Binning of data except from histcounts?

I am a matlab R2014b user. Using histcounts for binning a dataset is not giving me the exact result. Is there any alternative for this ?

 A = [0 0 1 1 1 0 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1];
N = histcounts(X,6)
>>N =

     6     0     2     0     0     2

This what I am getting every time.

A = [0 0 1 1 1 0 0 0 0 NaN NaN 1 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1];
C = categorical(A,[1 0 NaN],{'yes','no','undecided'})
[N,Categories] = histcounts(C)

And for the above I am getting this error.

Error using histcounts Expected input number 1, x, to be one of these types:

numeric

Instead its type was categorical.

Error in histcounts (line 96) validateattributes(x,{'numeric'},{'real'}, mfilename, 'x', 1)

Answer 1

[Bruno Luong]

ndivisions=4;
n = length(A);
partnum = floor(1+(0:n-1)/n*ndivisions);
n1 = accumarray(partnum(:),A(:)==1)

[Zara Khan]

Thank you for making this thing clear to me. I will give a try to your code.

[Walter Roberson]

In the case where the array is an exact multiple of ndivisions in length, then

sum(resize(A == 1, [], ndivisions),1)

[Bruno Luong]

resize -> reshape

[Walter Roberson]

Bruno is right, I should have written

sum(reshape(A == 1, [], ndivisions),1)

[Zara Khan]

Bruno Luong:You are code working well. how to get the size of the each part here?

[Bruno Luong]

"the size of the each part here"

? I'm not sure I understand your wording. Or do you mean how many elements of partnum that has a specific value? In this case

n = accumarray(partnum(:),1)

[Zara Khan]

Here we are diving the total data set in 8 equal parts. How can I calculate the size of the each part ?

[Bruno Luong]

Haven't try my answer?

[Zara Khan]

yes I have tried. now i want to get the size of the each part

[Bruno Luong]

Just give you, you have not read it or what?

[Zara Khan]

is this n = accumarray(partnum(:),1) ?

[Bruno Luong]

Yes.

[Zara Khan]

n = accumarray(partnum(:),1) this is giving the size of the each part then how can I count no of occurrence of 1's from each part ?

[Bruno Luong]

THAT IS ALREADY COMPUTED IN "N1", GRRRR

[Zara Khan]

Is this possible to plot and visualize ??

[Walter Roberson]

I have lost track of what would be plotted against what, and what kind of plot would be desired. But probably bar() would be useful.

[Zara Khan]

As each array is divided to 4 equal parts then from each part no of occrrence of 1's is being counted. Now I want to plot this total thing .

[Walter Roberson]

Sounds like you would calculate aa vector of the four counts and bar() that .

Answer 2

[Walter Roberson]

histcounts did not support categorical back then.

Use the second output of ismember to get the bin number, which you can then histc or histcounts or accumarray (most efficient)

[Zara Khan]

will you please explain this with an example?

[Walter Roberson]

[~, idx] = ismember(A, [1 0 nan]);
 N = accumarray(idx(:), 1);

[Zara Khan]

    A = [0 0 1 1 1 0 0 0 0 NaN NaN 1 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1];
>> [~, idx] = ismember(A, [1 0 nan]);
>>  N = accumarray(idx(:), 1);
Error using accumarray
First input SUBS must contain positive integer subscripts.

[Walter Roberson]

Interesting, I did not realize that ismember would not handle nan. It does make a kind of sense, in that nan have the oddity that

nan == nan

is false.

Answer 3

[Bruno Luong]

A = [0 0 1 1 1 0 0 0 0 NaN NaN 1 0 0 0 1 0 1 0 1 0 0 0 1 1 1 1];

[U,~,J] = unique(A);
inan = find(isnan(U),1,'first');
if inan
    U=U(1:inan); 
    J=min(J,inan); 
end
counts = accumarray(J(:), 1);
[U(:),counts]

ans =

     0    14
     1    11
   NaN     2

[Zara Khan]

I want to binning a dataset. what is this output giving ? and what will be the code for binning a dadaset when there no NaN's in the dataset ?

[Bruno Luong]

The output is

• 0 appears 14 times
• 1 appears 11 times
• Nan appears 2 timesin A

Plot the bar with it you'll get your "histogram".

The same code works with or without NaN. I can't see any bins in your question, and with an example with 0,1,NaN hard to see what kind of bins (beside [0,1,NaN]) is meaningful for you.

[Zara Khan]

Actually I am getting a linear array every and that is not a predefined size. I want to divide that array into 4 or 8 equal parts and wants to count occurence of 1's from each parts . How can I implement this ?

[Bruno Luong]

Sorry but this is not binning, binning means all the 1s fall at the same place: at the position 1.

If you use the wrong wording you should expect get the wrong answer.