Why does pandas.cut() behave differently in unique count in two similar cases?

Question

In the first case, I use a very simple DataFrame to try using pandas.cut() to count the number of unique values in one column within a range of another column. The code runs as expected:

However, in the following code, pandas.cut() counts the number of unique values wrong. I expect the first bin (1462320000, 1462406400] to have 5 unique values, and other bins including the last bin (1462752000, 1462838400] to have 0 unique values.

Instead, as shown in the result, the code returns 5 unique values in the last bin (1462752000, 1462838400], while the 2 highlighted values should not be counted because they are out of range.

So could anyone explain why pandas.cut() behaves so different in these 2 cases? And also, I would be really thankful if you can also tell me how can I correct the code to correctly count the number of unique values in one column within a range of value of another column.

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ADDITIONNAL INFO: (please import pandas and numpy to run the code, my pandas version is 0.19.2, and I am using python 2.7)

For your ready reference, I hereby post my DataFrame and the codes for you to reproduce my code:

Case 1:

df = pd.DataFrame({'No': [1,1.5,2,1,3,5,10], 'useragent': ['a', 'c', 'b', 'c', 'b','a','z']})
print type(df)
print df
df.groupby(pd.cut(df['No'], bins=np.arange(0,4,1))).useragent.nunique()

Case 2:

print type(df)
print len(df)
print df.time.nunique()
print df.hash.nunique()
print df[['time','hash']]
df.groupby(pd.cut(df['time'], bins =np.arange(1462320000,1462924800,86400))).hash.nunique()

Case 2's Data:

time      hash
1462328401 qo
1462328401 qQ
1462838401 q1
1462328401 q1
1462328401 qU
1462328401 qU
1462328401 qU
1462328401 qU
1462328401 qX
1462838401 qX

Answer 1

It's seems to be a bug.

On a simple example :

In [50]: df=pd.DataFrame({'atime': [28]*8+[38]*2, 'hash':randint(0,3,10)}
).sort_values('hash')
Out[50]: 
      atime  hash
1     28     0
3     28     0
4     28     0
5     28     0
8     38     0
2     28     1
6     28     1
0     28     2
7     28     2
9     38     2 

In [50bis;)]: df.groupby(pd.cut(df.atime,bins=arange(27,40,2))).hash.unique()
Out[50bis]: 
atime
(27, 29]                   [0, 1, 2]   # ok
(29, 31]                          []
(31, 33]                          []
(33, 35]                          []
(35, 37]                          []
(37, 39]                      [0, 2]
Name: hash, dtype: object

In [51]: df.groupby(pd.cut(df.atime,bins=arange(27,40,2))).hash.nunique()
Out[51]: 
atime
(27, 29]    2 # bug
(29, 31]    0
(31, 33]    0
(33, 35]    0
(35, 37]    0
(37, 39]    2
Name: hash, dtype: int64

Here seems to be a efficient workaround, converting the cut result in a list :

In [52]: df.groupby(pd.cut(df.atime,bins=arange(27,40,2)).tolist()
).hash.nunique()
Out[52]: 
atime
(27, 29]    3
(37, 39]    2
Name: hash, dtype: int64