12 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Bruno Luong am 10 Okt. 2018
Bearbeitet: Bruno Luong am 11 Okt. 2018
Akzeptierte Antwort: Bruno Luong
In MATLAB Online öffnen
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
>>
0 Kommentare -2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden
-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden
Melden Sie sich an, um zu kommentieren.
Melden Sie sich an, um diese Frage zu beantworten.
Akzeptierte Antwort
Bruno Luong am 10 Okt. 2018
Bearbeitet: Bruno Luong am 11 Okt. 2018
In MATLAB Online öffnen
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
2 Kommentare Keine anzeigenKeine ausblenden
Keine anzeigenKeine ausblenden
Bruno Luong am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620456
Bearbeitet: Bruno Luong am 10 Okt. 2018
Sorry, I must Accept my own answer but I think it's fair. Promise, I'll accept another solution if it's better.
Bruno Luong am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620466
Bearbeitet: Bruno Luong am 11 Okt. 2018
In MATLAB Online öffnen
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
Melden Sie sich an, um zu kommentieren.
Weitere Antworten (3)
Bruno Luong am 10 Okt. 2018
Bearbeitet: Bruno Luong am 10 Okt. 2018
In MATLAB Online öffnen
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
1 Kommentar -1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
-1 ältere Kommentare anzeigen-1 ältere Kommentare ausblenden
Stephen23 am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620386
Nice. This definitely wins the Answer of the Day award.
Perhaps the next step would be to make an enhancement request.
Melden Sie sich an, um zu kommentieren.
OCDER am 10 Okt. 2018
In MATLAB Online öffnen
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
5 Kommentare 3 ältere Kommentare anzeigen3 ältere Kommentare ausblenden
3 ältere Kommentare anzeigen3 ältere Kommentare ausblenden
Bruno Luong am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620400
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 am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620407
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 am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620422
"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 am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620447
In MATLAB Online öffnen
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 am 10 Okt. 2018
Direkter Link zu diesem Kommentar
https://de.mathworks.com/matlabcentral/answers/423252-histcounts-a-true-replacement-of-histc#comment_620459
Bearbeitet: Bruno Luong am 10 Okt. 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.
Melden Sie sich an, um zu kommentieren.
Bruno Luong am 10 Okt. 2018
Bearbeitet: Bruno Luong am 10 Okt. 2018
In MATLAB Online öffnen
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);
0 Kommentare -2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden
-2 ältere Kommentare anzeigen-2 ältere Kommentare ausblenden
Melden Sie sich an, um zu kommentieren.
Melden Sie sich an, um diese Frage zu beantworten.
Siehe auch
Kategorien
AI, Data Science, and StatisticsStatistics and Machine Learning ToolboxHypothesis Tests
Mehr zu Hypothesis Tests finden Sie in Help Center und File Exchange
Tags
- histogram
- histc
- histcounts
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Es ist ein Fehler aufgetreten
Da Änderungen an der Seite vorgenommen wurden, kann diese Aktion nicht abgeschlossen werden. Laden Sie die Seite neu, um sie im aktualisierten Zustand anzuzeigen.
Translated by 
Website auswählen
Wählen Sie eine Website aus, um übersetzte Inhalte (sofern verfügbar) sowie lokale Veranstaltungen und Angebote anzuzeigen. Auf der Grundlage Ihres Standorts empfehlen wir Ihnen die folgende Auswahl: .
Sie können auch eine Website aus der folgenden Liste auswählen:
Amerika
- América Latina (Español)
- Canada (English)
- United States (English)
Europa
- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)
- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- Deutsch
- English
- Français
- United Kingdom(English)
Asien-Pazifik
- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)
Kontakt zu Ihrer lokalen Niederlassung
