Custom sorting in pandas dataframe

Question

I have python pandas dataframe, in which a column contains month name.

How can I do a custom sort using a dictionary, for example:

custom_dict = {'March':0, 'April':1, 'Dec':3}  

Answer 1

import pandas as pd
custom_dict = {'March':0,'April':1,'Dec':3}

df = pd.DataFrame(...) # with columns April, March, Dec (probably alphabetically)

df = pd.DataFrame(df, columns=sorted(custom_dict, key=custom_dict.get))

returns a DataFrame with columns March, April, Dec

Answer 2

A bit late to the game, but here's a way to create a function that sorts pandas Series, DataFrame, and multiindex DataFrame objects using arbitrary functions.

I make use of the df.iloc[index] method, which references a row in a Series/DataFrame by position (compared to df.loc, which references by value). Using this, we just have to have a function that returns a series of positional arguments:

def sort_pd(key=None,reverse=False,cmp=None):
    def sorter(series):
        series_list = list(series)
        return [series_list.index(i) 
           for i in sorted(series_list,key=key,reverse=reverse,cmp=cmp)]
    return sorter

You can use this to create custom sorting functions. This works on the dataframe used in Andy Hayden's answer:

df = pd.DataFrame([
    [1, 2, 'March'],
    [5, 6, 'Dec'],
    [3, 4, 'April']], 
  columns=['a','b','m'])

custom_dict = {'March':0, 'April':1, 'Dec':3}
sort_by_custom_dict = sort_pd(key=custom_dict.get)

In [6]: df.iloc[sort_by_custom_dict(df['m'])]
Out[6]:
   a  b  m
0  1  2  March
2  3  4  April
1  5  6  Dec

This also works on multiindex DataFrames and Series objects:

months = ['Jan','Feb','Mar','Apr','May','Jun','Jul','Aug','Sep','Oct','Nov','Dec']

df = pd.DataFrame([
    ['New York','Mar',12714],
    ['New York','Apr',89238],
    ['Atlanta','Jan',8161],
    ['Atlanta','Sep',5885],
  ],columns=['location','month','sales']).set_index(['location','month'])

sort_by_month = sort_pd(key=months.index)

In [10]: df.iloc[sort_by_month(df.index.get_level_values('month'))]
Out[10]:
                 sales
location  month  
Atlanta   Jan    8161
New York  Mar    12714
          Apr    89238
Atlanta   Sep    5885

sort_by_last_digit = sort_pd(key=lambda x: x%10)

In [12]: pd.Series(list(df['sales'])).iloc[sort_by_last_digit(df['sales'])]
Out[12]:
2    8161
0   12714
3    5885
1   89238

To me this feels clean, but it uses python operations heavily rather than relying on optimized pandas operations. I haven't done any stress testing but I'd imagine this could get slow on very large DataFrames. Not sure how the performance compares to adding, sorting, then deleting a column. Any tips on speeding up the code would be appreciated!

Answer 3

Pandas 0.15 introduced Categorical Series, which allows a much clearer way to do this:

First make the month column a categorical and specify the ordering to use.

In [21]: df['m'] = pd.Categorical(df['m'], ["March", "April", "Dec"])

In [22]: df  # looks the same!
Out[22]:
   a  b      m
0  1  2  March
1  5  6    Dec
2  3  4  April

Now, when you sort the month column it will sort with respect to that list:

In [23]: df.sort("m")
Out[23]:
   a  b      m
0  1  2  March
2  3  4  April
1  5  6    Dec

Note: if a value is not in the list it will be converted to NaN.

---

An older answer for those interested...

You could create an intermediary series, and set_index on that:

df = pd.DataFrame([[1, 2, 'March'],[5, 6, 'Dec'],[3, 4, 'April']], columns=['a','b','m'])
s = df['m'].apply(lambda x: {'March':0, 'April':1, 'Dec':3}[x])
s.sort()

In [4]: df.set_index(s.index).sort()
Out[4]: 
   a  b      m
0  1  2  March
1  3  4  April
2  5  6    Dec

---

As commented, in newer pandas, Series has a replace method to do this more elegantly:

s = df['m'].replace({'March':0, 'April':1, 'Dec':3})

The slight difference is that this won't raise if there is a value outside of the dictionary (it'll just stay the same).