The full error message reads "Error using trainnet (line 46)
The number of mini-batch queue outputs (2) must match the number of network inputs plus
the number of network outputs (4)."
I’m using an arrayDatastore to pass the predictors (x2) and the targets(x2) to the transformer model. Both predictors have 410 features, one of the targets has 410 features and the other target is a scalar function.
Code to generate the dummy predictor and target data is pasted below:
%——————————————————————–
% data generation for encoder
numObs = 10;
seqLen = vocabSize;
x_enc = randi([1,10],[seqLen,numObs]);
y_enc = zeros(numObs,1);
for i = 1:numObs
idx = x_enc(1:2,i);
y_enc(i,:) = sum(x_enc(idx,i));
end
x_enc = num2cell(x_enc’,2);
y_enc = num2cell(y_enc)’;
x_1 = x_enc;
y_2 = y_enc’;
% data generation for decoder
x_series = randi([1,10],[seqLen,numObs]);
y_series = sin(rand([seqLen,numObs]));
x_dec = x_series(:,1:end)’;
y_dec = y_series(:,1:end)’;
x_dec = num2cell(x_dec,2); x_2 = x_dec;
y_dec = num2cell(y_dec,2); y_1 = y_dec;
cell_data = {}; cell_data = [cell_data x_1 x_2 y_1 y_2];
dstrain = arrayDatastore(cell_data,’OutputType’,’same’);
%——————————————————————-
cell_data is of the form:
cell_data

cell_data =

10×4 cell array

{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[10]}
{1×410 double} {1×410 double} {1×410 double} {[13]}
{1×410 double} {1×410 double} {1×410 double} {[20]}
{1×410 double} {1×410 double} {1×410 double} {[ 7]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[17]}
{1×410 double} {1×410 double} {1×410 double} {[ 6]}
{1×410 double} {1×410 double} {1×410 double} {[11]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}

If I were to use readall(dstrain) to read the datastore, I get the same format as cell_data:
fds = readall(dstrain)

fds =

10×4 cell array

{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[10]}
{1×410 double} {1×410 double} {1×410 double} {[13]}
{1×410 double} {1×410 double} {1×410 double} {[20]}
{1×410 double} {1×410 double} {1×410 double} {[ 7]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[17]}
{1×410 double} {1×410 double} {1×410 double} {[ 6]}
{1×410 double} {1×410 double} {1×410 double} {[11]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}

Finally, if I use minibatchqueue to create a minibatch of datastore ‘dstrain’, I get:
mbq = minibatchqueue(dstrain)

mbq =

minibatchqueue with 4 outputs and properties:

Mini-batch creation:
MiniBatchSize: 10
PartialMiniBatch: ‘return’
MiniBatchFcn: ‘collate’
PreprocessingEnvironment: ‘serial’

Outputs:
OutputCast: {‘single’ ‘single’ ‘single’ ‘single’}
OutputAsDlarray: [1 1 1 1]
MiniBatchFormat: {” ” ” ”}
OutputEnvironment: {‘auto’ ‘auto’ ‘auto’ ‘auto’}

As you can see, there are four outputs for the minibatch, which appears to contradict the original error message that there are only two minibatchqueue outputs
Also to confirm, i double checked the transformer input output structure:
net

net =

dlnetwork with properties:

Layers: [64×1 nnet.cnn.layer.Layer]
Connections: [1714×2 table]
Learnables: [110×3 table]
State: [0×3 table]
InputNames: {‘in_enc’ ‘in_dec’}
OutputNames: {‘decoder_out’ ‘fc_13’}
Initialized: 1

View summary with summary.
which shows two inputs and two outputs.
Could someone point me to the mistake I’m making here (likely with the datastore format) – it seems that during batching, the model is only choosing two of the cell columns from cell_data/dstrain for the input and output, rather than all four and its not clear why…thanks in advance for your help!
CGThe full error message reads "Error using trainnet (line 46)
The number of mini-batch queue outputs (2) must match the number of network inputs plus
the number of network outputs (4)."
I’m using an arrayDatastore to pass the predictors (x2) and the targets(x2) to the transformer model. Both predictors have 410 features, one of the targets has 410 features and the other target is a scalar function.
Code to generate the dummy predictor and target data is pasted below:
%——————————————————————–
% data generation for encoder
numObs = 10;
seqLen = vocabSize;
x_enc = randi([1,10],[seqLen,numObs]);
y_enc = zeros(numObs,1);
for i = 1:numObs
idx = x_enc(1:2,i);
y_enc(i,:) = sum(x_enc(idx,i));
end
x_enc = num2cell(x_enc’,2);
y_enc = num2cell(y_enc)’;
x_1 = x_enc;
y_2 = y_enc’;
% data generation for decoder
x_series = randi([1,10],[seqLen,numObs]);
y_series = sin(rand([seqLen,numObs]));
x_dec = x_series(:,1:end)’;
y_dec = y_series(:,1:end)’;
x_dec = num2cell(x_dec,2); x_2 = x_dec;
y_dec = num2cell(y_dec,2); y_1 = y_dec;
cell_data = {}; cell_data = [cell_data x_1 x_2 y_1 y_2];
dstrain = arrayDatastore(cell_data,’OutputType’,’same’);
%——————————————————————-
cell_data is of the form:
cell_data

cell_data =

10×4 cell array

{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[10]}
{1×410 double} {1×410 double} {1×410 double} {[13]}
{1×410 double} {1×410 double} {1×410 double} {[20]}
{1×410 double} {1×410 double} {1×410 double} {[ 7]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[17]}
{1×410 double} {1×410 double} {1×410 double} {[ 6]}
{1×410 double} {1×410 double} {1×410 double} {[11]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}

