Using while loops in matrices

2022 12 17
1 个回答
更新时间：2022 12 17
1 次查看（30 天）

0 个投票

I am trying to use while loop to change the diagonal entries of a square matrix rand(10) to 1, and other entries to zero
This code below is changing the whole entries to 1, i am stucked.
m= 1:10
n= 1:10
A = rand(10)
B = size (A)
while m==n
A(m,n) = 1;
if not (m==n)
A(m,n) = 0;
end
break
end
A

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回答（1 个）

Pin-Hao Cheng
Pin-Hao Cheng 2022-12-17
Ran in:

0 个投票

Ran in:
Let me know if this is what you are looking for. Happy to answer any further questions!
A = rand(10)
A = 10×10
0.1406 0.9234 0.5213 0.0271 0.2317 0.3967 0.3776 0.4847 0.3080 0.0222 0.4363 0.4473 0.2888 0.3199 0.6071 0.3528 0.3311 0.5584 0.4040 0.2190 0.2656 0.6738 0.6987 0.3731 0.8456 0.9905 0.1035 0.4882 0.4701 0.2163 0.0655 0.9590 0.1978 0.6575 0.6074 0.2159 0.2553 0.3445 0.9210 0.6419 0.8856 0.9335 0.8339 0.5396 0.2503 0.0700 0.4051 0.7461 0.3052 0.1634 0.4934 0.4743 0.6264 0.2365 0.6042 0.4961 0.4183 0.7427 0.0465 0.3843 0.2964 0.5497 0.6846 0.4194 0.2852 0.9456 0.7047 0.9608 0.4107 0.6736 0.7785 0.3134 0.1016 0.5672 0.2764 0.7794 0.3809 0.2773 0.2288 0.1847 0.6817 0.5795 0.6303 0.4710 0.6249 0.1779 0.3139 0.5706 0.5860 0.4254 0.6862 0.1720 0.4387 0.9392 0.0916 0.3381 0.4475 0.3649 0.5224 0.7542
for m = 1:10 % loop through rows
for n = 1:10 % loop through columns
if m == n % check if it's diagonal el
A(m,n) = 1;
else
A(m,n) = 0;
end
end
end
A
A = 10×10
1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1

7 个评论
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Olabayo
Olabayo 2022-12-17
This is using the FOR loop. I want to use the WHILE loop for achive this same result
Askic V
Askic V 2022-12-17
编辑：Askic V 2022-12-17
Ran in:
Ran in:
How about this:
% Initialize matrix A with random values
N = 5;
A = rand(N)
A = 5×5
10 2 6 7 8 6 2 7 9 7 3 6 6 2 4 2 6 6 4 9 1 7 3 1 1
% processing the elements
while N > 0
A(N,:) = 0;
A(N,N) = 1;
N = N-1;
end
A
A = 5×5
1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
Walter Roberson
Walter Roberson 2022-12-17
Ran in:
Ran in:
A = rand(10)
A = 10×10
0.5267 0.0485 0.6881 0.0751 0.5088 0.9623 0.4144 0.0369 0.5932 0.1203 0.0743 0.4432 0.1701 0.8212 0.2084 0.4752 0.3201 0.4127 0.1457 0.4379 0.4294 0.8825 0.5937 0.8722 0.1412 0.2457 0.6319 0.1385 0.0125 0.1849 0.1902 0.2588 0.2948 0.1180 0.3554 0.0058 0.1091 0.1462 0.6565 0.7121 0.3901 0.4582 0.6801 0.5591 0.1280 0.3830 0.4597 0.2188 0.6096 0.7307 0.7630 0.4357 0.6976 0.8166 0.7730 0.5614 0.6620 0.1153 0.8800 0.3424 0.9485 0.5221 0.1301 0.3194 0.6058 0.7353 0.1995 0.8348 0.8918 0.8874 0.3259 0.7321 0.2832 0.4322 0.1251 0.5902 0.5059 0.1353 0.5748 0.5123 0.2665 0.2283 0.9713 0.8249 0.1909 0.8411 0.5445 0.3293 0.1103 0.8057 0.1126 0.5050 0.2498 0.4914 0.5326 0.4910 0.3327 0.8345 0.2003 0.1004
m = 1;
while m <= 10 % loop through rows
n = 1;
while n <= 10 % loop through columns
if m == n % check if it's diagonal el
A(m,n) = 1;
else
A(m,n) = 0;
end
n = n + 1;
end
m = m + 1;
end
A
A = 10×10
1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1
Pin-Hao Cheng
Pin-Hao Cheng 2022-12-17
Was gonna comment exactly what @Walter Roberson showed. That should solve your problem :)
Olabayo
Olabayo 2022-12-17
Thank you for this.. this solves it
Jan
Jan 2022-12-17
The pattern:
if m == n
A(m,n) = 1;
else
A(m,n) = 0;
end
can be abbreviated in general to:
A(m, n) = (m == n);
Walter Roberson
Walter Roberson 2022-12-17
The whole thing abbreviates to a call to eye() and size()

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Olabayo
Olabayo
2022-12-17

评论：

Walter Roberson
Walter Roberson
2022-12-17

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