Getting Lat & Lang within 1000Ft radius of a central Lat Lang
I am working with a pandas data frame where I have 4 columns of lats and langs. You can call these as pickup and drop off points. I have identify lat and lang of a central location and figured the radius (1000FT)
of the circle. I want to select all the pickup and drop off points within this circle. Potentially create a column in the data frame with a flag.
(Pdb) x.dtypes
Pickup_longitude float64
Pickup_latitude float64
Dropoff_longitude float64
Dropoff_latitude float64
dtype: object
Pickup_longitude,Pickup_latitude,Dropoff_longitude,Dropoff_latitude
-73.97948455810547,40.68495559692383,-73.97943115234375,40.685020446777344
-74.01079559326173,40.91221618652344,-74.01078033447266,40.91221237182617
-73.92140960693358,40.76670837402344,-73.91441345214845,40.764686584472656
-73.92138671875,40.76667785644531,-73.93142700195312,40.771583557128906
-73.95548248291014,40.71404647827149,-73.94441223144531,40.71472930908203
-73.94529724121094,40.80818557739258,-73.93766784667969,40.821197509765625
-73.89087677001955,40.7464256286621,-73.87692260742188,40.75630569458008
-73.94670104980467,40.79732131958008,-73.9376449584961,40.80451583862305
-73.96315002441406,40.69382858276367,-73.956787109375,40.680530548095696
-73.89682006835938,40.74612808227539,-73.88862609863281,40.752723693847656
-73.82991790771484,40.713768005371094,-73.83494567871094,40.70729064941406
-73.9055404663086,40.77252578735352,-73.8953628540039,40.76896286010742
-73.941650390625,40.81829452514648,-73.950927734375,40.82603073120117
-73.93252563476562,40.85680389404297,-73.93136596679686,40.856117248535156
-73.95282745361328,40.808353424072266,-73.94914245605469,40.80290985107422
-73.904052734375,40.87870788574219,-73.89696502685547,40.90113067626953
-74.02134704589844,40.647010803222656,-74.00405883789062,40.65459442138672
-73.95098876953125,40.68049621582031,-73.9539566040039,40.69042205810547
-73.84967041015625,40.72400665283203,-73.83512878417969,40.70719909667969
-73.9466781616211,40.80628967285156,-73.93624877929686,40.79925155639648
-73.98682403564453,40.70245742797852,-73.97278594970702,40.69290161132813
-73.91205596923827,40.7754020690918,-73.89364624023438,40.76850509643555
-73.9617462158203,40.71377944946289,-73.98593139648438,40.71804809570313
-73.94091033935547,40.69868087768555,-73.94107818603516,40.6828498840332
-73.84423828125,40.72149658203125,-73.8086929321289,40.73440170288085
-73.89108276367188,40.74691009521485,-73.87848663330078,40.7494010925293
-73.84487915039062,40.75502395629883,-73.87222290039062,40.77399063110352
-73.87079620361328,40.73346328735352,-73.87014770507811,40.73357391357422
-73.94847869873048,40.81386184692383,-73.94107055664062,40.81884002685547
** subset of the data
I found following question in stackoverflow but not really sure if thats the correct approach.
Haversine Formula in Python (Bearing and Distance between two GPS points)
Update: So far solution
import math
def getPoints(lon, lat):
radius = 1800 # in feet (not sure if this is the right approach)
N = 360
# generate points
circlePoints =
for k in range(N):
angle = math.pi*2*k/N
python pandas latitude-longitude
add a comment |
I am working with a pandas data frame where I have 4 columns of lats and langs. You can call these as pickup and drop off points. I have identify lat and lang of a central location and figured the radius (1000FT)
of the circle. I want to select all the pickup and drop off points within this circle. Potentially create a column in the data frame with a flag.
