Find a number and range of group of the same number

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Grega Mocnik
Grega Mocnik am 14 Nov. 2022
Kommentiert: Bruno Luong am 14 Nov. 2022
I have an array with numbers [1 1 1 1 1 1 1 2 2 2 2 3 3 3 3 1 1 2 2 1 1 3 3 3 3 3 1 1 1 2 2 2 ]
The array has the length: of 1x42502
I would like the number of how many repeats the same number, and the range of the specific repeat.
Output example:
number 1 has 4 repeats, range of the first repeat is: 1:6, range of the second repeat is: 15:16 , etc
number 2 has 3 repeats, range ...
number 3 has 2 repeats, range ...
How to do this?
Histcounts return sum of all repeat.

Antworten (2)

Jan
Jan am 14 Nov. 2022
Bearbeitet: Jan am 14 Nov. 2022
a = [1 1 1 1 1 1 1 2 2 2 2 3 3 3 3 1 1 2 2 1 1 3 3 3 3 3 1 1 1 2 2 2];
[b, n, idx] = RunLength(a);
idxR = [idx(:), idx(:) + n(:) - 1];
idxR = 9×2
1 7 8 11 12 15 16 17 18 19 20 21 22 26 27 29 30 32
ub = unique(b);
result = splitapply(@(c) {c}, idxR, b(:));
result = 3×1 cell array
{4×2 double} {3×2 double} {2×2 double}
result{ub(1)} % Ranges, where e.g. ub(1), which equals 1, occurs:
ans = 4×2
1 7 16 17 20 21 27 29
If you do not have a C-compiler, use RunLength_M from the same submission.

Bruno Luong
Bruno Luong am 14 Nov. 2022
Bearbeitet: Bruno Luong am 14 Nov. 2022
A= [1 1 1 1 1 1 1 2 2 2 2 3 3 3 3 1 1 2 2 1 1 3 3 3 3 3 1 1 1 2 2 2 ];
[u,~,G]=unique(A);
n = length(u);
for g=1:n
i = find(G==g);
j = find([true; diff(i)>1; true]);
start=i(j(1:end-1));
stop=i(j(2:end)-1);
m = length(start);
fprintf('value=%g, %d intervals\n', u(g), m);
fprintf('\t(%d:%d)\n', [start, stop]');
end
value=1, 4 intervals
(1:7) (16:17) (20:21) (27:29)
value=2, 3 intervals
(8:11) (18:19) (30:32)
value=3, 2 intervals
(12:15) (22:26)
  1 Kommentar
Bruno Luong
Bruno Luong am 14 Nov. 2022
I think my approach is not good if the number of groups n is large. In this case one should use Jan's runlength.

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