pandas 熊猫，数据框，groupby，std
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pandas, dataframe, groupby, std
提问by LetMeSOThat4U
New to pandas here. A (trivial) problem: hosts, operations, execution times. I want to group by host, then by host+operation, calculate std deviation for execution time per host, then by host+operation pair. Seems simple?
这里的Pandas新手。一个（微不足道的）问题：主机、操作、执行时间。我想按主机分组，然后按主机+操作，计算每个主机的执行时间的标准偏差，然后按主机+操作对。看起来很简单？
It works for grouping by a single column:
它适用于按单列分组：
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
...
Nice. Now:
好的。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'
Huh?? Why do I get this exception?
嗯？？为什么我会收到此异常？
More questions:
更多问题：
how do I calculate std deviation on
dataframe.groupby([several columns])?how can I limit calculation to a selected column? E.g. it obviously doesn't make sense to calculate std dev on dates/timestamps here.
我如何计算标准偏差
dataframe.groupby([several columns])？如何将计算限制为选定的列？例如，在这里计算日期/时间戳的 std dev 显然没有意义。
采纳答案by Roman Pekar
It's important to know your version of Pandas / Python. Looks like this exception could arise in Pandas version < 0.10 (see ValueError: Buffer dtype mismatch, expected 'float64_t' but got 'float'). To avoid this, you can cast your floatcolumns to float64:
了解您的 Pandas / Python 版本很重要。看起来这个异常可能出现在 Pandas 版本 < 0.10 中（参见ValueError: Buffer dtype mismatch, expected 'float64_t' but got 'float'）。为避免这种情况，您可以将float列转换为float64：
df.astype('float64')
To calculate std()on selected columns, just select columns :)
要计算std()选定的列，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
update
更新
As far as it goes, it looks like std()is calling aggregation()on the groupbyresult, and a subtle bug (see here - Python Pandas: Using Aggregate vs Apply to define new columns). To avoid this, you can use apply():
至于它去，它看起来像std()呼吁aggregation()的groupby结果，而个微妙的问题（见这里- Python的Pandas：使用聚集VS应用来定义新列）。为避免这种情况，您可以使用apply()：
byhostandop['time'].apply(lambda x: x.std())
