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}
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}
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).
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
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!