How to sort a dataFrame in python pandas by two or more columns?

Question

Suppose I have a dataframe with columns a, b and c, I want to sort the dataframe by column b in ascending order, and by column c in descending order, how do I do this?

Solution 1

As of the 0.17.0 release, the sort method was deprecated in favor of sort_values. sort was completely removed in the 0.20.0 release. The arguments (and results) remain the same:

df.sort_values(['a', 'b'], ascending=[True, False])

---

You can use the ascending argument of sort:

df.sort(['a', 'b'], ascending=[True, False])

For example:

In [11]: df1 = pd.DataFrame(np.random.randint(1, 5, (10,2)), columns=['a','b'])

In [12]: df1.sort(['a', 'b'], ascending=[True, False])
Out[12]:
   a  b
2  1  4
7  1  3
1  1  2
3  1  2
4  3  2
6  4  4
0  4  3
9  4  3
5  4  1
8  4  1

---

As commented by @renadeen

Sort isn't in place by default! So you should assign result of the sort method to a variable or add inplace=True to method call.

that is, if you want to reuse df1 as a sorted DataFrame:

df1 = df1.sort(['a', 'b'], ascending=[True, False])

or

df1.sort(['a', 'b'], ascending=[True, False], inplace=True)

Solution 2

As of pandas 0.17.0, DataFrame.sort() is deprecated, and set to be removed in a future version of pandas. The way to sort a dataframe by its values is now is DataFrame.sort_values

As such, the answer to your question would now be

df.sort_values(['b', 'c'], ascending=[True, False], inplace=True)

Solution 3

For large dataframes of numeric data, you may see a significant performance improvement via numpy.lexsort, which performs an indirect sort using a sequence of keys:

import pandas as pd
import numpy as np

np.random.seed(0)

df1 = pd.DataFrame(np.random.randint(1, 5, (10,2)), columns=['a','b'])
df1 = pd.concat([df1]*100000)

def pdsort(df1):
    return df1.sort_values(['a', 'b'], ascending=[True, False])

def lex(df1):
    arr = df1.values
    return pd.DataFrame(arr[np.lexsort((-arr[:, 1], arr[:, 0]))])

assert (pdsort(df1).values == lex(df1).values).all()

%timeit pdsort(df1)  # 193 ms per loop
%timeit lex(df1)     # 143 ms per loop

One peculiarity is that the defined sorting order with numpy.lexsort is reversed: (-'b', 'a') sorts by series a first. We negate series b to reflect we want this series in descending order.

Be aware that np.lexsort only sorts with numeric values, while pd.DataFrame.sort_values works with either string or numeric values. Using np.lexsort with strings will give: TypeError: bad operand type for unary -: 'str'.

Comments

[renadeen]

Sort isn't in place by default! So you should assign result of the sort method to a variable or add inplace=True to method call.

[Andy Hayden]

@renadeen very good point, I've updated by answer with that comment.

[Kyle Heuton]

I was surprised to learn today that sort has been deprecated! Based on some of the opinions in this meta post: meta.stackoverflow.com/questions/297404/… I decided to add a new answer rather than attempt an edit to yours

[Andy Hayden]

@Snoozer Yeah, I don't think sort's ever going to go away (mainly as it's used extensively in Wes' book), but there has been some big changes in calling sort. Thanks! .. I really need to automate going through all my 1000s of pandas answers for deprecations!