GeoPandas: Find nearest point in other dataframe

Figured it out:

def min_dist(point, gpd2):
    gpd2['Dist'] = gpd2.apply(lambda row:  point.distance(row.geometry),axis=1)
    geoseries = gpd2.iloc[gpd2['Dist'].argmin()]
    return geoseries

Of course some criticism is welcome. I'm not a fan of recalculating gpd2['Dist'] for every row of gpd1...

If you have large dataframes, I've found that scipy's cKDTree spatial index .query method returns very fast results for nearest neighbor searches. As it uses a spatial index it's orders of magnitude faster than looping though the dataframe and then finding the minimum of all distances. It is also faster than using shapely's nearest_points with RTree (the spatial index method available via geopandas) because cKDTree allows you to vectorize your search whereas the other method does not.

Here is a helper function that will return the distance and 'Name' of the nearest neighbor in gpd2 from each point in gpd1. It assumes both gdfs have a geometry column (of points).

import geopandas as gpd
import numpy as np
import pandas as pd

from scipy.spatial import cKDTree
from shapely.geometry import Point

gpd1 = gpd.GeoDataFrame([['John', 1, Point(1, 1)], ['Smith', 1, Point(2, 2)],
                         ['Soap', 1, Point(0, 2)]],
                        columns=['Name', 'ID', 'geometry'])
gpd2 = gpd.GeoDataFrame([['Work', Point(0, 1.1)], ['Shops', Point(2.5, 2)],
                         ['Home', Point(1, 1.1)]],
                        columns=['Place', 'geometry'])

def ckdnearest(gdA, gdB):
    nA = np.array(list(gdA.geometry.apply(lambda x: (x.x, x.y))))
    nB = np.array(list(gdB.geometry.apply(lambda x: (x.x, x.y))))
    btree = cKDTree(nB)
    dist, idx = btree.query(nA, k=1)
    gdf = pd.concat(
        [gdA.reset_index(drop=True), gdB.loc[idx, gdB.columns != 'geometry'].reset_index(drop=True),
         pd.Series(dist, name='dist')], axis=1)
    return gdf

ckdnearest(gpd1, gpd2)

And if you want to find the closest point to a LineString, here is a full working example:

import itertools
from operator import itemgetter

import geopandas as gpd
import numpy as np
import pandas as pd

from scipy.spatial import cKDTree
from shapely.geometry import Point, LineString

gpd1 = gpd.GeoDataFrame([['John', 1, Point(1, 1)],
                         ['Smith', 1, Point(2, 2)],
                         ['Soap', 1, Point(0, 2)]],
                        columns=['Name', 'ID', 'geometry'])
gpd2 = gpd.GeoDataFrame([['Work', LineString([Point(100, 0), Point(100, 1)])],
                         ['Shops', LineString([Point(101, 0), Point(101, 1), Point(102, 3)])],
                         ['Home',  LineString([Point(101, 0), Point(102, 1)])]],
                        columns=['Place', 'geometry'])


def ckdnearest(gdfA, gdfB, gdfB_cols=['Place']):
    A = np.concatenate(
        [np.array(geom.coords) for geom in gdfA.geometry.to_list()])
    B = [np.array(geom.coords) for geom in gdfB.geometry.to_list()]
    B_ix = tuple(itertools.chain.from_iterable(
        [itertools.repeat(i, x) for i, x in enumerate(list(map(len, B)))]))
    B = np.concatenate(B)
    ckd_tree = cKDTree(B)
    dist, idx = ckd_tree.query(A, k=1)
    idx = itemgetter(*idx)(B_ix)
    gdf = pd.concat(
        [gdfA, gdfB.loc[idx, gdfB_cols].reset_index(drop=True),
         pd.Series(dist, name='dist')], axis=1)
    return gdf

c = ckdnearest(gpd1, gpd2)

The answer by Gene didn't work for me. Finally I discovered that gpd2.geometry.unary_union resulted in a geometry that only contained about 30.000 of my total of roughly 150.000 points. For anyone else running into the same problem, here's how I solved it:

from shapely.ops import nearest_points
from shapely.geometry import MultiPoint

gpd2_pts_list = gpd2.geometry.tolist()
gpd2_pts = MultiPoint(gpd2_pts_list)
def nearest(point, gpd2_pts, gpd2=gpd2, geom_col='geometry', src_col='Place'):
     # find the nearest point
     nearest_point = nearest_points(point, gpd2_pts)[1]
     # return the corresponding value of the src_col of the nearest point
     value = gpd2[gpd2[geom_col] == nearest_point][src_col].get_values()[0]
     return value

gpd1['Nearest'] = gpd1.apply(lambda x: nearest(x.geometry, gpd2_pts), axis=1)

For anyone having indexing errors with their own data while using the excellent answer from @JHuw, my problem was that my indexes did not align. Resetting the index of gdfA and gdfB solved my issues, maybe this can help you as well @Shakedk.

import itertools
from operator import itemgetter

import geopandas as gpd
import numpy as np
import pandas as pd

from scipy.spatial import cKDTree
from shapely.geometry import Point, LineString

gpd1 = gpd.GeoDataFrame([['John', 1, Point(1, 1)],
                         ['Smith', 1, Point(2, 2)],
                         ['Soap', 1, Point(0, 2)]],
                        columns=['Name', 'ID', 'geometry'])
gpd2 = gpd.GeoDataFrame([['Work', LineString([Point(100, 0), Point(100, 1)])],
                         ['Shops', LineString([Point(101, 0), Point(101, 1), Point(102, 3)])],
                         ['Home',  LineString([Point(101, 0), Point(102, 1)])]],
                        columns=['Place', 'geometry'])


def ckdnearest(gdfA, gdfB, gdfB_cols=['Place']):
    # resetting the index of gdfA and gdfB here.
    gdfA = gdfA.reset_index(drop=True)
    gdfB = gdfB.reset_index(drop=True)
    A = np.concatenate(
        [np.array(geom.coords) for geom in gdfA.geometry.to_list()])
    B = [np.array(geom.coords) for geom in gdfB.geometry.to_list()]
    B_ix = tuple(itertools.chain.from_iterable(
        [itertools.repeat(i, x) for i, x in enumerate(list(map(len, B)))]))
    B = np.concatenate(B)
    ckd_tree = cKDTree(B)
    dist, idx = ckd_tree.query(A, k=1)
    idx = itemgetter(*idx)(B_ix)
    gdf = pd.concat(
        [gdfA, gdfB.loc[idx, gdfB_cols].reset_index(drop=True),
         pd.Series(dist, name='dist')], axis=1)
    return gdf

c = ckdnearest(gpd1, gpd2)

This solution is extremely inefficient but it should work for any and all geometry types (including mixed geometry type gdfs). I would only try this if your gdfs are small (my use case was a gdf with about 2000 rows for which I wanted to find the nearest feature from another gdf with about 15 rows, and it took a few seconds on a typical office laptop). Intersecting features may cause it to spaz out though, so be warned. It was originally based on @RedM's solution but will instead assign the index of the feature in gdf2 that is nearest to gdf1

gdf1["gdf2_idx"] = gdf1.apply(
    lambda row1: gdf2.apply(
        lambda row2: row1.geometry.distance(row2.geometry), axis="columns").idxmin(),
    axis="columns"
)