I'm trying to understand how to use the chi2gof function in matlab with a very simple test. Let's assume that I toss a coin 190 times and get 94 heads and 96 tails. The null hypothesis should be that i get 95h, 95t. As far as I understand the documentation, I should be able to test the hypothesis by running
[h,p,stats] = chi2gof([94,96], 'expected', [95,95])
However, this returns h = 1, which supposedly means that null hypothesis is rejected, which makes no sense. Another pecular thing is that the O parameter in stats returns as O: [0 2] - but shouldn't this be my input ([94,96])? What am I doing wrong?
What am I doing wrong?
The problem is that you are passing the cumulative outcome of your coin tosses to chi2gof. The goodness-of-fit test must be performed on the full sample. From the official documentation (reference here):
x = sample data for the hypothesis test, specified as a vector (the wrong part of your code)
Expected = expected counts for each bin (the correct part of your code)
Let's make an example using the correct variables:
ct = randsample([0 1],190,true,[0.49 0.51]);
[h,p,stats] = chi2gof(ct,'Expected',[95 95]);
The returned value of h is 0, which is absolutely correct.
Now, let's make an example that is supposed to fail:
ct = randsample([0 1],190,true,[0.05 0.95]);
[h,p,stats] = chi2gof(ct,'Expected',[95 95]);
As you can see, h returned from this second test will be equal to 1.
On a final note, don't forget to take a look at the second output argument, which is the p-value of the test and is an important element to evaluate the significance of the result.
Related
I am doing another coursera assignemnt, this time with aerial robotics. I have to program a pd controller using the matlab ode45 (ordinary diff. equation). And the file that has to contain this code gets called as follows:
pd_controller(~, s, s_des, params)
I searched around but couldn't find anthing that explain this to me and how it works.
In the main program the function is called with a time variable which I would need for my ODE:
controlhandle(t, s, s_des, params)
Where this controlhandle is the functionhandler for pd_controller.
So, what does this mean? And can I access whatever is behind ~?
Besides:
I found one example, but the other around. A function, let's call it function = f(a,b) was called with f(~, b) where a and b has been declared inside the function.
The symbol is called a tilde, and it signifies that you are ignoring that input argument.
See the documentation here: https://mathworks.com/help/matlab/matlab_prog/ignore-function-inputs.html
In your case, the function controlhandle will not be passed a t variable, and probably has (should have) some check for this and perhaps a default t if none is given.
This works the same with output arguments, for example if you want the index of a max in an array, but not the max itself, you would use
a = [pi, 3.6, 1];
[~, idx] = max(a); % idx = 2, we don't know what the max value is
It means you don't need pass this parameter in this function call. Also, you can use it in the output of some functions too. For example:
A = [1 4 2 2 41];
[~, B] = sort(A);
this means you don't need the second output, and you can ignore that.
In your case, when no value sent for the first parameter t, probably the function acts on a default value for t in his computation.
Also, you can find more about that in matlab documentation.
I should have mentioned that this post exists as an answer, but it might be here instead.
I have a homework assignment that tasks me with writing a MATLAB function and I'm worried that I've missed something in my current answer. The function returns the complete solution to a linear equation of the form Ax=b, where A is a square matrix, and b is a vector of the appropriate dimension. The first line of the function is
function [Bs, Ns] = a(A, b)
where Bs is the basic solution (a vector), and Ns is the null solution - a matrix whose columns are a basis of the null space of A. There are also a few considerations in terms of the code used:
Code will be marked based on a set of test cases.
Built-in functions may be used in the code but it must be my original work.
Code that produces an error or warning, such as for a singular matrix, will be assigned a failing mark.
It can be assumed the test set will contain matrices that are all zero, non-singular, and otherwise rank deficient (complete solution exists, but MATLAB will produce an error or warning).
The code I've written is below.
function [Bs, Ns] = a(A, b)
ncols = size(A, 2);
x = pinv(A)*b;
Bs = x;
if ncols == rank(A)
Ns = zeros(ncols,1);
else
Ns = null(A);
end
end
The simplicity of my function has me worried that I've missed something (assignment is worth 4% of final grade) - either in my interpretation of the listed considerations, or that there are test cases which will cause errors/warnings. Any input would be appreciated.
I'm trying to use the MATLAB function fzero properly but my program keeps returning an error message. This is my code (made up of two m-files):
friction_zero.m
function fric_zero = friction_zero(reynolds)
fric_zero = 0.25*power(log10(5.74/(power(reynolds,0.9))),-2);
flow.m
function f = flow(fric)
f = 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(power(10,4));
z = fzero(#flow,f_initial)
The goal is to return z as the root for the equation specified by f when flow.m is run.
