Help for a function which works like triu function in matlab

2020 3 1
1 个回答
更新时间：2020 3 4
4 次查看（30 天）

0 个投票

we have seen the function triu(A,k) which extracts upper triangular part of A. But i have to write my own function with the header U = myTriu(A,k) which does the same thing and apply it with A= (2 5 6 4 8 9 10 15 12),k) where k=-2,-1,1
Could anyone help me,please?

3 个评论
显示 1更早的评论 隐藏 1更早的评论

Steven Lord
Steven Lord 2020-3-1
This sounds like a homework assignment. If it is, show us the code you've written to try to solve the problem and ask a specific question about where you're having difficulty and we may be able to provide some guidance.
If you aren't sure where to start because you're not familiar with how to write MATLAB code, I suggest you start with the MATLAB Onramp tutorial (https://www.mathworks.com/support/learn-with-matlab-tutorials.html) to quickly learn the essentials of MATLAB.
If you aren't sure where to start because you're not familiar with the mathematics you'll need to solve the problem, I recommend asking your professor and/or teaching assistant for help.
clc
clear all
close all
function U = myTriu(A,k)
for i=1:10
for j=1:10
if (i==1|| j==1 || i=10|| j=10
fprintf('i');
else fprintd('j');
end
end
end
I have written this but it does not seem to be correct
clear all
close all
function U = myTriu(A,k)
for (i=0 ; i<row; i++
{
for (j=0; j<col ;j++)
{
if (i>j)
{
matrix[i][j]=0;
}
count << matrix[i][j] << '' '' ;
}
count << end1
}
}

请先登录，再进行评论。

请先登录，再回答此问题。

回答（1 个）

Sai Sri Pathuri
Sai Sri Pathuri 2020-3-4

0 个投票

I think the first code you posted is not relevant to this question and the second code is written in C++ for k = 0 case.
You may use the following code which has a slight modification (checking for i + k > j) for all values of k
function U = myTriu(A,k)
for i = 1: size(A,1)
for j = 1:size(A,2)
if i + k > j
A(i,j) = 0;
end
end
end
U = A;
end

类别

在 帮助中心 和 File Exchange 中查找有关 Mathematics 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by