Find least frequent value in an array

Hello,

Let's say I have an array:

H = [1 1 2 2 3 3 4 5 5 5]

and if I use mode(H) I would get 5 since it's the most frequent, but can I do the opposite and find the least frequent? i.e. 4 in this case

[Dyuman Joshi]

What if there are two (or more) values which occur the least frequent?

For e.g.

H = [1 2 3 1 2 2 3 4 3 3 5 3 5 6]

Here, 4 and 6 occur only once. What should be the output in here and in such cases?

Answer 1

[Steven Lord]

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

Answer 2

[Voss]

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

[Voss]

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

Answer 3

[Star Strider]

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

.