I believe I have the correct syntax as I have spent a couple of hours online looking at examples. What happens is that it returns the following error message:
"Undefined function or variable 'fric'."
(Of course it's undefined, it's the variable I'm trying to solve!)
Can someone point out to me what I've done wrong? Thanks
EDIT
Thanks to all who helped! You have assisted me to eventually figure out my problem.
I had to add another file. Here is a full summary of the completed code with output.
friction_zero.m
function fric_zero = friction_zero(re)
fric_zero = 0.25*power(log10(5.74/(power(re,0.9))),-2); %starting value for fric
flow.m
function z = flow(fric)
re = power(10,4);
z = 1/(sqrt(fric))-1.873*log10(re*sqrt(fric))-233/((re*sqrt(fric))^0.9)-0.2361;
flow2.m
f_initial = friction_zero(re); %arbitrary starting value (Reynolds)
x = #flow;
fric_root = fzero(x,f_initial)
This returns an output of:
fric_root = 0.0235
Which seems to be the correct answer (phew!)
I realised that (1) I didn't define reynolds (which is now just re) in the right place, and (2) I was trying to do too much and thus skipped out on the line x = #flow;, for some reason when I added the extra line in, MATLAB stopped complaining. Not sure why it wouldn't have just taken #flow straight into fzero().
Once again, thanks :)
You need to make sure that f is a function in your code. This is simply an expression with reynolds being a constant when it isn't defined. As such, wrap this as an anonymous function with fric as the input variable. Also, you need to make sure the output variable from your function is z, not f. Since you're solving for fric, you don't need to specify this as the input variable into flow. Also, you need to specify f as the input into fzero, not flow. flow is the name of your main function. In addition, reynolds in flow is not defined, so I'm going to assume that it's the same as what you specified to friction_zero. With these edits, try doing this:
function z = flow()
reynolds = power(10,4);
f = #(fric) 1/(sqrt(fric))-1.873*log10(reynolds*sqrt(fric))-233/((reynolds*sqrt(fric))^0.9)-0.2361;
f_initial = friction_zero(reynolds);
z = fzero(#f, f_initial); %// You're solving for `f`, not flow. flow is your function name
The reason that you have a problem is because flow is called without argument I think. You should read a little more about matlab functions. By the way, reynolds is not defined either.
I am afraid I cannot help you completely since I have not been doing fluid mechanics. However, I can tell you about functions.
A matlab function definition looks something like this:
function x0 = f(xGuess)
a = 2;
fcn =#(t) a*t.^3+t; % t must not be an input to f.
disp(fcn);
a = 3;
disp(fcn);
x0 = fsolve(fcn1,xGuess); % x0 is calculated here
The function can then ne called as myX0 = f(myGuess). When you define a matlab function with arguments and return values, you must tell matlab what to do with them. Matlab cannot guess that. In this function you tell matlab to use xGuess as an initial guess to fsolve, when solving the anonymous function fcn. Notice also that matlab does not assume that an undefined variable is an independent variable. You need to tell matlab that now I want to create an anonymous function fcn which have an independent variable t.
Observation 1: I use .^. This is since the function will take an argument an evaluate it and this argument can also be a vector. In this particulat case I want pointwise evaluation. This is not really necessary when using fsolve but it is good practice if f is not a matrix equation, since "vectorization" is often used in matlab.
Observation 2: notice that even if a changes its value the function does not change. This is since matlab passes the value of a variable when defining a function and not the variable itself. A c programmer would say that a variable is passed by its value and not by a pointer. This means that fcn is really defined as fcn = #(x) 2*t.^3+t;. Using the variable a is just a conveniance (constants can may also be complicated to find, but when found they are just a value).
Armed with this knowledge, you should be able to tackle the problem in front of you. Also, the recursive call to flow in your function will eventuallt cause a crash. When you write a function that calls itself like this you must have a stopping criterium, something to tell the program when to stop. As it is now, flow will call ifself in the last row, like z = fzero(#flow,f_initial) for 500 times and then crash. Alos it is possible as well to define functions with zero inputs:
function plancksConstant = h()
plancksConstant = 6.62606957e−34;
Where the call h or h() will return Plancks constant.
Good luck!
I have recently started learning MatLab, and wrote the following script today as part of my practice to see how we can generate a vector:
x = [];
n = 4;
for i = i:n
x = [x,i^2];
end
x
When I run this script I get what I expect, namely the following vector:
x = 0 1 4 9 16
However, if I run the script a second time right afterwards I only get the following output:
x = 16
What is the reason for this? How come I only get the last vector entry as output the second time I run the script, and not the vector in its entirety? If anyone can explain this to me, I would greatly appreciate it.
