To Nicolas,
question: >> What happens if B = {A + noisy points} (false positive)? answer: You probably missed the second part of my previous email, where Card(B)>Card(A) with noise: I copied here, see: --------------------------------------------------------------------------- #-----the realistic implementation----- N = 100 # A.x = rand(N) #set A.x A.y = rand(N) #set A.y: coordinate pairs B.x = shake(A.x,10%) #slightly repositions points = noisy positions B.y = shake(A.y,10%) # randomly with 10% move B.x = B.x+rand(N/10) #adds extra 10% rand points = extra noisy points B.y = B.y+rand(N/10) #Card(B)=1.1*Card(A) M = PositionAccuracy(A,B) # Score = M/N*100 #my score=normalized based on N #N=Card(A) --------------------------------------------------------------------------- the computed score is: score = M(=#concordances)/N(=Card(A))*100 which seems to be right answer. Back to the first example, if A=B the score will be 100%.[correct] applying your scoring method if A=B then the score is smaller than 1. [incorrect]! Anyway, I'm happy you have found your satisfactory answer. To Duane: Thanks for your message. Do you have any information about existing statistically best random generator? I appreciate your replies. To All: Dear everybody, Is there any more robust/strong/reliable/high performance random generator satisfying statistically and being computing friendly? How can we evaluate the randomness of such generators then? To myself: Should double check the literature for concerns in randomness. Best Regards, . Younes yfa.st...@ymail.com http://alghalandis.com ________________________________ ________________________________ From: Nicolas Maisonneuve <n.maisonne...@gmail.com> To: Younes Fadakar <yfa.st...@ymail.com> Cc: Ask Geostatisticians <ai-geostats@jrc.it> Sent: Sun, 6 March, 2011 7:25:38 PM Subject: Re: AI-GEOSTATS: Estimation of the position accuracy of 2 set of points with different cardinalities In your example Card(A Union B) is always = Card(A) =N and that's an issue. What happens if B = {A + noisy points} (false positive)? According to your calcul the score will be 1.0... and that's not right. Actually I think the answer is actually trivial. (but I didn't think to formulate the problem in algebra terms) score = Card(A Intersection B)/Card(A Union B) score = # corcordances/ (#discordances+#concordances) score = # corcordances/ (# omissions (=Card(elements in A not included in B))+ # false positives(=Card(elements in B not included in A))+#concordances) Best, Nicolas On Sun, Mar 6, 2011 at 3:33 AM, Younes Fadakar <yfa.st...@ymail.com> wrote: Dear Nicolas, > >Hope this can help you. > >Let have a look at my implementation: > >#-----the simplest implementation----- >N = 100 #number of ref points=Crad(A) >A.x = rand(N) #set A.x >A.y = rand(N) #set A.y: coordinate pairs >B.X = A.x[:-10] #set B = sampling >B.Y = A.y[:-10] # has 10 points less than A > # Card(B)-Card(A)=-10 >M = PositionAccuracy(A,B) #as you defined=#concordances > >Score = M/N*100 #my score=normalized based on N > # N=Card(A) > >So the Score will be always in [0,1], here is 0.9 or 90.00%. > >and > >#-----the realistic implementation----- >N = 100 # >A.x = rand(N) #set A.x >A.y = rand(N) #set A.y: coordinate pairs >B.x = shake(A.x,10%) #slightly repositions points >B.y = shake(A.y,10%) # randomly with 10% move >B.x = B.x+rand(N/10) #adds extra 10% rand points >B.y = B.y+rand(N/10) #Card(B)=1.1*Card(A) > >M = PositionAccuracy(A,B) # > >Score = M/N*100 #my score=normalized based on N > #N=Card(A) > >Again the Score will be always in [0,1]. >This is what I used to generate the previously sent figures. > > >Best Regards, > >Younes >yfa.st...@ymail.com >http://alghalandis.com >________________________________ > > > > > > ________________________________ From: Nicolas Maisonneuve <n.maisonne...@gmail.com> >To: Younes Fadakar <yfa.st...@ymail.com> >Cc: Ask Geostatisticians <ai-geostats@jrc.it> >Sent: Wed, 2 March, 2011 6:27:48 PM >Subject: Re: AI-GEOSTATS: Estimation of the position accuracy of 2 set of >points >with different cardinalities > > >Thanks for your support Younges > >my idea was inspired and adapted from the Kendall correlation coefficient >(http://en.wikipedia.org/wiki/Kendall_tau_rank_correlation_coefficient >) but with the pb of cardinality. > >- number of concordances (accurate observations) >- number of discordances(omission + false positive) >and do a sum and then a normalisation to get something like 1.0 = max >corcordance max 0.0 = max discordance. >but I am not sure how to normalize: >- the range of concordance [0, Card(A)] is smaller than the >discordance [0, Card(A+B)] so anormalisation should be something like >(2Card(A)+Card(B)) but I am not sure about that , and I am not sure >the whole idea is right.. > >How did you normalize in your calcul? > > > > >On Wed, Mar 2, 2011 at 5:50 AM, Younes Fadakar <yfa.st...@ymail.com> wrote: >> Dear Nicolas, >> >> This is not the answer to your question but a try to implement your idea and >> to have an experience with it. >> Please see the attached, the output. >> It seems the total score provided by the method is very dependent to the >> 'r', the radius of search for neighbors around each ref point (A). >> However, being able to define the right 'r', the score seems a realistic >> measure of accuracy to me. >> Of course, this is just a practical understanding hoping the community could >> provide the statistical references. >> Anyway, I liked the idea. >> >> Best Regards, >> . >> Younes >> yfa.st...@ymail.com >> http://alghalandis.com >> ________________________________ >> >> >> ________________________________ >> From: Nicolas Maisonneuve <n.maisonne...@gmail.com> >> To: ai-geostats@jrc.it >> Sent: Mon, 28 February, 2011 6:21:49 PM >> Subject: AI-GEOSTATS: Estimation of the position accuracy of 2 set of points >> with different cardinalities >> >> Hi everyone, >> >> A simple question: >> I have 1 set of 2D location points A that I use as reference. >> I have another set of location points B generated by observations. >> >> Is there any standard method/measure to estimate a kind of position >> accuracy error knowing that >> - A and B dont have the same cardinality of elements e.g. B could have >> more points than A? >> - a point in A should be associated to only one point in B. >> >> For the moment I created my own error measure using 3 estimations. >> for a given accuracy rate (<20 meters) I compute: >> - O: number of omissions (when there is no observation in B closed >> enough of a point in A) , >> - FP: number of false positive (when a B point has been observed but >> not closed to a A point - or already taken from another >> observation) >> - M: number of matching (when a B point is closed enought of a A point) >> and then I aggregate the result = M- (O+FP) to get an indicator.. >> >> I am pretty sure there are other more traditional ways to do that. >> >> Thanks in advance >> -NM >> + >> + To post a message to the list, send it to ai-geost...@jrc.ec.europa.eu >> + To unsubscribe, send email to majordomo@ jrc.ec.europa.eu with no subject >> and "unsubscribe ai-geostats" in the message body. 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