(Pdb) x.dtypes
Pickup_longitude float64
Pickup_latitude float64
Dropoff_longitude float64
Dropoff_latitude float64
dtype: object
Pickup_longitude,Pickup_latitude,Dropoff_longitude,Dropoff_latitude
-73.97948455810547,40.68495559692383,-73.97943115234375,40.685020446777344
-74.01079559326173,40.91221618652344,-74.01078033447266,40.91221237182617
-73.92140960693358,40.76670837402344,-73.91441345214845,40.764686584472656
-73.92138671875,40.76667785644531,-73.93142700195312,40.771583557128906
-73.95548248291014,40.71404647827149,-73.94441223144531,40.71472930908203
-73.94529724121094,40.80818557739258,-73.93766784667969,40.821197509765625
-73.89087677001955,40.7464256286621,-73.87692260742188,40.75630569458008
-73.94670104980467,40.79732131958008,-73.9376449584961,40.80451583862305
-73.96315002441406,40.69382858276367,-73.956787109375,40.680530548095696
-73.89682006835938,40.74612808227539,-73.88862609863281,40.752723693847656
-73.82991790771484,40.713768005371094,-73.83494567871094,40.70729064941406
-73.9055404663086,40.77252578735352,-73.8953628540039,40.76896286010742
-73.941650390625,40.81829452514648,-73.950927734375,40.82603073120117
-73.93252563476562,40.85680389404297,-73.93136596679686,40.856117248535156
-73.95282745361328,40.808353424072266,-73.94914245605469,40.80290985107422
-73.904052734375,40.87870788574219,-73.89696502685547,40.90113067626953
-74.02134704589844,40.647010803222656,-74.00405883789062,40.65459442138672
-73.95098876953125,40.68049621582031,-73.9539566040039,40.69042205810547
-73.84967041015625,40.72400665283203,-73.83512878417969,40.70719909667969
-73.9466781616211,40.80628967285156,-73.93624877929686,40.79925155639648
-73.98682403564453,40.70245742797852,-73.97278594970702,40.69290161132813
-73.91205596923827,40.7754020690918,-73.89364624023438,40.76850509643555
-73.9617462158203,40.71377944946289,-73.98593139648438,40.71804809570313
-73.94091033935547,40.69868087768555,-73.94107818603516,40.6828498840332
-73.84423828125,40.72149658203125,-73.8086929321289,40.73440170288085
-73.89108276367188,40.74691009521485,-73.87848663330078,40.7494010925293
-73.84487915039062,40.75502395629883,-73.87222290039062,40.77399063110352
-73.87079620361328,40.73346328735352,-73.87014770507811,40.73357391357422
-73.94847869873048,40.81386184692383,-73.94107055664062,40.81884002685547
** subset of the data
I found following question in stackoverflow but not really sure if thats the correct approach.
Haversine Formula in Python (Bearing and Distance between two GPS points)
Update: So far solution
import math
def getPoints(lon, lat):
radius = 1800 # in feet (not sure if this is the right approach)
N = 360
# generate points
circlePoints =
for k in range(N):
angle = math.pi*2*k/N
python pandas latitude-longitude
Checked this?
– Cheche
Nov 20 '18 at 22:07
Totally dependent on your problem. You could have an infinite number of coords in range; do you have any way of pruning down the options before you run the test?
– roganjosh
Nov 20 '18 at 22:08
@roganjosh May be I am missing something How can there be infinite. We use the Central point and from that point we define radius and then select all the points that are in above columns that are within the distance.
– Null-Hypothesis
Nov 20 '18 at 22:53
1
I think OP wants to select, form a set of already determined lat,long pairs, those that are within a certain distance of another lat,long point. It boils down to a bunch of pairwise distance queries, if you don't use a fancier GIS index data structure.
– interfect
Nov 20 '18 at 22:59
My point was that lat and long are continuous (you may limit to a certain number of decimal places) but without some kind of geofencing you can quickly become swamped by the number of pairs to test.
– roganjosh
Nov 21 '18 at 6:36
add a comment |
I am working with a pandas data frame where I have 4 columns of lats and langs. You can call these as pickup and drop off points. I have identify lat and lang of a central location and figured the radius (1000FT)
of the circle. I want to select all the pickup and drop off points within this circle. Potentially create a column in the data frame with a flag.