If I were to use readall(dstrain) to read the datastore, I get the same format as cell_data:
fds = readall(dstrain)

fds =

10×4 cell array

{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[10]}
{1×410 double} {1×410 double} {1×410 double} {[13]}
{1×410 double} {1×410 double} {1×410 double} {[20]}
{1×410 double} {1×410 double} {1×410 double} {[ 7]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[17]}
{1×410 double} {1×410 double} {1×410 double} {[ 6]}
{1×410 double} {1×410 double} {1×410 double} {[11]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}

Finally, if I use minibatchqueue to create a minibatch of datastore ‘dstrain’, I get:
mbq = minibatchqueue(dstrain)

mbq =

minibatchqueue with 4 outputs and properties:

Mini-batch creation:
MiniBatchSize: 10
PartialMiniBatch: ‘return’
MiniBatchFcn: ‘collate’
PreprocessingEnvironment: ‘serial’

Outputs:
OutputCast: {‘single’ ‘single’ ‘single’ ‘single’}
OutputAsDlarray: [1 1 1 1]
MiniBatchFormat: {” ” ” ”}
OutputEnvironment: {‘auto’ ‘auto’ ‘auto’ ‘auto’}

As you can see, there are four outputs for the minibatch, which appears to contradict the original error message that there are only two minibatchqueue outputs
Also to confirm, i double checked the transformer input output structure:
net

net =

dlnetwork with properties:

Layers: [64×1 nnet.cnn.layer.Layer]
Connections: [1714×2 table]
Learnables: [110×3 table]
State: [0×3 table]
InputNames: {‘in_enc’ ‘in_dec’}
OutputNames: {‘decoder_out’ ‘fc_13’}
Initialized: 1

View summary with summary.
which shows two inputs and two outputs.
Could someone point me to the mistake I’m making here (likely with the datastore format) – it seems that during batching, the model is only choosing two of the cell columns from cell_data/dstrain for the input and output, rather than all four and its not clear why…thanks in advance for your help!
CG The full error message reads "Error using trainnet (line 46)
The number of mini-batch queue outputs (2) must match the number of network inputs plus
the number of network outputs (4)."
I’m using an arrayDatastore to pass the predictors (x2) and the targets(x2) to the transformer model. Both predictors have 410 features, one of the targets has 410 features and the other target is a scalar function.
Code to generate the dummy predictor and target data is pasted below:
%——————————————————————–
% data generation for encoder
numObs = 10;
seqLen = vocabSize;
x_enc = randi([1,10],[seqLen,numObs]);
y_enc = zeros(numObs,1);
for i = 1:numObs
idx = x_enc(1:2,i);
y_enc(i,:) = sum(x_enc(idx,i));
end
x_enc = num2cell(x_enc’,2);
y_enc = num2cell(y_enc)’;
x_1 = x_enc;
y_2 = y_enc’;
% data generation for decoder
x_series = randi([1,10],[seqLen,numObs]);
y_series = sin(rand([seqLen,numObs]));
x_dec = x_series(:,1:end)’;
y_dec = y_series(:,1:end)’;
x_dec = num2cell(x_dec,2); x_2 = x_dec;
y_dec = num2cell(y_dec,2); y_1 = y_dec;
cell_data = {}; cell_data = [cell_data x_1 x_2 y_1 y_2];
dstrain = arrayDatastore(cell_data,’OutputType’,’same’);
%——————————————————————-
cell_data is of the form:
cell_data

cell_data =

10×4 cell array

{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[10]}
{1×410 double} {1×410 double} {1×410 double} {[13]}
{1×410 double} {1×410 double} {1×410 double} {[20]}
{1×410 double} {1×410 double} {1×410 double} {[ 7]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[17]}
{1×410 double} {1×410 double} {1×410 double} {[ 6]}
{1×410 double} {1×410 double} {1×410 double} {[11]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}

If I were to use readall(dstrain) to read the datastore, I get the same format as cell_data:
fds = readall(dstrain)

fds =

10×4 cell array

{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[10]}
{1×410 double} {1×410 double} {1×410 double} {[13]}
{1×410 double} {1×410 double} {1×410 double} {[20]}
{1×410 double} {1×410 double} {1×410 double} {[ 7]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}
{1×410 double} {1×410 double} {1×410 double} {[17]}
{1×410 double} {1×410 double} {1×410 double} {[ 6]}
{1×410 double} {1×410 double} {1×410 double} {[11]}
{1×410 double} {1×410 double} {1×410 double} {[ 8]}

