Question

您好，我有一个像这样的数据框

df = pd.DataFrame( {'Item':['A','A','A','B','B','C','C','C','C'], 
'b':[Tom,John,Paul,Tom,Frank,Tom, John, Richard, James]})
df 
Item Name
A    Tom
A    John
A    Paul
B    Tom 
B    Frank
C    Tom
C    John
C    Richard
C    James

对于每个人，我想要具有相同项目和时间的人员列表

df1 
Name              People                          Times
Tom     [John, Paul, Frank, Richard, James]       [2,1,1,1,1]
John    [Tom, Richard, James]                     [2,1,1]
Paul    [Tom, John]                               [1,1]
Frank   [Tom]                                     [1]
Richard [Tom, John, James]                        [1,1,1]
James   [Tom, John, Richard]                      [1,1,1]

到目前为止，我已经尝试过这个来计算不同人物的不同物品

df.groupby("Item").agg({ "Name": pd.Series.nunique})
      Name
Item    
A      3
B      2
C      4

和

df.groupby("Name").agg({ "Item": pd.Series.nunique})
        Item
Name    
Frank   1
James   1
John    2
Paul    1
Richard 1
Tom     3

Answer 1

您可以将numpy.unique用于列表中的计数项目：

print df
  Item     Name
0    A      Tom
1    A     John
2    A     Paul
3    B      Tom
4    B    Frank
5    C      Tom
6    C     John
7    C  Richard
8    C    James

#merge M:N by column Item
df1 = pd.merge(df, df, on=['Item'])

#remove duplicity - column Name_x == Name_y
df1 = df1[~(df1['Name_x'] == df1['Name_y'])]
#print df1

#create lists
df1 = df1.groupby('Name_x')['Name_y'].apply(lambda x: x.tolist()).reset_index()
print df1
    Name_x                                     Name_y
0    Frank                                      [Tom]
1    James                       [Tom, John, Richard]
2     John           [Tom, Paul, Tom, Richard, James]
3     Paul                                [Tom, John]
4  Richard                         [Tom, John, James]
5      Tom  [John, Paul, Frank, John, Richard, James]

#get count by np.unique
df1['People'] = df1['Name_y'].apply(lambda a: np.unique((a), return_counts =True)[0])
df1['times'] = df1['Name_y'].apply(lambda a: np.unique((a), return_counts =True)[1])
#remove column Name_y
df1 = df1.drop('Name_y', axis=1).rename(columns={'Name_x':'Name'})
print df1
      Name                               People            times
0    Frank                                [Tom]              [1]
1    James                 [John, Richard, Tom]        [1, 1, 1]
2     John          [James, Paul, Richard, Tom]     [1, 1, 1, 2]
3     Paul                          [John, Tom]           [1, 1]
4  Richard                   [James, John, Tom]        [1, 1, 1]
5      Tom  [Frank, James, John, Paul, Richard]  [1, 1, 2, 1, 1]

Answer 2

ct = pd.crosstab(df.Name, df.Item)

d = {Name: [(name, val) 
            for name, val in ct.loc[ct.index != Name, ct.ix[Name] == 1]
            .sum(axis=1).iteritems() if val] 
     for Name in df.Name.unique()}

>>> pd.DataFrame({'Name': d.keys(), 
                  'People': [[t[0] for t in d[name]] for name in d], 
                  'times': [[t[1] for t in d[name]] for name in d]})
      Name                               People            times
0  Richard                   [James, John, Tom]        [1, 1, 1]
1    James                 [John, Richard, Tom]        [1, 1, 1]
2      Tom  [Frank, James, John, Paul, Richard]  [1, 1, 2, 1, 1]
3    Frank                                [Tom]              [1]
4     Paul                          [John, Tom]           [1, 1]
5     John          [James, Paul, Richard, Tom]     [1, 1, 1, 2]

<强>说明

交叉表按项目类型为您提供每个名称的位置。

>>> ct
Item     A  B  C
Name            
Frank    0  1  0
James    0  0  1
John     1  0  1
Paul     1  0  0
Richard  0  0  1
Tom      1  1  1

然后缩小此表。对于正在构建密钥的每个名称，将从表中删除该名称，并且仅选择显示该名称的列。

使用＆＃39; John＆＃39;举个例子：

>>> ct.loc[ct.index != 'John', ct.ix['John'] == 1]
Item     A  C
Name         
Frank    0  0
James    0  1
Paul     1  0
Richard  0  1
Tom      1  1

然后将这个结果沿着行求和，得到John的结果：

Name
Frank      0
James      1
Paul       1
Richard    1
Tom        2
dtype: int64

然后迭代这些结果以将它们打包成元组对并移除值为零的情况（例如上面的Frank）。

>>> [(name, val) for name, val in 
     ct.loc[ct.index != 'John', ct.ix['John'] == 1].sum(axis=1).iteritems() if val]
[('James', 1), ('Paul', 1), ('Richard', 1), ('Tom', 2)]

使用词典理解为每个名称执行此操作。

>>> d
{'Frank': [('Tom', 1)],
 'James': [('John', 1), ('Richard', 1), ('Tom', 1)],
 'John': [('James', 1), ('Paul', 1), ('Richard', 1), ('Tom', 2)],
 'Paul': [('John', 1), ('Tom', 1)],
 'Richard': [('James', 1), ('John', 1), ('Tom', 1)],
 'Tom': [('Frank', 1), ('James', 1), ('John', 2), ('Paul', 1), ('Richard', 1)]}

然后使用这个字典创建所需的数据帧，使用嵌套列表解析来解包元组对。

在pandas数据帧中对条目进行分组

2 个答案: