Question

我正在为我的Pandafile工作，我仍然没有想出如何解决这个问题。

我有以下熊猫对象：

pandaFile = pd.DataFrame([{'var1': 'Restaurant A','var2':'4.5','var3':
['AA','BB','CC'],'var4':['User1','User2','User3'],'var5':['Review 1','Review 
2','Review 3']},{'var1': 'Restaurant B','var2':'5.0','var3':
['AA','BB','CC'],'var4':['User1','User2','User3'], 'var5':['Review 1','Review 
2','Review 3']}])
print(pandaFile)

它看起来像这样：

   var1            var2   var3           var4                  var5  
0  Restaurant A    4.5    [AA, BB, CC]  [User1, User2, User3] [Review 1, Review 2, Review 3]     
1  Restaurant B    5.0    [AA, BB, CC]  [User1, User2, User3] [Review 1, Review 2, Review 3]

我想得到以下输出：

         var1 var2          var3   var4      var5
0   Restaurant A  4.5  [AA, BB, CC]  User1  Review 1
1   Restaurant A  4.5  [AA, BB, CC]  User2  Review 2
2   Restaurant A  4.5  [AA, BB, CC]  User3  Review 3
3   Restaurant B  5.0  [AA, BB, CC]  User1  Review 1
4   Restaurant B  5.0  [AA, BB, CC]  User2  Review 2
5   Restaurant B  5.0  [AA, BB, CC]  User3  Review 3

但我得到以下输出：

        var1 var2          var3   var4      var5
0   Restaurant A  4.5  [AA, BB, CC]  User1  Review 1
1   Restaurant A  4.5  [AA, BB, CC]  User1  Review 2
2   Restaurant A  4.5  [AA, BB, CC]  User1  Review 3
3   Restaurant A  4.5  [AA, BB, CC]  User2  Review 1
4   Restaurant A  4.5  [AA, BB, CC]  User2  Review 2
5   Restaurant A  4.5  [AA, BB, CC]  User2  Review 3
6   Restaurant A  4.5  [AA, BB, CC]  User3  Review 1
7   Restaurant A  4.5  [AA, BB, CC]  User3  Review 2
8   Restaurant A  4.5  [AA, BB, CC]  User3  Review 3
9   Restaurant B  5.0  [AA, BB, CC]  User1  Review 1
10  Restaurant B  5.0  [AA, BB, CC]  User1  Review 2
11  Restaurant B  5.0  [AA, BB, CC]  User1  Review 3
12  Restaurant B  5.0  [AA, BB, CC]  User2  Review 1
13  Restaurant B  5.0  [AA, BB, CC]  User2  Review 2
14  Restaurant B  5.0  [AA, BB, CC]  User2  Review 3
15  Restaurant B  5.0  [AA, BB, CC]  User3  Review 1
16  Restaurant B  5.0  [AA, BB, CC]  User3  Review 2
17  Restaurant B  5.0  [AA, BB, CC]  User3  Review 3

获取用户和评论的多行是错误的。

我尝试使用以下代码解决此问题：

mva_cols = ['var4', 'var5']
counter = 0

for x in zip(mva_cols):
    pandaFile = pd.DataFrame({col:np.repeat(pandaFile[col].values, 
pandaFile[mva_cols[counter]].str.len()) for col in 
pandaFile.columns.difference([mva_cols[counter]])}).assign(**
{mva_cols[counter]:np.concatenate(pandaFile[mva_cols[counter]].values)})
[pandaFile.columns.tolist()]
    counter = counter + 1
    print(counter)
    print(str(pandaFile).encode('utf-8'))

Answer 1

或者你可以尝试

new_df=df.reindex(df.index.repeat(df.var5.str.len()))
new_df.assign(var4=df.var4.sum(),var5=df.var5.sum())
Out[1022]: 
           var1  var2          var3   var4      var5
0  Restaurant A   4.5  [AA, BB, CC]  User1  Review 1
0  Restaurant A   4.5  [AA, BB, CC]  User2  Review 2
0  Restaurant A   4.5  [AA, BB, CC]  User3  Review 3
1  Restaurant B   5.0  [AA, BB, CC]  User1  Review 1
1  Restaurant B   5.0  [AA, BB, CC]  User2  Review 2
1  Restaurant B   5.0  [AA, BB, CC]  User3  Review 3

Answer 2

这是一个解决方案：

import pandas as pd

df = pd.DataFrame([['Restaurant A', 4.5, ['AA', 'BB', 'CC'], ['User1', 'User2', 'User3'], ['Review 1', 'Review 2', 'Review 3']],    
                   ['Restaurant B', 5.0, ['AA', 'BB', 'CC'], ['User1', 'User2', 'User3'], ['Review 1', 'Review 2', 'Review 3']]],
                  columns=['var1', 'var2', 'var3', 'var4', 'var5'])

df['var6'] = list(tuple(zip(i, j)) for i, j in zip(df['var4'], df['var5']))
lens = [len(item) for item in df['var6']]

df_out = pd.DataFrame( {'var1' : np.repeat(df['var1'].values, lens), 
                        'var2' : np.repeat(df['var2'].values, lens),
                        'var3' : np.repeat(df['var3'].values, lens),
                        'var4' : np.hstack(df['var4']),
                        'var5' : np.hstack(df['var5'])
                        })

#            var1  var2          var3   var4      var5
# 0  Restaurant A   4.5  [AA, BB, CC]  User1  Review 1
# 1  Restaurant A   4.5  [AA, BB, CC]  User2  Review 2
# 2  Restaurant A   4.5  [AA, BB, CC]  User3  Review 3
# 3  Restaurant B   5.0  [AA, BB, CC]  User1  Review 1
# 4  Restaurant B   5.0  [AA, BB, CC]  User2  Review 2
# 5  Restaurant B   5.0  [AA, BB, CC]  User3  Review 3

使用多个变量转换Panda文档

2 个答案: