在数据帧中循环时计算行的出现次数

Question

我有一个数据帧，每行重复3次。在循环浏览时，如何确定之前是否看到一行，然后在循环中的第二次出现时打印某个内容？你知道吗

print df
       user     date
0      User001  2014-11-01
40     User001  2014-11-01
80     User001  2014-11-01
120    User001  2014-11-08
200    User001  2014-11-08
160    User001  2014-11-08
280    User001  2014-11-15
240    User001  2014-11-15
320    User001  2014-11-15
400    User001  2014-11-22
440    User001  2014-11-22
360    User001  2014-11-22
...    ......   ..........
...    ......   ..........
1300   User008  2014-11-22
1341   User008  2014-11-22
1360   User008  2014-11-22

for line in df.itertuples():
    user = line[1]
    date = line[2]

    print user, date
    #do something after second occurrence of tuple i.e. print "second occurrence"

('User001', '2014-11-01')
('User001',  '2014-11-01')
second occurrence
('User001',  '2014-11-01')
('User001',  '2014-11-08')
('User001',  '2014-11-08')
second occurrence
('User001',  '2014-11-08')
('User001',  '2014-11-15')
('User001',  '2014-11-15')
second occurrence
('User001',  '2014-11-15')
('User001',  '2014-11-22')
('User001',  '2014-11-22')
second occurrence
('User001',  '2014-11-22')
('User008',  '2014-11-22')
('User008',  '2014-11-22')
second occurrence
('User008',  '2014-11-22')

Answer 1

可以使用^{}查找第二次出现的所有索引：

mask = df.groupby(['user', 'date']).cumcount() == 1
idx = mask[mask].index
print (idx)
Int64Index([40, 200, 240, 440], dtype='int64')

for line in df.itertuples():
    print (line.user)
    print (line.date)
    if line.Index in idx:
        print ('second occurrence')

User001
2014-11-01
User001
2014-11-01
second occurrence
User001
2014-11-01
User001
2014-11-08
User001
2014-11-08
second occurrence
User001
2014-11-08
User001
2014-11-15
User001
2014-11-15
second occurrence
User001
2014-11-15
User001
2014-11-22
User001
2014-11-22
second occurrence
User001
2014-11-22

查找索引的另一个解决方案是：

idx = df[df.duplicated(['user', 'date']) & 
         df.duplicated(['user', 'date'], keep='last')].index
print (idx)
Int64Index([40, 200, 240, 440], dtype='int64')