Expected input number 2, edges, to be one of these types

Hello all,

Here I am doing a Chi-Square calculation

main

data = randi([0,10],1,20);
% dummy variable
% 10, 0.05, 'unif', 1.5, 2.9
n = 10;
alpha = 0.05;
dist = 'unif';
a1 = 1.5;
a2 = 2.9;
a3 = 3.0;

%Call to the function
[chi2, critical] = chi2test (data, n, alpha, dist, a1, a2);

The function is straigth forward, however the part I am have a issue is the prod, and the invcdf anonymous function. I do not have the tool kit for the cdf, so I did it by sprintf, using the histcounts. Here is where the issue lays, is there a way to get around this error?

function [chi2, critical] = chi2test (data, n, alpha, dist, a1, a2)
  prob = @(a,b) sprintf('cdf(''%s'', b, %10.10g, %10.10g) - cdf(''%s'', a, %10.10g, %10.10g)', dist, a1, a2, dist, a1, a2);
  invcdf = @(x) sprintf('icdf(''%s'', x, %10.10g, %10.10g)', dist, a1, a2);
        
  pi = (1/n) .* (0:n);
  intvls = invcdf(pi);
        
  % The error is thrown here
  o_freq = histcounts(data, intvls);
end

I recevie the following error

Error using histcounts
Expected input number 2, edges, to be one of these types:
double, single, uint8, uint16, uint32, uint64, int8, int16, int32, int64, logical

Thank you all for any help

Answer 1

[Steven Lord]

function [chi2, critical] = chi2test (data, n, alpha, dist, a1, a2)
        prob = @(a,b) sprintf('cdf(''%s'', b, %10.10g, %10.10g) - cdf(''%s'', a, %10.10g, %10.10g)', dist, a1, a2, dist, a1, a2);
        invcdf = @(x) sprintf('icdf(''%s'', x, %10.10g, %10.10g)', dist, a1, a2);

This anonymous function does not call icdf on its inputs. It create a string that contains the icdf call that you would execute to compute the icdf. [The same holds for prob and cdf.] Don't use sprintf.

invcdf = @(x) icdf(dist, x, a1, a2);

If you really need the %10.10g behavior, use round on a1 and a2.

[SugerCodeCude]

Thank you so much for the quick responce.

However, when I have done that in the past. I would get

Error using chi2test>@(x)icdf(dist,x,a1,a2)
'icdf' requires Statistics and Machine Learning Toolbox.

Is there another way to go about this? I dont have that tool and the same for applies for both the icdf and cdf.

[SugerCodeCude]

Thank you Steven Lord, the fix on the anonymous function is alot clean, then for the issue on the error I get a matlab with the statistics needed to run chi-square directly.

I was looking to write the chi-square equation, however as time is of the essence I shell do that at a later time.

Thank you again

Answer 2

[dpb]

W/O the Statistics TB, you'll have to resort to some other way to compute the wanted distribution quantities -- a quick look at File Exchange located <generalized-chi-square-distribution> that purports to do what you want. There may be others there.

[SugerCodeCude]

Here is why I am big advocate of Matlab and the Matlab community!! That is exactly what I am looking for!!

dpb, I hope management see how great asset you are to the Matlab team!! thank you again for your time and effort in helping me resolve the error!!

Thank you again

[dpb]

Answers is totally volunteer; there are a number like Steven who are TMW employees who contribute, but I am just a retired geezer... :)