Get closest coordinate in 2D array










2














coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]

xy = (-222.4, -204.5)


What is the best way so that a given value xy can be compared to a 2D list of coordinates, and return the index number of the closest coordinate?



In this example, xy would be compared to the coordinates list and thus return the closest coordinate (-225.0, -299.5) or, more ideally, the index number 0.



I've tried researching for a method with itertools or numpy, but couldn't seem to understand how to get the result I want in my example.










share|improve this question




























    2














    coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]

    xy = (-222.4, -204.5)


    What is the best way so that a given value xy can be compared to a 2D list of coordinates, and return the index number of the closest coordinate?



    In this example, xy would be compared to the coordinates list and thus return the closest coordinate (-225.0, -299.5) or, more ideally, the index number 0.



    I've tried researching for a method with itertools or numpy, but couldn't seem to understand how to get the result I want in my example.










    share|improve this question


























      2












      2








      2







      coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]

      xy = (-222.4, -204.5)


      What is the best way so that a given value xy can be compared to a 2D list of coordinates, and return the index number of the closest coordinate?



      In this example, xy would be compared to the coordinates list and thus return the closest coordinate (-225.0, -299.5) or, more ideally, the index number 0.



      I've tried researching for a method with itertools or numpy, but couldn't seem to understand how to get the result I want in my example.










      share|improve this question















      coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]

      xy = (-222.4, -204.5)


      What is the best way so that a given value xy can be compared to a 2D list of coordinates, and return the index number of the closest coordinate?



      In this example, xy would be compared to the coordinates list and thus return the closest coordinate (-225.0, -299.5) or, more ideally, the index number 0.



      I've tried researching for a method with itertools or numpy, but couldn't seem to understand how to get the result I want in my example.







      python list multidimensional-array coordinates itertools






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 12 at 8:00









      martineau

      65.6k988177




      65.6k988177










      asked Nov 12 at 7:32









      qbuffer

      416




      416






















          4 Answers
          4






          active

          oldest

          votes


















          2














          You can use min with a proper key function. Sth along the following lines for instance:



          coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
          xy = (-222.4, -204.5)

          dist = lambda x, y: (x[0]-y[0])**2 + (x[1]-y[1])**2
          min(coordinates, key=lambda co: dist(co, xy))
          # (-225.0, -299.5)





          share|improve this answer




















          • I would suggest using math.hypot() instead—it's probably faster.
            – martineau
            Nov 12 at 8:03










          • @martineau I don't quite see how the distance from the origin helps in this general distance case.
            – schwobaseggl
            Nov 12 at 8:13










          • min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
            – martineau
            Nov 12 at 8:44










          • @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
            – schwobaseggl
            Nov 12 at 8:46










          • Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
            – martineau
            Nov 12 at 19:19



















          1














          Your question is equivalent to : How do I sort a Python list using a custom method to define the sorting key. This can be done in raw python without using external libraries.



          When using the sorted() function of python you can pass a lambda to the key argument to get the sort done on a specific key.



          From there you just have to define the key as your own distance calculation method (here using the distances between the points):



          from math import *
          coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
          xy = (-222.4, -204.5)
          results = sorted(coordinates, key= lambda v: sqrt(pow((v[0] - xy[0]), 2) + pow((v[1] - xy[1]), 2)))
          # Output : [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]


          From there you can just take the first element of the list if you want the closer point, etc. If you still want to external module I think you can use some third parties function such as from scipy.spatial import distance as the key parameter of the sort.






          share|improve this answer




















          • Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
            – martineau
            Nov 12 at 19:30



















          0














          You could simply create a function that iterates through the list of coordinates and keep the index of the one in which the distance between two points is the smallest (using the Pythagorean theorem).



