Pandas 纬度-经度到连续行之间的距离
Pandas Latitude-Longitude to distance between successive rows
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
我想计算数据框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 .
尝试
This post看起来有点像，但它是计算固定点之间的距离。我需要连续点之间的距离。
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))
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]
此错误已根据 MaxU 的评论修复。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
根据：
- 这个online calculator：如果我使用 Latitude1 = 74.166061，
经度 1 = 30.512811，纬度 2 = 72.249672，经度 2 = 33.427724
然后我得到 233 公里
- 已找到正弦函数
here 作为：
print haversine(30.512811, 74.166061, 33.427724, 72.249672) 那么我
达到 232.55 公里
答案应该是 233 公里，但我的方法是 ~8000 公里。我认为我尝试在连续行之间迭代的方式有问题。
问题：
在 Pandas 中有没有办法做到这一点？或者我是否需要一次一行地遍历数据框？
附加信息：
要创建上述 DF，select 并将其复制到剪贴板。那么：
import pandas as pd
df = pd.read_clipboard()
print df
你可以使用 this great solution (c) @derricw（别忘了点赞 ;-):
# vectorized haversine function
def haversine(lat1, lon1, lat2, lon2, to_radians=True, earth_radius=6371):
"""
slightly modified version: of
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
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
我想计算数据框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 .
尝试
This post看起来有点像，但它是计算固定点之间的距离。我需要连续点之间的距离。
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))
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]
此错误已根据 MaxU 的评论修复。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
根据：
- 这个online calculator：如果我使用 Latitude1 = 74.166061， 经度 1 = 30.512811，纬度 2 = 72.249672，经度 2 = 33.427724 然后我得到 233 公里
- 已找到正弦函数
here 作为：
print haversine(30.512811, 74.166061, 33.427724, 72.249672)那么我 达到 232.55 公里
答案应该是 233 公里，但我的方法是 ~8000 公里。我认为我尝试在连续行之间迭代的方式有问题。
问题： 在 Pandas 中有没有办法做到这一点？或者我是否需要一次一行地遍历数据框？
附加信息：
要创建上述 DF，select 并将其复制到剪贴板。那么：
import pandas as pd
df = pd.read_clipboard()
print df
你可以使用 this great solution (c) @derricw（别忘了点赞 ;-):
# vectorized haversine function
def haversine(lat1, lon1, lat2, lon2, to_radians=True, earth_radius=6371):
"""
slightly modified version: of
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