HISTCOUNTS a true replacement of HISTC? (2024)

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Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 11 Oct 2018

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Since few latest releases we get the code warning "histc is not recommended. Use histcounts instead".

However how to replace HISTC with specified DIM argument, for example what is the command to get C1 and C2 in this example?

I hope you won't tell me I need a for-loop or some sort disguised loop.

>> A=reshape(linspace(0,1,20),[4 5])

A =

0 0.2105 0.4211 0.6316 0.8421

0.0526 0.2632 0.4737 0.6842 0.8947

0.1053 0.3158 0.5263 0.7368 0.9474

0.1579 0.3684 0.5789 0.7895 1.0000

>> b=linspace(0,1,3)

b =

0 0.5000 1.0000

>> c1=histc(A,b,1)

c1 =

4 4 2 0 0

0 0 2 4 3

0 0 0 0 1

>> c2=histc(A,b,2)

c2 =

3 2 0

3 2 0

2 3 0

2 2 1

>>

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Accepted Answer

Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 11 Oct 2018

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Third solution, may be the best in term of speed/memory

% replacement of [n,i] = histc(x,edges,dim);

[~,~,i] = histcounts(x, [edges,Inf]);

s = uint32(size(x));

nbins = uint32(length(edges));

p1 = uint32(prod(s(1:dim-1)));

p2 = uint32(prod(s(dim+1:end)));

if p1 == 1 % happens for dim==1

k = reshape(uint32(i),[s(dim),p2]);

l = nbins*(0:p2-1);

ilin = l + k;

else

j = reshape(1:p1,[p1,1,1]);

k = reshape(uint32(i-1)*p1,[p1,s(dim),p2]);

l = reshape((p1*nbins)*(0:p2-1),[1,1,p2]);

ilin = (j + l) + k;

end

ilin = ilin(:);

s(dim) = nbins;

n = accumarray(ilin(i~=0),1,[prod(s),1]);

n = reshape(n,s);

EDIT: slight improve code by (1) avoid reduce calculation for dim=1, (2) indexing cast to UINT32

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Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 10 Oct 2018

Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 11 Oct 2018

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Timing of the above algo for arrays of 100x100x100

Working along dimension #1

Time HISTC = 15.549 [ms]

Time HISTCOUNTS = 19.481 [ms]

Penalty factor = 1.3

Working along dimension #2

Time HISTC = 20.104 [ms]

Time HISTCOUNTS = 16.454 [ms]

Penalty factor = 0.8

Working along dimension #3

Time HISTC = 21.446 [ms]

Time HISTCOUNTS = 18.485 [ms]

Penalty factor = 0.9

The test code (in function) is as following:

x = reshape(linspace(0,1,1e6),[100 100 100]);

edges = linspace(0,1,100);

t = zeros(2,ndims(x));

for dim=1:ndims(x)

tic

[nn,~] = histc(x,edges,dim);

t(1,dim) = toc;

tic

[~,~,i] = histcounts(x, [edges,Inf]);

s = uint32(size(x));

nbins = uint32(length(edges));

p1 = uint32(prod(s(1:dim-1)));

p2 = uint32(prod(s(dim+1:end)));

if p1 == 1

k = reshape(uint32(i),[s(dim),p2]);

l = reshape(nbins*(0:p2-1),1,[]);

ilin = l + k;

else

j = reshape(1:p1,[],1,1);

k = reshape(uint32(i-1)*p1,[p1,s(dim),p2]);

l = reshape((p1*nbins)*(0:p2-1),1,1,[]);

ilin = (j + l) + k;

end

ilin = ilin(:);

s(dim) = nbins;

n = accumarray(ilin(i~=0),1,[prod(s),1]);

n = reshape(n,s);

t(2,dim) = toc;

end

t = t*1e3;

for dim=1:ndims(x)

fprintf('Working along dimension #%d\n', dim);

fprintf('\tTime HISTC = %1.3f [ms]\n', t(1,dim));

fprintf('\tTime HISTCOUNTS = %1.3f [ms]\n', t(2,dim));

fprintf('\tPenalty factor = %1.1f\n', t(2,dim)/t(1,dim));

end

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Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 10 Oct 2018

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OK, here is the answer of my own question.

