我是 pandas(ver 0.14.0)的新手，遇到了以下问题:
我正在尝试使用多索引对 pandas 数据帧进行切片。索引包含时间戳。如果仅使用日期作为时间戳进行切片，则效果很好。使用时间戳中的时间切片时，它不返回任何内容或返回异常。
分割包含日期和时间的时间戳的正确方法是什么？
更新: 切片时间戳和其他索引和列的正确方法是什么？
这是我的代码:
dates = pd.DatetimeIndex([datetime.datetime(2012,1,1,12,12,12)+datetime.timedelta(days = i) for i in range(6)])
freq = [1,2]
iterables = [dates, freq]
index = pd.MultiIndex.from_product(iterables, names=['date','frequency'])
df = pd.DataFrame(np.random.randn(6*2,4),index=index,columns=list('ABCD'))
print df.loc[(slice(None), slice(None)),:] # works
print df.loc[(slice(None), slice(1,1)),:] # works
df.loc[(slice('2012-01-01 12:12:12','2012-01-03 12:12:12'), slice(None)),:] # returns empty
返回:
A B C D
date frequency
2012-01-01 12:12:12 1 0.903078 -0.250419 0.191373 0.491633
2 -2.571769 1.906471 -0.712225 0.255760
2012-01-02 12:12:12 1 1.056798 -0.753387 0.509417 2.001925
2 -0.746595 0.435158 0.955275 -1.854974
2012-01-03 12:12:12 1 0.139800 -0.728467 -1.196661 0.201817
2 -0.006282 -0.644041 0.138642 -1.232355
2012-01-04 12:12:12 1 -0.895909 0.504779 -0.803993 1.306559
2 0.268643 -0.642575 -0.573525 0.914382
2012-01-05 12:12:12 1 0.608634 -2.650082 -0.404462 0.593098
2 -0.376576 -1.514299 -1.053566 0.130654
2012-01-06 12:12:12 1 0.658660 -0.575514 0.665777 -1.282307
2 0.162896 0.302550 1.609635 -2.146004
A B C D
date frequency
2012-01-01 12:12:12 1 0.903078 -0.250419 0.191373 0.491633
2012-01-02 12:12:12 1 1.056798 -0.753387 0.509417 2.001925
2012-01-03 12:12:12 1 0.139800 -0.728467 -1.196661 0.201817
2012-01-04 12:12:12 1 -0.895909 0.504779 -0.803993 1.306559
2012-01-05 12:12:12 1 0.608634 -2.650082 -0.404462 0.593098
2012-01-06 12:12:12 1 0.658660 -0.575514 0.665777 -1.282307
Empty DataFrame
Columns: [A, B, C, D]
Index: []
或者，如果我尝试以下操作，它会返回一个错误:
df.loc[(slice(dates[0],dates[2]), slice(None)),:]
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-126-016ed3a2c8ff> in <module>()
----> 1 df.loc[(slice(dates[0],dates[2]), slice(None)),:]
2 #print df.loc[(slice(pd.to_datetime(datetime.datetime(2013, 1, 2, 2, 3, 40)),pd.to_datetime(datetime.datetime(2013, 1, 3, 2, 3, 40))), 1),:]
C:\Anaconda\lib\site-packages\pandas\core\indexing.pyc in __getitem__(self, key)
1125 def __getitem__(self, key):
1126 if type(key) is tuple:
-> 1127 return self._getitem_tuple(key)
1128 else:
1129 return self._getitem_axis(key, axis=0)
C:\Anaconda\lib\site-packages\pandas\core\indexing.pyc in _getitem_tuple(self, tup)
643 def _getitem_tuple(self, tup):
644 try:
--> 645 return self._getitem_lowerdim(tup)
646 except IndexingError:
647 pass
C:\Anaconda\lib\site-packages\pandas\core\indexing.pyc in _getitem_lowerdim(self, tup)
751 # we may have a nested tuples indexer here
752 if self._is_nested_tuple_indexer(tup):
