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Error finding matrix indices where elements obey a condition

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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 Commenti
madhan ravi
madhan ravi il 22 Mar 2019
When you use vpa() or vpasolve() the class of the variable becomes symbolic.

Accedi per commentare.

Risposta accettata

Luna
Luna il 22 Mar 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)

Più risposte (1)

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

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R2018b

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