假设我有这个数据帧，我希望每个唯一的用户ID都有自己的基于日期戳的排名值：In [93]:

df = pd.DataFrame({

'userid':['a', 'a', 'a', 'a', 'b', 'b', 'b', 'b'],

'date_stamp':['2016-02-01', '2016-02-01', '2016-02-04', '2016-02-08', '2016-02-04', '2016-02-10', '2016-02-10', '2016-02-12'],

'tie_breaker':[1,2,3,4,1,2,3,4]})

df['date_stamp'] = df['date_stamp'].map(lambda x: datetime.datetime.strptime(x, "%Y-%m-%d"))

df['rank'] = df.groupby(['userid'])['date_stamp'].rank(ascending=True, method='min')

df

Out[93]:

date_stamp tie_breaker userid rank

0 2016-02-01 1 a 1

1 2016-02-01 2 a 1

2 2016-02-04 3 a 3

3 2016-02-08 4 a 4

4 2016-02-04 1 b 1

5 2016-02-10 2 b 2

6 2016-02-10 3 b 2

7 2016-02-12 4 b 4

好吧，但是如果我想在有两个相同日期的情况下添加另一个字段作为打平比分的工具呢？我希望有些事情可以简单到：

^{pr2}$

但那没用-有什么想法吗？在

理想输出：date_stamp tie_breaker userid rank

0 2/1/16 1 a 1

1 2/1/16 2 a 2

2 2/4/16 3 a 3

3 2/8/16 4 a 4

4 2/4/16 1 b 1

5 2/10/16 2 b 2

6 2/10/16 3 b 3

7 2/12/16 4 b 4

编辑以获得真实数据

看起来这里的顶级解决方案不能正确处理tie峎u breaker字段中的零-知道发生了什么吗？在df = pd.DataFrame({

'userid':['10010012083198581013', '10010012083198581013', '10010012083198581013', '10010012083198581013','10010012083198581013'],

'date_stamp':['2015-12-26 13:24:37', '2015-11-25 11:24:13', '2015-10-25 12:13:59', '2015-02-16 22:59:58','2015-08-17 11:43:43'],

'tie_breaker':[460000156735858, 460000152444239, 460000147374709, 11083155016444116916,0]})

df['date_stamp'] = df['date_stamp'].map(lambda x: datetime.datetime.strptime(x, "%Y-%m-%d %H:%M:%S"))

df['userid'] = df['userid'].astype(str)

df['tie_breaker'] = df['tie_breaker'].astype(str)

def myrank(g):

return pd.DataFrame(1 + np.lexsort((g['tie_breaker'].rank(),

g['date_stamp'].rank())),

index=g.index)

df['rank']=df.groupby(['userid']).apply(myrank)

df.sort('date_stamp')

Out[101]:

date_stamp tie_breaker userid rank

3 2015-02-16 11083155016444116916 10010012083198581013 2

4 2015-08-17 0 10010012083198581013 1

2 2015-10-25 460000147374709 10010012083198581013 3

1 2015-11-25 460000152444239 10010012083198581013 5

0 2015-12-26 460000156735858 10010012083198581013 4