17

What is the best way to get a random sample of the elements of a groupby? As I understand it, a groupby is just an iterable over groups.

The standard way I would do this for an iterable, if I wanted to select N = 200 elements is:

rand = random.sample(data, N)

If you attempt the above where data is a 'grouped' the elements of the resultant list are tuples for some reason.

I found the below example for randomly selecting the elements of a single key groupby, however this does not work with a multi-key groupby. From, How to access pandas groupby dataframe by key

create groupby object

grouped = df.groupby('some_key')

pick N dataframes and grab their indices

sampled_df_i = random.sample(grouped.indices, N)

grab the groups using the groupby object 'get_group' method

df_list = map(lambda df_i: grouped.get_group(df_i),sampled_df_i)

optionally - turn it all back into a single dataframe object

sampled_df = pd.concat(df_list, axis=0, join='outer')
yoav_aaa
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sfortney
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  • For me had to add `import random` before the 2nd step and use `list(df_list)` after map in 3rd step to put it as `pd.concat(list(df_list))` finally – MItrajyoti Aug 29 '22 at 14:19

3 Answers3

16

You can take a randoms sample of the unique values of df.some_key.unique(), use that to slice the df and finally groupby on the resultant:

In [337]:

df = pd.DataFrame({'some_key': [0,1,2,3,0,1,2,3,0,1,2,3],
                   'val':      [1,2,3,4,1,5,1,5,1,6,7,8]})
In [338]:

print df[df.some_key.isin(random.sample(df.some_key.unique(),2))].groupby('some_key').mean()
               val
some_key          
0         1.000000
2         3.666667

If there are more than one groupby keys:

In [358]:

df = pd.DataFrame({'some_key1':[0,1,2,3,0,1,2,3,0,1,2,3],
                   'some_key2':[0,0,0,0,1,1,1,1,2,2,2,2],
                   'val':      [1,2,3,4,1,5,1,5,1,6,7,8]})
In [359]:

gby = df.groupby(['some_key1', 'some_key2'])
In [360]:

print gby.mean().ix[random.sample(gby.indices.keys(),2)]
                     val
some_key1 some_key2     
1         1            5
3         2            8

But if you are just going to get the values of each group, you don't even need to groubpy, MultiIndex will do:

In [372]:

idx = random.sample(set(pd.MultiIndex.from_product((df.some_key1, df.some_key2)).tolist()),
                    2)
print df.set_index(['some_key1', 'some_key2']).ix[idx]
                     val
some_key1 some_key2     
2         0            3
3         1            5
CT Zhu
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4

I feel like lower-level numpy operations are cleaner:

import pandas as pd
import numpy as np

df = pd.DataFrame(
    {
        "some_key": [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3],
        "val": [1, 2, 3, 4, 1, 5, 1, 5, 1, 6, 7, 8],
    }
)

ids = df["some_key"].unique()
ids = np.random.choice(ids, size=2, replace=False)
ids

# > array([3, 2])

df.loc[df["some_key"].isin(ids)]

# >     some_key  val
# 2          2    3
# 3          3    4
# 6          2    1
# 7          3    5
# 10         2    7
# 11         3    8
jsta
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-1

Although this question was asked and answered long ago, I think the following is cleaner:

import pandas as pd


df = pd.DataFrame(
    {
        "some_key1": [0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3],
        "some_key2": [0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2],
        "val": [1, 2, 3, 4, 1, 5, 1, 5, 1, 6, 7, 8]
    }
)

# Set the number of samples by group
n_samples_by_group = 1
samples_by_group = df \
    .groupby(by=["some_key1", "some_key2"]) \
    .sample(n_samples_by_group)
Esteban Jiménez
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