我有一个数据框,如下所示:
ID Date
16911 2017-04-15
16911 2017-04-25
16911 2017-04-27
16911 2017-05-08
16911 2017-05-20
16911 2017-05-25
16911 2017-08-08
16911 2017-08-11
16911 2017-08-24
16912 2017-04-15
16912 2017-04-25
16812 2017-04-27
16812 2017-05-08
16812 2017-05-20
16812 2017-05-25
16812 2017-08-08
16812 2017-08-11
日期已排序,我想找到时间戳之间的差异并找到每个ID的平均值。
另外,
假设对于ID-16911,我想要日期差异列表例如 - >列出a;
16911 2017-04-15
16911 2017-04-25
difference between the above two dates is 10, so a is
a = [10]
16911 2017-04-25
16911 2017-04-27
difference between the above two dates is 2, so a is
a=[10,2]
16911 2017-04-27
16911 2017-05-08
difference between the above two dates is 11(assuming), so a is
a=[10,2,11]
所以最终的输出应该是:
ID Average_Day Diff
16911 3 days [10,2,11]
答案 0 :(得分:4)
df = df.groupby('ID')['Date'].apply(lambda x: x.diff().mean()).reset_index()
print (df)
ID Date
0 16812 21 days 04:48:00
1 16911 16 days 09:00:00
2 16912 10 days 00:00:00
如果需要转换timedeltas,例如到days
:
df = df.groupby('ID')['Date'].apply(lambda x: x.diff().mean().days).reset_index()
print (df)
ID Date
0 16812 21
1 16911 16
2 16912 10
编辑:
#create difference column per ID
df['new'] = df.groupby('ID')['Date'].diff().dt.days
#remove NaT rows (first for each group)
df = df.dropna(subset=['new'])
#convert to integers
df['new'] = df['new'].astype(int)
#aggreagte lists and mean
df = df.groupby('ID', sort=False)['new'].agg([('val', lambda x: x.tolist()),('avg', 'mean')])
print (df)
ID
16911 [10, 2, 11, 12, 5, 75, 3, 13] 16.375
16912 [10] 10.000
16812 [11, 12, 5, 75, 3] 21.200