熊猫经纬度到连续行之间的距离

Question

我在Python 2.7中的Pandas DataFrame中具有以下内容：

Ser_Numb        LAT      LONG
       1  74.166061 30.512811
       2  72.249672 33.427724
       3  67.499828 37.937264
       4  84.253715 69.328767
       5  72.104828 33.823462
       6  63.989462 51.918173
       7  80.209112 33.530778
       8  68.954132 35.981256
       9  83.378214 40.619652
       10 68.778571 6.607066

我正在寻找计算数据帧中连续行之间的距离。输出应如下所示：

Ser_Numb          LAT        LONG   Distance
       1    74.166061   30.512811          0
       2    72.249672   33.427724          d_between_Ser_Numb2 and Ser_Numb1
       3    67.499828   37.937264          d_between_Ser_Numb3 and Ser_Numb2
       4    84.253715   69.328767          d_between_Ser_Numb4 and Ser_Numb3
       5    72.104828   33.823462          d_between_Ser_Numb5 and Ser_Numb4
       6    63.989462   51.918173          d_between_Ser_Numb6 and Ser_Numb5
       7    80.209112   33.530778   .
       8    68.954132   35.981256   .
       9    83.378214   40.619652   .
       10   68.778571   6.607066    .

我尝试如下进行调整：

df['LAT_rad'], df['LON_rad'] = np.radians(df['LAT']), np.radians(df['LONG'])
df['dLON'] = df['LON_rad'] - np.radians(df['LON_rad'].shift(1))
df['dLAT'] = df['LAT_rad'] - np.radians(df['LAT_rad'].shift(1))
df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin(df['dLAT']/2)**2 + math.cos(df['LAT_rad'].astype(float).shift(-1)) * np.cos(df['LAT_rad']) * np.sin(df['dLON']/2)**2))

但是，出现以下错误：

Traceback (most recent call last):
  File "C:\Python27\test.py", line 115, in <module>
    df['distance'] = 6367 * 2 * np.arcsin(np.sqrt(np.sin(df['dLAT']/2)**2 + math.cos(df['LAT_rad'].astype(float).shift(-1)) * np.cos(df['LAT_rad']) * np.sin(df['dLON']/2)**2))
  File "C:\Python27\lib\site-packages\pandas\core\series.py", line 78, in wrapper
    "{0}".format(str(converter)))
TypeError: cannot convert the series to <type 'float'>
[Finished in 2.3s with exit code 1]

此错误已通过MaxU的注释修复。修复后，此计算的输出没有意义-距离近8000 km：

   Ser_Numb        LAT       LONG   LAT_rad   LON_rad      dLON      dLAT     distance
0         1  74.166061  30.512811  1.294442  0.532549       NaN       NaN          NaN
1         2  72.249672  33.427724  1.260995  0.583424  0.574129  1.238402  8010.487211
2         3  67.499828  37.937264  1.178094  0.662130  0.651947  1.156086  7415.364469
3         4  84.253715  69.328767  1.470505  1.210015  1.198459  1.449943  9357.184623
4         5  72.104828  33.823462  1.258467  0.590331  0.569212  1.232802  7992.087820
5         6  63.989462  51.918173  1.116827  0.906143  0.895840  1.094862  7169.812123
6         7  80.209112  33.530778  1.399913  0.585222  0.569407  1.380421  8851.558260
7         8  68.954132  35.981256  1.203477  0.627991  0.617777  1.179044  7559.609520
8         9  83.378214  40.619652  1.455224  0.708947  0.697986  1.434220  9194.371978
9        10  68.778571   6.607066  1.200413  0.115315  0.102942  1.175014          NaN

根据：

• 这个在线计算器：如果我使用Latitude1 = 74.166061，Longitude1 = 30.512811，Latitude2 = 72.249672，Longitude2 = 33.427724，那么我得到233公里
• 在这里发现haversine函数 为：print haversine(30.512811, 74.166061, 33.427724, 72.249672)然后我得到232.55公里

答案应该是233公里，但我的方法是给出约8000公里。我认为我试图在连续的行之间进行迭代存在问题。

问题： 在熊猫中有没有办法做到这一点？还是我需要一次遍历数据帧一行？

附加信息：

要创建上述DF，请选择它并复制到剪贴板。然后：

import pandas as pd
df = pd.read_clipboard()
print df

Answer 1

您可以使用此出色的解决方案（c）@derricw（不要忘记对其进行投票;-)：

# vectorized haversine function
def haversine(lat1, lon1, lat2, lon2, to_radians=True, earth_radius=6371):
    """
    slightly modified version: of http://stackoverflow.com/a/29546836/2901002

    Calculate the great circle distance between two points
    on the earth (specified in decimal degrees or in radians)

    All (lat, lon) coordinates must have numeric dtypes and be of equal length.

    """
    if to_radians:
        lat1, lon1, lat2, lon2 = np.radians([lat1, lon1, lat2, lon2])

    a = np.sin((lat2-lat1)/2.0)**2 + \
        np.cos(lat1) * np.cos(lat2) * np.sin((lon2-lon1)/2.0)**2

    return earth_radius * 2 * np.arcsin(np.sqrt(a))


df['dist'] = \
    haversine(df.LAT.shift(), df.LONG.shift(),
                 df.loc[1:, 'LAT'], df.loc[1:, 'LONG'])

结果：

In [566]: df
Out[566]:
   Ser_Numb        LAT       LONG         dist
0         1  74.166061  30.512811          NaN
1         2  72.249672  33.427724   232.549785
2         3  67.499828  37.937264   554.905446
3         4  84.253715  69.328767  1981.896491
4         5  72.104828  33.823462  1513.397997
5         6  63.989462  51.918173  1164.481327
6         7  80.209112  33.530778  1887.256899
7         8  68.954132  35.981256  1252.531365
8         9  83.378214  40.619652  1606.340727
9        10  68.778571   6.607066  1793.921854

更新： 这将有助于理解逻辑：

In [573]: pd.concat([df['LAT'].shift(), df.loc[1:, 'LAT']], axis=1, ignore_index=True)
Out[573]:
           0          1
0        NaN        NaN
1  74.166061  72.249672
2  72.249672  67.499828
3  67.499828  84.253715
4  84.253715  72.104828
5  72.104828  63.989462
6  63.989462  80.209112
7  80.209112  68.954132
8  68.954132  83.378214
9  83.378214  68.778571