添加等級字段通過獨特的羣體和多列排序，以大熊貓數據幀

Question

說我有這個數據幀，我想每一個唯一的用戶ID基於日期戳有它自己的等級值：

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 

所以這是很好，但是如果我想在有兩個相同的日期時添加另一個場來作爲決勝局，那該怎麼辦呢？我希望事情會是一樣簡單：

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

但是，這並不工作 - 任何想法？

理想輸出：

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 

編輯以有真實數據
貌似頂部的解決方案在這裏正確tie_breaker場不處理零 - 任何想法，這是怎麼回事？

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 

Answer 1

有了這個數據幀：

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]}) 

一種方式來做到這一點是：

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) 

輸出：

date_stamp tie_breaker userid rank 
0 2016-02-01   1  a  1 
1 2016-02-01   2  a  2 
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  3 
7 2016-02-12   4  b  4