Find least frequent value in an array

H = [1 1 2 2 3 3 4 5 5 5 6]
H = 1×11
     1     1     2     2     3     3     4     5     5     5     6
[counts, values] = histcounts(H, [unique(H) Inf])
counts = 1×6
     2     2     2     1     3     1
values = 1×7
     1     2     3     4     5     6   Inf
allMinimumCountsLocations = counts == min(counts)
allMinimumCountsLocations = 1×6 logical array
   0   0   0   1   0   1
values(allMinimumCountsLocations)
ans = 1×2
     4     6

I added Inf at the end of the list of unique values so that the last bin contained only the last value from unique(H) rather than the last two. If I had omitted it, the last bin would have contained both 5 and 6 (both of the last two edges) rather than just 6. This is not a bug; see the description of the edges input argument or the BinEdges name-value argument on the histcounts documentation page.

[counts, values, bins] = histcounts(H, unique(H))
counts = 1×5
     2     2     2     1     4
values = 1×6
     1     2     3     4     5     6
bins = 1×11
     1     1     2     2     3     3     4     5     5     5     5
valuesInLastBin = H(bins==max(bins))
valuesInLastBin = 1×4
     5     5     5     6

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Answer 2

Voss 2023-9-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2026447-find-least-frequent-value-in-an-array#answer_1320172

编辑：Voss 2023-9-27

在 MATLAB Online 中打开

H = [1 1 2 2 3 3 4 5 5 5];
[uu,ii] = unique(sort(H(:)));
[~,idx] = min(diff([ii; numel(H)+1]));
result = uu(idx)
result = 4
% another example:
H = randi(5,10,10)
H = 10×10
     1     5     4     1     3     3     3     5     2     3
     2     1     5     1     2     3     2     1     2     4
     2     3     5     5     4     5     2     5     3     2
     1     1     1     2     4     5     5     2     1     1
     1     3     2     5     2     3     2     5     4     3
     2     3     3     5     2     4     2     4     3     2
     5     2     4     1     1     2     5     5     4     1
     1     4     4     1     1     3     1     4     2     3
     2     3     5     2     4     3     4     4     3     2
     4     1     4     3     1     2     1     1     4     5
for i = 1:5
    fprintf('%d appears %d times\n',i,nnz(H == i))
end
1 appears 22 times
2 appears 24 times
3 appears 19 times
4 appears 18 times
5 appears 17 times
[uu,ii] = unique(sort(H(:)));
[~,idx] = min(diff([ii; numel(H)+1]));
result = uu(idx)
result = 5

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Voss 2023-9-27

在 MATLAB Online 中打开

Note that in case more than one element of H appears the least often, this method returns the lowest valued one, just like mode() does with the most frequently appearing value.

H = [2 2 3 3 4 5 5 5 0 1 1 1];
mode(H) % returns 1, not 5
ans = 1
[uu,ii] = unique(sort(H(:)));
[~,idx] = min(diff([ii; numel(H)+1]));
result = uu(idx) % returns 0, not 4
result = 0

But you can modify it:

[uu,ii] = unique(sort(H(:)));
d = diff([ii; numel(H)+1]);
result = uu(d == min(d)) % returns 0 and 4
result = 2×1
     0
     4

And you can use a similar approach to get multiple most-frequent elements in case there are more than one:

[uu,ii] = unique(sort(H(:)));
d = diff([ii; numel(H)+1]);
result = uu(d == max(d)) % returns 1 and 5
result = 2×1
     1
     5

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Answer 3

Star Strider 2023-9-27

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2026447-find-least-frequent-value-in-an-array#answer_1320177

编辑：Star Strider 2023-9-27

在 MATLAB Online 中打开

One approach —

H = [1 1 2 2 3 3 4 5 5 5];
[Hu,~,uidx] = unique(H, 'stable');
[Out,idx] = min(accumarray(uidx,(1:numel(uidx)).', [], @(x)numel(H(x))));
Result = Hu(idx)
Result = 4

EDIT — (27 Sep 2023 at 17:02)

Addressing Dyuman Joshi’s Comment, my code changes to —

H = [1 2 3 1 2 2 3 4 3 3 5 3 5 6];
[Hu,~,uidx] = unique(H, 'stable');
Out = accumarray(uidx,(1:numel(uidx)).', [], @(x)numel(H(x)));
Result = Hu(find(Out == min(Out)))
Result = 1×2
     4     6