Have any matrix's column same number?
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Hello. I have a question. If you can help me, i am pleasure.
Question: Have any matrix's column same number?
In my idea, my code just find number of diagonal.
clc;
clear;
tic
sum=0;
A=[1 2 3;4 5 3;7 8 9];
B=size(A);
m=B(1,1);
n=B(1,2);
for i=1:m
for j=1:n
if((A(i,j) == A(j,j) || A(j,i) == A(i,i)) && i ~= j )
sum=sum+1;
end
end
end
if sum==0
disp('different')
else
disp('same')
end
A
toc
7 件のコメント
James Tursa
2015 年 5 月 19 日
編集済み: James Tursa
2015 年 5 月 19 日
I don't understand the question completely. Is your current code not producing the result you want? Are you trying to find out if any diagonal elements are duplicated in the same column? Are you trying to generalize your current code? Or what? Can you give a numeric example?
Mehmet Gurturk
2015 年 5 月 19 日
編集済み: Mehmet Gurturk
2015 年 5 月 19 日
Stephen23
2015 年 5 月 19 日
@Mehmet Gurturk: the explanation is not completely clear. Can you please give us an exact example, with numbers, like this:
A = [1,2,3;4,5,6;1,8,9]
what should happen? What output do you want?
Mehmet Gurturk
2015 年 5 月 19 日
編集済み: Image Analyst
2015 年 5 月 19 日
Mehmet Gurturk
2015 年 5 月 19 日
Joseph Cheng
2015 年 5 月 19 日
Stephen he is trying to say that in the code he has presented it marks the second example as "different" due to the fact that his code only checks the diagonals.
回答 (3 件)
Joseph Cheng
2015 年 5 月 19 日
編集済み: Joseph Cheng
2015 年 5 月 19 日
you can use the function unique() to determine if there are repeated values within the matrix
x = randi(10,4,4);
disp(x);
for ind = 1:size(x,2)
if numel(unique(x(:,ind)))<numel(x(:,ind))
disp('same')
else
disp('different')
end
end
1 件のコメント
Joseph Cheng
2015 年 5 月 19 日
if you only want 1 answer then
x = randi(10,4,4);
disp(x);
same = 0;
for ind = 1:size(x,2)
if numel(unique(x(:,ind)))<numel(x(:,ind))
same = same+1;
end
end
if same>0
disp('same');
else
disp('different');
end
Here is a very fast and simple solution that does not require any loops (i.e. fully vectorized code). An output of 0 indicates different, and output of 1 indicates same.
>> A = [1,2,3; 4,5,6; 7,8,9]
A =
1 2 3
4 5 6
7 8 9
>> any(any(0==diff(sort(A))))
ans =
0
>> A = [1,2,3; 4,5,6; 1,8,9]
A =
1 2 3
4 5 6
1 8 9
>> any(any(0==diff(sort(A))))
ans =
1
Here it is on your second example matrix:
>> A = [1,2,3; 4,5,3; 7,8,9]
A =
1 2 3
4 5 3
7 8 9
>> any(any(0==diff(sort(A))))
ans =
1
And if you want to know which column contains repeated values then simply remove one of the any commands:
>> any(0==diff(sort(A)))
ans =
0 0 1
This code works very simply: it sorts the columns, then uses diff to find if any adjacent values are the same. You can use the all(..) to control the if statement, like this:
if all(...)
disp('same')
else
disp('different')
end
1 件のコメント
Joseph Cheng
2015 年 5 月 19 日
very nice.. I like the use of sort.
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