(Pdb) x.dtypes
Pickup_longitude float64
Pickup_latitude float64
Dropoff_longitude float64
Dropoff_latitude float64
dtype: object
Pickup_longitude,Pickup_latitude,Dropoff_longitude,Dropoff_latitude
-73.97948455810547,40.68495559692383,-73.97943115234375,40.685020446777344
-74.01079559326173,40.91221618652344,-74.01078033447266,40.91221237182617
-73.92140960693358,40.76670837402344,-73.91441345214845,40.764686584472656
-73.92138671875,40.76667785644531,-73.93142700195312,40.771583557128906
-73.95548248291014,40.71404647827149,-73.94441223144531,40.71472930908203
-73.94529724121094,40.80818557739258,-73.93766784667969,40.821197509765625
-73.89087677001955,40.7464256286621,-73.87692260742188,40.75630569458008
-73.94670104980467,40.79732131958008,-73.9376449584961,40.80451583862305
-73.96315002441406,40.69382858276367,-73.956787109375,40.680530548095696
-73.89682006835938,40.74612808227539,-73.88862609863281,40.752723693847656
-73.82991790771484,40.713768005371094,-73.83494567871094,40.70729064941406
-73.9055404663086,40.77252578735352,-73.8953628540039,40.76896286010742
-73.941650390625,40.81829452514648,-73.950927734375,40.82603073120117
-73.93252563476562,40.85680389404297,-73.93136596679686,40.856117248535156
-73.95282745361328,40.808353424072266,-73.94914245605469,40.80290985107422
-73.904052734375,40.87870788574219,-73.89696502685547,40.90113067626953
-74.02134704589844,40.647010803222656,-74.00405883789062,40.65459442138672
-73.95098876953125,40.68049621582031,-73.9539566040039,40.69042205810547
-73.84967041015625,40.72400665283203,-73.83512878417969,40.70719909667969
-73.9466781616211,40.80628967285156,-73.93624877929686,40.79925155639648
-73.98682403564453,40.70245742797852,-73.97278594970702,40.69290161132813
-73.91205596923827,40.7754020690918,-73.89364624023438,40.76850509643555
-73.9617462158203,40.71377944946289,-73.98593139648438,40.71804809570313
-73.94091033935547,40.69868087768555,-73.94107818603516,40.6828498840332
-73.84423828125,40.72149658203125,-73.8086929321289,40.73440170288085
-73.89108276367188,40.74691009521485,-73.87848663330078,40.7494010925293
-73.84487915039062,40.75502395629883,-73.87222290039062,40.77399063110352
-73.87079620361328,40.73346328735352,-73.87014770507811,40.73357391357422
-73.94847869873048,40.81386184692383,-73.94107055664062,40.81884002685547
** subset of the data
I found following question in stackoverflow but not really sure if thats the correct approach.
Haversine Formula in Python (Bearing and Distance between two GPS points)
Update: So far solution
import math
def getPoints(lon, lat):
radius = 1800 # in feet (not sure if this is the right approach)
N = 360
# generate points
circlePoints =
for k in range(N):
angle = math.pi*2*k/N
python pandas latitude-longitude
I am working with a pandas data frame where I have 4 columns of lats and langs. You can call these as pickup and drop off points. I have identify lat and lang of a central location and figured the radius (1000FT)
of the circle. I want to select all the pickup and drop off points within this circle. Potentially create a column in the data frame with a flag.
