Error finding matrix indices where elements obey a condition
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ferda sonmez
el 22 de Mzo. de 2019
Comentada: Luna
el 28 de Mzo. de 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 comentarios
madhan ravi
el 22 de Mzo. de 2019
When you use vpa() or vpasolve() the class of the variable becomes symbolic.
Respuesta aceptada
Luna
el 22 de Mzo. de 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)
2 comentarios
Más respuestas (1)
Walter Roberson
el 23 de Mzo. de 2019
B = isAlways(A>0);
will give you a logical result. Or you could
B = logical(A>0);
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