Finally, if I use minibatchqueue to create a minibatch of datastore ‘dstrain’, I get:
mbq = minibatchqueue(dstrain)

mbq =

minibatchqueue with 4 outputs and properties:

Mini-batch creation:
MiniBatchSize: 10
PartialMiniBatch: ‘return’
MiniBatchFcn: ‘collate’
PreprocessingEnvironment: ‘serial’

Outputs:
OutputCast: {‘single’ ‘single’ ‘single’ ‘single’}
OutputAsDlarray: [1 1 1 1]
MiniBatchFormat: {” ” ” ”}
OutputEnvironment: {‘auto’ ‘auto’ ‘auto’ ‘auto’}

As you can see, there are four outputs for the minibatch, which appears to contradict the original error message that there are only two minibatchqueue outputs
Also to confirm, i double checked the transformer input output structure:
net

net =

dlnetwork with properties:

Layers: [64×1 nnet.cnn.layer.Layer]
Connections: [1714×2 table]
Learnables: [110×3 table]
State: [0×3 table]
InputNames: {‘in_enc’ ‘in_dec’}
OutputNames: {‘decoder_out’ ‘fc_13’}
Initialized: 1

View summary with summary.
which shows two inputs and two outputs.
Could someone point me to the mistake I’m making here (likely with the datastore format) – it seems that during batching, the model is only choosing two of the cell columns from cell_data/dstrain for the input and output, rather than all four and its not clear why…thanks in advance for your help!
CG deep learning, minibatch, datastore MATLAB Answers — New Questions

​

I’m trying to initialize a variable (MyHandle) that I want to be a handle pointing to some object but which can’t currently be assigned to said object. I have code that needs to behave differently depending on if MyHandle is assigned or not yet and I haven’t found a good way to test if MyHandle has been assigned to a valid object or not.
The function ‘isvalid’ seems like the right choice to test this, however, isvalid throws the error "Incorrect number or types of inputs or outputs for function isvalid." when run on MyHandle in its unassigned state. I’ve tried different ways to initialize it like "handle([])" and "[]", but they all return that same error.
I’ve found some bandaid solutions like initializing MyHandle to an instance of some arbitrary handle subclass and then deleting that initialized object, or assigning MyHandle to be a value of type double and then using "isnumeric" as my test function instead of "isvalid," but I was wondering if there is a more "correct" way to do it for the sake of best practices.
Below is a sample class that recreates the behavior described above. To restate my question in terms of the class below: Is there an initialization for MyHandle I can use in the properties block that will allow IsHandleAssigned to return false before the first time AssignHandle is run with a valid handle and return true afterwards? OR is there a different function besides isvalid that I should be using in IsHandleAssigned to achieve that behavior?
classdef ExampleClass < handle
properties
MyHandle %Will be a handle pointing to an object of an arbitrary class
% MyHandle = handle([]) < This initialization returns an error
% MyHandle = [] < This initialization returns an error
% MyHandle = 0 < This works if I replace ‘isvalid’ with ‘isnumeric’
% but that seems like poor practice to do since it obfuscates the
% purpose of MyHandle
end

methods
function obj = AssignHandle(obj,handleIn)
%This method assigns the handle of whatever class it happens to
%be to my property.
obj.MyHandle = handleIn;
end
function HandleIsAssigned = IsHandleAssigned(obj)
%This method tests whether or not my handle was assigned. Is
%there a better function than isvalid to use?
HandleIsAssigned = isvalid(obj.MyHandle);
end
end
endI’m trying to initialize a variable (MyHandle) that I want to be a handle pointing to some object but which can’t currently be assigned to said object. I have code that needs to behave differently depending on if MyHandle is assigned or not yet and I haven’t found a good way to test if MyHandle has been assigned to a valid object or not.
The function ‘isvalid’ seems like the right choice to test this, however, isvalid throws the error "Incorrect number or types of inputs or outputs for function isvalid." when run on MyHandle in its unassigned state. I’ve tried different ways to initialize it like "handle([])" and "[]", but they all return that same error.
I’ve found some bandaid solutions like initializing MyHandle to an instance of some arbitrary handle subclass and then deleting that initialized object, or assigning MyHandle to be a value of type double and then using "isnumeric" as my test function instead of "isvalid," but I was wondering if there is a more "correct" way to do it for the sake of best practices.
Below is a sample class that recreates the behavior described above. To restate my question in terms of the class below: Is there an initialization for MyHandle I can use in the properties block that will allow IsHandleAssigned to return false before the first time AssignHandle is run with a valid handle and return true afterwards? OR is there a different function besides isvalid that I should be using in IsHandleAssigned to achieve that behavior?
classdef ExampleClass < handle
properties
MyHandle %Will be a handle pointing to an object of an arbitrary class
% MyHandle = handle([]) < This initialization returns an error
% MyHandle = [] < This initialization returns an error
% MyHandle = 0 < This works if I replace ‘isvalid’ with ‘isnumeric’
% but that seems like poor practice to do since it obfuscates the
% purpose of MyHandle
end