Beginning with a fresh workspace, i will simply be the complex number 1i (as in x^2=-1). I imagine you got this warning on the first run:
Warning: Colon operands must be real scalars.
So the for statement basically loops over for i = real(1i):4. Note that real(1i)=0.
When you rerun the script again with the variables already initialized (assuming you didn't clear the workspace), i will refer to a variable containing the last value of 4, shadowing the builtin function i with the same name, and the for-loop executes:
x=[];
for i=4:4
x = [x, i^2]
end
which iterates only one time, thus you end up with x=16
you forget to initialize i.
after first execution i is 4 and remains 4.
then you initialize x as an empty vector but because i is 4 the loop runs only once.
clear your workspace and inspect it before and after first execution.
Is it possibly a simple typo?
for i = i:n
and should actually mean
for i = 1:n
as i is (probably) uninitialized in the first run, and therefore 0, it works just fine.
The second time, i is still n (=4), and only runs once.
Also, as a performance-tip: in every iteration of your loop you increase the size of your vector, the more efficient (and more matlaboid) way would be to create the vector with the basevalues first, for example with
x = 1:n
and then square each value by
x = x^2
In Matlab, using vector-operations (or matrix-operations on higher dimensions) should be prefered over iterative loop approaches, as it gives matlab the opportunity to do optimised operations. It is also often more readable that way.
i have two problems in mathematica and want to do them in matlab:
measure := RandomReal[] - 0.5
m = 10000;
data = Table[measure, {m}];
fig1 = ListPlot[data, PlotStyle -> {PointSize[0.015]}]
Histogram[data]
matlab:
measure =# (m) rand(1,m)-0.5
m=10000;
for i=1:m
data(:,i)=measure(:,i);
end
figure(1)
plot(data,'b.','MarkerSize',0.015)
figure(2)
hist(data)
And it gives me :
??? The following error occurred
converting from function_handle to
double: Error using ==> double
If i do :
measure =rand()-0.5
m=10000;
data=rand(1,m)-0.5
then, i get the right results in plot1 but in plot 2 the y=axis is wrong.
Also, if i have this in mathematica :
steps[m_] := Table[2 RandomInteger[] - 1, {m}]
steps[20]
Walk1D[n_] := FoldList[Plus, 0, steps[n]]
LastPoint1D[n_] := Fold[Plus, 0, steps[n]]
ListPlot[Walk1D[10^4]]
I did this :
steps = # (m) 2*randint(1,m,2)-1;
steps(20)
Walk1D =# (n) cumsum(0:steps(n)) --> this is ok i think
LastPointold1D= # (n) cumsum(0:steps(n))
LastPoint1D= # (n) LastPointold1D(end)-->but here i now i must take the last "folding"
Walk1D(10)
LastPoint1D(10000)
plot(Walk1D(10000),'b')
and i get an empty matrix and no plot..
Since #Itamar essentially answered your first question, here is a comment on the second one. You did it almost right. You need to define
Walk1D = # (n) cumsum(steps(n));
since cumsum is a direct analog of FoldList[Plus,0,your-list]. Then, the plot in your code works fine. Also, notice that, either in your Mathematica or Matlab code, it is not necessary to define LastPoint1D separately - in both cases, it is the last point of your generated list (vector) steps.
EDIT:
Expanding a bit on LastPoint1D: my guess is that you want it to be a last point of the walk computed by Walk1D. Therefore, it would IMO make sense to just make it a function of a generated walk (vector), that returns its last point. For example:
lastPoint1D = #(walk) (walk(end));
Then, you use it as:
walk = Walk1D(10000);
lastPoint1D(walk)
HTH
You have a few errors/mistakes translating your code to Matlab:
If I am not wrong, the line data = Table[measure, {m}]; creates m copies of measure, which in your case will create a random vector of size (1,m). If that is true, in Matlab it would simply be data = measure(m);
The function you define gets a single argument m, therefor it makes no sense using a matrix notation (the :) when calling it.
Just as a side-note, if you insert data into a matrix inside a for loop, it will run much faster if you allocate the matrix in advance, otherwise Matlab will re-allocate memory to resize the matrix in each iteration. You do this by data = zeros(1,m);.
What do you mean by "in plot 2 the y=axis is wrong"? What do you expect it to be?
EDIT
Regarding your 2nd question, it would be easier to help you if you describe in words what you want to achieve, rather than trying to read your (error producing) code. One thing which is clearly wrong is using expression like 0:steps(n), since you use m:n with two scalars m and n to produce a vector, but steps(n) produces a vector, not a scalar. You probably get an empty matrix since the first value in the vector returned by steps(n) might be -1, and 0:-1 produces an empty vector.