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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>
>
