Custom sorting with Pandas

One approach would be to use a custom dict to create a 'rank' column, we then use to sort with and then drop the column after sorting:

In [17]:
custom_dict = {'Critical':0, 'High':1, 'Medium':2, 'Low':3}  
df['rank'] = df['Criticality'].map(custom_dict)
df

Out[17]:

  Name Criticality  rank
0  baz        High     1
1  foo    Critical     0
2  baz         Low     3
3  foo      Medium     2
4  bar        High     1
5  bar         Low     3
6  bar      Medium     2

[7 rows x 3 columns]

In [19]:
# now sort by 'Name' and 'rank', it will first sort by 'Name' column first and then 'rank'
df.sort(columns=['Name', 'rank'],inplace=True)
df

Out[19]:

  Name Criticality  rank
4  bar        High     1
6  bar      Medium     2
5  bar         Low     3
0  baz        High     1
2  baz         Low     3
1  foo    Critical     0
3  foo      Medium     2

[7 rows x 3 columns]

In [21]:
# now drop the 'rank' column
df.drop(labels=['rank'],axis=1)

Out[21]:

  Name Criticality
4  bar        High
6  bar      Medium
5  bar         Low
0  baz        High
2  baz         Low
1  foo    Critical
3  foo      Medium

[7 rows x 2 columns]

Based on the answer of EdChum, this worked for me :

custom_dict = {'Critical':0, 'High':1, 'Medium':2, 'Low':3}  

df['rank'] = df['Criticality'].map(custom_dict)

# now sort by 'Name' and 'rank', it will first sort by 'Name' column first and then 'rank'
df.sort_values(by=['Name', 'rank'],inplace=True) 

# now drop the 'rank' column
df.drop(labels=['rank'],axis=1)

Basically, I used "sort_values" instead of "sort" and 'by" instead of "columns"

I works for me using pd.Categorical

In [114]: df = pd.DataFrame({
          'Name' : ["baz","foo","baz","foo","bar","bar","bar"],
          'Criticality' : ["hi", "crt", "lo", "med", "hi", "lo", "med"]
          })

     ...: df['Criticality'] = pd.Categorical(df['Criticality'], ["crt","hi", "med", "lo"])

     ...: df.sort_values(['Name','Criticality'])
Out[114]: 
  Name Criticality
4  bar          hi
6  bar         med
5  bar          lo
0  baz          hi
2  baz          lo
1  foo         crt
3  foo         med