          However, if you need something fast given by an external module rather than writing your own, I am not aware of libraries I already used that already have that function, so I'm not helpful here.






          share|improve this answer




















          • I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
            – qbuffer
            Nov 12 at 7:46










          • @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
            – schwobaseggl
            Nov 12 at 7:50











          • If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
            – marcioz98
            Nov 12 at 7:55


















          0














          Using scipy.spatial.KDTree:



          from scipy import spatial
          import numpy as np
          coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
          x = [(-222.4, -204.5)]
          distance,index = spatial.KDTree(coordinates).query(x)
          print(distance)
          print(index)


          The kd-tree method is O(N*log(N)) and is much faster than Brute Force method that takes O(N**2) time for large enough N.






          share|improve this answer




















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            4 Answers
            4






            active

            oldest

            votes








            4 Answers
            4






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            You can use min with a proper key function. Sth along the following lines for instance:



            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)

            dist = lambda x, y: (x[0]-y[0])**2 + (x[1]-y[1])**2
            min(coordinates, key=lambda co: dist(co, xy))
            # (-225.0, -299.5)





            share|improve this answer




















            • I would suggest using math.hypot() instead—it's probably faster.
              – martineau
              Nov 12 at 8:03










            • @martineau I don't quite see how the distance from the origin helps in this general distance case.
              – schwobaseggl
              Nov 12 at 8:13










            • min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
              – martineau
              Nov 12 at 8:44










            • @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
              – schwobaseggl
              Nov 12 at 8:46










            • Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
              – martineau
              Nov 12 at 19:19
















            2














            You can use min with a proper key function. Sth along the following lines for instance:



            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)

            dist = lambda x, y: (x[0]-y[0])**2 + (x[1]-y[1])**2
            min(coordinates, key=lambda co: dist(co, xy))
            # (-225.0, -299.5)





            share|improve this answer




















            • I would suggest using math.hypot() instead—it's probably faster.
              – martineau
              Nov 12 at 8:03










            • @martineau I don't quite see how the distance from the origin helps in this general distance case.
              – schwobaseggl
              Nov 12 at 8:13










            • min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
              – martineau
              Nov 12 at 8:44










            • @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
              – schwobaseggl
              Nov 12 at 8:46










            • Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
              – martineau
              Nov 12 at 19:19














            2












            2








            2






            You can use min with a proper key function. Sth along the following lines for instance:



            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)

            dist = lambda x, y: (x[0]-y[0])**2 + (x[1]-y[1])**2
            min(coordinates, key=lambda co: dist(co, xy))
            # (-225.0, -299.5)





            share|improve this answer












            You can use min with a proper key function. Sth along the following lines for instance:



            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)

            dist = lambda x, y: (x[0]-y[0])**2 + (x[1]-y[1])**2
            min(coordinates, key=lambda co: dist(co, xy))
            # (-225.0, -299.5)






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 12 at 7:46









            schwobaseggl

            36.1k32340




            36.1k32340











            • I would suggest using math.hypot() instead—it's probably faster.
              – martineau
              Nov 12 at 8:03










            • @martineau I don't quite see how the distance from the origin helps in this general distance case.
              – schwobaseggl
              Nov 12 at 8:13










            • min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
              – martineau
              Nov 12 at 8:44










            • @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
              – schwobaseggl
              Nov 12 at 8:46










            • Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
              – martineau
              Nov 12 at 19:19

















            • I would suggest using math.hypot() instead—it's probably faster.
              – martineau
              Nov 12 at 8:03










            • @martineau I don't quite see how the distance from the origin helps in this general distance case.
              – schwobaseggl
              Nov 12 at 8:13










            • min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
              – martineau
              Nov 12 at 8:44










            • @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
              – schwobaseggl
              Nov 12 at 8:46










            • Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
              – martineau
              Nov 12 at 19:19
















            I would suggest using math.hypot() instead—it's probably faster.
            – martineau
            Nov 12 at 8:03




            I would suggest using math.hypot() instead—it's probably faster.
            – martineau
            Nov 12 at 8:03












            @martineau I don't quite see how the distance from the origin helps in this general distance case.
            – schwobaseggl
            Nov 12 at 8:13




            @martineau I don't quite see how the distance from the origin helps in this general distance case.
            – schwobaseggl
            Nov 12 at 8:13












            min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
            – martineau
            Nov 12 at 8:44




            min(coordinates, key=lambda p: math.hypot(p[0]-xy[0], p[1]-xy[1]))
            – martineau
            Nov 12 at 8:44












            @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
            – schwobaseggl
            Nov 12 at 8:46




            @martineau So you are adding an extra function call and still doing calculations in pure Python? Not too convincing....
            – schwobaseggl
            Nov 12 at 8:46












            Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
            – martineau
            Nov 12 at 19:19





            Timing the execution is the only definitive way to tell. Doing so, I found that for larger numbers of points, your way is about 19% slower than using hypot() with Python 3.7.1. Here's the benchmark used to determine this (results are at the very end).
            – martineau
            Nov 12 at 19:19














            1














            Your question is equivalent to : How do I sort a Python list using a custom method to define the sorting key. This can be done in raw python without using external libraries.



