这里有熊猫新手。一个（平凡的）问题：主机，操作，执行时间。我想按主机分组，然后按主机+操作，计算每个主机执行时间的std偏差，然后按主机+操作对计算。看似简单吗？
适用于按单列分组：
df
Out[360]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 132564 entries, 0 to 132563
Data columns (total 9 columns):
datespecial 132564 non-null values
host 132564 non-null values
idnum 132564 non-null values
operation 132564 non-null values
time 132564 non-null values
...
dtypes: float32(1), int64(2), object(6)
byhost = df.groupby('host')
byhost.std()
Out[362]:
datespecial idnum time
host
ahost1.test 11946.961952 40367.033852 0.003699
host1.test 15484.975077 38206.578115 0.008800
host10.test NaN 37644.137631 0.018001
...
尼斯。现在：
byhostandop = df.groupby(['host', 'operation'])
byhostandop.std()
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
<ipython-input-364-2c2566b866c4> in <module>()
----> 1 byhostandop.std()
/home/username/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in std(self, ddof)
386 # todo, implement at cython level?
387 if ddof == 1:
--> 388 return self._cython_agg_general('std')
389 else:
390 f = lambda x: x.std(ddof=ddof)
/home/username/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in _cython_agg_general(self, how, numeric_only)
1615
1616 def _cython_agg_general(self, how, numeric_only=True):
-> 1617 new_blocks = self._cython_agg_blocks(how, numeric_only=numeric_only)
1618 return self._wrap_agged_blocks(new_blocks)
1619
/home/username/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in _cython_agg_blocks(self, how, numeric_only)
1653 values = com.ensure_float(values)
1654
-> 1655 result, _ = self.grouper.aggregate(values, how, axis=agg_axis)
1656
1657 # see if we can cast the block back to the original dtype
/home/username/anaconda/lib/python2.7/site-packages/pandas/core/groupby.pyc in aggregate(self, values, how, axis)
838 if is_numeric:
839 result = lib.row_bool_subset(result,
--> 840 (counts > 0).view(np.uint8))
841 else:
842 result = lib.row_bool_subset_object(result,
/home/username/anaconda/lib/python2.7/site-packages/pandas/lib.so in pandas.lib.row_bool_subset (pandas/lib.c:16540)()
ValueError: Buffer dtype mismatch, expected 'float64_t' but got 'float'
咦??为什么我会得到这个例外？
更多问题：
如何计算dataframe.groupby([several columns])上的标准偏差？
如何限制所选列的计算？例如。在这里计算日期/时间戳的std dev显然没有意义。
答案 0 :(得分：2)
了解您的Pandas / Python版本非常重要。看起来这个例外可能出现在Pandas版本＆lt; 0.10（见ValueError: Buffer dtype mismatch, expected 'float64_t' but got 'float'）。为避免这种情况，您可以将float列投射到float64：
df.astype('float64')
要计算所选列的std()，只需选择列：）
>>> df = pd.DataFrame({'a':range(10), 'b':range(10,20), 'c':list('abcdefghij'), 'g':[1]*3 + [2]*3 + [3]*4})
>>> df
a b c g
0 0 10 a 1
1 1 11 b 1
2 2 12 c 1
3 3 13 d 2
4 4 14 e 2
5 5 15 f 2
6 6 16 g 3
7 7 17 h 3
8 8 18 i 3
9 9 19 j 3
>>> df.groupby('g')[['a', 'b']].std()
a b
g
1 1.000000 1.000000
2 1.000000 1.000000
3 1.290994 1.290994
就目前而言，看起来std()正在aggregation()结果上调用groupby，并且有一个微妙的错误（请参阅此处 - Python Pandas: Using Aggregate vs Apply to define new columns）。为避免这种情况，您可以使用apply()：
byhostandop['time'].apply(lambda x: x.std())