How can i count how many time a number appears and for how long in a vector?

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salvatore vergine
salvatore vergine 2021 年 1 月 20 日
コメント済み: salvatore vergine 2021 年 1 月 21 日
I have a vector like this
v=[+1 +1 +1 0 0 0 +1 +1 -1 -1 +1 0 0 -1 -1 +1 +1]
I want to count how many times each number appears and for how long. i mean, for example let's consider +1. it appears indivigually only 1 time, then it appears in couple 2 times, then it appear in triplet only 1 time and so on. Could someone help me? thank you!

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Jan
Jan 2021 年 1 月 20 日
編集済み: Jan 2021 年 1 月 20 日
Start with a run-length-encoding, e.g. by FileExchange: RunLength . If you do not have a C-Compiler installed, use RunLength_M of this submission. Then:
v = [+1 +1 +1 0 0 0 +1 +1 -1 -1 +1 0 0 -1 -1 +1 +1];
[Num, Len] = RunLength_M(v)
Now you have the values and the lengths of the repetitions. Nur you can use histcounts to count, how often each sequence appears. If this is not time-ciritical, a simple loop does it also:
uniqNum = unique(Num);
for aNum = uniqNum(:).'
fprintf('Number: %d\n', aNum);
LenList = Len(Num == aNum);
uniqLenList = unique(LenList);
for aLen = uniqLenList(:)'
fprintf(' Block length: %d', aLen);
fprintf(' : %d times.\n', sum(LenList == aLen));
end
end
  5 件のコメント
Jan
Jan 2021 年 1 月 21 日
I do not reliably recognize the relation between the inputs and outputs.
salvatore vergine
salvatore vergine 2021 年 1 月 21 日
I try to explain it better. In each vector each column represents a pattern. For example, let s consider the number 1 and the input vector [0 1 1 2 1 1 1]. The result is the vector v1=[0 1 1 0 0 0 0] in which column 1 indicates how many times number 1 appears individually in the input vector (zero times in this case). Column 2 indicates how many times number 1 appears coupled in the input vector (only 1 as you can see). Column 3 indicates how many times number 1 appears as a triplet (only 1 too) and so on.
Another example with the input vector [0 1 1 0 1 1 0 1 2]. In this case the outpur vectors are:
v0=[3 0 0 0 0 0 0 0 0]
v1=[1 2 0 0 0 0 0 0 0]
v2=[1 0 0 0 0 0 0 0 0]
I hope I expressed myself better.

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