How to select array elements which are within a range and get their indices?

24 Ansichten (letzte 30 Tage)
Hi, I would like to get the value and index of elements from a cell array. For instant, I wanted to get the values and indices of the elements which falls within the range 0.95< x <1.15 from the row 'data{1,1}(:,5)'. Then I need to calculate the average of these elements.
I have thousands of elements in a row and I need to repeat the process for hundreds times. How do I do it efficiently?
Thank you in adv.
  2 Kommentare
the cyclist
the cyclist am 28 Aug. 2017
Bearbeitet: the cyclist am 28 Aug. 2017
Note that your statement
0.95 < x < 0.15
would exclude all x. It looks like Image Analyst and I interpreted what you meant in different ways, so be careful about how you implement our code (if you do).
Jo 5
Jo 5 am 29 Aug. 2017
Sorry typo! Should be 0.95 < x < 1.15. Appreciate your help!

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Image Analyst
Image Analyst am 28 Aug. 2017
Try this:
% Get the whole row
row = data(...........whatever...
% Find values of row in the range "0.95< x <0.15"
% where x is either below 0.15 OR more than 0.95 but not in between.
logicalIndexes = row < 0.15 | row > 0.95;
% Get means where row is in range
meanValue = mean(row(logicalIndexes));
% You don't need the actual index numbers but if you want them, you can get them with find():
linearIndexes = find(logicalIndexes);

Weitere Antworten (1)

the cyclist
the cyclist am 28 Aug. 2017
Bearbeitet: the cyclist am 28 Aug. 2017
I am not certain I understood exactly what you wanted, but this solution is probably close enough that you can figure it out:
% Make up some pretend data
rng default
C{1} = rand(1,50);
C{2} = rand(1,100);
% For each element of the cell array, calculated the mean of the vector elements that lie strictly within the range (0.15,0.95)
meanWithinRange = cellfun(@(x)mean(x(x>0.15&x<0.95)),C)

Kategorien

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

Community Treasure Hunt

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

Start Hunting!

Translated by