# Sum of highest length of consecutive lowest values from a array.

조회 수: 16(최근 30일)
편집: Roger Stafford 2018년 4월 11일
A = [5 1 1 1 6 1 1 1 1 1 1 1 7 1 1 1 7]
I would like to sum the highest consecutive numbers those are less than 2. In this case, there three 1 at the beginning and at the end but those will not be counted because there are 7 consecutive 1 at the middle. And I am looking for highest consecutive numbers those are less than 2 that's why the answer would be 1+1+1+1+1+1+1=7.

댓글을 달려면 로그인하십시오.

### 답변(3개)

Image Analyst 2018년 4월 9일
You could do this:
A = [5 1 1 1 6 1 1 1 1 1 1 1 7 1 1 1 7] % Assume A is all integers.
% Get unique integers in A
ua = unique(A)
for k = 1 : length(ua)
thisNumber = ua(k)
% Measure lengths of all regions comprised of this number.
props = regionprops(A==thisNumber, A, 'Area')
% Get maximum length for this number in A.
maxAreas(k) = max([props.Area])
end
maxAreas % Show in command window.
##### 댓글 수: 1표시숨기기 없음
Image Analyst 2018년 4월 11일
My code still works with your new, floating point numbers:
% A = [5 1 1 1 6 1 1 1 1 1 1 1 7 1 1 1 7] % Assume A is all integers.
A = [5 1 1 1 6 1 .5 .8 .55 .65 .95 1 7 1 1 1 7]
% Get unique numbers in A
ua = unique(A)
for k = 1 : length(ua)
thisNumber = ua(k);
% Measure lengths of all regions comprised of this number.
props = regionprops(A==thisNumber, A, 'Area');
% Get maximum length for this number in A.
maxAreas(k) = max([props.Area]);
fprintf('%.2f shows up %d times.\n', thisNumber, maxAreas(k));
end
I added an fprintf() to show you the results:
0.50 shows up 1 times.
0.55 shows up 1 times.
0.65 shows up 1 times.
0.80 shows up 1 times.
0.95 shows up 1 times.
1.00 shows up 3 times.
5.00 shows up 1 times.
6.00 shows up 1 times.
7.00 shows up 1 times.
Note the count (area) is only the length of the longest run, not a count of all the times the number appears. Thus 1 gives 3, not 8.

댓글을 달려면 로그인하십시오.

David Fletcher 2018년 4월 9일
편집: David Fletcher 2018년 4월 9일
A = [5 1 1 1 6 1 1 1 1 1 1 1 7 1 1 1 7]
B=int32(A==min(A));
[startInd lastInd]=regexp(char(B+48),'[1]+');
counts=lastInd-startInd+1;
sumContinuous=max(counts)
##### 댓글 수: 4표시숨기기 이전 댓글 수: 3
David Fletcher 2018년 4월 10일
I've just had another look at your 'revised' question, and I'm struggling to make sense of it. When I first saw it, I assumed that you now wanted the sum of all elements <=1 (rather than the largest contiguous length of the lowest value elements that you originally requested). However, I now see that in the list of elements you want summed together, there are only two values of 1 listed. Either this is a mistake, or there is some other weird exclusion factor that you are applying to all the other values of 1. So, in short, I haven't a clue what you now want.

댓글을 달려면 로그인하십시오.

Roger Stafford 2018년 4월 11일
편집: Roger Stafford 2018년 4월 11일
L = min(A);
f = find(diff([false,A==L,false])~=0);
h = L*max(f(2:2:end)-f(1:2:end-1));
h is the highest sum of consecutive lowest values.
[Note: You should be careful how your fractions are generated. You might have, say, two consecutive values which appear equal to the minimum value, .2, .2, but which are not exactly equal. These would not be detected as part of a consecutive sequence of least values.]

댓글을 달려면 로그인하십시오.

### 범주

Find more on Logical in Help Center and File Exchange

### Community Treasure Hunt

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

Start Hunting!

Translated by