pandas 熊猫经纬度到连续行之间的距离
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原文地址: http://stackoverflow.com/questions/40452759/
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Pandas Latitude-Longitude to distance between successive rows
提问by edesz
I have the following in a Pandas DataFrame in Python 2.7:
我在 Python 2.7 的 Pandas DataFrame 中有以下内容：
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
I am looking to calculate the distance between successive rows in the dataframe. The output should look something like this:
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 .
Attempt
试图
This postlooks somewhat similar but it is calculating the distance between fixed points. I need the distance between successive points.
这篇文章看起来有些相似，但它正在计算固定点之间的距离。我需要连续点之间的距离。
I tried to adapt this as follows:
我尝试将其调整如下：
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))
However, I get the following error:
但是，我收到以下错误：
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]
This error was fixed from MaxU's comment. With the fix, the output of this calculation is not making sense - the distance is nearly 8000 km:
此错误已从 MaxU 的评论中修复。修复后，此计算的输出没有意义 - 距离接近 8000 公里：
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
According to:
根据：
- this online calculator: If I use Latitude1 = 74.166061, Longitude1 = 30.512811, Latitude2 = 72.249672, Longitude2 = 33.427724 then I get 233 km
- haversine function found
hereas:
print haversine(30.512811, 74.166061, 33.427724, 72.249672)then I get 232.55 km
- 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
print haversine(30.512811, 74.166061, 33.427724, 72.249672)找到的半正弦函数 为：然后我得到 232.55 公里
The answer should be 233 km, but my approach is giving ~8000 km. I think there is something wrong with how I am trying to iterate between successive rows.
答案应该是 233 公里，但我的方法是给出 ~8000 公里。我认为我尝试在连续行之间进行迭代的方式有问题。
Question:Is there a way to do this in Pandas? Or do I need to loop through the dataframe one row at a time?
问题：有没有办法在 Pandas 中做到这一点？或者我是否需要一次遍历数据帧一行？
Additional Information:
附加信息：
To create the above DF, select it and copy to clipboard. Then:
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
回答by MaxU
you can use this great solution (c) @derricw(don't forget to upvote it ;-):
你可以使用这个很棒的解决方案（c）@derricw（不要忘记给它点赞；-）：
# 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'])
Result:
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
UPDATE:this will help to understand the logic:
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