The replacement of

[n,i] = histc(x, edges, dim);

is

[~,~,i] = histcounts(x, edges);

s = size(x);

nd = length(s);

idx = arrayfun(@(n) 1:n, s, 'unif', 0);

[idx{:}] = ndgrid(idx{:});

idx{dim} = i;

s(dim) = length(edges);

idx = cat(nd+1,idx{:});

idx = reshape(idx,[],nd);

n = accumarray(idx(i~=0,:),1,s);

That's a lot of code. If an authority of TMW can shed a light if there is a shorter, faster, cleaner solution, I'm in.

NOTE: there is still a subtle difference caused by data right at the outer edge

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Stephen23 on 10 Oct 2018

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Nice. This definitely wins the Answer of the Day award.

Perhaps the next step would be to make an enhancement request.

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OCDER on 10 Oct 2018

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Interesting finding that histcounts doesn't have the Dim option. You should ask TMW to add that feature - they must have overlooked that.

"I hope you won't tell me I need a for-loop or some sort disguised loop."

But, there really is no need to avoid a simpler for-loop solution. Also, your answer has a "disguised loop" inside the arrayfun - arrayfun uses an internal for loop.

A =[ 0 0.2105 0.4211 0.6316 0.8421;

0.0526 0.2632 0.4737 0.6842 0.8947;

0.1053 0.3158 0.5263 0.7368 0.9474;

0.1579 0.3684 0.5789 0.7895 1.0000];

Bin = linspace(0, 1, 3);

C1 = histc(A, Bin, 1);

C2 = myHistC(A, Bin, 1);

D1 = histc(A, Bin, 2);

D2 = myHistC(A, Bin, 2);

assert(isequal(C1, C2) && isequal(D1, D2), 'Not the same thing');

function C = myHistC(A, Bin, Dim)

if nargin < 3; Dim = 1; end

if Dim == 1

C = zeros(numel(Bin), size(A, 2));

for j = 1:size(A, 2)

C(:, j) = histcounts(A(:, j), [Bin(:); Inf]);

end

else

C = zeros(size(A, 1), numel(Bin));

for j = 1:size(A, 1)

C(j, :) = histcounts(A(j, :), [Bin(:); Inf]);

end

end

end

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Bruno Luong on 10 Oct 2018

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What I try to avoid is the loop on HISTCOUNTS, the small parasite loops I don't mind.

BTW your code can't handle nd array as it is.

OCDER on 10 Oct 2018

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Oh, I see. Yeah, TMW should step in and suggest a better solution (if any). Plus, my solution also cannot handle the variable outputs generated by histc/histcounts. I'm curious as to what's the alternative to histc.

Guillaume on 10 Oct 2018

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"they must have overlooked that"

I'm not speaking for Mathworks, but I'd say they're very much aware of it. You just have to look at Replace Discouraged Instances of hist and histc which clearly states under Code Updates for histc:

"histcounts treats the input matrix as a single tall vector and calculates the bin counts for the entire matrix [...] Use a for-loop to calculate bin counts over each column. [...] If performance is a problem due to a large number of columns in the matrix, then consider continuing to use histc for the column-wise bin counts."

OCDER on 10 Oct 2018

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I see. But if the suggestion is to use histc, I'm wondering why histcounts doesn't implement this directly using a name-value input like

histcounts(..., 'dim', N) ?

Overall, I find it confusing that they acknowledged the shortcoming of histcounts, offered a workaround code in the example but do not implement this within histcounts, and then, made histc generate warning messages to use histcounts instead, which has no replacement for histc's column/row-wise binning. It's an interesting decision.

From the website:

Use a for-loop to calculate bin counts over each column.

