yes I read too quickly  :)
anyway thanks for your help Younes

On Sun, Mar 6, 2011 at 12:51 PM, Younes Fadakar <yfa.st...@ymail.com> wrote:

> 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
>> > +
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