(Pdb) x.dtypes
Pickup_longitude float64
Pickup_latitude float64
Dropoff_longitude float64
Dropoff_latitude float64
dtype: object
Pickup_longitude,Pickup_latitude,Dropoff_longitude,Dropoff_latitude
-73.97948455810547,40.68495559692383,-73.97943115234375,40.685020446777344
-74.01079559326173,40.91221618652344,-74.01078033447266,40.91221237182617
-73.92140960693358,40.76670837402344,-73.91441345214845,40.764686584472656
-73.92138671875,40.76667785644531,-73.93142700195312,40.771583557128906
-73.95548248291014,40.71404647827149,-73.94441223144531,40.71472930908203
-73.94529724121094,40.80818557739258,-73.93766784667969,40.821197509765625
-73.89087677001955,40.7464256286621,-73.87692260742188,40.75630569458008
-73.94670104980467,40.79732131958008,-73.9376449584961,40.80451583862305
-73.96315002441406,40.69382858276367,-73.956787109375,40.680530548095696
-73.89682006835938,40.74612808227539,-73.88862609863281,40.752723693847656
-73.82991790771484,40.713768005371094,-73.83494567871094,40.70729064941406
-73.9055404663086,40.77252578735352,-73.8953628540039,40.76896286010742
-73.941650390625,40.81829452514648,-73.950927734375,40.82603073120117
-73.93252563476562,40.85680389404297,-73.93136596679686,40.856117248535156
-73.95282745361328,40.808353424072266,-73.94914245605469,40.80290985107422
-73.904052734375,40.87870788574219,-73.89696502685547,40.90113067626953
-74.02134704589844,40.647010803222656,-74.00405883789062,40.65459442138672
-73.95098876953125,40.68049621582031,-73.9539566040039,40.69042205810547
-73.84967041015625,40.72400665283203,-73.83512878417969,40.70719909667969
-73.9466781616211,40.80628967285156,-73.93624877929686,40.79925155639648
-73.98682403564453,40.70245742797852,-73.97278594970702,40.69290161132813
-73.91205596923827,40.7754020690918,-73.89364624023438,40.76850509643555
-73.9617462158203,40.71377944946289,-73.98593139648438,40.71804809570313
-73.94091033935547,40.69868087768555,-73.94107818603516,40.6828498840332
-73.84423828125,40.72149658203125,-73.8086929321289,40.73440170288085
-73.89108276367188,40.74691009521485,-73.87848663330078,40.7494010925293
-73.84487915039062,40.75502395629883,-73.87222290039062,40.77399063110352
-73.87079620361328,40.73346328735352,-73.87014770507811,40.73357391357422
-73.94847869873048,40.81386184692383,-73.94107055664062,40.81884002685547
** subset of the data
I found following question in stackoverflow but not really sure if thats the correct approach.
Haversine Formula in Python (Bearing and Distance between two GPS points)
Update: So far solution
import math
def getPoints(lon, lat):
radius = 1800 # in feet (not sure if this is the right approach)
N = 360
# generate points
circlePoints =
for k in range(N):
angle = math.pi*2*k/N
python pandas latitude-longitude
python pandas latitude-longitude
edited Nov 21 '18 at 1:47
Null-Hypothesis
asked Nov 20 '18 at 22:04
Null-HypothesisNull-Hypothesis
5,76132103174
5,76132103174
Checked this?
– Cheche
Nov 20 '18 at 22:07
Totally dependent on your problem. You could have an infinite number of coords in range; do you have any way of pruning down the options before you run the test?
– roganjosh
Nov 20 '18 at 22:08
@roganjosh May be I am missing something How can there be infinite. We use the Central point and from that point we define radius and then select all the points that are in above columns that are within the distance.
– Null-Hypothesis
Nov 20 '18 at 22:53
1
I think OP wants to select, form a set of already determined lat,long pairs, those that are within a certain distance of another lat,long point. It boils down to a bunch of pairwise distance queries, if you don't use a fancier GIS index data structure.
– interfect
Nov 20 '18 at 22:59
My point was that lat and long are continuous (you may limit to a certain number of decimal places) but without some kind of geofencing you can quickly become swamped by the number of pairs to test.
– roganjosh
Nov 21 '18 at 6:36
add a comment |
Checked this?
– Cheche
Nov 20 '18 at 22:07
Totally dependent on your problem. You could have an infinite number of coords in range; do you have any way of pruning down the options before you run the test?
– roganjosh
Nov 20 '18 at 22:08
@roganjosh May be I am missing something How can there be infinite. We use the Central point and from that point we define radius and then select all the points that are in above columns that are within the distance.
– Null-Hypothesis
Nov 20 '18 at 22:53
1
I think OP wants to select, form a set of already determined lat,long pairs, those that are within a certain distance of another lat,long point. It boils down to a bunch of pairwise distance queries, if you don't use a fancier GIS index data structure.
– interfect
Nov 20 '18 at 22:59
My point was that lat and long are continuous (you may limit to a certain number of decimal places) but without some kind of geofencing you can quickly become swamped by the number of pairs to test.
