从纬度经度坐标以矢量化方式(无循环)构建距离矩阵

问题描述

我想提出一种更快的方法来在所有纬度对之间创建距离矩阵.此 QA 解决了使用标准线性代数，但不包含纬度Lon Lon坐标.

I would like to come up with a faster way to create a distance matrix between all lat lon pairs. This QA addresses doing a vectorized way with standard Linear Algebra, but without Lat Lon coordinates.

在我的情况下，这些经久不衰的农场是农场.这是我的Python代码，对于完整的数据集(4000(拉特，lon))，至少需要五分钟.有什么想法吗?

In my case these lat longs are farms. Here is my Python code, which for the full data set (4000 (lat, lon)'s) takes at least five minutes. Any ideas?

> def slowdistancematrix(df, distance_calc=True, sparse=False, dlim=100):
    """
    inputs: df

    returns:
    1.) distance between all farms in miles
    2.) distance^2

    """

    from scipy.spatial import distance_matrix
    from geopy.distance import geodesic

    unique_farms = pd.unique(df.pixel)
    df_unique = df.set_index('pixel')
    df_unique = df_unique[~df_unique.index.duplicated(keep='first')] # only keep unique index values
    distance = np.zeros((unique_farms.size,unique_farms.size))

    for i in range(unique_farms.size):
        lat_lon_i = df_unique.Latitude.iloc[i],df_unique.Longitude.iloc[i]
        for j in range(i):
            lat_lon_j = df_unique.Latitude.iloc[j],df_unique.Longitude.iloc[j]
            if distance_calc == True:
                distance[i,j] = geodesic(lat_lon_i, lat_lon_j).miles
                distance[j,i] = distance[i,j] # make use of symmetry

    return distance, np.power(distance, 2)

推荐答案

我的解决方案是:

import numpy as np

def dist(v):
    v = np.radians(v)

    dlat = v[:, 0, np.newaxis] - v[:, 0]
    dlon = v[:, 1, np.newaxis] - v[:, 1]

    a = np.sin(dlat / 2.0) ** 2 + np.cos(v[:, 0]) * np.cos(v[:, 0]) * np.sin(dlon / 2.0) ** 2

    c = 2 * np.arcsin(np.sqrt(a))
    result = 3956 * c

    return result

但是，您将需要使用属性values将数据框转换为numpy数组.例如:

However you will need to convert your dataframe to a numpy array, using the attribute values. For example:

df = pd.read_csv('some_csv_file.csv')
distances = dist(df[['lat', 'lng']].values)

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