            When using the sorted() function of python you can pass a lambda to the key argument to get the sort done on a specific key.



            From there you just have to define the key as your own distance calculation method (here using the distances between the points):



            from math import *
            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)
            results = sorted(coordinates, key= lambda v: sqrt(pow((v[0] - xy[0]), 2) + pow((v[1] - xy[1]), 2)))
            # Output : [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]


            From there you can just take the first element of the list if you want the closer point, etc. If you still want to external module I think you can use some third parties function such as from scipy.spatial import distance as the key parameter of the sort.






            share|improve this answer




















            • Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
              – martineau
              Nov 12 at 19:30
















            1














            Your question is equivalent to : How do I sort a Python list using a custom method to define the sorting key. This can be done in raw python without using external libraries.



            When using the sorted() function of python you can pass a lambda to the key argument to get the sort done on a specific key.



            From there you just have to define the key as your own distance calculation method (here using the distances between the points):



            from math import *
            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)
            results = sorted(coordinates, key= lambda v: sqrt(pow((v[0] - xy[0]), 2) + pow((v[1] - xy[1]), 2)))
            # Output : [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]


            From there you can just take the first element of the list if you want the closer point, etc. If you still want to external module I think you can use some third parties function such as from scipy.spatial import distance as the key parameter of the sort.






            share|improve this answer




















            • Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
              – martineau
              Nov 12 at 19:30














            1












            1








            1






            Your question is equivalent to : How do I sort a Python list using a custom method to define the sorting key. This can be done in raw python without using external libraries.



            When using the sorted() function of python you can pass a lambda to the key argument to get the sort done on a specific key.



            From there you just have to define the key as your own distance calculation method (here using the distances between the points):



            from math import *
            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)
            results = sorted(coordinates, key= lambda v: sqrt(pow((v[0] - xy[0]), 2) + pow((v[1] - xy[1]), 2)))
            # Output : [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]


            From there you can just take the first element of the list if you want the closer point, etc. If you still want to external module I think you can use some third parties function such as from scipy.spatial import distance as the key parameter of the sort.






            share|improve this answer












            Your question is equivalent to : How do I sort a Python list using a custom method to define the sorting key. This can be done in raw python without using external libraries.



            When using the sorted() function of python you can pass a lambda to the key argument to get the sort done on a specific key.



            From there you just have to define the key as your own distance calculation method (here using the distances between the points):



            from math import *
            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            xy = (-222.4, -204.5)
            results = sorted(coordinates, key= lambda v: sqrt(pow((v[0] - xy[0]), 2) + pow((v[1] - xy[1]), 2)))
            # Output : [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]


            From there you can just take the first element of the list if you want the closer point, etc. If you still want to external module I think you can use some third parties function such as from scipy.spatial import distance as the key parameter of the sort.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 12 at 7:49









            Anto

            1062




            1062











            • Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
              – martineau
              Nov 12 at 19:30

















            • Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
              – martineau
              Nov 12 at 19:30
















            Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
            – martineau
            Nov 12 at 19:30





            Sorting a list with a custom key function involves quite a bit more processing than simply finding the list element that produces the minimum when passed to that functiont—so, no, the question isn't equivalent to that. If it were, I would suggest using the math.hypot() function because it's faster that the way you're doing it. See this comment about that.
            – martineau
            Nov 12 at 19:30












            0














            You could simply create a function that iterates through the list of coordinates and keep the index of the one in which the distance between two points is the smallest (using the Pythagorean theorem).



            However, if you need something fast given by an external module rather than writing your own, I am not aware of libraries I already used that already have that function, so I'm not helpful here.






            share|improve this answer




















            • I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
              – qbuffer
              Nov 12 at 7:46










            • @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
              – schwobaseggl
              Nov 12 at 7:50











            • If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
              – marcioz98
              Nov 12 at 7:55















            0














            You could simply create a function that iterates through the list of coordinates and keep the index of the one in which the distance between two points is the smallest (using the Pythagorean theorem).



