Replace numbers in a matrix depending on if statements from arrays

3 ビュー (過去 30 日間)
Askeladden2
Askeladden2 2020 年 6 月 10 日
コメント済み: Askeladden2 2020 年 6 月 10 日
Dear all Community members,
I have a 5x5 matrix and two arrays (1x5 and 5x1) that contains only numerical values. I want to remove values below a variable threshold matrix and add the sum of these values to the next horisontal element (row wise) that is above the threshold.
My input matrix is as follows:
A=[1 6 7 8 13; 1 3 11 8 15; 1 2 4 13 19; 1 3 4 8 11; 1 3 4 5 11];
h=[0.5;1;2;2.5;3];
t=[1 2 3 4 5];
A=[1 6 7 8 13; 1 3 11 8 15; 1 2 4 13 19; 1 3 4 8 11; 1 3 4 5 11];
for i = 1:size(h)
for j = 1:size(t)
thres(i,j)=3.1*sqrt(h(i))
end
end
Resulting threshold matrix will be:
thresh=[2.2 2.2 2.2 2.2 2.2; 3.1 3.1 3.1 3.1 3.1; 4.4 4.4 4.4 4.4 4.4; 4.9 4.9 4.9 4.9 4.9; 5.4 5.4 5.4 5.4 5.4];
I.e. I want to replace all values <=thresh with zeros and add the sum of these to the next horisontal element that is above the threshold. So the output matrix shall be:
B=[0 7 7 8 13; 0 0 15 8 15; 0 0 0 20 19; 0 0 0 16 11; 0 0 0 0 24];
Can anyone assist me?
Thanks in advance.

採用された回答

madhan ravi
madhan ravi 2020 年 6 月 10 日
編集済み: madhan ravi 2020 年 6 月 10 日
thresh = 3.1 * sqrt(h);
ix = A <= thresh;
s = sum(A .* ix,2);
ix1 = cumsum(~ix,2)==1;
Wanted = cumsum(ix1,2) .* ...
(ix1 .* s + A)
  1 件のコメント
Askeladden2
Askeladden2 2020 年 6 月 10 日
Dear madhan ravi,
Thank you for your help

サインインしてコメントする。

その他の回答 (2 件)

Ameer Hamza
Ameer Hamza 2020 年 6 月 10 日
編集済み: Ameer Hamza 2020 年 6 月 10 日
This is one of the solutions. A more direct solution is probably possible,
A=[1 6 7 8 13; 1 3 11 8 15; 1 2 4 13 19; 1 3 4 8 11; 1 3 4 5 11];
h=[0.5;1;2;2.5;3];
thresh = 3.1*sqrt(h);
mask1 = A > thresh;
mask2 = ~[zeros(size(A,1), 1) mask1(:, 1:end-1)];
mask = mask1.*mask2;
B = cumsum(A.*mask2, 2).*mask;
A_new = (A.*~mask + B).*mask1;

rajkumar k
rajkumar k 2020 年 6 月 10 日
Try this code:
A=[1 6 7 8 13; 1 3 11 8 15; 1 2 4 13 19; 1 3 4 8 11; 1 3 4 5 11];
h=[0.5;1;2;2.5;3];
t=[1 2 3 4 5];
[ta tb]=size(t);
for i = 1:size(h)
for j = 1:tb
thres(i,j)=3.1*sqrt(h(i))
end
end
for i=1:size(h)
b=0;
for j=1:tb
if A(i,j)<=thres(i,j)
b=A(i,j)+b;
B(i,j)=0;
else
B(i,j)=b+A(i,j);
end
end
end

カテゴリ

Help Center および File ExchangeMultidimensional Arrays についてさらに検索

製品


リリース

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by