sort sorts inplace so returns nothing:

In [54]:
df = pd.DataFrame({'A':[1,1,3,2,6,2,8]})
a = df['A'].unique()
a.sort()
a

Out[54]:
array([1, 2, 3, 6, 8], dtype=int64)

So you have to call print a again after the call to sort.

Eg.:

In [55]:
df = pd.DataFrame({'A':[1,1,3,2,6,2,8]})
a = df['A'].unique()
a.sort()
print(a)

[1 2 3 6 8]

I would suggest using numpy’s sort, as it is anyway what pandas is doing in background:

import numpy as np
np.sort(df.A.unique())

But doing all in pandas is valid as well.

You can also use the drop_duplicates() instead of unique()

df = pd.DataFrame({'A':[1,1,3,2,6,2,8]})
a = df['A'].drop_duplicates()
a.sort()
print a

I prefer the oneliner:

print(sorted(df['Column Name'].unique()))

Came across the question myself today. I think the reason that your code returns ‘None’ (exactly what I got by using the same method) is that

a.sort()

is calling the sort function to mutate the list a. In my understanding, this is a modification command. To see the result you have to use print(a).

My solution, as I tried to keep everything in pandas:

pd.Series(df['A'].unique()).sort_values()

Another way is using set data type.

Some characteristic of Sets: Sets are unordered, can include mixed data types, elements in a set cannot be repeated, are mutable.

Solving your question:

df = pd.DataFrame({'A':[1,1,3,2,6,2,8]})
sorted(set(df.A))

The answer in List type:

[1, 2, 3, 6, 8]

Fastest code

for large data frames:

df['A'].drop_duplicates().sort_values()

Surprised no one suggested this:

df['A'].sort_values().unique()