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split the text by comma and append rows pandas

0

I have a dataframe a like below:

df = pd.DataFrame(improve this question

asked Mar 6 at 17:24

Kumar AKKumar AK

173213

0

I have a dataframe a like below:

df = pd.DataFrame(improve this question

asked Mar 6 at 17:24

Kumar AKKumar AK

173213

df = pd.DataFrame( 
 
 

2 Answers
2

active

oldest

votes

1

May be this:

import ast
df[['Text1','Text2']]=df.pop('Text').str.split(":",expand=True)
df.Text2=df.Text2.replace("}","",regex=True).apply(ast.literal_eval)
df.Text1=df.Text1.replace("W",'',regex=True)

s=pd.DataFrame('B':np.concatenate(df.Text2.values),index=df.index.repeat(df.Text2.str.len()))
df.join(s).drop('Text2',1).rename(columns='B':'Text2')

Output

 User Text1 Text2
0 101 y 8
0 101 y 8 plus
1 102 x 7
2 103 x 7
2 103 x 7+
2 103 x 7++
3 104 x 7

share|improve this answer

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks but the code is. It dynamic. Suppose we have 3 or 4 values.it will throw error. It only works 2 vlaues.
  
  – Kumar AK
  Mar 7 at 6:43
  

  
  
  
  

add a comment | 

1

Assuming the first dictionary is missing the closing bracket (}). You could use ast.literal_eval:

import ast
import pandas as pd

df = pd.DataFrame('User': ['101', '102', '103', '104'],
 'Text': [""""y":["8","8 plus"]""", """"x":["7"]""", """"x":["7","7+","7++"]""",
 """"x":["7"]"""])

# convert to dictionary and drop the text column
dictionaries = df.assign(D=df.Text.apply(ast.literal_eval)).drop('Text', axis=1)

# convert each row to multiple ones (given by the values of each dictionary)
tuples = [(u, k, v) for u, r in dictionaries.values for k, vs in r.items() for v in vs]

result = pd.DataFrame(tuples, columns=['User', 'Text1', 'Text2'])
print(result)

Output

 User Text1 Text2
0 101 y 8
1 101 y 8 plus
2 102 x 7
3 103 x 7
4 103 x 7+
5 103 x 7++
6 104 x 7

share|improve this answer

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks but it's a string not dictionary.
  
  – Kumar AK
  Mar 7 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KumarAK Is a string representing a dictionary right?
  
  – Daniel Mesejo
  Mar 7 at 10:01
  

  
  
  
  

add a comment | 

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1

May be this:

import ast
df[['Text1','Text2']]=df.pop('Text').str.split(":",expand=True)
df.Text2=df.Text2.replace("}","",regex=True).apply(ast.literal_eval)
df.Text1=df.Text1.replace("W",'',regex=True)

s=pd.DataFrame('B':np.concatenate(df.Text2.values),index=df.index.repeat(df.Text2.str.len()))
df.join(s).drop('Text2',1).rename(columns='B':'Text2')

Output

 User Text1 Text2
0 101 y 8
0 101 y 8 plus
1 102 x 7
2 103 x 7
2 103 x 7+
2 103 x 7++
3 104 x 7

share|improve this answer

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks but the code is. It dynamic. Suppose we have 3 or 4 values.it will throw error. It only works 2 vlaues.
  
  – Kumar AK
  Mar 7 at 6:43
  

  
  
  
  

add a comment | 

1

May be this:

import ast
df[['Text1','Text2']]=df.pop('Text').str.split(":",expand=True)
df.Text2=df.Text2.replace("}","",regex=True).apply(ast.literal_eval)
df.Text1=df.Text1.replace("W",'',regex=True)

s=pd.DataFrame('B':np.concatenate(df.Text2.values),index=df.index.repeat(df.Text2.str.len()))
df.join(s).drop('Text2',1).rename(columns='B':'Text2')

Output

 User Text1 Text2
0 101 y 8
0 101 y 8 plus
1 102 x 7
2 103 x 7
2 103 x 7+
2 103 x 7++
3 104 x 7

share|improve this answer

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks but the code is. It dynamic. Suppose we have 3 or 4 values.it will throw error. It only works 2 vlaues.
  
  – Kumar AK
  Mar 7 at 6:43
  

  
  
  
  

add a comment | 

May be this:

import ast
df[['Text1','Text2']]=df.pop('Text').str.split(":",expand=True)
df.Text2=df.Text2.replace("}","",regex=True).apply(ast.literal_eval)
df.Text1=df.Text1.replace("W",'',regex=True)

s=pd.DataFrame('B':np.concatenate(df.Text2.values),index=df.index.repeat(df.Text2.str.len()))
df.join(s).drop('Text2',1).rename(columns='B':'Text2')

