Question

说我有这个数据框，我希望每个唯一的用户ID都有基于日期戳的自己的排名值：

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

这样很好，但如果我想在有两个相同的日期时添加另一个字段作为打破平局怎么办？我希望有些事情会像以下一样简单：

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

但这不起作用 - 任何想法？

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

修改了真实数据
看起来这里的顶级解决方案并没有正确处理tie_breaker字段中的零 - 任何想法会发生什么？

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

使用此数据框：

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]})

一种方法是：

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)

输出：

   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

按唯一组向pandas数据框添加排名字段，并按多列排序

1 个答案: