How to remove columns from matrix?
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Sahar abdalah
el 15 de Abr. de 2015
I have a matrix A and I want to remove from this matrix each column that contain the value -1.
A= [-1.192 -1.020 -1 -1.050 -1 -1 -1.070;
-1.213 -1.096 -1 -1.045 -1 -1 -1.102;
-1.036 -1.061 -1 -1.085 -1 -1 -1.137;
-1.176 -1.120 -1 -1.004 -1 -1 -1.115;
-1.124 -1.034 -1 -1.073 -1 -1 -1.134;
-1.202 -1.145 -1 -1.078 -1 -1 -1.057;
-1.127 -1.023 -1 -1.056 -1 -1 -1.066]
the size of A is 7*7 and I want to have a matrix 7*4 after removing colums? How can I do please?
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Star Strider
el 15 de Abr. de 2015
A more general solution:
A = reshape(A(~arrayfun(@isequal, A, -ones(size(A)))),size(A,1),[]);
It first finds a logical array of vectors that are equal to -1 using arrayfun, takes the values of the matrix that are not those values and creates a new vector from them. It then takes that vector and reshapes it to the row size of the original matrix to create the new ‘A’ matrix.
5 comentarios
Star Strider
el 15 de Abr. de 2015
Did you try it yourself to see what it does? (My code does a version of what you intend with that.)
BTW, I always test my code to be sure it works before I post it (or clearly label it ‘untested code’ if I can’t). It saves time and enhances credibility.
pfb
el 15 de Abr. de 2015
I did test it. Here is what I get
>> A = [-1 2 -1 3 -1; 2 -1 -1 0 1];
>> A==-ones(size(A))
ans =
1 0 1 0 1
0 1 1 0 0
>> arrayfun(@isequal, A, -ones(size(A)))
ans =
1 0 1 0 1
0 1 1 0 0
It seems to me that they are equivalent, at least for this minimal example. I jotted down the matrix randomly, making sure that at least one column was composed of -1 alone. Maybe I was lucky. I have to admit I did not embark in systematic testing. Anyway, that is why I asked.
Más respuestas (3)
N/A
el 15 de Abr. de 2015
Editada: N/A
el 16 de Abr. de 2015
% A(:, any(A == -1)) = [] %% Please try this.
% revised 'any' to 'all'. 'any' means the columns in which there is any '-1'; 'all'means the columns in which they are all '-1'.
A(:, all(A == -1)) = []
2 comentarios
James Tursa
el 15 de Abr. de 2015
"all" instead of "any" to achieve the goal "... remove columns that entirely contain the value -1 ..."
Sahar abdalah
el 15 de Abr. de 2015
3 comentarios
Star Strider
el 15 de Abr. de 2015
What do you mean by ‘remove the index’? The resulting matrix will be smaller, so you will have to address its columns differently than you did in the original matrix. You will have to rewrite your code to do that.
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