Apply a lambda and call sample with param frac:

In [2]:
df = pd.DataFrame({'a': [1,2,3,4,5,6,7],
                   'b': [1,1,1,0,0,0,0]})
​
grouped = df.groupby('b')
grouped.apply(lambda x: x.sample(frac=0.3))

Out[2]:
     a  b
b        
0 6  7  0
1 2  3  1

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

Convert sample_info to dictionary. Group population by Group ID. Pass the sample size values to DataFrame.sample using the dictionary.

mapper = sample_info.set_index('Group ID')['Sample Size'].to_dict()

population.groupby('Group ID').apply(lambda x: x.sample(n=mapper.get(x.name))).reset_index(drop = True)

You can do with shuffle and ngroup

g = df.groupby(['col1', 'col2'])

a=np.arange(g.ngroups)
np.random.shuffle(a)

df[g.ngroup().isin(a[:2])]# change 2 to what you need :-)