I think you're almost there, try removing the extra square brackets around the lst's (Also you don't need to specify the column names when you're creating a dataframe from a dict like this):

import pandas as pd
lst1 = range(100)
lst2 = range(100)
lst3 = range(100)
percentile_list = pd.DataFrame(
    {'lst1Title': lst1,
     'lst2Title': lst2,
     'lst3Title': lst3
    })

percentile_list
    lst1Title  lst2Title  lst3Title
0          0         0         0
1          1         1         1
2          2         2         2
3          3         3         3
4          4         4         4
5          5         5         5
6          6         6         6
...

If you need a more performant solution you can use np.column_stack rather than zip as in your first attempt, this has around a 2x speedup on the example here, however comes at bit of a cost of readability in my opinion:

import numpy as np
percentile_list = pd.DataFrame(np.column_stack([lst1, lst2, lst3]), 
                               columns=['lst1Title', 'lst2Title', 'lst3Title'])

Starting from Pandas 0.25.0, there is internal method DataFrame.explode(), which was designed just for that:

res = df.explode("b")

output

In [98]: res
Out[98]:
   a  b
0  1  1
0  1  2
1  2  2
1  2  3
1  2  4
2  3  5

Solution for Pandas versions < 0.25: generic vectorized approach - will work also for multiple columns DFs:

assuming we have the following DF:

In [159]: df
Out[159]:
   a          b  c
0  1     [1, 2]  5
1  2  [2, 3, 4]  6
2  3        [5]  7

Solution:

In [160]: lst_col = 'b'

In [161]: pd.DataFrame({
     ...:     col:np.repeat(df[col].values, df[lst_col].str.len())
     ...:     for col in df.columns.difference([lst_col])
     ...: }).assign(**{lst_col:np.concatenate(df[lst_col].values)})[df.columns.tolist()]
     ...:
Out[161]:
   a  b  c
0  1  1  5
1  1  2  5
2  2  2  6
3  2  3  6
4  2  4  6
5  3  5  7

Setup:

df = pd.DataFrame({
    "a" : [1,2,3],
    "b" : [[1,2],[2,3,4],[5]],
    "c" : [5,6,7]
})

Vectorized NumPy approach:

In [124]: pd.DataFrame({'a':np.repeat(df.a.values, df.b.str.len()),
                        'b':np.concatenate(df.b.values)})
Out[124]:
   a  b
0  1  1
1  1  2
2  2  2
3  2  3
4  2  4
5  3  5

OLD answer:

Try this:

In [89]: df.set_index('a', append=True).b.apply(pd.Series).stack().reset_index(level=[0, 2], drop=True).reset_index()
Out[89]:
   a    0
0  1  1.0
1  1  2.0
2  2  2.0
3  2  3.0
4  2  4.0
5  3  5.0

Or bit nicer solution provided by @Boud:

In [110]: df.set_index('a').b.apply(pd.Series).stack().reset_index(level=-1, drop=True).astype(int).reset_index()
Out[110]:
   a  0
0  1  1
1  1  2
2  2  2
3  2  3
4  2  4
5  3  5

You can call DataFrame constructor after zipping both lists, where A, B represents column names and a, b are lists

df = pd.DataFrame(columns=['A','B'], data=zip(a, b))

If lists are of uneven lengths

from itertools import zip_longest
df = pd.DataFrame(columns=['A','B'], data=zip_longest(a, b)

import itertools
import pandas as pd
import numpy as np

def melt_series(s):
    lengths = s.str.len().values
    flat = [i for i in itertools.chain.from_iterable(s.values.tolist())]
    idx = np.repeat(s.index.values, lengths)
    return pd.Series(flat, idx, name=s.name)


melt_series(df.type).to_frame().join(df.drop('type', 1)).reindex_axis(df.columns, 1)

setup

df = pd.DataFrame({'key':[1,2,3],
                   'type':[[1,3],[1,2,3],[1,2]],
                   'value':[5,1,8]})
df

We can solve this using pandas.DataFrame.explode function which was introduced in version 0.25.0 if you have same or higher version, you can use below code.
explode function reference: https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.explode.html

import pandas as pd
import ast

data = {
    'Location_City': ['Los Angeles','Texas'],
    'Location_State': ['CA','TX'],
    'Name': ['John','Jack'],
    'hobbies': ["['Music', 'Running']", "['Swimming', 'Trekking']"]
}
df = pd.DataFrame(data)

# Converting a string representation of a list into an actual list object

list_eval = lambda x: ast.literal_eval(x)
df['hobbies'] = df['hobbies'].apply(list_eval)

# Exploding the list
df = df.explode('hobbies')

print(df)

  Location_City Location_State  Name   hobbies
0   Los Angeles             CA  John     Music
0   Los Angeles             CA  John   Running
1         Texas             TX  Jack  Swimming
1         Texas             TX  Jack  Trekking

You can use itertools.chain to flatten each list, construct a dictionary with the flattened lists and cast it to a DataFrame:

from itertools import chain
A, B, C = [list(chain.from_iterable(lst)) for lst in [List_a, List_b, List_c]]
out = pd.DataFrame({'A': A, 'B': B, 'C': C})

Output:

     A   B   C
0    1  16  31
1    2  17  32
2    3  18  33
3    4  19  34
4    5  20  35
5    6  21  36
6    7  22  37
7    8  23  38
8    9  24  39
9   10  25  40
10  11  26  41
11  12  27  42
12  13  28  43
13  14  29  44
14  15  30  45

You can use pandas for this:

import pandas as pd
import io

data = '''ID,dep_id
EMP1,"1,2,3"
EMP2,"4,5,6"'''

df = pd.read_csv(io.StringIO(data)) # insert your csv here
df['dep_id'] = ('['+df['dep_id']+ ']').apply(pd.eval) # turn string to list
df.explode('dep_id')df.explode('dep_id').reset_index(drop=True)

Output:

Here you go, just add column 'C' and you are sorted. Just to note, you must be seeking a pandas solution because merge is a pandas command and you're asking for a Python solution.

import pandas as pd

a = ["house","garden", "living room", "dog","cat"]
b= ["cat","dog", "chicken"]
df = pd.DataFrame(a, columns = ['a'])
df2 = pd.DataFrame(b, columns = ['b'])
dfa = df['a'].value_counts()
dfa.columns = ['a']
dfb = df2['b'].value_counts()
dfb.columns = ['b']
dfa = dfa.to_frame()
dfb = dfb.to_frame()
df3 = dfa.join(dfb).replace(np.nan, 0).astype(int)
print (df3)

Output

             a  b
house        1  0
garden       1  0
living room  1  0
dog          1  1
cat          1  1

and if you want to remove the 'a' column

print (df3.drop('a', axis=1))

             b
house        0
garden       0
living room  0
dog          1
cat          1

Notes The key command here is value_counts and makes a frequency plot. Its output is a Series rather than a DataFrame so it needs converting.

If you want to keep all inputs the commands is

df3 = pd.concat([dfa, dfb]).replace(np.nan, 0).astype(int)

Alternatively,

a = ["house","garden", "living room", "dog","cat"]
b= ["cat","dog", "chicken"]
df = pd.DataFrame(a).value_counts().to_frame('a')
df2 = pd.DataFrame(b).value_counts().to_frame('b')
df3 = df.join(df2).replace(np.nan, 0).astype(int).rename_axis(None)
print (df3.drop('a', axis=1))