Have any matrix's column same number?

1 Ansicht (letzte 30 Tage)
Mehmet Gurturk
Mehmet Gurturk am 19 Mai 2015
Beantwortet: Mehmet Gurturk am 20 Mai 2015
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 Kommentare
Joseph Cheng
Joseph Cheng am 19 Mai 2015
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.
Stephen23
Stephen23 am 19 Mai 2015
Bearbeitet: Stephen23 am 19 Mai 2015
@Joseph Cheng: thank you for the clarification, I was a little bit confused about the meaning.

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Antworten (3)

Joseph Cheng
Joseph Cheng am 19 Mai 2015
Bearbeitet: Joseph Cheng am 19 Mai 2015
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 Kommentar
Joseph Cheng
Joseph Cheng am 19 Mai 2015
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

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Stephen23
Stephen23 am 19 Mai 2015
Bearbeitet: Stephen23 am 19 Mai 2015
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

Mehmet Gurturk
Mehmet Gurturk am 20 Mai 2015
Thank you so much :)

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