1

我想计算两个gps坐标之间的距离（每个tripId的第一个和最后一个），以获得每次旅行的距离我的数据框看起来像这样

   tripId   latitude    longitude   timestamp
0   1817603 53.155273   8.207176    2021-05-24 00:29:22
1   1817603 53.155271   8.206898    2021-05-24 00:29:38
2   1817603 53.155213   8.206314    2021-05-24 00:29:44
3   1817603 53.155135   8.206429    2021-05-24 00:29:50
4   1817603 53.154950   8.206565    2021-05-24 00:29:56
... ... ... ... ...
195 1817888 53.092805   8.212095    2021-05-24 08:27:54
196 1817888 53.093024   8.211756    2021-05-24 08:27:59
197 1817888 53.093305   8.211383    2021-05-24 08:28:05
198 1817888 53.093594   8.211026    2021-05-24 08:28:10
199 1817888 53.093853   8.210708    2021-05-24 08:28:15

我对每个步骤都这样做了，s = pd.Series(haversine_vector(df, df.shift(),Unit.KILOMETERS), index=df.index, name='distance_K') 但我需要知道每个 ID 的整个行程的距离我用这个作为测试并且它有效，但我需要知道每次行程的确切持续时间（最终持续时间）

for i in range(1,df.shape[0]-1):
    if df['tripId'][i]==df['tripId'][i+1]:
        df['distance'][i]=df['distance'][i-1]+df['distance_K'][i]
    else:
        df['distance'][i]=df['distance_K'][i]
4

2 回答 2

4

用于groupby_apply计算每次行程的偏正距离：

# Inspired by https://stackoverflow.com/a/4913653/15239951
def haversine_series(sr):
    lon1 = sr['longitude']
    lat1 = sr['latitude']
    lon2 = sr['longitude'].shift(fill_value=sr['longitude'].iloc[0])
    lat2 = sr['latitude'].shift(fill_value=sr['latitude'].iloc[0])
    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 = 6371 * c
    return km

df['distance_K'] = df.groupby('tripId').apply(haversine_series).droplevel(0)

注意：我想您的数据框已经按timestamp列排序。

此时，您的数据框如下所示：

>>> df
      tripId   latitude  longitude            timestamp  distance_K
0    1817603  53.155273   8.207176  2021-05-24 00:29:22    0.000000
1    1817603  53.155271   8.206898  2021-05-24 00:29:38    0.018538
2    1817603  53.155213   8.206314  2021-05-24 00:29:44    0.039470
3    1817603  53.155135   8.206429  2021-05-24 00:29:50    0.011577
4    1817603  53.154950   8.206565  2021-05-24 00:29:56    0.022481
195  1817888  53.092805   8.212095  2021-05-24 08:27:54    0.000000
196  1817888  53.093024   8.211756  2021-05-24 08:27:59    0.033248
197  1817888  53.093305   8.211383  2021-05-24 08:28:05    0.039958
198  1817888  53.093594   8.211026  2021-05-24 08:28:10    0.040012
199  1817888  53.093853   8.210708  2021-05-24 08:28:15    0.035781

现在很容易得到每次旅行的总距离和时间groupby_agg：

>>> df.groupby('tripId') \
      .agg(total_distance=('distance_K', 'sum'), 
           total_time=('timestamp', lambda x: x.max()-x.min())) \
      .reset_index()

    tripId  total_distance      total_time
0  1817603        0.092066 0 days 00:00:34
1  1817888        0.148999 0 days 00:00:21
于 2021-11-04T22:25:49.653 回答
2

您可以使用

from haversine import haversine_vector

df = df.groupby('tripId').apply(
    lambda g: g.assign(distance=lambda g: [0, *haversine_vector(
        g.iloc[:-1][['latitude', 'longitude']].values,
        g.iloc[1:][['latitude', 'longitude']].values,
    )])
).droplevel(0)
df
#     tripId   latitude  longitude                timestamp  distance
# 0  1817603  53.155273   8.207176      2021-05-24 00:29:22  0.000000
# 1  1817603  53.155271   8.206898      2021-05-24 00:29:38  0.018538
# 2  1817603  53.155213   8.206314      2021-05-24 00:29:44  0.039470
# 3  1817603  53.155135   8.206429      2021-05-24 00:29:50  0.011577
# 4  1817603  53.154950   8.206565      2021-05-24 00:29:56  0.022481
# 5  1817888  53.092805   8.212095      2021-05-24 08:27:54  0.000000
# 6  1817888  53.093024   8.211756      2021-05-24 08:27:59  0.033248
# 7  1817888  53.093305   8.211383      2021-05-24 08:28:05  0.039958
# 8  1817888  53.093594   8.211026      2021-05-24 08:28:10  0.040012
# 9  1817888  53.093853   8.210708      2021-05-24 08:28:15  0.035781

并获得总时间和距离

df.groupby('tripId').agg(
    {
        'timestamp': lambda g: g.iloc[-1] - g.iloc[0],
        'distance':'sum'
    }
)
#               timestamp  distance
# tripId                           
# 1817603 0 days 00:00:34  0.092066
# 1817888 0 days 00:00:21  0.148999
于 2021-11-04T22:36:25.397 回答