How to normalize values in a matrix to be between 0 and 1?

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Sahar abdalah
Sahar abdalah 2015 年 4 月 8 日
コメント済み: Tubi 2018 年 3 月 22 日
I have a matrix Ypred that contain negative values and I want to normalize this matrix between 0 and 1
Ypred=[-0.9630 -1.0107 -1.0774
-1.2075 -1.4164 -1.2135
-1.0237 -1.0082 -1.0714
-1.0191 -1.3686 -1.2105];
I'm new in matlab, please help me, there is a matlab function or toolbox that can do this? thanks
  2 件のコメント
James Tursa
James Tursa 2015 年 4 月 8 日
編集済み: James Tursa 2015 年 4 月 8 日
What do you mean by "normalize"? Divide by the max value in the matrix and make all values positive? Do this by columns or rows? Divide by norm of columns or rows? Or what?
Sahar abdalah
Sahar abdalah 2015 年 4 月 8 日
I want to divide by norm of row to make positive values

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採用された回答

Jos (10584)
Jos (10584) 2015 年 4 月 8 日
編集済み: Jos (10584) 2015 年 4 月 8 日
This can be simply done in a two step process
  1. subtract the minimum
  2. divide by the new maximum
normA = A - min(A(:))
normA = normA ./ max(normA(:)) % *
note that A(:) makes A into a long list of values. Otherwise min(A) would not return a single value ... Try fro yourself!
  • Edited after comment ...
  2 件のコメント
Sahar abdalah
Sahar abdalah 2015 年 4 月 8 日
Thanks for the code, I just tried this and it not normalised between 0 and 1
Ypred =
-0.9630 -1.0107 -1.0774
-1.2075 -1.4164 -1.2135
-1.0237 -1.0082 -1.0714
-1.0191 -1.3686 -1.2105
normA =
0.4534 0.4057 0.3390
0.2089 0 0.2029
0.3927 0.4082 0.3450
0.3973 0.0478 0.2059
normA =
1.0000 1.0495 1.1188
1.2539 1.4708 1.2601
1.0630 1.0469 1.1126
1.0583 1.4212 1.2570
Jos (10584)
Jos (10584) 2015 年 4 月 8 日
Sorry! The second line of code is wrong ;-) It should read
normA = normA ./ max(normA(:))

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その他の回答 (3 件)

James Tursa
James Tursa 2015 年 4 月 8 日
NormRows = sqrt(sum(Ypred.*Ypred,2));
Ynorm = bsxfun(@rdivide,abs(Ypred),NormRows);
  3 件のコメント
John D'Errico
John D'Errico 2018 年 3 月 10 日
編集済み: John D'Errico 2018 年 3 月 10 日
Why would you need a citation? If you truly want a citation, just cite MATLAB itself, or perhaps the doc page for bsxfun, but it seems a bit silly to need a citation for a simple code fragment.
Tubi
Tubi 2018 年 3 月 22 日
Many Thanks John, I believe you are right in your suggestion since they are mostly common MATLAB commands and functions.

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Sahar abdalah
Sahar abdalah 2015 年 4 月 9 日
thank you for your answers. I used both codes and I found two different result. what is the result that I can use?
normA = Ypred - min(Ypred(:))
normA = normA ./ max(normA(:))
normA =
1.0000 0.8948 0.7477
0.4607 0 0.4475
0.8661 0.9003 0.7609
0.8763 0.1054 0.4541
NormRows = sqrt(sum(Ypred.*Ypred,2));
Ynorm = bsxfun(@rdivide,abs(Ypred),NormRows);
Ynorm =
0.5461 0.5731 0.6110
0.5435 0.6375 0.5462
0.5712 0.5625 0.5978
0.4871 0.6542 0.5786
  2 件のコメント
James Tursa
James Tursa 2015 年 4 月 9 日
Use whichever is appropriate for your problem. Jos and I are both trying to interpret what you want, but without any background about the problem you are solving it is a bit of a guessing game at our end.
Sahar abdalah
Sahar abdalah 2015 年 4 月 9 日
ok thanks

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c4ndc
c4ndc 2017 年 8 月 12 日
Hello, What is the name of this norm in the accepted answer? (Euclidean, Frobenius etc.)
  1 件のコメント
Jan
Jan 2017 年 8 月 12 日
Please post comments to answers in the section for the comments. You message is not an answer.
The accepted answer does not contain a norm at all, but a "normalization". A matrix norm would reply a scalar, the normalization replies a matrix with the same size, but with shifted and scaled values.

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