快速的Haversine逼近(Python/Pandas)

Question

Pandas数据帧中的每一行包含2个点的lat/lng坐标.使用下面的Python代码,计算许多(数百万)行的这两个点之间的距离需要很长时间!

考虑到2点相距不到50英里并且准确性不是很重要,是否可以更快地进行计算？

from math import radians, cos, sin, asin, sqrt
def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(radians, [lon1, lat1, lon2, lat2])
    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = sin(dlat/2)**2 + cos(lat1) * cos(lat2) * sin(dlon/2)**2
    c = 2 * asin(sqrt(a)) 
    km = 6367 * c
    return km


for index, row in df.iterrows():
    df.loc[index, 'distance'] = haversine(row['a_longitude'], row['a_latitude'], row['b_longitude'], row['b_latitude'])

Answer 1

这是相同功能的矢量化numpy版本:

import numpy as np

def haversine_np(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees)

    All args must be of equal length.    

    """
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])

    dlon = lon2 - lon1
    dlat = lat2 - lat1

    a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2

    c = 2 * np.arcsin(np.sqrt(a))
    km = 6367 * c
    return km

输入都是值的数组,它应该能够立即完成数百万个点.要求是输入是ndarray但是pandas表的列将起作用.

例如,随机生成的值:

>>> import numpy as np
>>> import pandas
>>> lon1, lon2, lat1, lat2 = np.random.randn(4, 1000000)
>>> df = pandas.DataFrame(data={'lon1':lon1,'lon2':lon2,'lat1':lat1,'lat2':lat2})
>>> km = haversine_np(df['lon1'],df['lat1'],df['lon2'],df['lat2'])

在python中循环遍历数据数组非常慢.Numpy提供了对整个数据数组进行操作的函数,可以避免循环并大幅提高性能.

这是矢量化的一个例子.

Answer 2

纯粹是为了一个说明性的例子,我numpy从@ballsdotballs的答案中获取了该版本,并且还通过调用了一个伴随的C实现ctypes.由于这numpy是一个高度优化的工具,我的C代码几乎不可能有效,但它应该有点接近.这里的一大优势是通过运行C类型的示例,它可以帮助您了解如何将自己的个人C函数连接到Python而不会产生太多开销.当你只想通过在一些C源而不是Python中编写那个小块来优化一小块更大的计算时,这是特别好的.简单地使用numpy将在大多数情况下解决问题,但是对于那些你真的不需要全部numpy并且你不想numpy在一些代码中添加耦合以要求使用数据类型的情况,知道如何使用它是非常方便的.下载到内置ctypes库并自己动手.

首先让我们创建我们的C源文件,名为haversine.c:

#include <stdlib.h>
#include <stdio.h>
#include <math.h>

int haversine(size_t n, 
              double *lon1, 
              double *lat1, 
              double *lon2, 
              double *lat2,
              double *kms){

    if (   lon1 == NULL 
        || lon2 == NULL 
        || lat1 == NULL 
        || lat2 == NULL
        || kms == NULL){
        return -1;
    }

    double km, dlon, dlat;
    double iter_lon1, iter_lon2, iter_lat1, iter_lat2;

    double km_conversion = 2.0 * 6367.0; 
    double degrees2radians = 3.14159/180.0;

    int i;
    for(i=0; i < n; i++){
        iter_lon1 = lon1[i] * degrees2radians;
        iter_lat1 = lat1[i] * degrees2radians;
        iter_lon2 = lon2[i] * degrees2radians;
        iter_lat2 = lat2[i] * degrees2radians;

        dlon = iter_lon2 - iter_lon1;
        dlat = iter_lat2 - iter_lat1;

        km = pow(sin(dlat/2.0), 2.0) 
           + cos(iter_lat1) * cos(iter_lat2) * pow(sin(dlon/2.0), 2.0);

        kms[i] = km_conversion * asin(sqrt(km));
    }

    return 0;
}

// main function for testing
int main(void) {
    double lat1[2] = {16.8, 27.4};
    double lon1[2] = {8.44, 1.23};
    double lat2[2] = {33.5, 20.07};
    double lon2[2] = {14.88, 3.05};
    double kms[2]  = {0.0, 0.0};
    size_t arr_size = 2;

    int res;
    res = haversine(arr_size, lon1, lat1, lon2, lat2, kms);
    printf("%d\n", res);

    int i;
    for (i=0; i < arr_size; i++){
        printf("%3.3f, ", kms[i]);
    }
    printf("\n");
}

请注意,我们正在努力遵守C约定.通过引用显式传递数据参数,使用size_t大小变量,并期望我们的haversine函数通过改变其中一个传递的输入来工作,这样它将在退出时包含预期的数据.该函数实际上返回一个整数,这是一个成功/失败标志,可供该函数的其他C级使用者使用.

我们需要找到一种方法来处理Python内部所有这些特定于C的小问题.