--> 753 return self._getitem_nested_tuple(tup)
754
755 # we maybe be using a tuple to represent multiple dimensions here
C:\Anaconda\lib\site-packages\pandas\core\indexing.pyc in _getitem_nested_tuple(self, tup)
823
824 current_ndim = obj.ndim
--> 825 obj = getattr(obj, self.name)._getitem_axis(key, axis=axis, validate_iterable=True)
826 axis += 1
827
C:\Anaconda\lib\site-packages\pandas\core\indexing.pyc in _getitem_axis(self, key, axis, validate_iterable)
1254 return self._getitem_iterable(key, axis=axis)
1255 elif _is_nested_tuple(key, labels):
-> 1256 locs = labels.get_locs(key)
1257 indexer = [ slice(None) ] * self.ndim
1258 indexer[axis] = locs
C:\Anaconda\lib\site-packages\pandas\core\index.pyc in get_locs(self, tup)
3580 np.logical_or,[ _convert_indexer(self._get_level_indexer(x, level=i)
3581 ) for x in k ]))
-> 3582 elif k == slice(None):
3583 # include all from this level
3584 pass
C:\Anaconda\lib\site-packages\pandas\tslib.pyd in pandas.tslib._Timestamp.__richcmp__ (pandas\tslib.c:13056)()
TypeError: Cannot compare type 'Timestamp' with type 'NoneType'
这也失败了:
df.loc[(slice(pd.Timestamp('2012-01-01 12:12:12'),pd.Timestamp('2012-01-03 12:12:12')),slice(1,1)), slice('A','B')]
更新 以下工作但仍然无法一步完成:
df_temp = df.loc[(slice(pd.Timestamp('2012-01-01 12:12:12'),pd.Timestamp('2012-01-03 12:12:12'))), slice('A','B')]
df_temp.loc[(slice(None),slice(1,1)),:]
A B
date frequency
2012-01-01 12:12:12 1 0.840330 -0.051184
2012-01-02 12:12:12 1 -0.468037 -0.012381
2012-01-03 12:12:12 1 -0.061229 0.613407
最佳答案
您可以在时间戳而不是字符串上切片:
In [11]: df.loc[(slice(pd.Timestamp('2012-01-01 12:12:12'),pd.Timestamp('2012-01-03 12:12:12')))]
Out[11]:
A B C D
date frequency
2012-01-01 12:12:12 1 0.796501 -0.914335 1.205684 0.707926
2 0.659782 -0.823599 0.786772 -1.265034
2012-01-02 12:12:12 1 0.907892 1.248585 -0.037800 -0.893048
2 -0.595936 -0.286499 0.595300 -0.359440
2012-01-03 12:12:12 1 0.145403 0.621906 0.865768 -0.228813
2 1.169412 0.213809 0.551384 0.870852
In [12]: df.loc[(slice(pd.Timestamp('2012-01-01 12:12:12'),pd.Timestamp('2012-01-03 12:12:12')), slice(None))]
Out[12]:
A B C D
date frequency
2012-01-01 12:12:12 1 0.796501 -0.914335 1.205684 0.707926
2 0.659782 -0.823599 0.786772 -1.265034
2012-01-02 12:12:12 1 0.907892 1.248585 -0.037800 -0.893048
2 -0.595936 -0.286499 0.595300 -0.359440
2012-01-03 12:12:12 1 0.145403 0.621906 0.865768 -0.228813
2 1.169412 0.213809 0.551384 0.870852
我认为字符串用于切片的事实非常疯狂!
话虽如此，我似乎无法通过以下方法对两者进行切片:
df.loc[(slice(pd.Timestamp('2012-01-01 12:12:12'),pd.Timestamp('2012-01-03 12:12:12')), slice(1, 1))]
KeyError: 'start bound [1] is not the [columns]'
关于python - 使用 datetime 数据类型切片 pandas multiindex，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/24152509/