What is the best strategy for determining a GPS tracked user's location, once they have stopped moving for a reasonable period?
What is the best strategy for determining a user's location from a series of GPS fixes, once they are considered not to be moving?
When tracking a user, if they should stop moving there will subsequently be a sequence of fixes roughly in the same location.
If possible I would like to not just use the last fix, but also take into consideration previous fixes so as to calculate a more accurate position for them.
Factors that I would have thought need to be considered:
- The best way to determine a user is stationary (from experience speed from the GPS fixes is not sufficiently reliable)
- Each fix has an accuracy, how can this be factored in?
- Are there well established algorithms/libraries that could be used?
Any suggestions greatly appreciated
geolocation statistics gps geospatial geostatistics
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
add a comment |
What is the best strategy for determining a user's location from a series of GPS fixes, once they are considered not to be moving?
When tracking a user, if they should stop moving there will subsequently be a sequence of fixes roughly in the same location.
If possible I would like to not just use the last fix, but also take into consideration previous fixes so as to calculate a more accurate position for them.
Factors that I would have thought need to be considered:
- The best way to determine a user is stationary (from experience speed from the GPS fixes is not sufficiently reliable)
- Each fix has an accuracy, how can this be factored in?
- Are there well established algorithms/libraries that could be used?
Any suggestions greatly appreciated
geolocation statistics gps geospatial geostatistics
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
add a comment |
What is the best strategy for determining a user's location from a series of GPS fixes, once they are considered not to be moving?
When tracking a user, if they should stop moving there will subsequently be a sequence of fixes roughly in the same location.
If possible I would like to not just use the last fix, but also take into consideration previous fixes so as to calculate a more accurate position for them.
Factors that I would have thought need to be considered:
- The best way to determine a user is stationary (from experience speed from the GPS fixes is not sufficiently reliable)
- Each fix has an accuracy, how can this be factored in?
- Are there well established algorithms/libraries that could be used?
Any suggestions greatly appreciated
geolocation statistics gps geospatial geostatistics
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
What is the best strategy for determining a user's location from a series of GPS fixes, once they are considered not to be moving?
When tracking a user, if they should stop moving there will subsequently be a sequence of fixes roughly in the same location.
If possible I would like to not just use the last fix, but also take into consideration previous fixes so as to calculate a more accurate position for them.
Factors that I would have thought need to be considered:
- The best way to determine a user is stationary (from experience speed from the GPS fixes is not sufficiently reliable)
- Each fix has an accuracy, how can this be factored in?
- Are there well established algorithms/libraries that could be used?
Any suggestions greatly appreciated
geolocation statistics gps geospatial geostatistics
geolocation statistics gps geospatial geostatistics
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
edited Nov 13 '18 at 8:39
Anony-Mousse
58.6k797162
58.6k797162
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
asked Nov 12 '18 at 21:19
Mark MellingMark Melling
1,1571013
1,1571013
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
What's wrong with taking the average?
If you want to take uncertainty into account, use a weighted average. Or a trimmed one, discarding those measurements that deviate most.
But it is a known fact that other factors such as reflections on buildings, can have a much larger effect on the accuracy.
Nevertheless, this is not so much a programming question, but one that needs GPS expertise. You are better off reading expert literature than asking random internet users for their opinion.
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
add a comment |
Assuming you only have access to the position and accuracy (and in this case we can assume the chipset is already giving you the best fix available, and factoring in the previous positions, normally through a Kalman filter) , and no access to the raw measurements given by the GPS receiver, you can try the following experience:
Calculate the weighted mean, where the weight is given by the inverse of the variance (let's assume your accuracy value equals the standard deviation);
Calculate the variance of the weighted mean through Error Propagation (Wikipedia will help you);
Evaluate the behaviour of the variance. If you're getting a variance with a lower value than the square of any of the original accuracies, then you're onto something. If no, stick with the position fix with the highest accuracy (lowest standard deviation).
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
What's wrong with taking the average?
If you want to take uncertainty into account, use a weighted average. Or a trimmed one, discarding those measurements that deviate most.
But it is a known fact that other factors such as reflections on buildings, can have a much larger effect on the accuracy.
Nevertheless, this is not so much a programming question, but one that needs GPS expertise. You are better off reading expert literature than asking random internet users for their opinion.
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
add a comment |
What's wrong with taking the average?
If you want to take uncertainty into account, use a weighted average. Or a trimmed one, discarding those measurements that deviate most.
But it is a known fact that other factors such as reflections on buildings, can have a much larger effect on the accuracy.
Nevertheless, this is not so much a programming question, but one that needs GPS expertise. You are better off reading expert literature than asking random internet users for their opinion.
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
add a comment |
What's wrong with taking the average?