methods
function obj = AssignHandle(obj,handleIn)
%This method assigns the handle of whatever class it happens to
%be to my property.
obj.MyHandle = handleIn;
end
function HandleIsAssigned = IsHandleAssigned(obj)
%This method tests whether or not my handle was assigned. Is
%there a better function than isvalid to use?
HandleIsAssigned = isvalid(obj.MyHandle);
end
end
end I’m trying to initialize a variable (MyHandle) that I want to be a handle pointing to some object but which can’t currently be assigned to said object. I have code that needs to behave differently depending on if MyHandle is assigned or not yet and I haven’t found a good way to test if MyHandle has been assigned to a valid object or not.
The function ‘isvalid’ seems like the right choice to test this, however, isvalid throws the error "Incorrect number or types of inputs or outputs for function isvalid." when run on MyHandle in its unassigned state. I’ve tried different ways to initialize it like "handle([])" and "[]", but they all return that same error.
I’ve found some bandaid solutions like initializing MyHandle to an instance of some arbitrary handle subclass and then deleting that initialized object, or assigning MyHandle to be a value of type double and then using "isnumeric" as my test function instead of "isvalid," but I was wondering if there is a more "correct" way to do it for the sake of best practices.
Below is a sample class that recreates the behavior described above. To restate my question in terms of the class below: Is there an initialization for MyHandle I can use in the properties block that will allow IsHandleAssigned to return false before the first time AssignHandle is run with a valid handle and return true afterwards? OR is there a different function besides isvalid that I should be using in IsHandleAssigned to achieve that behavior?
classdef ExampleClass < handle
properties
MyHandle %Will be a handle pointing to an object of an arbitrary class
% MyHandle = handle([]) < This initialization returns an error
% MyHandle = [] < This initialization returns an error
% MyHandle = 0 < This works if I replace ‘isvalid’ with ‘isnumeric’
% but that seems like poor practice to do since it obfuscates the
% purpose of MyHandle
end

methods
function obj = AssignHandle(obj,handleIn)
%This method assigns the handle of whatever class it happens to
%be to my property.
obj.MyHandle = handleIn;
end
function HandleIsAssigned = IsHandleAssigned(obj)
%This method tests whether or not my handle was assigned. Is
%there a better function than isvalid to use?
HandleIsAssigned = isvalid(obj.MyHandle);
end
end
end handles MATLAB Answers — New Questions

​

Is there some sort of dual fuel engine model already accessible? If not, are there any sort of forums or videos i could use as help to build Such a model. My simulink skills are very basic at the moment.Is there some sort of dual fuel engine model already accessible? If not, are there any sort of forums or videos i could use as help to build Such a model. My simulink skills are very basic at the moment. Is there some sort of dual fuel engine model already accessible? If not, are there any sort of forums or videos i could use as help to build Such a model. My simulink skills are very basic at the moment. dual fuel, lpg, engine MATLAB Answers — New Questions

​

I am trying to replicate finding the optimized parameters of a Lorentz model defined in the paper titled, "EXTRACTION OF EFFECTIVE METAMATERIAL PARAMETERS BY PARAMETER FITTING OF DISPERSIVE MODEL" (linked here). I’ve provided the equation and table of paramters below:

I’ve tried the curve fitting tool and get different results especially when I try to change the bounds. Sometimes I am able to get paramters that match the table above but my confidence in repeatbility isn’t high. I’ve also tried scripting code to use the "fittype", "lsqnonlin", or "fminsearch" but also get different answers as well. In the code below, I’m comparing the fit to the real and imaginary parts of the Lorentz model. The fit to the real data is way off and even though the fit to the imaginary values looks qualitatively good. For the fit to the imaginary part, I get different values for the estimated parameters compared to what’s reported in the paper.

%% Parameters in Table
e_inf = 1.62;
wp = 2*pi*14.63*1e9; % GHz
vc = 30.69*1e6; %MHz
mu_s = 1.26;
mu_inf = 1.12;
wo = 2*pi*9.67*1e9; % GHz
delta = 1.24*1e9; % GHz

%%
fo = [7:0.01:12].*1e9; % GHz

w = 2*pi*fo;

mu_eff = mu_inf + ((mu_s – mu_inf)*wo.^2)./(wo.^2 + 1i*w*delta – w.^2); % equation in paper / Lorentz model

real_mu_eff = real(mu_eff);

imag_mu_eff = imag(mu_eff);

x = fo;
y = real_mu_eff;
x2 = fo;
y2 = imag_mu_eff;

myfittype = fittype("real(a+((b-a)*(2*pi*1e9*c).^2)./((2*pi*1e9*c).^2+1i*2*pi*x*d*1e9-x.^2))",…
dependent="y",independent="x",…
coefficients=["a" "b" "c" "d"])

myfit = fit(x’,y’,myfittype)

figure
plot(myfit,x,y)

myfittype2 = fittype("imag(a2+((b2-a2)*(2*pi*1e9*c2).^2)./((2*pi*1e9*c2).^2+1i*2*pi*x2*d2*1e9-x2.^2))",…
dependent="y2",independent="x2",…
coefficients=["a2" "b2" "c2" "d2"])

myfit2 = fit(x2′,y2′,myfittype2)

figure
plot(myfit2,x2,y2)

