How to get a subarray from a matrix
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Hi. I am looking to solve this Question: Write a function called top_right that takes two inputs: a matrix N and a scalar non-negative integer n, in that order, where each dimension of N is greater than or equal to n. The function returns the n-by-n square array at the top right corner of N.
So I am looking to take the top right portion of matrix N to get a subarray that is n by n units. I thought the code might look like this:
function M = top_right(N,n)
M = N(end-n+1:end, 1:n);
However, this gives me the bottom left of the matrix and i want the top right.
6 件のコメント
Vijay Ramanathan
2017 年 10 月 2 日
I was trying M= [N(1:n,end-n+1:end)] with M= [N(1:n,end-n:end)] before i saw your answer. I thought it wont apply for all but later it did :) Thanks Karthik and Yasin
回答 (3 件)
James Tursa
2016 年 8 月 5 日
編集済み: James Tursa
2016 年 8 月 5 日
Switch your row and column indexing. E.g.
M = N(1:n,end-n+1:end);
0 件のコメント
Star Strider
2016 年 8 月 5 日
That’s different than the wording of your earlier Question.
As far as the indexing code, see if this does what you want:
N = randi(99,8,6) % Argument Matrix
[R,C] = size(N);
min_dim = min(R,C);
n = randi(min_dim-1) % Argument Parameter
M = N(1:n, C-n+1:C) % Return Top Right Corner Square Matrix
N =
79 65 20 86 17 8
10 49 4 80 97 52
26 78 74 58 71 10
34 71 50 19 50 81
68 90 48 24 47 81
14 89 90 88 6 72
72 34 61 3 68 15
11 70 62 49 5 66
n =
3
M =
86 17 8
80 97 52
58 71 10
0 件のコメント
Aakriti Mittal
2018 年 5 月 30 日
function N=top_right(M,n) T = M(1:n,1:n); N=T; Try this code...Hopefully it will work.. Aakriti
1 件のコメント
Guillaume
2018 年 5 月 30 日
- please learn to format your answer
- You're answering a two year old question. Hopefylly, the OP is still not waiting for an answer
- Your function returns the top left portion of the matrix, not the top right
- What is the point of
T = ...;
N = T;
? Why not write directly
N = ...;
and not have an intermediate variable that serves zero purpose?
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