Replace diagonals in a matrix

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Hasan Hassoun
Hasan Hassoun 2021-1-19
评论： Image Analyst 2021-1-23
Hello every one,
How to replace the upper and lower part of a n*n matrix with zeros (The upper part starts from "diagonal+2" until n while the lower part starts from "diagonal-2" until n)?
  1 个评论
Image Analyst
Image Analyst 2021-1-23
Original question is below in case he deletes it like he's done before:
Replace diagonals in a matrix
Hello every one,
How to replace the upper and lower part of a n*n matrix with zeros (The upper part starts from "diagonal+2" until n while the lower part starts from "diagonal-2" until n)?

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采纳的回答

Matt J
Matt J 2021-1-19
Ran in:
For example,
A=rand(10),
A = 10×10
0.3632 0.5683 0.2698 0.2133 0.4828 0.8948 0.0478 0.7711 0.5088 0.7301 0.8024 0.1100 0.1475 0.6313 0.8957 0.8629 0.1435 0.8356 0.1381 0.8203 0.8949 0.0767 0.3939 0.2073 0.3055 0.5721 0.7656 0.1443 0.3270 0.6013 0.5722 0.1502 0.8491 0.7124 0.9746 0.7291 0.9774 0.9558 0.4083 0.0934 0.1948 0.7491 0.7331 0.3686 0.4528 0.2436 0.2373 0.4254 0.9163 0.6997 0.4224 0.4079 0.6316 0.8894 0.8130 0.2013 0.0054 0.5217 0.1405 0.7939 0.0219 0.5942 0.3860 0.2447 0.6384 0.9649 0.3000 0.2048 0.2233 0.1450 0.4361 0.2666 0.2646 0.3323 0.6236 0.6866 0.9302 0.3413 0.4398 0.0531 0.3450 0.6491 0.0609 0.1894 0.8414 0.7489 0.1080 0.4159 0.2769 0.3304 0.0411 0.9652 0.9593 0.3298 0.9979 0.3895 0.2349 0.0686 0.9871 0.3078
mask=tril( triu( true(size(A)), -2 ), +2);
B=A.*mask
B = 10×10
0.3632 0.5683 0.2698 0 0 0 0 0 0 0 0.8024 0.1100 0.1475 0.6313 0 0 0 0 0 0 0.8949 0.0767 0.3939 0.2073 0.3055 0 0 0 0 0 0 0.1502 0.8491 0.7124 0.9746 0.7291 0 0 0 0 0 0 0.7331 0.3686 0.4528 0.2436 0.2373 0 0 0 0 0 0 0.8894 0.8130 0.2013 0.0054 0.5217 0 0 0 0 0 0 0.6384 0.9649 0.3000 0.2048 0.2233 0 0 0 0 0 0 0.6866 0.9302 0.3413 0.4398 0.0531 0 0 0 0 0 0 0.1080 0.4159 0.2769 0.3304 0 0 0 0 0 0 0 0.0686 0.9871 0.3078

更多回答（1 个）

David Goodmanson
David Goodmanson 2021-1-19
编辑：David Goodmanson 2021-1-19
Hi Hasan,
here is one way
r = rand(7,7)
n = size(r,1);
m = (n-1)/2;
a = (-m:m)-(-m:m)';
r(abs(a)>1)=0
assuming the main diagonal is number 0, and the zeros start on the +-2nd diagonal, otherwise adjust the '1' on the last line of code accordingly

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