matrix where each row is the sum of one more colum then the last.

Basically, it would look like this:

How would I go about doing that?

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Dyuman Joshi 2023-11-9

3
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2044897-how-to-apply-the-same-operation-to-an-ever-increasing-number-of-columns#answer_1349297

在 MATLAB Online 中打开

C = cumsum(A.*B, 2)

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 2

Onesimus Hewett 2023-11-9

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2044897-how-to-apply-the-same-operation-to-an-ever-increasing-number-of-columns#answer_1349272

在 MATLAB Online 中打开

The easiest method is to use a for loop.

C = zeros(20, 7);
for i = 1:7
    C(:, i) = B(:, 1:i)*A(1:i)';
end
C

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

Answer 3

William Rose 2023-11-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2044897-how-to-apply-the-same-operation-to-an-ever-increasing-number-of-columns#answer_1349292

在 MATLAB Online 中打开

@Zulfiqar Alibhai

[I moved my suggestion from comment section to answer section.]

a=[1:7]; b=zeros(20,7); c=b;
for j=1:7, b(:,j)=10*[1:20]'+j; end
disp(b(1:3,:))
    11    12    13    14    15    16    17
    21    22    23    24    25    26    27
    31    32    33    34    35    36    37
for j=1:7
    for i=1:j
        c(:,j)=c(:,j)+b(:,i)*a(i);
    end
end
disp(c(1:3,1:3))
    11    35    74
    21    65   134
    31    95   194