Question

我有一本字典，我想将其变成一个数据框，然后将该数据框的 some 列合并为一列。

我的字典看起来像这样：

mydict = {'Participants': {'source': ['1', '2', '3'],
                           'name': ['A', 'B', 'C'],
                           'Entry (1)': ['Address1', 'Address2', 'Address3'],
                           'Entry (2)': ['Number1', 'Number2', 'Number2'],
                           'Entry (3)': ['Start1', 'Start2', 'Start3']},
            'Countries': {'DK': ['1', '2', '3'],
                      'UK': ['1', '3', '2'],
                      'CDN': ['3', '2', '1'],
                      'FR': ['1', '2', '3']}}

，结果数据帧如下所示： df = pd.DataFrame(mydict)

df：

           Countries                    Participants
CDN        [3, 2, 1]                             NaN
DK         [1, 2, 3]                             NaN
Entry (1)        NaN  [Address1, Address2, Address3]
Entry (2)        NaN     [Number1, Number2, Number2]
Entry (3)        NaN        [Start1, Start2, Start3]
FR         [1, 2, 3]                             NaN
UK         [1, 3, 2]                             NaN
name             NaN                       [A, B, C]
source           NaN                       [1, 2, 3]

我有多个“条目（n）”列，其中包含每个参与者（df['Participants']['name']）的“地址，号码和开始”信息。现在，我需要一个附加的“条目”列，该列为每行合并Entry (1)，Entry (2)和Entry(3)的信息。由于条目数（Entry (n)随数据源的不同而变化，我需要获取如下条目数：

entries = re.findall(r'Entry \(\d\)', str(mydict['Participants'].keys()))

这给我留下了所有条目的列表：['Entry (1)', 'Entry (2)', 'Entry (3)']。

最后我想拥有一个像这样的数据框：

           Countries                    Participants
CDN        [3, 2, 1]                             NaN
DK         [1, 2, 3]                             NaN
Entry (1)        NaN  [Address1, Address2, Address3]
Entry (2)        NaN  [Number1, Number2, Number2]
Entry (3)        NaN  [Start1, Start2, Start3]
Entries          Nan  ['Address1\nNumber1\Start1', 'Address2\nNumber2\Start2', 'Address3\nNumber3\nStart3']  <<-- I need this
FR         [1, 2, 3]                             NaN
UK         [1, 3, 2]                             NaN
name             NaN                       [A, B, C]
source           NaN                       [1, 2, 3]

有人可以告诉我如何实现这一目标的熊猫特有方式吗？

Answer 1

您似乎需要

s=pd.DataFrame(df.filter(like='Entry',axis=0).Participants.tolist()).apply('/n'.join).tolist()
df.loc['Entries','Participants']=s
df
Out[64]: 
                                                Participants  Countries
CDN                                                      NaN  [3, 2, 1]
DK                                                       NaN  [1, 2, 3]
Entry (1)                     [Address1, Address2, Address3]        NaN
Entry (2)                        [Number1, Number2, Number2]        NaN
Entry (3)                           [Start1, Start2, Start3]        NaN
FR                                                       NaN  [1, 2, 3]
UK                                                       NaN  [1, 3, 2]
name                                               [A, B, C]        NaN
source                                             [1, 2, 3]        NaN
Entries    [Address1/nNumber1/nStart1, Address2/nNumber2/...        NaN

请注意，您可以在末尾添加sort_index

Answer 2

让我们尝试一下：

~~df.at ['Entries'，'Participants'] = ['\ n'.join（i）for in in（zip（* df.loc [['Entry（1）'，'Entry （2）'，'条目（3）']，'参与者']））]]~~

使用过滤器而不是索引列表从@ W-B解决方案中借来：

df.at['Entries','Participants'] = ['\n'.join(i) for i in (zip(*df.filter(like='Entry', axis=0)['Participants']))]
df.sort_index()

输出：

                                                Participants  Countries
CDN                                                      NaN  [3, 2, 1]
DK                                                       NaN  [1, 2, 3]
Entries    [Address1\nNumber1\nStart1, Address2\nNumber2\...        NaN
Entry (1)                     [Address1, Address2, Address3]        NaN
Entry (2)                        [Number1, Number2, Number2]        NaN
Entry (3)                           [Start1, Start2, Start3]        NaN
FR                                                       NaN  [1, 2, 3]
UK                                                       NaN  [1, 3, 2]
name                                               [A, B, C]        NaN
source                                             [1, 2, 3]        NaN

熊猫从多列（条目为列表）的行中获得新列

2 个答案: