Find the unique values in a column and then sort them

sorted(iterable): Return a new sorted list from the items in iterable.

CODE

import pandas as pd
df = pd.DataFrame({'A':[1,1,3,2,6,2,8]})
a = df['A'].unique()
print(sorted(a))

OUTPUT

[1, 2, 3, 6, 8]

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]

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

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()