Count instances of subarray inside array

2 Ansichten (letzte 30 Tage)
Mitsu
Mitsu am 4 Nov. 2021
Bearbeitet: Turlough Hughes am 7 Nov. 2021
For an array
A = [2,2,2,2];
I would like to count how many times [2,2] appears in A, using each value only in one possible pair.
For example,
numel(strfind(A,[2,2]))
returns 3.
But if each "2" is used only once, then we have only 2 instances, instead of 3.
A is not necessarily of all same elements or even-number size, it could be e.g. A = [1,2,3].
Is there a simple way to do that?
  2 Kommentare
KSSV
KSSV am 4 Nov. 2021
Try reshape into n*2 array and use ismember with rows option.
Mitsu
Mitsu am 4 Nov. 2021
I mean for this approach to work regardless of the size and elements, such as A = [1,2,3]. I edited the question to clarify.

Melden Sie sich an, um zu kommentieren.

Antworten (1)

Turlough Hughes
Turlough Hughes am 4 Nov. 2021
You could do the following:
b = A == 2; % binary indicating where A equal to 2
count = 0;
ii = 1;
while ii < numel(A)
if b(ii+1) && b(ii)
count = count + 1;
ii = ii + 1;
end
ii = ii + 1;
end
When there is a match, the loop variable, ii, increments an extra time so that each "2" is only used once.
  3 Kommentare
Mitsu
Mitsu am 4 Nov. 2021
Bearbeitet: Mitsu am 4 Nov. 2021
Thank you. There might be no other way around but to do it following the logic in your answer. For now I will wait to see if there are other answers with simpler, maybe one-line, approach.
Turlough Hughes
Turlough Hughes am 4 Nov. 2021
Bearbeitet: Turlough Hughes am 7 Nov. 2021
Another way would be to get the length of each group of twos (based on this answer by @Image Analyst) and then round down each group's length to the nearest whole number divisible by two:
measurements = regionprops(A == 2, 'Area');
allLengths = [measurements.Area];
count = sum(floor(allLengths/2))
(if you have the image processing toolbox)

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Matrices and Arrays finden Sie in Help Center und File Exchange

Tags

Produkte


Version

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by