Re: AI-GEOSTATS: Estimation of the position accuracy of 2 set of points with different cardinalities
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.comwrote:
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
Re: AI-GEOSTATS: Estimation of the position accuracy of 2 set of points with different cardinalities
sorry I read defintively too quickly:
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]!
1/ my scoring method = Card(A inter B)/ Card(A Union B) = # corcordances /
(# omissions + # false positives+#concordances)
if A = B = score= Card(A) / (0 + Card(A) + 0) = 1.0 so it works actually.
2/ sorry but I maintain that in the case of B= A union Noisy points, the
fact you divise by Card(A) and not by Card(A Union B) is an issue .
In this context:
your score= #concordance / Card(a)= Card(a)/Card(a) = 1.0
my score = #concordance / (#omission+ #false positives + #concordances) =
Card(A)/ Card(False positive)+Card(A)) 1.0 , the right behavior.
Anyway again thanks for the exchange,
Best
Nicolas
On Sun, Mar 6, 2011 at 2:16 PM, Nicolas Maisonneuve n.maisonne...@gmail.com
wrote:
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.comwrote:
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.comwrote:
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)
