我是python的新手，不知道如何处理此任务： 我需要从点数据框中找到每个点的2个数据框，从轨迹数据框中找到2个最近的点
轨迹数据框：
datetime lon_deg lat_deg
2501 28.03.2018 11:58 13.35994653 48.59990204
2502 28.03.2018 11:58 13.35880586 48.60004335
2503 28.03.2018 11:59 13.35766636 48.600205100000004
2504 28.03.2018 11:59 13.35653218 48.60039648
2505 28.03.2018 12:00 13.35539451 48.60058775
2506 28.03.2018 12:00 13.35426064 48.60079647
2507 28.03.2018 12:01 13.3531299 48.60096096
2508 28.03.2018 12:01 13.352004 48.60099219
点数据框：
datetime lon_deg lat_deg
2018-01-29 08:08:59.000 13.359284659333333 48.600108882
29.01.2018 8:09 13.358371081166666 48.60023545666667
2018-01-29 08:09:19.000 13.358347605833334 48.600238692333335
29.01.2018 8:09 13.358324105166666 48.600241913333335
2018-01-29 08:09:20.000 13.358300611666667 48.600245154666666
29.01.2018 8:09 13.358277134 48.600248416
2018-01-29 08:09:21.000 13.358253648166666 48.60025165216667
2018-01-29 08:09:54.000 13.356701967 48.60046564733333
29.01.2018 8:09 13.356678427 48.6004688765
2018-01-29 08:09:55.000 13.356654635 48.6004718285
29.01.2018 8:09 13.356443313166666 48.600502414833336
2018-01-29 08:10:00.000 13.356419901333334 48.60050610933333
29.01.2018 8:10 13.356396262666667 48.600509612
2018-01-29 08:10:09.000 13.355999669 48.6005754975
29.01.2018 8:10 13.355976287333334 48.600579365
2018-01-29 08:10:10.000 13.355952748166667 48.60058305983333
29.01.2018 8:10 13.355929286666667 48.600586781666664
2018-01-29 08:10:11.000 13.355905869 48.6005904815
29.01.2018 8:10 13.355882745166667 48.60059446966667
2018-01-29 08:10:12.000 13.355859396333333 48.600598258666665
29.01.2018 8:10 13.3558361535 48.600602143
2018-01-29 08:10:13.000 13.355812639 48.600605769
29.01.2018 8:10 13.355789295666666 48.60060949333333
2018-01-29 08:10:14.000 13.355765727833333 48.60061298866667
29.01.2018 8:10 13.355742236833333 48.60061659483333
2018-01-29 08:10:15.000 13.3557187615 48.60062014216667
29.01.2018 8:10 13.355695496166666 48.60062391466667
2018-01-29 08:10:16.000 13.35567225 48.600627667833336
29.01.2018 8:10 13.355649023166666 48.600631406
2018-01-29 08:10:17.000 13.355625505 48.60063494533333
29.01.2018 8:10 13.3556019655 48.60063844983333
2018-01-29 08:10:18.000 13.355578551333334 48.60064199316667
29.01.2018 8:10 13.355461117166668 48.60065928433333
2018-01-29 08:10:21.000 13.355437626833334 48.600662660333334
2018-01-29 08:10:24.000 13.3552968655 48.600682845166666
29.01.2018 8:10 13.3552734295 48.600686212333336
2018-01-29 08:10:25.000 13.355249975 48.600689552333336
2018-01-29 08:10:29.000 13.355062269 48.6007157075
29.01.2018 8:10 13.355038871833333 48.60071868083333
2018-01-29 08:10:30.000 13.355015400166666 48.6007218995
29.01.2018 8:10 13.354991943833333 48.60072502533333
2018-01-29 08:10:31.000 13.354968547333334 48.60072815216667
29.01.2018 8:10 13.353912527 48.60085315883333
2018-01-29 08:10:54.000 13.353889066666667 48.60085595533333
2018-01-29 08:11:00.000 13.353607144333333 48.60088610016667
我将不胜感激！
答案 0 :(得分：0)
我想这很大程度上取决于您的数据大小。
暴力破解方法类似于：
import numpy as np
points_dataframe = np.random.rand(20,2)
trajecotry_dataframe = np.random.rand(5,2)
print('points_dataframe:')
print(points_dataframe)
print('\n\ntrajecotry_dataframe:')
print(trajecotry_dataframe)
print('\n\n')
for index_points, (x1, y1) in enumerate(points_dataframe):
distance_list = []
for index_trajecotry, (x2, y2) in enumerate(trajecotry_dataframe):
distance_list.append(np.sqrt((x1-x2)**2 + (y1-y2)**2))
sorted_list = np.sort(distance_list)
print('+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++')
print(f'for element {index_points} in the points_dataframe the two closest points are:')
point0 = np.where(distance_list==sorted_list[0])[0][0]
print(f'element {point0} from the trajecotry_dataframe')
point1 = np.where(distance_list==sorted_list[1])[0][0]
print(f'element {point1} from the trajecotry_dataframe')
但是当数据集更大或您不得不更频繁地重复计算时， 也许您应该考虑将数据保存在经过地理编码的数据库中。
答案 1 :(得分：0)
以下是一些用Matlab编写的代码，可能会有所帮助。如果有用，则必须将它们转换为Python。这种方法是蛮力的，而不是最优雅的方法。但是，我尝试包括近似的坐标系转换，这些转换将地球的形状解释为椭球。如果假设地球是一个球体，那么事情就可以简化一点。另外，为了提高精度（尽管很有可能忽略不计），可以通过球体表面（在给定点最接近椭球体的球体）局部近似椭圆体的表面，并使用球体代替欧几里得几何。
可能会有一些错别字或错误，但是也许您可以了解坐标，转换和方法的概念。
使用以下两个功能，您可以转换为：
到欧几里得坐标的点long_lat0 = [long0, lat0]附近的大地坐标（即经度）是WGS84地球椭球上实际，真实大地坐标的一阶线性近似
相反，您可以从欧几里得坐标转换回大地纬度
long_lat0 = [long0, lat0]; % a point from dataset 2
long_lat % the n x 2 matrix of points from dataset 1 (or a chunk of it)
%center of approximate Euclidean coordinate system is point long_lat0
% with long_lat coordinates and the scaling coefficient
% a of longitude and b of latitude,
% which equalizes longitude and latitude distance at point long_lat0, is
function [x, a, b] = convert_to_local_Eucl(long_lat, long_lat0)
% long_lat0 = [long_0, lat_0] is the origin of the local coordinate system
% long_lat = [long_1, lat_1;
% long_2, lat_2;
% ............