The results I was expecting were: a=1.12, b=1.26, c=9.67, and d=1.24. I ‘ve atttached my attempts with using "lsqnonlin", "lsqcurvefit", and "fminsearch"; however I didn’t have success in using those methods either. What is best way to find the parameters of the Lorentz model to get the values in the table above?I am trying to replicate finding the optimized parameters of a Lorentz model defined in the paper titled, "EXTRACTION OF EFFECTIVE METAMATERIAL PARAMETERS BY PARAMETER FITTING OF DISPERSIVE MODEL" (linked here). I’ve provided the equation and table of paramters below:

I’ve tried the curve fitting tool and get different results especially when I try to change the bounds. Sometimes I am able to get paramters that match the table above but my confidence in repeatbility isn’t high. I’ve also tried scripting code to use the "fittype", "lsqnonlin", or "fminsearch" but also get different answers as well. In the code below, I’m comparing the fit to the real and imaginary parts of the Lorentz model. The fit to the real data is way off and even though the fit to the imaginary values looks qualitatively good. For the fit to the imaginary part, I get different values for the estimated parameters compared to what’s reported in the paper.

%% Parameters in Table
e_inf = 1.62;
wp = 2*pi*14.63*1e9; % GHz
vc = 30.69*1e6; %MHz
mu_s = 1.26;
mu_inf = 1.12;
wo = 2*pi*9.67*1e9; % GHz
delta = 1.24*1e9; % GHz

%%
fo = [7:0.01:12].*1e9; % GHz

w = 2*pi*fo;

mu_eff = mu_inf + ((mu_s – mu_inf)*wo.^2)./(wo.^2 + 1i*w*delta – w.^2); % equation in paper / Lorentz model

real_mu_eff = real(mu_eff);

imag_mu_eff = imag(mu_eff);

x = fo;
y = real_mu_eff;
x2 = fo;
y2 = imag_mu_eff;

myfittype = fittype("real(a+((b-a)*(2*pi*1e9*c).^2)./((2*pi*1e9*c).^2+1i*2*pi*x*d*1e9-x.^2))",…
dependent="y",independent="x",…
coefficients=["a" "b" "c" "d"])

myfit = fit(x’,y’,myfittype)

figure
plot(myfit,x,y)

myfittype2 = fittype("imag(a2+((b2-a2)*(2*pi*1e9*c2).^2)./((2*pi*1e9*c2).^2+1i*2*pi*x2*d2*1e9-x2.^2))",…
dependent="y2",independent="x2",…
coefficients=["a2" "b2" "c2" "d2"])

myfit2 = fit(x2′,y2′,myfittype2)

figure
plot(myfit2,x2,y2)

The results I was expecting were: a=1.12, b=1.26, c=9.67, and d=1.24. I ‘ve atttached my attempts with using "lsqnonlin", "lsqcurvefit", and "fminsearch"; however I didn’t have success in using those methods either. What is best way to find the parameters of the Lorentz model to get the values in the table above? I am trying to replicate finding the optimized parameters of a Lorentz model defined in the paper titled, "EXTRACTION OF EFFECTIVE METAMATERIAL PARAMETERS BY PARAMETER FITTING OF DISPERSIVE MODEL" (linked here). I’ve provided the equation and table of paramters below:

I’ve tried the curve fitting tool and get different results especially when I try to change the bounds. Sometimes I am able to get paramters that match the table above but my confidence in repeatbility isn’t high. I’ve also tried scripting code to use the "fittype", "lsqnonlin", or "fminsearch" but also get different answers as well. In the code below, I’m comparing the fit to the real and imaginary parts of the Lorentz model. The fit to the real data is way off and even though the fit to the imaginary values looks qualitatively good. For the fit to the imaginary part, I get different values for the estimated parameters compared to what’s reported in the paper.

%% Parameters in Table
e_inf = 1.62;
wp = 2*pi*14.63*1e9; % GHz
vc = 30.69*1e6; %MHz
mu_s = 1.26;
mu_inf = 1.12;
wo = 2*pi*9.67*1e9; % GHz
delta = 1.24*1e9; % GHz