            However, if you need something fast given by an external module rather than writing your own, I am not aware of libraries I already used that already have that function, so I'm not helpful here.






            share|improve this answer




















            • I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
              – qbuffer
              Nov 12 at 7:46










            • @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
              – schwobaseggl
              Nov 12 at 7:50











            • If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
              – marcioz98
              Nov 12 at 7:55













            0












            0








            0






            You could simply create a function that iterates through the list of coordinates and keep the index of the one in which the distance between two points is the smallest (using the Pythagorean theorem).



            However, if you need something fast given by an external module rather than writing your own, I am not aware of libraries I already used that already have that function, so I'm not helpful here.






            share|improve this answer












            You could simply create a function that iterates through the list of coordinates and keep the index of the one in which the distance between two points is the smallest (using the Pythagorean theorem).



            However, if you need something fast given by an external module rather than writing your own, I am not aware of libraries I already used that already have that function, so I'm not helpful here.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 12 at 7:40









            marcioz98

            1




            1











            • I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
              – qbuffer
              Nov 12 at 7:46










            • @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
              – schwobaseggl
              Nov 12 at 7:50











            • If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
              – marcioz98
              Nov 12 at 7:55
















            • I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
              – qbuffer
              Nov 12 at 7:46










            • @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
              – schwobaseggl
              Nov 12 at 7:50











            • If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
              – marcioz98
              Nov 12 at 7:55















            I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
            – qbuffer
            Nov 12 at 7:46




            I thought about this, but wouldn't the Pythagorean theorem not work if I have negative and positive coordinates in my list? e.g. if coordinates = [(-255.0,-255.0), (255.0,255.0)] -> it wouldn't know which coordinate is closer?
            – qbuffer
            Nov 12 at 7:46












            @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
            – schwobaseggl
            Nov 12 at 7:50





            @Ben There is no issue with negative coordinates, Pythagoras will still take good care of the correct distance. You just have to decide on a tie breaker if in deed, two sets of coordinates havce the same distance.
            – schwobaseggl
            Nov 12 at 7:50













            If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
            – marcioz98
            Nov 12 at 7:55




            If you have negative and positive coordinates it will surely work, the formula doesn't care if you have positive or negative coordinates. link If the distance from a specific point from two other points is the same, it will take the index of the first point that appears on the list.
            – marcioz98
            Nov 12 at 7:55











            0














            Using scipy.spatial.KDTree:



            from scipy import spatial
            import numpy as np
            coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
            x = [(-222.4, -204.5)]
            distance,index = spatial.KDTree(coordinates).query(x)
            print(distance)
            print(index)


            The kd-tree method is O(N*log(N)) and is much faster than Brute Force method that takes O(N**2) time for large enough N.






            share|improve this answer

























              0














              Using scipy.spatial.KDTree:



              from scipy import spatial
              import numpy as np
              coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
              x = [(-222.4, -204.5)]
              distance,index = spatial.KDTree(coordinates).query(x)
              print(distance)
              print(index)


              The kd-tree method is O(N*log(N)) and is much faster than Brute Force method that takes O(N**2) time for large enough N.






              share|improve this answer























                0












                0








                0






                Using scipy.spatial.KDTree:



                from scipy import spatial
                import numpy as np
                coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
                x = [(-222.4, -204.5)]
                distance,index = spatial.KDTree(coordinates).query(x)
                print(distance)
                print(index)


                The kd-tree method is O(N*log(N)) and is much faster than Brute Force method that takes O(N**2) time for large enough N.






                share|improve this answer












                Using scipy.spatial.KDTree:



                from scipy import spatial
                import numpy as np
                coordinates = [(-225.0, -299.5), (-150.0, 75.5), (0.0, 0.0), (225.0, 300.5)]
                x = [(-222.4, -204.5)]
                distance,index = spatial.KDTree(coordinates).query(x)
                print(distance)
                print(index)


                The kd-tree method is O(N*log(N)) and is much faster than Brute Force method that takes O(N**2) time for large enough N.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 at 8:10









                Abhinav Vajpeyi

                163




                163



























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