How do i count same consecutive occurrences
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Consider I have an array of occurrences
A=[1,1,1,1,1,2,2,2,2,3,3,3,3,2,2,2,1,1,1,1]
I want to find out how many 2 are on each occurrence. The answer should be 4 starting 6th position and 3 starting 14 positions.
Is it possible to do it in a wise Matlab way without many loop complications?
Réponses (3)
madhan ravi
le 23 Oct 2018
Modifié(e) : madhan ravi
le 23 Oct 2018
0 votes
Daniel Pare
le 5 Nov 2019
I wanted to count how many time in a row Head or Tail will occure from a random draw and I came up with this.
Let (x) be your vector of observation like x = [1 1 0 1 0 0 0 1]
The result will be: s = [ 2 1 1 3 1]
s=0 ;
j=2;
i=1;
while i < nb_it+1
c=1;
if i == nb_it
s = [ s c];
break
else
while x(i) == x(i+1)
c = c+1;
i = i+1;
if i == nb_it
break
end
end
end
s = [ s c];
i = i+1;
end
s = s(2:end); % to remove the first zero
sum(s) % the sum should be equal to the number of element in (x)
max(s) % This is the maximum of consecutive ocurence from the draw
5 commentaires
Guillaume
le 5 Nov 2019
Or you could do:
s = diff([0, find(diff(x)), numel(x)]);
and be done...
Or similarly:
>> s = diff(find([1,diff(x),1]))
s =
2 1 1 3 1
Daniel Pare
le 5 Nov 2019
Awesome, and this way takes way less time than doing many loop.
Thanks
giannit
le 23 Avr 2020
That one line command is amazing, many thanks!
gummiyummi
le 3 Août 2020
I get an error: Error using horzcat. Dimensions of arrays being concatenated are not consistent.
Can anybody help resolve this?
Bruno Luong
le 3 Août 2020
Modifié(e) : Bruno Luong
le 16 Déc 2021
Example:
A=[1,1,1,1,1,2,2,2,2,3,3,3,3,2,2,2,1,1,1,1]
Code
d = diff([0, A==2, 0]);
startidx = find(d==1)
lgt = find(d==-1)-startidx % EDIT error
Result
startidx =
6 14
lgt =
4 3
4 commentaires
gummiyummi
le 3 Août 2020
I still get the same error...
Bruno Luong
le 3 Août 2020
Modifié(e) : Bruno Luong
le 3 Août 2020
@As Sc: Can you tell if your A is a row vector? column vector? Something else?
Please type
class(A)
size(A)
then report the result. A is your array.
Ayush Meena
le 15 Déc 2021
@Bruno can you please tell me what is idx here?
Bruno Luong
le 16 Déc 2021
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