1

从以下开始

df = pd.DataFrame( {'Item':['A','A','A','B','B','C','C','C','C'], 
    'Name': ['Tom','John','Paul','Tom','Frank','Tom', 'John', 'Richard', 'James'],
    'Total':[3,3,3,2,2,4,4,4,4]})
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]

我有一个如下的数据框：

print df 
      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]

我想为每个Name考虑People为条形和times值的堆积条形图。

我想做这样的事情

sub_df = df.groupby(['Name','People'])['Times'].sum().unstack()
sub_df.plot(kind='bar',stacked=True)

但它返回

TypeError：不可散列的类型：'numpy.ndarray'

4

1 回答 1

0

您必须使用 apply 之后的灵活类型的“agg” groupby：

df1['People'] = df1['Name_y'].apply(lambda x: tuple(x))
df1['Times'] = df1['Name_y'].apply(lambda x: [x.count(name) for name in list(set(x))])
s = df1.groupby(['Name_x','People']).apply(lambda x: sum(x.iloc[0]['Times']))

然后你得到以下

Name_x   People                                   
Frank    (Tom,)                                       1
James    (Tom, John, Richard)                         3
John     (Tom, Paul, Tom, Richard, James)             5
Paul     (Tom, John)                                  2
Richard  (Tom, John, James)                           3
Tom      (John, Paul, Frank, John, Richard, James)    6
dtype: int64

你可以随心所欲地绘制

s.plot(kind='bar', stacked=True)
于 2016-03-03T23:39:19.137 回答