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Need help understanding variable assignment, pointers, ...

The following is reproducible.

import pandas as pd

df = pd.DataFrame(
 'listData': [
 ['c', 'f', 'd', 'a', 'e', 'b'], 
 [5, 2, 1, 4, 3]
 ])

df['listDataSort'] = df['listData']

gives:

 listData listDataSort
0 [c, f, d, a, e, b] [c, f, d, a, e, b]
1 [5, 2, 1, 4, 3] [5, 2, 1, 4, 3]

If I only want to sort the lists in the listDataSort column, I might try:

df['listDataSort'].apply(lambda l: l.sort())
df

However, that sorts the lists in both columns, in-place.

 listData listDataSort
0 [a, b, c, d, e, f] [a, b, c, d, e, f]
1 [1, 2, 3, 4, 5] [1, 2, 3, 4, 5]

I can fix this by instead doing:

df = pd.DataFrame(
 'listData': [
 ['c', 'f', 'd', 'a', 'e', 'b'], 
 [5, 2, 1, 4, 3]
 ])

df['listDataSort'] = df['listData'].apply(sorted)

giving:

 listData listDataSort
0 [c, f, d, a, e, b] [a, b, c, d, e, f]
1 [5, 2, 1, 4, 3] [1, 2, 3, 4, 5]

Assigning df to a different variable, say df2 still changes everything back to the original source list. Furthermore, how do I create a new dataframe based on an existing dataframe so I can make changes to the new dataframe without making the same changes to the existing dataframe?

df = pd.DataFrame(
 'listData': [
 ['c', 'f', 'd', 'a', 'e', 'b'], 
 [5, 2, 1, 4, 3]
 ])

df2 = df
print('ndfn', df)
print('ndf2n', df2)

df2['listDataSort'] = df2['listData']
print('ndfn', df)
print('ndf2n', df2)

df2['listDataSort'].apply(lambda l: l.sort())
print('ndfn', df)
print('ndf2n', df2)

prints:

df
 listData
0 [c, f, d, a, e, b]
1 [5, 2, 1, 4, 3]

df2
 listData
0 [c, f, d, a, e, b]
1 [5, 2, 1, 4, 3]

df
 listData listDataSort
0 [c, f, d, a, e, b] [c, f, d, a, e, b]
1 [5, 2, 1, 4, 3] [5, 2, 1, 4, 3]

df2
 listData listDataSort
0 [c, f, d, a, e, b] [c, f, d, a, e, b]
1 [5, 2, 1, 4, 3] [5, 2, 1, 4, 3]

df
 listData listDataSort
0 [a, b, c, d, e, f] [a, b, c, d, e, f]
1 [1, 2, 3, 4, 5] [1, 2, 3, 4, 5]

df2
 listData listDataSort
0 [a, b, c, d, e, f] [a, b, c, d, e, f]
1 [1, 2, 3, 4, 5] [1, 2, 3, 4, 5]

also:

df = pd.DataFrame(
 'listData': [
 ['c', 'f', 'd', 'a', 'e', 'b'], 
 [5, 2, 1, 4, 3]
 ])
print('ndfn', df)

df3 = df
df3['listDataSort'] = df3['listData'].apply(sorted)
print('ndfn', df)
print('ndf3n', df3)

prints:

df
 listData
0 [c, f, d, a, e, b]
1 [5, 2, 1, 4, 3]

df
 listData listDataSort
0 [c, f, d, a, e, b] [a, b, c, d, e, f]
1 [5, 2, 1, 4, 3] [1, 2, 3, 4, 5]

df3
 listData listDataSort
0 [c, f, d, a, e, b] [a, b, c, d, e, f]
1 [5, 2, 1, 4, 3] [1, 2, 3, 4, 5]

When you run

df['listDataSort'] = df['listData']

All you do is copy the references of the lists to new columns. This means only a shallow copy is performed and both columns reference the same lists. So any change to one column will likely affect another.

You can use a list comprehension with sorted which returns a copy of the data. This should be the easiest option for you.

df['listDataSort'] = [sorted(x) for x in df['listDataSort']]
df

 listData listDataSort
0 [c, f, d, a, e, b] [a, b, c, d, e, f]
1 [5, 2, 1, 4, 3] [1, 2, 3, 4, 5]

Now, when it comes to the problem of making a copy of the entire DataFrame, things are a little more complicated. I would recommend deepcopy:

import copy
df2 = df.apply(copy.deepcopy)