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Error finding matrix indices where elements obey a condition

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ferda sonmez
ferda sonmez am 22 Mär. 2019
Kommentiert: Luna am 28 Mär. 2019
Hi,
I have a (30,1) matrix A filled with both positive and negative double values with high precisions.
I do the B= A>0 expecting a resulting matrix filled with 0s and 1s based on the value of original matrix such as
[0,
0,
1,
1,
0,
....]
However, when I display I get a matrix like below
0.0 < -60.80089912054130723383673691361028456512560544254638307756720395046146077078219605027697980403900146484375
0.0 < -40.47441497086561319883182687840919573785507473004455923328153712191512791918057700968347489833831787109375
0.0 < 11.9150901590727927897158716535914255438713276913571208738565018227240077663964257226325571537017822265625
0.0 < 50.24337655630030618369864852262179095255359883940103235308691394245261818696235422976315021514892578125
0.0 < -12.788094842081277874683606597932106962408104528383932295293478643982698628178695798851549625396728515625
0.0 < -19.7994345791724984686511623375698392780606832842132285447736268595153585891921466100029647350311279296875
0.0 < 44.20960440529531749044620672799080999950637846265719961065220598196712220584458918892778456211090087890625
0.0 < -38.4752715808701180299125340913012275096601820744906678455217984013980725421788520179688930511474609375
0.0 < -57.1261686699021779161260124218737353314343618323505877827184033790341999292650143615901470184326171875
0.0 < -15.50675798085333259859508957067109961395543474125240585269404109021451620975540208746679127216339111328125 0
Later when use matrix B in a matrix multiplication I get a lot of NaN values. Where am I doing wrong? I will be very appreciated if I can find a solution dfor this . Cause I am struggling due to this issue for a few days..
Best Regards,
Ferda
  7 Kommentare
madhan ravi
madhan ravi am 22 Mär. 2019
When you use vpa() or vpasolve() the class of the variable becomes symbolic.
Luna
Luna am 25 Mär. 2019
Ah now I understand, thx :)

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Akzeptierte Antwort

Luna
Luna am 22 Mär. 2019
Hi Ferda,
Please see my comments below:
% first convert it to double array:
A = double(input_of_hidden_layer2);
% indexing your conditions:
B = A > 0;
% get the values which are greater than zero from that index:
A(B)

Weitere Antworten (1)

Walter Roberson
Walter Roberson am 23 Mär. 2019
B = isAlways(A>0);
will give you a logical result. Or you could
B = logical(A>0);

Produkte


Version

R2018b

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