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I am new to python, and I can't figure out how to find the minimum distance from a given lat/lon point (which is not given from the grid, but selected by me) to a find the closest indices of a lat/lon point on a grid.

Basically , I am reading in an ncfile that contains 2D coordinates:

coords = 'coords.nc'
fh = Dataset(coords,mode='r')
lons = fh.variables['latitudes'][:,:]
lats = fh.variables['longitudes'][:,:]
fh.close()

>>> lons.shape
(94, 83)
>>> lats.shape
(94, 83)

I want to find the indices in the above grid for the nearest lat lon to the below values: 

sel_lat=71.60556
sel_lon=-161.458611

I tried to make lat/lon pairs in order to use the scipy.spatial.distance function, but I still am having problems because I did not set up the input arrays to the format it wants, but I don't understand how to do that:

latLon_pairsGrid = np.vstack(([lats.T],[lons.T])).T

>>> latLon_pairsGrid.shape
(94, 83, 2)

distance.cdist([sel_lat,sel_lon],latLon_pairsGrid,'euclidean')

Any help or hints would be appreciated

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  • 1
    show us what lons and lats look like and what are their types. Even better: gives us an example that we can run in our computers and the expected output. Commented Oct 12, 2016 at 23:10
  • @cd98 I'm not sure what you mean by an example, but maybe I wasn't clear enough in my question. The sel_lat, sel_lon values are given above, and those are the coordinates I need to find the nearest corresponding indices for on the grid from the 'coords.nc' file. The lons.shape = (93,83) and same with the lats.shape. And so that means the latLon_pairsGrid I made is a (93,83,2). I hope this is clear enough, I'm not sure what else to add Commented Oct 12, 2016 at 23:28
  • "does not have the format it wants" can you print the error please Commented Oct 12, 2016 at 23:36
  • also can you try to do lats.T.as_matrix() and lons.T.as_matrix() to your arrays inside of the np.vstack to convert them to pure numpy arrays instead of pandas stuff Commented Oct 12, 2016 at 23:38
  • @Julius I'm not using pandas, when I do lats.T.as_matrix() it gives the error ` >>> lats.T.as_matrix() Traceback (most recent call last): File "<stdin>", line 1, in <module> AttributeError: 'numpy.ndarray' object has no attribute 'as_matrix' ` For the error received in the distance.cdist command in the last line of my question it returns: ` 2027, in cdist raise ValueError('XA must be a 2-dimensional array.') ` Commented Oct 12, 2016 at 23:49

2 Answers 2

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Checkout the pyresample package. It provides spatial nearest neighbour search using a fast kdtree approach:

import pyresample
import numpy as np

# Define lat-lon grid
lon = np.linspace(30, 40, 100)
lat = np.linspace(10, 20, 100)
lon_grid, lat_grid = np.meshgrid(lon, lat)
grid = pyresample.geometry.GridDefinition(lats=lat_grid, lons=lon_grid)

# Generate some random data on the grid
data_grid = np.random.rand(lon_grid.shape[0], lon_grid.shape[1])

# Define some sample points
my_lons = np.array([34.5, 36.5, 38.5])
my_lats = np.array([12.0, 14.0, 16.0])
swath = pyresample.geometry.SwathDefinition(lons=my_lons, lats=my_lats)

# Determine nearest (w.r.t. great circle distance) neighbour in the grid.
_, _, index_array, distance_array = pyresample.kd_tree.get_neighbour_info(
    source_geo_def=grid, target_geo_def=swath, radius_of_influence=50000,
    neighbours=1)

# get_neighbour_info() returns indices in the flattened lat/lon grid. Compute
# the 2D grid indices:
index_array_2d = np.unravel_index(index_array, grid.shape)

print "Indices of nearest neighbours:", index_array_2d
print "Longitude of nearest neighbours:", lon_grid[index_array_2d]
print "Latitude of nearest neighbours:", lat_grid[index_array_2d]
print "Great Circle Distance:", distance_array

There is also a shorthand method for directly obtaining the data values at the nearest grid points:

data_swath = pyresample.kd_tree.resample_nearest(
    source_geo_def=grid, target_geo_def=swath, data=data_grid,
    radius_of_influence=50000)
print "Data at nearest grid points:", data_swath
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Comments

1

I think I found an answer, but it is a workaround that avoids calculating distance between the chosen lat/lon and the lat/lons on the grid. This doesn't seem completely accurate because I am never calculating distances, just the closest difference between lat/lon values themselves.

I used the answer to the question find (i,j) location of closest (long,lat) values in a 2D array

a = abs(lats-sel_lat)+abs(lons-sel_lon)
i,j = np.unravel_index(a.argmin(),a.shape)

Using those returned indices i,j, I can then find on the grid the coordinates that correspond most closely to my selected lat, lon value:

>>> lats[i,j]
71.490295
>>> lons[i,j]
-161.65045
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1 Comment

You're minimizing an L1 metric, also called Manhattan distance, which is not totally unreasonable and cheaper to compute than the usual L2 metric a=sqrt(dx**2+dy**2). And like you said, you're computing distance in the lat/lon domain, not geodesic or Cartesian distances, which would require coordinate transformations. If you're going to query multiple points and speed is an issue, definitely check out @Funkensieper's answer or other tree-based approaches.

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