Output

 User Text1 Text2
0 101 y 8
0 101 y 8 plus
1 102 x 7
2 103 x 7
2 103 x 7+
2 103 x 7++
3 104 x 7

share|improve this answer

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

import ast
df[['Text1','Text2']]=df.pop('Text').str.split(":",expand=True)
df.Text2=df.Text2.replace("}","",regex=True).apply(ast.literal_eval)
df.Text1=df.Text1.replace("W",'',regex=True)

s=pd.DataFrame('B':np.concatenate(df.Text2.values),index=df.index.repeat(df.Text2.str.len()))
df.join(s).drop('Text2',1).rename(columns='B':'Text2')

 User Text1 Text2
0 101 y 8
0 101 y 8 plus
1 102 x 7
2 103 x 7
2 103 x 7+
2 103 x 7++
3 104 x 7

share|improve this answer

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

answered Mar 6 at 17:36

anky_91anky_91

8,1122721

1

Assuming the first dictionary is missing the closing bracket (}). You could use ast.literal_eval:

import ast
import pandas as pd

df = pd.DataFrame('User': ['101', '102', '103', '104'],
 'Text': [""""y":["8","8 plus"]""", """"x":["7"]""", """"x":["7","7+","7++"]""",
 """"x":["7"]"""])

# convert to dictionary and drop the text column
dictionaries = df.assign(D=df.Text.apply(ast.literal_eval)).drop('Text', axis=1)

# convert each row to multiple ones (given by the values of each dictionary)
tuples = [(u, k, v) for u, r in dictionaries.values for k, vs in r.items() for v in vs]

result = pd.DataFrame(tuples, columns=['User', 'Text1', 'Text2'])
print(result)

Output

 User Text1 Text2
0 101 y 8
1 101 y 8 plus
2 102 x 7
3 103 x 7
4 103 x 7+
5 103 x 7++
6 104 x 7

share|improve this answer

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks but it's a string not dictionary.
  
  – Kumar AK
  Mar 7 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KumarAK Is a string representing a dictionary right?
  
  – Daniel Mesejo
  Mar 7 at 10:01
  

  
  
  
  

add a comment | 

1

Assuming the first dictionary is missing the closing bracket (}). You could use ast.literal_eval:

import ast
import pandas as pd

df = pd.DataFrame('User': ['101', '102', '103', '104'],
 'Text': [""""y":["8","8 plus"]""", """"x":["7"]""", """"x":["7","7+","7++"]""",
 """"x":["7"]"""])

# convert to dictionary and drop the text column
dictionaries = df.assign(D=df.Text.apply(ast.literal_eval)).drop('Text', axis=1)

# convert each row to multiple ones (given by the values of each dictionary)
tuples = [(u, k, v) for u, r in dictionaries.values for k, vs in r.items() for v in vs]

result = pd.DataFrame(tuples, columns=['User', 'Text1', 'Text2'])
print(result)

Output

 User Text1 Text2
0 101 y 8
1 101 y 8 plus
2 102 x 7
3 103 x 7
4 103 x 7+
5 103 x 7++
6 104 x 7

share|improve this answer

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Thanks but it's a string not dictionary.
  
  – Kumar AK
  Mar 7 at 6:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KumarAK Is a string representing a dictionary right?
  
  – Daniel Mesejo
  Mar 7 at 10:01
  

  
  
  
  

add a comment | 

Assuming the first dictionary is missing the closing bracket (}). You could use ast.literal_eval:

import ast
import pandas as pd

df = pd.DataFrame('User': ['101', '102', '103', '104'],
 'Text': [""""y":["8","8 plus"]""", """"x":["7"]""", """"x":["7","7+","7++"]""",
 """"x":["7"]"""])

# convert to dictionary and drop the text column
dictionaries = df.assign(D=df.Text.apply(ast.literal_eval)).drop('Text', axis=1)

# convert each row to multiple ones (given by the values of each dictionary)
tuples = [(u, k, v) for u, r in dictionaries.values for k, vs in r.items() for v in vs]

result = pd.DataFrame(tuples, columns=['User', 'Text1', 'Text2'])
print(result)

Output

 User Text1 Text2
0 101 y 8
1 101 y 8 plus
2 102 x 7
3 103 x 7
4 103 x 7+
5 103 x 7++
6 104 x 7

share|improve this answer

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433

import ast
import pandas as pd

df = pd.DataFrame('User': ['101', '102', '103', '104'],
 'Text': [""""y":["8","8 plus"]""", """"x":["7"]""", """"x":["7","7+","7++"]""",
 """"x":["7"]"""])

# convert to dictionary and drop the text column
dictionaries = df.assign(D=df.Text.apply(ast.literal_eval)).drop('Text', axis=1)

# convert each row to multiple ones (given by the values of each dictionary)
tuples = [(u, k, v) for u, r in dictionaries.values for k, vs in r.items() for v in vs]

result = pd.DataFrame(tuples, columns=['User', 'Text1', 'Text2'])
print(result)

 User Text1 Text2
0 101 y 8
1 101 y 8 plus
2 102 x 7
3 103 x 7
4 103 x 7+
5 103 x 7++
6 104 x 7

share|improve this answer

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433

answered Mar 6 at 17:51

Daniel MesejoDaniel Mesejo

18.7k21433