| Re: A better way to find a dip (local minimum with certain conditions) [message #69037] |
Wed, 16 December 2009 15:58  |
Giorgio
Messages: 31
Registered: March 2008
|
Member |
|
|
On Dec 16, 1:39 pm, chris <rog...@googlemail.com> wrote:
> Hi David,
> as i understood the following might work (untested):
>
> FUNCTION finddip,array,minim,range=range
> range=keyword_set(range)? 0>range<n_elements(array) : 4
> minimum = (min(array-minim,wheremin,/nan))[0] ; [0]-> to get only
> the first match
> if wheremin ne -1 then begin
> ar = array[-1>(wheremin-range):(wheremin+range)<(n_elements
> (array)-1)]
> if max(((arr=ar[sort(ar)]-ar))) eq min(arr) then return,wheremin
> endif else return,-1
> end
>
> Maybe it works and maybe that the principle behind gives you a hint
> for solving the problem
>
> Regards
>
> CR
Hi,
Following CR, this is a small program to find all the local minima.
I did not test it yet.
function findip, array,minim
range = 4
nel = n_elements(array)
; look for candidates
candidates = where(array GT minim)
mins = candidates
; look which of them is a local minima
FOR i =0, n_elements(candidates)-1 do begin
localmin = where(array[((candidates[i]-range)>0):((candidates[i]
+range)<(nel-1) GT array[candidates[i]])
IF localmin GE 0 THEN mins[i] = 0
endfor
return, where(mins NE 0)
end
|
|
|
|
|
|
| Re: A better way to find a dip (local minimum with certain conditions) [message #69104 is a reply to message #69040] |
Mon, 21 December 2009 10:44  |
DavidPS
Messages: 10
Registered: December 2009
|
Junior Member |
|
|
Thanks to all!! I've seen all your suggestions and you have me
pointed me through the way I wanted.
Finally, what I've got from Jim P and James version and adding stuff
here and there is:
;===================== CODE=========================
FUNCTION finddip5,array,minim=minim,range=range
range=KEYWORD_SET(range) ? 1>range<(n_elements(array)/2-1) : 4
; Find the negative gradients. We're only interested in the sign of
the gradient, not the magnitude.
c1 = fix(convol(array, [-1, 1]) GT 0)
; Find the positive gradients, and invert the sign.
c2 = -fix(convol(array, [1, -1]) GT 0)
; Combine the two Filter for four negative gradients, a zero, and
four positive gradients.
minima = where(fix(convol(c1+c2, [replicate(-1,range+1),replicate
(1,range)]) eq 2*range+1))
;Look for those greater than minim if set.
IF (minima[0] NE -1) AND (KEYWORD_SET(minim)) THEN BEGIN
lab = where(array[minima] ge minim,nlab)
minima=(nlab GT 0) ? minima[lab] : -1
ENDIF
RETURN,minima
END
;=================== END CODE=======================
Which is very clever, scalable and a lot faster than what I had.
However... I still have a small problem, which I've been trying to
solve using the same method... but I have not found the way to do it.
In my first example I was interested also on those dips where on the
decreasing or increasing part of it there's repeated values. For
example consider these arrays:
[-5,-4,18,12,12,3,1,-0.5,7,11,13,30,29,5] or
[-5,-4,18,16,12,3,1,-0.5,7,11,11,30,29,5]
or a mix of both:
[-5,-4,18,12,12,3,1,-0.5,7,11,11,30,29,5]
I've managed to get when there's on one side or another of the dip,
but not in both... making ones in C1 and C1 everything bigger than
-0.0001 and then making 1 the values gt 2 on c1+c2.
does someone have any clue?
Thank you to everybody again!!
David
|
|
|
|
comp.lang.idl-pvwave archive
Members
Search
Help
Login
Home