– roganjosh
Nov 21 '18 at 6:36
Checked this?
– Cheche
Nov 20 '18 at 22:07
Checked this?
– Cheche
Nov 20 '18 at 22:07
Totally dependent on your problem. You could have an infinite number of coords in range; do you have any way of pruning down the options before you run the test?
– roganjosh
Nov 20 '18 at 22:08
Totally dependent on your problem. You could have an infinite number of coords in range; do you have any way of pruning down the options before you run the test?
– roganjosh
Nov 20 '18 at 22:08
@roganjosh May be I am missing something How can there be infinite. We use the Central point and from that point we define radius and then select all the points that are in above columns that are within the distance.
– Null-Hypothesis
Nov 20 '18 at 22:53
@roganjosh May be I am missing something How can there be infinite. We use the Central point and from that point we define radius and then select all the points that are in above columns that are within the distance.
– Null-Hypothesis
Nov 20 '18 at 22:53
1
1
I think OP wants to select, form a set of already determined lat,long pairs, those that are within a certain distance of another lat,long point. It boils down to a bunch of pairwise distance queries, if you don't use a fancier GIS index data structure.
– interfect
Nov 20 '18 at 22:59
I think OP wants to select, form a set of already determined lat,long pairs, those that are within a certain distance of another lat,long point. It boils down to a bunch of pairwise distance queries, if you don't use a fancier GIS index data structure.
– interfect
Nov 20 '18 at 22:59
My point was that lat and long are continuous (you may limit to a certain number of decimal places) but without some kind of geofencing you can quickly become swamped by the number of pairs to test.
– roganjosh
Nov 21 '18 at 6:36
My point was that lat and long are continuous (you may limit to a certain number of decimal places) but without some kind of geofencing you can quickly become swamped by the number of pairs to test.
– roganjosh
Nov 21 '18 at 6:36
add a comment |
1 Answer
1
active
oldest
votes
How important is accuracy, especially near the poles of Earth?
The easy thing to do would be to work out the number of feet per degree of latitude and longitude at your central point. Then for each point, you calculate the latitude and longitude differences from your central point in degrees, an convert them to feet using the conversion factor that is correct at the central point. Then you compute the distance in feet as sqrt(latFeet^2 + longFeet^2)
.
The latitude/foot conversion factor will be constant everywhere (1 Earth circumference / 360 degrees of latitude), but the longitude/foot conversion factor will change as your latitude changes. If you draw a diagram and do some trigonometry, you can see that the distance around the (assumed spherical) Earth at a given latitude is 2 * Pi * Earth Radius * sin(90 degrees - abs(latitude))
, and dividing that by 360 degrees of latitude gives you your distance per degree.
The approximation that 1 degree of latitude or longitude is the same distance in feet over the area you are working in breaks down at the poles, but if you don't have to operate near the poles, and if you don't care about the oblateness of the Earth, it should hold relatively well over distances of 1000 feet.
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53402280%2fgetting-lat-lang-within-1000ft-radius-of-a-central-lat-lang%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
How important is accuracy, especially near the poles of Earth?
The easy thing to do would be to work out the number of feet per degree of latitude and longitude at your central point. Then for each point, you calculate the latitude and longitude differences from your central point in degrees, an convert them to feet using the conversion factor that is correct at the central point. Then you compute the distance in feet as sqrt(latFeet^2 + longFeet^2)
.
The latitude/foot conversion factor will be constant everywhere (1 Earth circumference / 360 degrees of latitude), but the longitude/foot conversion factor will change as your latitude changes. If you draw a diagram and do some trigonometry, you can see that the distance around the (assumed spherical) Earth at a given latitude is 2 * Pi * Earth Radius * sin(90 degrees - abs(latitude))
, and dividing that by 360 degrees of latitude gives you your distance per degree.
The approximation that 1 degree of latitude or longitude is the same distance in feet over the area you are working in breaks down at the poles, but if you don't have to operate near the poles, and if you don't care about the oblateness of the Earth, it should hold relatively well over distances of 1000 feet.
add a comment |
How important is accuracy, especially near the poles of Earth?