If you want to take uncertainty into account, use a weighted average. Or a trimmed one, discarding those measurements that deviate most.
But it is a known fact that other factors such as reflections on buildings, can have a much larger effect on the accuracy.
Nevertheless, this is not so much a programming question, but one that needs GPS expertise. You are better off reading expert literature than asking random internet users for their opinion.
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
What's wrong with taking the average?
If you want to take uncertainty into account, use a weighted average. Or a trimmed one, discarding those measurements that deviate most.
But it is a known fact that other factors such as reflections on buildings, can have a much larger effect on the accuracy.
Nevertheless, this is not so much a programming question, but one that needs GPS expertise. You are better off reading expert literature than asking random internet users for their opinion.
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
answered Nov 13 '18 at 8:43
Anony-MousseAnony-Mousse
58.6k797162
58.6k797162
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
add a comment |
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
There may be nothing wrong, with taking an average or weighted average, although I would say you still need a way to determine which fixes to include (they were moving prior to stopping in one location), some clustering algorithm perhaps? I'm not convinced that GPS expertise really helps in this case, I would say that this 'expertise' is encapsulated in the accuracy figure provided in a fix. I wasn't sure if this was the right site to post this to, possibly Data Science might be more appropriate?
– Mark Melling
Nov 13 '18 at 11:28
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
No, migrating the question to GIS is more reasonable. Clustering is not reliable enough. K-means will split a very obvious uniform distribution right in the middle. Don't assume clustering can magically solve whatever you toss at it - it won't. Instead. It is fragile and needs to be carefully guided. So you may as well just do the median latitude and longitude, remove the farthest 10%, and do the resulting trimmed mean. It's just that all the GPS experts will tell you that your accuracy probably does not improve by using more fixes. Because that reduces only variance, not bias.
– Anony-Mousse
Nov 13 '18 at 22:08
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
Thanks, I'll do some experimenting.
– Mark Melling
Nov 14 '18 at 16:16
add a comment |
Assuming you only have access to the position and accuracy (and in this case we can assume the chipset is already giving you the best fix available, and factoring in the previous positions, normally through a Kalman filter) , and no access to the raw measurements given by the GPS receiver, you can try the following experience:
Calculate the weighted mean, where the weight is given by the inverse of the variance (let's assume your accuracy value equals the standard deviation);
Calculate the variance of the weighted mean through Error Propagation (Wikipedia will help you);
Evaluate the behaviour of the variance. If you're getting a variance with a lower value than the square of any of the original accuracies, then you're onto something. If no, stick with the position fix with the highest accuracy (lowest standard deviation).
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
add a comment |
Assuming you only have access to the position and accuracy (and in this case we can assume the chipset is already giving you the best fix available, and factoring in the previous positions, normally through a Kalman filter) , and no access to the raw measurements given by the GPS receiver, you can try the following experience:
Calculate the weighted mean, where the weight is given by the inverse of the variance (let's assume your accuracy value equals the standard deviation);
Calculate the variance of the weighted mean through Error Propagation (Wikipedia will help you);
Evaluate the behaviour of the variance. If you're getting a variance with a lower value than the square of any of the original accuracies, then you're onto something. If no, stick with the position fix with the highest accuracy (lowest standard deviation).
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
add a comment |
Assuming you only have access to the position and accuracy (and in this case we can assume the chipset is already giving you the best fix available, and factoring in the previous positions, normally through a Kalman filter) , and no access to the raw measurements given by the GPS receiver, you can try the following experience:
Calculate the weighted mean, where the weight is given by the inverse of the variance (let's assume your accuracy value equals the standard deviation);
Calculate the variance of the weighted mean through Error Propagation (Wikipedia will help you);
Evaluate the behaviour of the variance. If you're getting a variance with a lower value than the square of any of the original accuracies, then you're onto something. If no, stick with the position fix with the highest accuracy (lowest standard deviation).
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
Assuming you only have access to the position and accuracy (and in this case we can assume the chipset is already giving you the best fix available, and factoring in the previous positions, normally through a Kalman filter) , and no access to the raw measurements given by the GPS receiver, you can try the following experience:
Calculate the weighted mean, where the weight is given by the inverse of the variance (let's assume your accuracy value equals the standard deviation);
Calculate the variance of the weighted mean through Error Propagation (Wikipedia will help you);
Evaluate the behaviour of the variance. If you're getting a variance with a lower value than the square of any of the original accuracies, then you're onto something. If no, stick with the position fix with the highest accuracy (lowest standard deviation).
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
answered Nov 14 '18 at 1:57
lusitanicalusitanica
523211
523211
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
add a comment |
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
I'll do some experiment using your suggestions.
– Mark Melling
Nov 14 '18 at 16:17
add a comment |
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