Return order of unique values in a pandas column

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I am trying to find a more efficient way to return the index of unique values in a pandas df

For the df below I want to return the index of the first time a unique value occurs. 

import pandas as pd
import numpy as np

d = ({
    'Day' : ['Mon','Mon','Tues','Mon','Tues','Wed'],                                
     })

df = pd.DataFrame(data=d)

I can manually counti the index of unique value and return below:

first = df.iloc[0].Location
second = df.iloc[2].Location 
third = df.iloc[5].Location

I was thinking of doing something like 

first = (df['Day'] == 'Mon')

But I still have to change this to find the 2nd, 3rd unique value. Is there a more efficient method?

pandas unique python
2 Answers

If want filter all unique index values use drop_duplicates with keep=False:

print (df['Day'].drop_duplicates(keep=False))
5    Wed
Name: Day, dtype: object

print (df['Day'].drop_duplicates(keep=False).index)
Int64Index([5], dtype='int64')

Or:

print (df.index[~df['Day'].duplicated(keep=False)])
Int64Index([5], dtype='int64')

If want filter first unique values use only drop_duplicates:

print (df['Day'].drop_duplicates())
0     Mon
2    Tues
5     Wed
Name: Day, dtype: object

print (df['Day'].drop_duplicates().index)
Int64Index([0, 2, 5], dtype='int64')
user_29010020

IIUC, you could use:

df['Day'].mask(df['Day'].duplicated(keep=False)).dropna().index

Update

Or update thanks to @Josh Friedlanders suggestion:

df[~df['Day'].duplicated(keep=False)].index
user_102015800
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