如何通过密钥按数据组访问熊猫

问题:如何通过密钥按数据组访问熊猫

如何通过密钥访问groupby对象中的相应groupby数据帧?

通过以下groupby:

rand = np.random.RandomState(1)
df = pd.DataFrame({'A': ['foo', 'bar'] * 3,
                   'B': rand.randn(6),
                   'C': rand.randint(0, 20, 6)})
gb = df.groupby(['A'])

我可以遍历它来获取密钥和组:

In [11]: for k, gp in gb:
             print 'key=' + str(k)
             print gp
key=bar
     A         B   C
1  bar -0.611756  18
3  bar -1.072969  10
5  bar -2.301539  18
key=foo
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

我希望能够通过其键访问组:

In [12]: gb['foo']
Out[12]:  
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

但是当我尝试这样做时,gb[('foo',)]我得到了这个奇怪的pandas.core.groupby.DataFrameGroupBy对象,似乎没有任何与我想要的DataFrame相对应的方法。

我能想到的最好的是:

In [13]: def gb_df_key(gb, key, orig_df):
             ix = gb.indices[key]
             return orig_df.ix[ix]

         gb_df_key(gb, 'foo', df)
Out[13]:
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14  

但是,考虑到这些事情上熊猫通常是多么好,这有点令人讨厌。
这样做的内置方式是什么?

How do I access the corresponding groupby dataframe in a groupby object by the key?

With the following groupby:

rand = np.random.RandomState(1)
df = pd.DataFrame({'A': ['foo', 'bar'] * 3,
                   'B': rand.randn(6),
                   'C': rand.randint(0, 20, 6)})
gb = df.groupby(['A'])

I can iterate through it to get the keys and groups:

In [11]: for k, gp in gb:
             print 'key=' + str(k)
             print gp
key=bar
     A         B   C
1  bar -0.611756  18
3  bar -1.072969  10
5  bar -2.301539  18
key=foo
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

I would like to be able to access a group by its key:

In [12]: gb['foo']
Out[12]:  
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

But when I try doing that with gb[('foo',)] I get this weird pandas.core.groupby.DataFrameGroupBy object thing which doesn’t seem to have any methods that correspond to the DataFrame I want.

The best I could think of is:

In [13]: def gb_df_key(gb, key, orig_df):
             ix = gb.indices[key]
             return orig_df.ix[ix]

         gb_df_key(gb, 'foo', df)
Out[13]:
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14  

but this is kind of nasty, considering how nice pandas usually is at these things.
What’s the built-in way of doing this?


回答 0

您可以使用以下get_group方法:

In [21]: gb.get_group('foo')
Out[21]: 
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

注意:这不需要为每个组创建一个中间字典/每个子数据帧的副本,因此与使用来创建朴素的字典相比,其内存效率更高dict(iter(gb))。这是因为它使用了groupby对象中已经可用的数据结构。


您可以使用groupby切片选择不同的列:

In [22]: gb[["A", "B"]].get_group("foo")
Out[22]:
     A         B
0  foo  1.624345
2  foo -0.528172
4  foo  0.865408

In [23]: gb["C"].get_group("foo")
Out[23]:
0     5
2    11
4    14
Name: C, dtype: int64

You can use the get_group method:

In [21]: gb.get_group('foo')
Out[21]: 
     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

Note: This doesn’t require creating an intermediary dictionary / copy of every subdataframe for every group, so will be much more memory-efficient that creating the naive dictionary with dict(iter(gb)). This is because it uses data-structures already available in the groupby object.


You can select different columns using the groupby slicing:

In [22]: gb[["A", "B"]].get_group("foo")
Out[22]:
     A         B
0  foo  1.624345
2  foo -0.528172
4  foo  0.865408

In [23]: gb["C"].get_group("foo")
Out[23]:
0     5
2    11
4    14
Name: C, dtype: int64

回答 1

Python for Data Analysis中的Wes McKinney(熊猫的作者)提供了以下配方:

groups = dict(list(gb))

它返回一个字典,其键是您的组标签,其值是DataFrames,即

groups['foo']

将产生您想要的东西:

     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

Wes McKinney (pandas’ author) in Python for Data Analysis provides the following recipe:

groups = dict(list(gb))

which returns a dictionary whose keys are your group labels and whose values are DataFrames, i.e.

groups['foo']

will yield what you are looking for:

     A         B   C
0  foo  1.624345   5
2  foo -0.528172  11
4  foo  0.865408  14

回答 2

而不是

gb.get_group('foo')

我更喜欢使用 gb.groups

df.loc[gb.groups['foo']]

因为这样您也可以选择多个列。例如:

df.loc[gb.groups['foo'],('A','B')]

Rather than

gb.get_group('foo')

I prefer using gb.groups

df.loc[gb.groups['foo']]

Because in this way you can choose multiple columns as well. for example:

df.loc[gb.groups['foo'],('A','B')]

回答 3

gb = df.groupby(['A'])

gb_groups = grouped_df.groups

如果要查找选择性的groupby对象,请执行:gb_groups.keys(),然后将所需的密钥输入到以下key_list中。

gb_groups.keys()

key_list = [key1, key2, key3 and so on...]

for key, values in gb_groups.iteritems():
    if key in key_list:
        print df.ix[values], "\n"
gb = df.groupby(['A'])

gb_groups = grouped_df.groups

If you are looking for selective groupby objects then, do: gb_groups.keys(), and input desired key into the following key_list..

gb_groups.keys()

key_list = [key1, key2, key3 and so on...]

for key, values in gb_groups.iteritems():
    if key in key_list:
        print df.ix[values], "\n"

回答 4

我正在寻找对GroupBy obj的几个成员进行抽样的方法-必须解决发布的问题才能完成此任务。

创建分组对象

grouped = df.groupby('some_key')

选择N个数据框并获取其索引

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

抢团体

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

可选-将所有内容重新转换为单个dataframe对象

sampled_df = pd.concat(df_list, axis=0, join='outer')

I was looking for a way to sample a few members of the GroupBy obj – had to address the posted question to get this done.

create groupby object

grouped = df.groupby('some_key')

pick N dataframes and grab their indicies

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

grab the groups

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')