接下来让我们将我们numpy的函数版本以及一些导入和一些测试数据放入一个名为的文件中haversine.py:

import time
import ctypes
import numpy as np
from math import radians, cos, sin, asin, sqrt

def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])
    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = (np.sin(dlat/2)**2 
         + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2)
    c = 2 * np.arcsin(np.sqrt(a)) 
    km = 6367 * c
    return km

if __name__ == "__main__":
    lat1 = 50.0 * np.random.rand(1000000)
    lon1 = 50.0 * np.random.rand(1000000)
    lat2 = 50.0 * np.random.rand(1000000)
    lon2 = 50.0 * np.random.rand(1000000)

    t0 = time.time()
    r1 = haversine(lon1, lat1, lon2, lat2)
    t1 = time.time()
    print t1-t0, r1

我选择制作在0到50之间随机选择的lats和lons(以度为单位),但这对于这个解释并不重要.

接下来我们需要做的是以可以由Python动态加载的方式编译我们的C模块.我正在使用Linux系统(您可以在Google上轻松找到其他系统的示例),因此我的目标是编译haversine.c成共享对象,如下所示:

gcc -shared -o haversine.so -fPIC haversine.c -lm

我们还可以编译成可执行文件并运行它以查看C程序的main功能显示的内容:

> gcc haversine.c -o haversine -lm
> ./haversine
0
1964.322, 835.278, 

现在我们已经编译了共享对象haversine.so,我们可以使用ctypes它来加载它,我们需要提供文件的路径来执行此操作:

lib_path = "/path/to/haversine.so" # Obviously use your real path here.
haversine_lib = ctypes.CDLL(lib_path)

现在haversine_lib.haversine几乎就像Python函数一样,除了我们可能需要做一些手动类型封送以确保输入和输出被正确解释.

numpy实际上为此提供了一些很好的工具,我将在这里使用的工具numpy.ctypeslib.我们将构建一个指针类型,它允许我们传递numpy.ndarrays给这些ctypes加载的函数,就像它们是指针一样.这是代码:

arr_1d_double = np.ctypeslib.ndpointer(dtype=np.double, 
                                       ndim=1, 
                                       flags='CONTIGUOUS')

haversine_lib.haversine.restype = ctypes.c_int
haversine_lib.haversine.argtypes = [ctypes.c_size_t,
                                    arr_1d_double, 
                                    arr_1d_double,
                                    arr_1d_double,
                                    arr_1d_double,
                                    arr_1d_double] 

请注意,我们告诉haversine_lib.haversine函数代理根据我们想要的类型解释其参数.

现在,要从Python中测试它剩下的只是创建一个大小变量,以及一个将被变异的数组(就像在C代码中一样)来包含结果数据,然后我们可以调用它:

size = len(lat1)
output = np.empty(size, dtype=np.double)
print "====="
print output
t2 = time.time()
res = haversine_lib.haversine(size, lon1, lat1, lon2, lat2, output)
t3 = time.time()
print t3 - t2, res
print type(output), output

全部放在一起的__main__块haversine.py,整个文件现在看起来是这样的:

import time
import ctypes
import numpy as np
from math import radians, cos, sin, asin, sqrt

def haversine(lon1, lat1, lon2, lat2):
    """
    Calculate the great circle distance between two points 
    on the earth (specified in decimal degrees)
    """
    # convert decimal degrees to radians 
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])
    # haversine formula 
    dlon = lon2 - lon1 
    dlat = lat2 - lat1 
    a = (np.sin(dlat/2)**2 
         + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2)**2)
    c = 2 * np.arcsin(np.sqrt(a)) 
    km = 6367 * c
    return km

if __name__ == "__main__":
    lat1 = 50.0 * np.random.rand(1000000)
    lon1 = 50.0 * np.random.rand(1000000)
    lat2 = 50.0 * np.random.rand(1000000)
    lon2 = 50.0 * np.random.rand(1000000)

    t0 = time.time()
    r1 = haversine(lon1, lat1, lon2, lat2)
    t1 = time.time()
    print t1-t0, r1

    lib_path = "/home/ely/programming/python/numpy_ctypes/haversine.so"
    haversine_lib = ctypes.CDLL(lib_path)
    arr_1d_double = np.ctypeslib.ndpointer(dtype=np.double, 
                                           ndim=1, 
                                           flags='CONTIGUOUS')

    haversine_lib.haversine.restype = ctypes.c_int
    haversine_lib.haversine.argtypes = [ctypes.c_size_t,
                                        arr_1d_double, 
                                        arr_1d_double,
                                        arr_1d_double,
                                        arr_1d_double,
                                        arr_1d_double]

    size = len(lat1)
    output = np.empty(size, dtype=np.double)
    print "====="
    print output
    t2 = time.time()
    res = haversine_lib.haversine(size, lon1, lat1, lon2, lat2, output)
    t3 = time.time()
    print t3 - t2, res
    print type(output), output

要运行它,它将分别运行和定时Python和ctypes版本并打印一些结果,我们可以这样做

python haversine.py

显示:

0.111340045929 [  231.53695005  3042.84915093   169.5158946  ...,  1359.2656769
  2686.87895954  3728.54788207]
=====
[  6.92017600e-310   2.97780954e-316   2.97780954e-316 ...,
   3.20676686e-001   1.31978329e-001   5.15819721e-001]
0.148446083069 0
<type 'numpy.ndarray'> [  231.53675618  3042.84723579   169.51575588 ...,  1359.26453029
  2686.87709456  3728.54493339]

正如预期的那样,numpy版本稍微快一些(长度为1百万的向量为0.11秒),但我们快速而肮脏的ctypes版本并不懈怠:相同数据上的0.148秒可观.

