熊猫数据帧矢量化采样

Question

I have a simple df forming a pivot_table:

    d = {'one' : ['A', 'B', 'B', 'C', 'C', 'C'], 'two' : [6., 5., 4., 3., 2., 1.],     'three' : [6., 5., 4., 3., 2., 1.], 'four' : [6., 5., 4., 3., 2., 1.]}
    df = pd.DataFrame(d)
    pivot = pd.pivot_table(df,index=['one','two'])

I would like to randomly sample 1 row from each different element from column 'one' of the resulting pivot object. (In this example, 'A' will always be sampled while there are more options for 'B' and 'C'.) I just began using the 0.18.0 version of pandas and am aware of the .sample method. I messed with the .groupby method applying a sampling function something like this:

    grouped = pivot.groupby('one').apply(lambda x: x.sample(n=1, replace=False))

I raise a KeyError when I tried variations on that theme so I thought it was time for some fresh perspective on this seemingly simple question...

Thanks for any assistance!

Answer 1

The KeyError is raised since 'one' is not a column in pivot but the name of an index:

In [11]: pivot
Out[11]:
         four  three
one two
A   6.0   6.0    6.0
B   4.0   4.0    4.0
    5.0   5.0    5.0
C   1.0   1.0    1.0
    2.0   2.0    2.0
    3.0   3.0    3.0

You have to use the level argument:

In [12]: pivot.groupby(level='one').apply(lambda x: x.sample(n=1, replace=False))
Out[12]:
             four  three
one one two
A   A   6.0   6.0    6.0
B   B   4.0   4.0    4.0
C   C   1.0   1.0    1.0

This isn't quite right since the index is repeated! It's slightly better with as_index=False :

In [13]: pivot.groupby(level='one', as_index=False).apply(lambda x: x.sample(n=1))
Out[13]:
           four  three
  one two
0 A   6.0   6.0    6.0
1 B   4.0   4.0    4.0
2 C   2.0   2.0    2.0

Note: This picks a random row each time.

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As an alternative, a potentially more performant variant (that pulls out a subframe:

In [21]: df.iloc[[np.random.choice(x) for x in g.indices.values()]]
Out[21]:
   four one  three  two
1   5.0   B    5.0  5.0
3   3.0   C    3.0  3.0
0   6.0   A    6.0  6.0