Help on a Function

Greetings!

I would like your help on a function I'd like to make on Matlab.

I want to make a function called R,

[Roots] = R(X,Y,eps) and this function will be returning a table with 2 columns showing the X and Y of the points where abs(Y)<eps

Thank you in advance.

[dpb]

OK, simple enough. What have you got so far and where, specifically did you get stuck?

[Dhmhgr]

I'm new to Matlab and I saw this interesting exercise online and there is no solution given. I don't know what to do, but I'd like to know how this type of exercises are getting solved.

[dpb]

I suggest reading the Getting Started tutorial sections in the online documentation...it outlines basic syntax and operations that will get you started. There are lots of examples of writing simple functions with the doc for function

Logical addressing is the trick for such questions as this...see what

eps=0.5;
x=rand(10,1);
x(x<eps)

does.

[Dhmhgr]

I see, but it didn't help me.

Answer 1

[Image Analyst]

Try this (untested)

% This function will be returning a table with 2 columns showing the X and Y of the points where abs(Y)<eps
function [Roots] = R(X,Y,eps) 
Roots = []; % Initialize to no values.
indexes = abs(Y) < eps;
if isempty(indexes)
    % No Y satisfies the criteria.
    return;
end
xSmall = reshape(X(indexes), [], 1); % Extract indexes and make sure they're in a column vector.
ySmall = reshape(Y(indexes), [], 1); % Extract indexes and make sure they're in a column vector.
Roots = table(xSmall, ySmall);
end

[Dhmhgr]

I forgot to say that X and Y are row vectors. Is this solution ok if the X and Y are row vectors?

I made them like that: X=linspace(-2,2)

[Walter Roberson]

Yes, that code will work with row vectors. Image Analyst took care of the detail that the values passed into table() should normally be column vectors.

[dpb]

Continuing w/ Walter's explanation, because if row vectors,

>> X=1:10;Y=randn(size(X));
>> table(X,Y)
ans =
  1×2 table
          X                Y      
    _____________    _____________
    [1x10 double]    [1x10 double]
>> 

you see you get a row vector for each of the two variables as the one record in the table...this probably is not what is wanted.

[Dhmhgr]

So, what should the correct answer be? The first one by the person "Image Analyst"?

[Image Analyst]

Of course. dpb was merely showing what would happen if I didn't use reshape to turn the data into a column vector before passing in to the table() function. It was not a separate answer in itself, and I did use reshape() so I did it correctly.

Answer 2

[Dhmhgr]

??

[Image Analyst]

I didn't notice at first that we were using eps. That is probably a bad thing to use (unless it's a trick question) because no value of Y can be less than eps unless it's pure zero.

Try this (with a variable named minValue):

X = rand(1, 100);
Y = rand(1, 100);
minValue = 0.1;
Roots = R(X,Y,minValue) 

% This function will be returning a table with 2 columns showing the X and Y of the points where abs(Y)<eps
function [Roots] = R(X,Y,minValue) 
Roots = []; % Initialize to no values.
indexes = abs(Y) < minValue;
if isempty(indexes)
    % No Y satisfies the criteria.
    return;
end
xSmall = reshape(X(indexes), [], 1); % Extract indexes and make sure they're in a column vector.
ySmall = reshape(Y(indexes), [], 1); % Extract indexes and make sure they're in a column vector.
Roots = table(xSmall, ySmall);
end

You'll see

Roots =

11×2 table

xSmall ySmall

_________________ ___________________

0.970592781760616 0.0838213779969326

0.792207329559554 0.0781755287531837

0.849129305868777 0.00463422413406744

0.743132468124916 0.0844358455109103

0.489764395788231 0.0759666916908419

0.679702676853675 0.0496544303257421

0.890903252535798 0.0964545251683886

0.257508254123736 0.0597795429471558

0.243524968724989 0.0154034376515551

0.929263623187228 0.0430238016578078

0.567821640725221 0.0811257688657853

Answer 3

[Dhmhgr]

Okay, function made correctly.

I have an other problem regarding the same exercise.

I have:

x=linspace(-2,2,100) y=x.^4-3*x.^3+x.^2+8*x

I want to give eps a number so that by calling the R function I will be able to find the points where abs(y)<eps.

What should the eps number be?

[Walter Roberson]

What should the eps number be?

42.

[Image Analyst]

eps, which remember I said must be renamed, can be anything. Whatever you want. Whatever you use, R will find the points, if any exists, or null if they don't.