I have a pandas dataframe. I want to print the unique values of one of its columns in ascending order. This is how I am doing it:

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

The problem is that I am getting a None for the output.

8 Answers

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 

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() 

Fastest code

for large data frames:

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

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.

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]