如何在groupby数据框上应用功能
给出数据框df。
userid trip_id lat long
141.0 1.0 39.979547 116.306813
141.0 1.0 39.979558 116.306823
141.0 1.0 39.979575 116.306835
141.0 1.0 39.979587 116.306847
141.0 2.0 39.979603 116.306852
141.0 2.0 39.979612 116.306867
141.0 2.0 39.979627 116.306877
141.0 2.0 39.979635 116.306888
141.0 3.0 39.979645 116.306903
141.0 3.0 39.979657 116.306913
141.0 3.0 39.979670 116.306920
141.0 3.0 39.979682 116.306920
我想为每组数据框计算Vincenty距离。数据框分为两列，即（userid，trip_id）
我可以通过给定的语句计算整个数据帧的Vincenty距离
from geopy.distance import vincenty
df['lat_next'] = df['lat'].shift(-1)
df['long_next'] = df['long'].shift(-1)
df['Vincenty_distance'] = df.dropna().apply(lambda x: vincenty((x['lat'], x['long']), (x['lat_next'], x['long_next'])).meters, axis = 1)
df = df.drop(['lat_next','long_next'], axis=1)
我想将此功能应用于每个组，我尝试使用此语句但出现错误。
df['Vincenty_distance'] = df.dropna().groupby(['userid','trip_id']).apply(lambda x: vincenty((x['lat'], x['long']), (x['lat_next'], x['long_next'])).meters,axis=1)
我期待以下结果。
userid trip_id lat long Vincenty_distance
141.0 1.0 39.979547 116.306813 2.563812
141.0 1.0 39.979558 116.306823 2.956183
141.0 1.0 39.979575 116.306835 2.332577
141.0 1.0 39.979587 116.306847 Nan
141.0 2.0 39.979603 116.306852 2.334821
141.0 2.0 39.979612 116.306867 2.332577
141.0 2.0 39.979627 116.306877 1.695449
141.0 2.0 39.979635 116.306888 Nan
141.0 3.0 39.979645 116.306903 1.871784
141.0 3.0 39.979657 116.306913 1.982752
141.0 3.0 39.979670 116.306920 2.220685
141.0 3.0 39.979682 116.306920 Nan
答案 0 :(得分：1)
我认为您首先需要DataFrameGroupBy.shift来进行next列的每组转换，因此groupby和vincenty并不是必需的：
df = df.join(df.groupby(['userid','trip_id'])[['lat','long']].shift(-1).add_suffix('_next'))
print (df)
userid trip_id lat long lat_next long_next
0 141.0 1.0 39.979547 116.306813 39.979558 116.306823
1 141.0 1.0 39.979558 116.306823 39.979575 116.306835
2 141.0 1.0 39.979575 116.306835 39.979587 116.306847
3 141.0 1.0 39.979587 116.306847 NaN NaN
4 141.0 2.0 39.979603 116.306852 39.979612 116.306867
5 141.0 2.0 39.979612 116.306867 39.979627 116.306877
6 141.0 2.0 39.979627 116.306877 39.979635 116.306888
7 141.0 2.0 39.979635 116.306888 NaN NaN
8 141.0 3.0 39.979645 116.306903 39.979657 116.306913
9 141.0 3.0 39.979657 116.306913 39.979670 116.306920
10 141.0 3.0 39.979670 116.306920 39.979682 116.306920
11 141.0 3.0 39.979682 116.306920 NaN NaN
f = lambda x: vincenty((x['lat'], x['long']), (x['lat_next'], x['long_next'])).meters
df['Vincenty_distance'] = df.dropna().apply(f, axis = 1)
df = df.drop(['lat_next','long_next'], axis=1)
print (df)
userid trip_id lat long Vincenty_distance
0 141.0 1.0 39.979547 116.306813 1.490437
1 141.0 1.0 39.979558 116.306823 2.147940
2 141.0 1.0 39.979575 116.306835 1.681071
3 141.0 1.0 39.979587 116.306847 NaN
4 141.0 2.0 39.979603 116.306852 1.624902
5 141.0 2.0 39.979612 116.306867 1.871784
6 141.0 2.0 39.979627 116.306877 1.293017
7 141.0 2.0 39.979635 116.306888 NaN
8 141.0 3.0 39.979645 116.306903 1.582706
9 141.0 3.0 39.979657 116.306913 1.562388
10 141.0 3.0 39.979670 116.306920 1.332411
11 141.0 3.0 39.979682 116.306920 NaN
答案 1 :(得分：0)
查看此示例：
>>>
>>> d=pd.DataFrame([[1,2,3],[1,2,1],[2,3,4],[2,3,2],[3,4,5],[3,4,3]],columns=['a
','b','c'])
>>> d
a b c
0 1 2 3
1 1 2 1
2 2 3 4
3 2 3 2
4 3 4 5
5 3 4 3
>>> def gr(grp):
... grp['c_next']=grp['c'].shift(-1)
... grp.fillna(0, inplace=True)
... ####You can have your own operation here
... grp['c_dist']=grp['c_next']-grp['c']
... return grp
...
>>> d.groupby(['a','b']).apply(gr)
a b c c_next c_dist
0 1 2 3 1.0 -2.0
1 1 2 1 0.0 -1.0
2 2 3 4 2.0 -2.0
3 2 3 2 0.0 -2.0
4 3 4 5 3.0 -2.0
5 3 4 3 0.0 -3.0
>>>