Python: Split pandas dataframe column containing a list and a value into two columns

Question

I have a Pandas data frame with a column that contains a list and a value: ([z, z, z, z, m, ., c, l, u, b, .], 0.0)

How do I split this column into two columns that I add to the data frame? The output I want: one column will contain the list, the other column will contain the value. For example:

[z, z, z, z, m, ., c, l, u, b, .] and 0.0

I have tried str.split(...,expand=True,) but the output is just a column of NaN . I can't use the comma delimiter and ], both produce one column of NaN rather than a column of lists and a column of values.

Here's 4 rows of the column of my Pandas data frame that I'm trying to manipulate.

X['set']
1                  ([z, z, z, z, m, ., c, l, u, b, .], 0.0)
2                  ([z, z, z, z, g, ., c, l, u, b, .], 0.0)
3              ([z, z, z, z, cy, s, ., l, o, a, n, .], 0.0)
4                        ([z, z, z, x, c, ., u, s, .], 0.0)

Answer 1

I was able to figure it out based on deduction using the answers of other users.

pd.DataFrame(X['set'].tolist(), index=df.index)

Related post: how to split column of tuples in pandas dataframe?

Answer 2

can you try making the delimiter ], ?

Answer 3

You just need a bit of string gymnastics:

def separate(x):
    closing_bracket_index = x.index(']')
    list_vals = x[:closing_bracket_index+1]
    val = x[closing_bracket_index+3:]

    return pd.Series([list_vals, val], index=['list', 'val'])

X['set'].apply(separate)

Answer 4

Hope this works

import numpy as np
import pandas as pd

a = (['g','f'],0.0)
b = (['d','e'],0.1)
df = pd.DataFrame({'col':[a,b]})
df

Out[1]: 
             col
0  ([g, f], 0.0)
1  ([d, e], 0.1)

def split_val(col):
    list_val = col[0]
    value    = col[1]
    return pd.Series([list_val, value], index=['list', 'val'])


df[['list_val','value']] = df['col'].apply(split_val) 
df

Out[2]: 
             col list_val  value
0  [[g, f], 0.0]   [g, f]    0.0
1  [[d, e], 0.1]   [d, e]    0.1