%%
fo = [7:0.01:12].*1e9; % GHz

w = 2*pi*fo;

mu_eff = mu_inf + ((mu_s – mu_inf)*wo.^2)./(wo.^2 + 1i*w*delta – w.^2); % equation in paper / Lorentz model

real_mu_eff = real(mu_eff);

imag_mu_eff = imag(mu_eff);

x = fo;
y = real_mu_eff;
x2 = fo;
y2 = imag_mu_eff;

myfittype = fittype("real(a+((b-a)*(2*pi*1e9*c).^2)./((2*pi*1e9*c).^2+1i*2*pi*x*d*1e9-x.^2))",…
dependent="y",independent="x",…
coefficients=["a" "b" "c" "d"])

myfit = fit(x’,y’,myfittype)

figure
plot(myfit,x,y)

myfittype2 = fittype("imag(a2+((b2-a2)*(2*pi*1e9*c2).^2)./((2*pi*1e9*c2).^2+1i*2*pi*x2*d2*1e9-x2.^2))",…
dependent="y2",independent="x2",…
coefficients=["a2" "b2" "c2" "d2"])

myfit2 = fit(x2′,y2′,myfittype2)

figure
plot(myfit2,x2,y2)

The results I was expecting were: a=1.12, b=1.26, c=9.67, and d=1.24. I ‘ve atttached my attempts with using "lsqnonlin", "lsqcurvefit", and "fminsearch"; however I didn’t have success in using those methods either. What is best way to find the parameters of the Lorentz model to get the values in the table above? optimization, curve fitting, physics, electrical, material characterization, debye, lorentz MATLAB Answers — New Questions

​

Hello everyone,

I’m a user of MPT3(multi-parametric toolbox 3), and my computer is a Mac M1 Pro, my Matlab version is 2024b.
When I installed the MPT3 on my computer, the following issue happens:
Error using tbxmanager>main_install (line 468)
Toolbox "cddmex" is not available for your platform.

Error using tbxmanager (line 150)
Cannot continue, see message above.

Error in install_mpt3 (line 55)
tbxmanager install mpt mptdoc cddmex fourier glpkmex hysdel lcp sedumi yalmip
^^^^^^^^^^

Actually, this does not only happen for tool "cddmex", other toolboxes, including "fourier glpkmex hysdel lcp sedumi", also has this issue.
However, when I installed it for my matlab2021b, it was successful.

Does anyone know the way to fix it?Hello everyone,

I’m a user of MPT3(multi-parametric toolbox 3), and my computer is a Mac M1 Pro, my Matlab version is 2024b.
When I installed the MPT3 on my computer, the following issue happens:
Error using tbxmanager>main_install (line 468)
Toolbox "cddmex" is not available for your platform.

Error using tbxmanager (line 150)
Cannot continue, see message above.

Error in install_mpt3 (line 55)
tbxmanager install mpt mptdoc cddmex fourier glpkmex hysdel lcp sedumi yalmip
^^^^^^^^^^

Actually, this does not only happen for tool "cddmex", other toolboxes, including "fourier glpkmex hysdel lcp sedumi", also has this issue.
However, when I installed it for my matlab2021b, it was successful.

Does anyone know the way to fix it? Hello everyone,

I’m a user of MPT3(multi-parametric toolbox 3), and my computer is a Mac M1 Pro, my Matlab version is 2024b.
When I installed the MPT3 on my computer, the following issue happens:
Error using tbxmanager>main_install (line 468)
Toolbox "cddmex" is not available for your platform.

Error using tbxmanager (line 150)
Cannot continue, see message above.

Error in install_mpt3 (line 55)
tbxmanager install mpt mptdoc cddmex fourier glpkmex hysdel lcp sedumi yalmip
^^^^^^^^^^

Actually, this does not only happen for tool "cddmex", other toolboxes, including "fourier glpkmex hysdel lcp sedumi", also has this issue.
However, when I installed it for my matlab2021b, it was successful.

Does anyone know the way to fix it? mpt3, toolbox, control, optimization MATLAB Answers — New Questions

​

Previously posted but not answered: https://www.mathworks.com/matlabcentral/answers/1667539-how-do-i-display-zeros-in-confusion-chart
Is there a way to display 0’s in a confusionchart?

i have searched through MATLAB document for confusionchart to no avail, and a Google search brought me to the first question but no other answers. 🙁

I am attaching the code that I used to get the confusion chart, and any guidance will be appreciated! Otherwise manually adding a 0’s in power point will be my option. Thanks!

path = ‘X:GC_Experiments2MelioDataMelio_Clean_6_predictions.csv’;
data = readtable(path);
group1 = table2array(data(:,2));
group2 = table2array(data(:,3));
D = unique(group1);
labels = [D(1,1),D(2,1),D(3,1),D(4,1),D(5,1),D(6,1),D(7,1),D(8,1),D(9,1)];
C = confusionmat(group1,group2,’Order’,labels);

figure;
confusionchart(C,’OffDiagonalColor’,’b’);Previously posted but not answered: https://www.mathworks.com/matlabcentral/answers/1667539-how-do-i-display-zeros-in-confusion-chart
Is there a way to display 0’s in a confusionchart?

i have searched through MATLAB document for confusionchart to no avail, and a Google search brought me to the first question but no other answers. 🙁

I am attaching the code that I used to get the confusion chart, and any guidance will be appreciated! Otherwise manually adding a 0’s in power point will be my option. Thanks!

