列之间的计算（经度/纬度）非常慢

Question

我有两个独立的数据集，df 和 df2，每个数据集都有 longitude 和 latitude 列。我要做的是在df 中找到最接近df2 中的点，并在km 中计算它们之间的距离，并将每个值附加到df2 中的新列中。

我想出了一个解决方案，但请记住，df 有 +700,000 行，df2 有大约 60,000 行，所以我的解决方案需要很长时间才能计算出来。我能想出的唯一解决方案是使用双 for 循环...

def compute_shortest_dist(df, df2):
    # array to store all closest distances
    shortest_dist = []

    # radius of earth (used for calculation)
    R = 6373.0
    for i in df2.index:
        # keeps track of current minimum distance
        min_dist = -1

        # latitude and longitude from df2
        lat1 = df2.ix[i]['Latitude']
        lon1 = df2.ix[i]['Longitude']

        for j in df.index:

            # the following is just the calculation necessary
            # to calculate the distance between each point in km
            lat2 = df.ix[j]['Latitude']
            lon2 = df.ix[j]['Longitude']
            dlon = lon2 - lon1
            dlat = lat2 - lat1
            a = sin(dlat / 2)**2 + cos(lat1) * cos(lat2) * sin(dlon / 2)**2
            c = 2 * atan2(sqrt(a), sqrt(1 - a))
            distance = R * c

            # store new shortest distance
            if min_dist == -1 or distance > min_dist:
                min_dist = distance
        # append shortest distance to array
        shortest_dist.append(min_dist)

这个函数的计算时间太长，我知道必须有更有效的方法，但我不太擅长pandas 语法。

感谢您的帮助。

Answer 1

您可以在numpy 中编写内部循环，这应该会大大加快速度：

import numpy as np

def compute_shortest_dist(df, df2):
    # array to store all closest distances
    shortest_dist = []

    # radius of earth (used for calculation)
    R = 6373.0
    lat1 = df['Latitude']
    lon1 = df['Longitude']
    for i in df2.index:
        # the following is just the calculation necessary
        # to calculate the distance between each point in km
        lat2 = df2.loc[i, 'Latitude']
        dlat = lat1 - lat2
        dlon = lon1 - df2.loc[i, 'Longitude']
        a = np.sin(dlat / 2)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon / 2)**2
        distance = 2* R * np.arctan2(np.sqrt(a), np.sqrt(1 - a))

        # append shortest distance to array
        shortest_dist.append(distance.min())
    return shortest_dist