Chendi Lin
MathWorks
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I am a Software Engineer in Simulink Code Inspector at MathWorks. My main responsibilities include feature development and codebase maintenance for Simulink Code Inspector.
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Storing multiple matrices from a loop into a single variable without cell function
Hi Ilker, If I understand your question correctly, A is a 3x3 matrix. For each i in the iteration, you want to store A withou...
Storing multiple matrices from a loop into a single variable without cell function
Hi Ilker, If I understand your question correctly, A is a 3x3 matrix. For each i in the iteration, you want to store A withou...
3 years 前 | 1
已回答
Returning values based on range of variable values
Hi Kevin, The index of MATLAB starts from 1. Have you tried U(x,1)? Best, CD
Returning values based on range of variable values
Hi Kevin, The index of MATLAB starts from 1. Have you tried U(x,1)? Best, CD
3 years 前 | 0
已回答
isempty function does not work for me
Hi Nelson, I believe the correct way to use "isempty" is to call "isempty(out1)". Here is the sample code: >> a = zeros(0,2)...
isempty function does not work for me
Hi Nelson, I believe the correct way to use "isempty" is to call "isempty(out1)". Here is the sample code: >> a = zeros(0,2)...
3 years 前 | 0
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Differentiating a Symbolic Function
Hi Ammar, Without explicitly defining the differentiation variable, "diff" uses the default variable, which is "x" in your ca...
Differentiating a Symbolic Function
Hi Ammar, Without explicitly defining the differentiation variable, "diff" uses the default variable, which is "x" in your ca...
3 years 前 | 1
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how to subtract the mean value of the network output values
To normalize the net I think you can simply do advantage_Net = advantage_Net - mean(advantage_Net) Please let me know if I un...
how to subtract the mean value of the network output values
To normalize the net I think you can simply do advantage_Net = advantage_Net - mean(advantage_Net) Please let me know if I un...
3 years 前 | 0
已回答
Problem is unbounded in linear programming
Hi Klaus, Because all your design variables are non-negative, you will have AT = [1 4 2 3; 4 5 3 1]; u = [300 400]; g...
Problem is unbounded in linear programming
Hi Klaus, Because all your design variables are non-negative, you will have AT = [1 4 2 3; 4 5 3 1]; u = [300 400]; g...
3 years 前 | 3