Finding series of values within array

1 visualización (últimos 30 días)
Oscar
Oscar el 18 de Nov. de 2015
Editada: Stephen23 el 12 de En. de 2016
Hi, I am up for a problem which I cannot totally solve.
Imagine I have an array:
testtimes=[0.2 1.2 1.202 1.205 1.209 1.8 2.1 2.6 2.604 2.606 3.601 3.603 3.605];
Now I need to find all series of values where the difference between values is less than 0.01. So for now I did this:
% reference = 0.01
% isitesttimes=diff(testtimes); belowBIthreshold=(isitesttimes<reference);
% belowBIthreshold = [false, belowBIthreshold, false];
% edges = diff(belowBIthreshold);
% rising = find(edges==1); falling = find(edges==-1); spanWidth = falling - rising; wideEnough = spanWidth >= 3;
% startPos = rising(wideEnough);
% endPos = falling(wideEnough)-1;
% allInSpan = cell2mat(arrayfun(@(x,y) x:1:y, startPos, endPos, 'uni', false))
However, the answer now is
allInSpan = 2 3 4
Which means that on position 2,3,4 in testtimes the values are postioned which have a difference less than 0.01. However, the correct answer in this example would be 2,3,4,5 (since also position 5 has a difference of less than 0.01 compared to the one on position 4) (apart from that 8,9,10,11,12,13 should also be in the answer but that has to do (I guess) with the cut-off/Span-Width of 3).
Can somebody help me out?
Regards

Respuestas (1)

Stephen23
Stephen23 el 12 de En. de 2016
Editada: Stephen23 el 12 de En. de 2016
There is no need to use find when logical indexing is much faster. The following code extracts all values from X where the difference with an adjacent element is less than R:
>> V = [0.2,1.2,1.202,1.205,1.209,1.8,2.1,2.6,2.604,2.606,3.601,3.603,3.605];
>> R = 0.01;
>> D = diff(V)<R;
>> X = [false,D] | [D,false];
>> V(X)
ans =
1.2000 1.2020 1.2050 1.2090 2.6000 2.6040 2.6060 3.6010 3.6030 3.6050
But of course you can still use find if you need to:
>> find(X)
ans =
2 3 4 5 8 9 10 11 12 13

Categorías

Más información sobre Mathematics en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by