how to sort descending an alphanumeric pandas index.

Question

I have an pandas data frame that looks like:

df = DataFrame({\'id\':[\'a132\',\'a132\',\'b5789\',\'b5789\',\'c1112\',\'c1112\'], \'value\':[0,0,0,0,0,0,]}) 
         


        

Answer 1

Categoricals provide a reasonably easy way to define an arbitrary ordering

In [35]: df['id'] = df['id'].astype('category')

In [39]: df['id'] = (df['id'].cat.reorder_categories(
                         sorted(df['id'].cat.categories, key = lambda x: int(x[1:]), reverse=True)))
In [40]: df.groupby('id').sum()
Out[40]: 
       value
id          
b5789      0
c1112      0
a132       0

Answer 2

A simple solution is to:

• split the index to extract the number in a temporary key column
• sort by this column descending
• drop the temporary key column

---

df = DataFrame({'id':['a132','a132','b5789','b5789','c1112','c1112'], 'value':[0,0,0,0,0,0,]}) 

df = df.groupby('id').sum()

df['key'] = df.index
df['key'] = df['key'].str.split('(\d+)').str[1].astype(int)
df = df.sort('key', ascending=False).drop('key', axis=1)

# Result
       value
id          
b5789      0
c1112      0
a132       0