让我们将它与Python中的一个天真的for循环解决方案进行比较:

from math import radians, cos, sin, asin, sqrt

def slow_haversine(lon1, lat1, lon2, lat2):
    n = len(lon1)
    kms = np.empty(n, dtype=np.double)
    for i in range(n):
       lon1_v, lat1_v, lon2_v, lat2_v = map(
           radians, 
           [lon1[i], lat1[i], lon2[i], lat2[i]]
       )

       dlon = lon2_v - lon1_v 
       dlat = lat2_v - lat1_v 
       a = (sin(dlat/2)**2 
            + cos(lat1_v) * cos(lat2_v) * sin(dlon/2)**2)
       c = 2 * asin(sqrt(a)) 
       kms[i] = 6367 * c
    return kms

当我把它放在与其他文件相同的Python文件中并将它计入相同的百万元素数据时,我一直在机器上看到大约2.65秒的时间.

因此,通过快速切换到ctypes我们将速度提高了大约18倍.对于许多可以从访问裸,连续数据中获益的计算,您经常会看到比这更高的收益.

只是要非常清楚,我并不认为这是一个比使用更好的选择numpy.这正是为解决问题numpy而构建的问题,因此ctypes无论何时(a)numpy在您的应用程序中合并数据类型是有意义的(b)存在一种将代码映射到numpy等效代码的简单方法,因此自行编写自己的代码.不是很有效率.

但是,如果你喜欢用C语言写一些东西但是用Python调用它,或者依赖于numpy不实用的情况(例如在一个numpy无法安装的嵌入式系统中),知道如何做到这一点仍然非常有用..

Answer 3

如果允许使用scikit-learn,我会给出以下机会:

from sklearn.neighbors import DistanceMetric
dist = DistanceMetric.get_metric('haversine')

# example data
lat1, lon1 = 36.4256345, -5.1510261
lat2, lon2 = 40.4165, -3.7026
lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])

X = [[lat1, lon1],
     [lat2, lon2]]
kms = 6367
print(kms * dist.pairwise(X))

Answer 4

对@derricw向量化解决方案的一个简单扩展，您可以使用它numba来将性能提高约 2倍，而几乎不更改您的代码。对于纯数值计算，这可能应该用于基准测试/测试，而不是可能更有效的解决方案。

from numba import njit

@njit
def haversine_nb(lon1, lat1, lon2, lat2):
    lon1, lat1, lon2, lat2 = np.radians(lon1), np.radians(lat1), np.radians(lon2), np.radians(lat2)
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2
    return 6367 * 2 * np.arcsin(np.sqrt(a))

基准测试与 Pandas 函数：

%timeit haversine_pd(df['lon1'], df['lat1'], df['lon2'], df['lat2'])
# 1 loop, best of 3: 1.81 s per loop

%timeit haversine_nb(df['lon1'].values, df['lat1'].values, df['lon2'].values, df['lat2'].values)
# 1 loop, best of 3: 921 ms per loop

完整的基准测试代码：

import pandas as pd, numpy as np
from numba import njit

def haversine_pd(lon1, lat1, lon2, lat2):
    lon1, lat1, lon2, lat2 = map(np.radians, [lon1, lat1, lon2, lat2])
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2
    return 6367 * 2 * np.arcsin(np.sqrt(a))

@njit
def haversine_nb(lon1, lat1, lon2, lat2):
    lon1, lat1, lon2, lat2 = np.radians(lon1), np.radians(lat1), np.radians(lon2), np.radians(lat2)
    dlon = lon2 - lon1
    dlat = lat2 - lat1
    a = np.sin(dlat/2.0)**2 + np.cos(lat1) * np.cos(lat2) * np.sin(dlon/2.0)**2
    return 6367 * 2 * np.arcsin(np.sqrt(a))

np.random.seed(0)
lon1, lon2, lat1, lat2 = np.random.randn(4, 10**7)
df = pd.DataFrame(data={'lon1':lon1,'lon2':lon2,'lat1':lat1,'lat2':lat2})
km = haversine_pd(df['lon1'], df['lat1'], df['lon2'], df['lat2'])
km_nb = haversine_nb(df['lon1'].values, df['lat1'].values, df['lon2'].values, df['lat2'].values)

assert np.isclose(km.values, km_nb).all()

%timeit haversine_pd(df['lon1'], df['lat1'], df['lon2'], df['lat2'])
# 1 loop, best of 3: 1.81 s per loop

%timeit haversine_nb(df['lon1'].values, df['lat1'].values, df['lon2'].values, df['lat2'].values)
# 1 loop, best of 3: 921 ms per loop