The easy thing to do would be to work out the number of feet per degree of latitude and longitude at your central point. Then for each point, you calculate the latitude and longitude differences from your central point in degrees, an convert them to feet using the conversion factor that is correct at the central point. Then you compute the distance in feet as sqrt(latFeet^2 + longFeet^2)
.
The latitude/foot conversion factor will be constant everywhere (1 Earth circumference / 360 degrees of latitude), but the longitude/foot conversion factor will change as your latitude changes. If you draw a diagram and do some trigonometry, you can see that the distance around the (assumed spherical) Earth at a given latitude is 2 * Pi * Earth Radius * sin(90 degrees - abs(latitude))
, and dividing that by 360 degrees of latitude gives you your distance per degree.
The approximation that 1 degree of latitude or longitude is the same distance in feet over the area you are working in breaks down at the poles, but if you don't have to operate near the poles, and if you don't care about the oblateness of the Earth, it should hold relatively well over distances of 1000 feet.
add a comment |
How important is accuracy, especially near the poles of Earth?
The easy thing to do would be to work out the number of feet per degree of latitude and longitude at your central point. Then for each point, you calculate the latitude and longitude differences from your central point in degrees, an convert them to feet using the conversion factor that is correct at the central point. Then you compute the distance in feet as sqrt(latFeet^2 + longFeet^2)
.
The latitude/foot conversion factor will be constant everywhere (1 Earth circumference / 360 degrees of latitude), but the longitude/foot conversion factor will change as your latitude changes. If you draw a diagram and do some trigonometry, you can see that the distance around the (assumed spherical) Earth at a given latitude is 2 * Pi * Earth Radius * sin(90 degrees - abs(latitude))
, and dividing that by 360 degrees of latitude gives you your distance per degree.
The approximation that 1 degree of latitude or longitude is the same distance in feet over the area you are working in breaks down at the poles, but if you don't have to operate near the poles, and if you don't care about the oblateness of the Earth, it should hold relatively well over distances of 1000 feet.
How important is accuracy, especially near the poles of Earth?
The easy thing to do would be to work out the number of feet per degree of latitude and longitude at your central point. Then for each point, you calculate the latitude and longitude differences from your central point in degrees, an convert them to feet using the conversion factor that is correct at the central point. Then you compute the distance in feet as sqrt(latFeet^2 + longFeet^2)
.
The latitude/foot conversion factor will be constant everywhere (1 Earth circumference / 360 degrees of latitude), but the longitude/foot conversion factor will change as your latitude changes. If you draw a diagram and do some trigonometry, you can see that the distance around the (assumed spherical) Earth at a given latitude is 2 * Pi * Earth Radius * sin(90 degrees - abs(latitude))
, and dividing that by 360 degrees of latitude gives you your distance per degree.
The approximation that 1 degree of latitude or longitude is the same distance in feet over the area you are working in breaks down at the poles, but if you don't have to operate near the poles, and if you don't care about the oblateness of the Earth, it should hold relatively well over distances of 1000 feet.
answered Nov 20 '18 at 23:10
interfectinterfect
1,4791021
1,4791021
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53402280%2fgetting-lat-lang-within-1000ft-radius-of-a-central-lat-lang%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Checked this?
– Cheche
Nov 20 '18 at 22:07
Totally dependent on your problem. You could have an infinite number of coords in range; do you have any way of pruning down the options before you run the test?
– roganjosh
Nov 20 '18 at 22:08
@roganjosh May be I am missing something How can there be infinite. We use the Central point and from that point we define radius and then select all the points that are in above columns that are within the distance.
– Null-Hypothesis
Nov 20 '18 at 22:53
1
I think OP wants to select, form a set of already determined lat,long pairs, those that are within a certain distance of another lat,long point. It boils down to a bunch of pairwise distance queries, if you don't use a fancier GIS index data structure.
– interfect
Nov 20 '18 at 22:59
My point was that lat and long are continuous (you may limit to a certain number of decimal places) but without some kind of geofencing you can quickly become swamped by the number of pairs to test.
– roganjosh
Nov 21 '18 at 6:36