% long_n, lat_n] is an n x 2 array of points in lat and long coordinates
% on the Earth's ellipsoid
% x = [x_1, y_1;
% x_2, y_2;
% ..........
% x_n, y_n]
% is the n x 2 matrix of Euclidean coordinates with origin the point long_lat0
% a is a number, correction factor of longitude coordinate
% b is a number, correction factor of latitude
R = 6378137.0 %in meters;
e_2 = ( R^2 - (6356752.314245)^2 ) / R^2;
a = R * (1-e_2) * cosd(long_lat0(2)) / (1 - e_2*sind(long_lat0(2))^(1/2)); % dlong
b = R * (1-e_2) / (1 - e_2*sind(long_lat0(2))^(3/2); %dlat
% a and b are correcting/rescaling coefficients
% that correct the longitude-latitude coordinates of all points
% near point long_lat0 in geodetic coordinates of WGS84.
x = long_lat .- long_lat0; % subtract the long_lat0 from the coordinates of every point
% from the list long_lat, i.e. for each j = 1...n
% x(j, 1) = long_lat(j, 1) - long_lat0(1);
% x(j, 2) = long_lat(j, 2) - long_lat0(2);
x = [ a * x(:,1), b * x(:, 2)];
% multiply the first column of coordinates by the scaling factor a and
% multiply the second column of coordinates by the scaling factor b
% these coordinates are first order linear Euclidean approximation
% of the real geodetic coordinates of WGS84.
% Near the point long_lat0
% the error is negligible, especially within a couple of kilometers.
% The farther you go from that point, the error slowly increases,
% but then it doesn't matter since such points are not the closest anyway.
end
function long_lat = convert_to_long_lat(x, long_lat0, a, b)
% from Euclidean coordinates x = [x(1), x(2)] of a point near long_lat0 go back to
% long_lat = [long, lat] coordinates of that points. a and b are the scaling
% coefficients at point long_lat0
long_lat = [long_lat0(1) + x(1)/a, long_lat0(2) + x(2)/b];
end
对于数据集2中的每个点long_lat0 = [long0, lat0]，首先从大地纬向经纬度转换为long_lat0处的近似欧几里得坐标
数据集1（第二和第三列）的整个long_lat列表（或大块）：
x = convert_2_local_Eucl(long_lat, long_lat0);
然后计算所有2D行向量的大小（即长度）
x(j,:) = [x(j,1), x(j,1)]
来自数据集x
magnitudes = norm(x); %you have to either find this function or write one yourself
之后，找到x的索引和元素的最小值：
[j, min] = min(magnitudes);
然后为两对：
x1 = x(j,:) and x2 = x(j+1,:)
和
x1 = x(j,:) and x2 = x(j-1,:)
使用下面的函数计算最接近的点：
function [dist, long_lat] = dist_point_to_reference(x1, x2, long_lat0, a, b)
% calculates the shortest distance dist from the point long_lat0
% to the closest point on the segment between x1 and x2
% and then obtain the long_lat coordinates of this closest point
dist = dot(x1, x1) * dot(x2 - x1, x2 - x1) - dot(x1, x2 - x1)^2 ; % dot is dot product
dist = sqrt( dist / ( dot(x2 - x1, x2 - x1)^2) );
% dist is the distance from the point at the origin [0, 0]
% to the straight Euclidean interval between
% the points x1 = [x1(1), x1(2)] and x2 = [x2(1), x2(2)]
if dot(x1, x2 - x1) > 0 % if the height of the triangle is outside, on the side of x1
dist = sqrt( dot(x1, x1) );
long_lat = x1;
elseif dot(x2, x1 - x2) > 0 % if the height of the triangle is outside, on the side of x2
dist = sqrt( dot(x2, x2) );
long_lat = x1;
else
long_lat(1) = - x2(2) + x1(2);
long_lat(2) = x2(1) - x1(1);
long_lat = long_lat / sqrt(dot(long_lat, long_lat));
long_lat = - dot(x1, long_lat) * long_lat; % despite the name, these are Eucldean coordinates
end
long_lat = convert_to_long_lat(long_lat, a, b); % finally, geodetic coordinates
end