How to remove columns from matrix?

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Sahar abdalah
Sahar abdalah 2015 年 4 月 15 日
編集済み: N/A 2015 年 4 月 16 日
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?

採用された回答

Star Strider
Star Strider 2015 年 4 月 15 日
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 件のコメント
Star Strider
Star Strider 2015 年 4 月 15 日
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
pfb 2015 年 4 月 15 日
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.

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その他の回答 (3 件)

N/A
N/A 2015 年 4 月 15 日
編集済み: N/A 2015 年 4 月 16 日
% 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 件のコメント
James Tursa
James Tursa 2015 年 4 月 15 日
"all" instead of "any" to achieve the goal "... remove columns that entirely contain the value -1 ..."
N/A
N/A 2015 年 4 月 16 日
編集済み: N/A 2015 年 4 月 16 日
Yeah. Just a while after I shut down my PC last night, I realized it should be 'all'.
A(:, all(A == -1)) = []
Thanks!

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pfb
pfb 2015 年 4 月 15 日
編集済み: pfb 2015 年 4 月 15 日
for that particular matrix you want
B = A(:, [1 2 4 7]);
If this is a more general problem, that's a bit more tricky. Should the column consist entirely of -1 or just one single -1 is sufficient for eliminating it?

Sahar abdalah
Sahar abdalah 2015 年 4 月 15 日
My goal is to remove the columns that entirely contain the value -1. the problem is more general, I have a matrix A: size m * n with m number of test samples and n is the label of the class . The size of A in my case is 1000 * 1181. I want to remove the column and also their index.
  3 件のコメント
Sahar abdalah
Sahar abdalah 2015 年 4 月 15 日
yes it work but I want to remove the index also
Star Strider
Star Strider 2015 年 4 月 15 日
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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