Find least frequent value in an array

6 次查看（过去 30 天）
Marcus Johnson
Marcus Johnson 2023-9-27
回答： Steven Lord 2023-9-27
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
  1 个评论
Dyuman Joshi
Dyuman Joshi 2023-9-27
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?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（3 个）

Steven Lord
Steven Lord 2023-9-27
Ran in:
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
Voss
Voss 2023-9-27
编辑：Voss 2023-9-27
Ran in:
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
  1 个评论
Voss
Voss 2023-9-27
Ran in:
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

请先登录，再进行评论。

Star Strider
Star Strider 2023-9-27
编辑：Star Strider 2023-9-27
Ran in:
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
.

类别

在 Help Center 和 File Exchange 中查找有关 Environment and Settings 的更多信息

标签

产品


版本

R2023a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by