How to split lists over a range of column?

Question

I have a dataframe with several columns that contains a list inside. I want to split this list to different columns. I currently found this question here in stackoverflow, but it seem that it is only splitting the list inside 1 column, which I want to apply to multiple columns containing unequal number of objects in the list.

My df looks something like this:

     ID |  value_0  |  value_1  |  value_2  | value_3   | value_4
0   1001|[1001,1002]|   None    |   None    |   None    |  None 
1   1010|[1010,2001]|[2526,1000]|   None    |   None    |  None  
2   1100|[1234,5678]|[9101,1121]|[3141,5161]|[1718,1920]|[2122,2324]

I want to transform it to:

     ID | 0  | 1  |  2   |  3   | 4
0   1001|1001|1002| None | None | None 
1   1010|1010|2001| 2526 | 1000 | None  
2   1100|1234|5678| 9101 | 1121 | 3141 ....etc.

Currently this is my code but it only outputs a dataframe containing "None" value. I'm not sure how to fix it cause it seem that it is only getting the last column and not really splitting the list.

length = len(list(df.columns.values))-1

for i in range(length):
    temp = "value_" + str(i)
    x = df[temp]
    new_df = pd.DataFrame(df[temp].values.tolist())

The result the new_df that I got is:

   | 0
  0| None
  1| None
  2| [2122,2324]

However if I just focus of only 1 column (ie. value_0) it splits the list just fine.

new_df = pd.DataFrame(df['value_0'].values.tolist())

Any help is very much appreciated

jezrael · Accepted Answer · 2019-06-06 07:53:14Z

0

Idea is reshape values by DataFrame.stack for remove None values, so possible use DataFrame constructor, then reshape back by Series.unstack, sorting column and set default columns names:

import ast
#if strings in columns instead lists
#df.iloc[:, 1:] = df.iloc[:, 1:].applymap(ast.literal_eval)

s = df.set_index('ID', append=True).stack()

df = pd.DataFrame(s.values.tolist(), index=s.index).unstack().sort_index(axis=1, level=1)
df.columns = np.arange(len(df.columns))

df = df.reset_index(level=1)
print (df)
     ID       0       1       2       3       4       5       6       7  \
0  1001  1001.0  1002.0     NaN     NaN     NaN     NaN     NaN     NaN   
1  1010  1010.0  2001.0  2526.0  1000.0     NaN     NaN     NaN     NaN   
2  1100  1234.0  5678.0  9101.0  1121.0  3141.0  5161.0  1718.0  1920.0   

        8       9  
0     NaN     NaN  
1     NaN     NaN  
2  2122.0  2324.0

Solution for pandas 0.24+ for missing values with integers:

df = df.astype('Int64').reset_index(level=1)
print (df)
     ID     0     1     2     3     4     5     6     7     8     9
0  1001  1001  1002   NaN   NaN   NaN   NaN   NaN   NaN   NaN   NaN
1  1010  1010  2001  2526  1000   NaN   NaN   NaN   NaN   NaN   NaN
2  1100  1234  5678  9101  1121  3141  5161  1718  1920  2122  2324

edited Jun 6, 2019 at 7:53

answered Jun 6, 2019 at 7:47

jezrael

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2 Comments

Funky Over a year ago

Hi when I tried to view the whole dataframe, there seem to be NaN values in between columns that have a value. For example there would be a value in column 0 and 1, then 'NaN' value in column 3-4, then there's value again in 5-6. How can I remove this NaN values in between?

jezrael Over a year ago

@Funky - Sorry, I was offline. So added answer now. Btw, accepted answer is not recommended - check this. Only if performance is not important or small DataFrame it does not matter.

iamklaus · Accepted Answer · 2019-06-06 08:00:20Z

0

First using pd.concat and pd.Seriesto expand the list into separate columns and append to the original df,then just dropping the original columns

for i in df.columns:
    df = pd.concat([df, df[i].apply(pd.Series)], axis=1)

df.drop(['ID','value_0','value_1','value_2','value_3','value_4'], axis=1, inpalce=True)

Output

          0     0     1       0       1       0       1       0       1  \
   0   1001  1001  1002     NaN     NaN     NaN     NaN     NaN     NaN   
   1   1010  1010  2001  2526.0  1000.0     NaN     NaN     NaN     NaN   
   2   1100  1234  5678  9101.0  1121.0  3141.0  5161.0  1718.0  1920.0   

           0       1  
   0     NaN     NaN  
   1     NaN     NaN  
   2  2122.0  2324.0

answered Jun 6, 2019 at 8:00

iamklaus

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