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

102 visualizzazioni (ultimi 30 giorni)
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 Commenti
James Tursa
James Tursa il 8 Apr 2015
Modificato: James Tursa il 8 Apr 2015
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 il 8 Apr 2015
I want to divide by norm of row to make positive values

Accedi per commentare.

Risposta accettata

Jos (10584)
Jos (10584) il 8 Apr 2015
Modificato: Jos (10584) il 8 Apr 2015
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 Commenti
Sahar abdalah
Sahar abdalah il 8 Apr 2015
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) il 8 Apr 2015
Sorry! The second line of code is wrong ;-) It should read
normA = normA ./ max(normA(:))

Accedi per commentare.

Più risposte (3)

James Tursa
James Tursa il 8 Apr 2015
NormRows = sqrt(sum(Ypred.*Ypred,2));
Ynorm = bsxfun(@rdivide,abs(Ypred),NormRows);
  3 Commenti
John D'Errico
John D'Errico il 10 Mar 2018
Modificato: John D'Errico il 10 Mar 2018
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 il 22 Mar 2018
Many Thanks John, I believe you are right in your suggestion since they are mostly common MATLAB commands and functions.

Accedi per commentare.


Sahar abdalah
Sahar abdalah il 9 Apr 2015
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 Commenti
James Tursa
James Tursa il 9 Apr 2015
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.

Accedi per commentare.


c4ndc
c4ndc il 12 Ago 2017
Hello, What is the name of this norm in the accepted answer? (Euclidean, Frobenius etc.)
  1 Commento
Jan
Jan il 12 Ago 2017
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.

Accedi per commentare.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by