Binning of data except from histcounts?

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Zara Khan
Zara Khan 2018-10-9
评论： Walter Roberson 2018-11-14
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)

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采纳的回答

Bruno Luong
Bruno Luong 2018-10-9
ndivisions=4;
n = length(A);
partnum = floor(1+(0:n-1)/n*ndivisions);
n1 = accumarray(partnum(:),A(:)==1)
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Zara Khan
Zara Khan 2018-11-14
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
Walter Roberson 2018-11-14
Sounds like you would calculate aa vector of the four counts and bar() that .

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更多回答（2 个）

Walter Roberson
Walter Roberson 2018-10-9
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)
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Zara Khan
Zara Khan 2018-10-9
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
Walter Roberson 2018-10-9
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.

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Bruno Luong
Bruno Luong 2018-10-9
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
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Zara Khan
Zara Khan 2018-10-9
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
Bruno Luong 2018-10-9
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.

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