A = randn(100,10);

nbins = 10;

N = zeros(nbins, size(A,2));

for k = 1:size(A,2)

N(:,k) = histcounts(A(:,k),nbins);

end

Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 10 Oct 2018

+1 OCDER

And the example in the document merely deals with a workaround for 2D case. It would be a nightmare for someone who is not familiar with MATLAB and find a decend alternative for nd-array case.

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Bruno Luong on 10 Oct 2018

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Edited: Bruno Luong on 10 Oct 2018

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Another solution of replacement, still long, but consumes less memory than my first solution at the cost of array permutations:

% replacement of [n,i] = histc(x,edges,dim);

s = size(x);

nd = length(s);

p = 1:nd;

p([1 dim]) = p([dim 1]);

q(1:nd) = p;

xp = permute(x, p);

sp = s(p);

[~,~,i] = histcounts(xp, [edges, Inf]);

m = prod(sp(2:end));

sp1 = sp(1);

nbins = length(edges);

j = repmat(nbins*(0:m-1),[sp1,1]);

keep = i~=0;

linidx = i(keep) + j(keep);

sp(1) = nbins;

n = accumarray(linidx(:),1,[prod(sp) 1]);

n = permute(reshape(n,sp),q);

i = permute(i,q);

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HISTCOUNTS a true replacement of HISTC? (14)

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HISTCOUNTS a true replacement of HISTC? (2024)

FAQs

What is histc in Matlab? ›

histc calculates the bin counts for each column of input data. For an input matrix of size m -by- n , histc returns a matrix of bin counts of size length(edges) -by- n .

Keep Reading ›
What is bin count in Matlab? ›

bincounts = histc( x , binranges ) counts the number of values in x that are within each specified bin range. The input, binranges , determines the endpoints for each bin. The output, bincounts , contains the number of elements from x in each bin.

Tell Me More ›
What does histcount mean in MATLAB? ›

The histcounts function uses an automatic binning algorithm that returns uniform bins chosen to cover the range of elements in X and reveal the underlying shape of the distribution. example. [ N , edges ] = histcounts( X , nbins ) uses a number of bins specified by the scalar, nbins .

Keep Reading ›
How to use hist in MATLAB? ›

hist( x , xbins ) sorts x into bins with intervals or categories determined by the vector xbins .
  1. If xbins is a vector of evenly spaced values, then hist uses the values as the bin centers.
  2. If xbins is a vector of unevenly spaced values, then hist uses the midpoints between consecutive values as the bin edges.

See More ›
What is bin count? ›

The bincount() method of the NumPy module is used to find the frequency of each element in a NumPy array of positive integers. The element's index in the frequency array or bin is stored as the element's count. Because each bin value represents an instance of its index, we can adjust the bin size accordingly.

View More ›
What are bins in a histogram? ›

A histogram displays numerical data by grouping data into "bins" of equal width. Each bin is plotted as a bar whose height corresponds to how many data points are in that bin. Bins are also sometimes called "intervals", "classes", or "buckets".

See More ›
How many bins can a histogram have MATLAB? ›

If you specify BinWidth , then Histogram can use a maximum of 65,536 bins (or 216). If the specified bin width requires more bins, then histogram uses a larger bin width corresponding to the maximum number of bins.

Keep Reading ›
What does Histeq mean in MATLAB? ›

histeq performs histogram equalization. It enhances the contrast of images by transforming the values in an intensity image so that the histogram of the output image approximately matches a specified histogram (uniform distribution by default). adapthisteq performs contrast-limited adaptive histogram equalization.

Get More Info ›
What does the function hist return? ›

The function returns a tuple that contains the frequencies of the histogram bins, the edges of the bins, and the respective patches that create the histogram.

See Details ›
How to show the histogram of an image in MATLAB? ›

The imhist function displays the histogram, by default. The histogram shows a peak at around 100, corresponding to the dark gray background in the image. figure; imhist(I);

Get More Info ›

What is the bin of a histogram? ›

A histogram displays numerical data by grouping data into "bins" of equal width. Each bin is plotted as a bar whose height corresponds to how many data points are in that bin. Bins are also sometimes called "intervals", "classes", or "buckets".

See Details ›
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