path = ‘X:GC_Experiments2MelioDataMelio_Clean_6_predictions.csv’;
data = readtable(path);
group1 = table2array(data(:,2));
group2 = table2array(data(:,3));
D = unique(group1);
labels = [D(1,1),D(2,1),D(3,1),D(4,1),D(5,1),D(6,1),D(7,1),D(8,1),D(9,1)];
C = confusionmat(group1,group2,’Order’,labels);

figure;
confusionchart(C,’OffDiagonalColor’,’b’); Previously posted but not answered: https://www.mathworks.com/matlabcentral/answers/1667539-how-do-i-display-zeros-in-confusion-chart
Is there a way to display 0’s in a confusionchart?

i have searched through MATLAB document for confusionchart to no avail, and a Google search brought me to the first question but no other answers. 🙁

I am attaching the code that I used to get the confusion chart, and any guidance will be appreciated! Otherwise manually adding a 0’s in power point will be my option. Thanks!

path = ‘X:GC_Experiments2MelioDataMelio_Clean_6_predictions.csv’;
data = readtable(path);
group1 = table2array(data(:,2));
group2 = table2array(data(:,3));
D = unique(group1);
labels = [D(1,1),D(2,1),D(3,1),D(4,1),D(5,1),D(6,1),D(7,1),D(8,1),D(9,1)];
C = confusionmat(group1,group2,’Order’,labels);

figure;
confusionchart(C,’OffDiagonalColor’,’b’); confusion matrix, display, plot MATLAB Answers — New Questions

​

Hello everyone!
I created artificial neural network to compute answers for my diagnosis. Now, I would like to compare the amount of true positive, true negative, false positive and false negative. I was thinking about creating a loop and count every values.
Thank you for your help.Hello everyone!
I created artificial neural network to compute answers for my diagnosis. Now, I would like to compare the amount of true positive, true negative, false positive and false negative. I was thinking about creating a loop and count every values.
Thank you for your help. Hello everyone!
I created artificial neural network to compute answers for my diagnosis. Now, I would like to compare the amount of true positive, true negative, false positive and false negative. I was thinking about creating a loop and count every values.
Thank you for your help. neural networks, confusion, count MATLAB Answers — New Questions

​

I’m trying to plot 4 different sets of data with a linear regression for each. I used the fitlm command to find hte linear regression. My problem arises with the plot all being the same color for each set of data, blue for the points and red for the regression line. I need them to be different colors for each data set for clarity sake.
data=readmatrix(‘straingage_caldata.csv’);
c1F=data(:,1);
c1V=data(:,2);
c3F=data(:,3);
c3V=data(:,4);
c4F=data(:,5);
c4V=data(:,6);
c5F=data(:,7);
c5V=data(:,8);

c1LR=fitlm(c1F,c1V)
c3LR=fitlm(c3F,c3V)
c4LR=fitlm(c4F,c4V)
c5LR=fitlm(c5F,c5V)

figure
hold on
plot(c1LR,"r");
plot(c3LR);
plot(c4LR);
plot(c5LR);
legend(‘Crane 1′,’Crane 1 Lin Reg’,’Crane 3′,’Crane 3 Lin Reg’,’Crane 4′,’Crane 4 Lin Reg’,’Crane 5′,’Crane 5 Lin Reg’,"Location","SouthEast")

thanksI’m trying to plot 4 different sets of data with a linear regression for each. I used the fitlm command to find hte linear regression. My problem arises with the plot all being the same color for each set of data, blue for the points and red for the regression line. I need them to be different colors for each data set for clarity sake.
data=readmatrix(‘straingage_caldata.csv’);
c1F=data(:,1);
c1V=data(:,2);
c3F=data(:,3);
c3V=data(:,4);
c4F=data(:,5);
c4V=data(:,6);
c5F=data(:,7);
c5V=data(:,8);

c1LR=fitlm(c1F,c1V)
c3LR=fitlm(c3F,c3V)
c4LR=fitlm(c4F,c4V)
c5LR=fitlm(c5F,c5V)

figure
hold on
plot(c1LR,"r");
plot(c3LR);
plot(c4LR);
plot(c5LR);
legend(‘Crane 1′,’Crane 1 Lin Reg’,’Crane 3′,’Crane 3 Lin Reg’,’Crane 4′,’Crane 4 Lin Reg’,’Crane 5′,’Crane 5 Lin Reg’,"Location","SouthEast")

thanks I’m trying to plot 4 different sets of data with a linear regression for each. I used the fitlm command to find hte linear regression. My problem arises with the plot all being the same color for each set of data, blue for the points and red for the regression line. I need them to be different colors for each data set for clarity sake.
data=readmatrix(‘straingage_caldata.csv’);
c1F=data(:,1);
c1V=data(:,2);
c3F=data(:,3);
c3V=data(:,4);
c4F=data(:,5);
c4V=data(:,6);
c5F=data(:,7);
c5V=data(:,8);

c1LR=fitlm(c1F,c1V)
c3LR=fitlm(c3F,c3V)
c4LR=fitlm(c4F,c4V)
c5LR=fitlm(c5F,c5V)

figure
hold on
plot(c1LR,"r");
plot(c3LR);
plot(c4LR);
plot(c5LR);
legend(‘Crane 1′,’Crane 1 Lin Reg’,’Crane 3′,’Crane 3 Lin Reg’,’Crane 4′,’Crane 4 Lin Reg’,’Crane 5′,’Crane 5 Lin Reg’,"Location","SouthEast")

thanks fitlm, plot MATLAB Answers — New Questions

​

Hi!
I am setting up my Raspberry Pi (model 4B) for deep learning deployment. I am following the set up steps of the Support Package for Raspberry Pi. The Pi has been flashed with the image "mathworks_raspbian_DL_R22.2.0.img".
When I tried to install the OpenCV library by executing "./installOpenCVLib", I encountered some errors, as shown in the attached screenshot. Apparently the file "opencv_raspberrypi_10.zip" could not be downloaded, although the 3 packages seems to be succesfully installed. Indeed, when I tried to connect to the link to the Mathworks site on my computer, the link was not accessible.
Can you please suggest any alternative ways to install opencv?
Thank you for your help!Hi!
I am setting up my Raspberry Pi (model 4B) for deep learning deployment. I am following the set up steps of the Support Package for Raspberry Pi. The Pi has been flashed with the image "mathworks_raspbian_DL_R22.2.0.img".
When I tried to install the OpenCV library by executing "./installOpenCVLib", I encountered some errors, as shown in the attached screenshot. Apparently the file "opencv_raspberrypi_10.zip" could not be downloaded, although the 3 packages seems to be succesfully installed. Indeed, when I tried to connect to the link to the Mathworks site on my computer, the link was not accessible.
Can you please suggest any alternative ways to install opencv?
Thank you for your help! Hi!
I am setting up my Raspberry Pi (model 4B) for deep learning deployment. I am following the set up steps of the Support Package for Raspberry Pi. The Pi has been flashed with the image "mathworks_raspbian_DL_R22.2.0.img".
When I tried to install the OpenCV library by executing "./installOpenCVLib", I encountered some errors, as shown in the attached screenshot. Apparently the file "opencv_raspberrypi_10.zip" could not be downloaded, although the 3 packages seems to be succesfully installed. Indeed, when I tried to connect to the link to the Mathworks site on my computer, the link was not accessible.
Can you please suggest any alternative ways to install opencv?
Thank you for your help! opencv, raspberry pi, deep learning MATLAB Answers — New Questions

​

I am trying to convert an array into numpy, but I keep getting this issue:
Python Error: ModuleNotFoundError: No module named ‘libmwbuffer’
Python function ” might not be able to accept input argument at position 1. The function may require a specific
data type that you can construct from the MATLAB array. For more information, see the documentation for Python
function ” and working with Python arrays.

I checked and I have this pyversion, which is supposed to be compatible with Matlab R2024a. what’s the issue?
version: ‘3.9’
executable: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3python.EXE’
library: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3python39.dll’
home: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3’
isloaded: 1

I tried multiple different versions of numpy save
py.numpy.save(output_filename, data);
or
py.numpy.save(strcat(output_filename, ‘.npy’), py.numpy.array(data));
or
py.numpy.save(strcat(output_filename, ‘.npy’), data);I am trying to convert an array into numpy, but I keep getting this issue:
Python Error: ModuleNotFoundError: No module named ‘libmwbuffer’
Python function ” might not be able to accept input argument at position 1. The function may require a specific
data type that you can construct from the MATLAB array. For more information, see the documentation for Python
function ” and working with Python arrays.

I checked and I have this pyversion, which is supposed to be compatible with Matlab R2024a. what’s the issue?
version: ‘3.9’
executable: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3python.EXE’
library: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3python39.dll’
home: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3’
isloaded: 1

I tried multiple different versions of numpy save
py.numpy.save(output_filename, data);
or
py.numpy.save(strcat(output_filename, ‘.npy’), py.numpy.array(data));
or
py.numpy.save(strcat(output_filename, ‘.npy’), data); I am trying to convert an array into numpy, but I keep getting this issue:
Python Error: ModuleNotFoundError: No module named ‘libmwbuffer’
Python function ” might not be able to accept input argument at position 1. The function may require a specific
data type that you can construct from the MATLAB array. For more information, see the documentation for Python
function ” and working with Python arrays.

I checked and I have this pyversion, which is supposed to be compatible with Matlab R2024a. what’s the issue?
version: ‘3.9’
executable: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3python.EXE’
library: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3python39.dll’
home: ‘C:Program FilesArcGISProbinPythonenvsarcgispro-py3’
isloaded: 1

I tried multiple different versions of numpy save
py.numpy.save(output_filename, data);
or
py.numpy.save(strcat(output_filename, ‘.npy’), py.numpy.array(data));
or
py.numpy.save(strcat(output_filename, ‘.npy’), data); numpy, python, libmwbuffer MATLAB Answers — New Questions

​