Sorting a Python list by two fields [duplicate]

Question

I have the following list created from a sorted csv

list1 = sorted(csv1, key=operator.itemgetter(1))

I would actually like to sort the list by two criteria: first by the value in field 1 and then by the value in field 2. How do I do this?

Answer 1

No need to import anything when using lambda functions.
The following sorts list by the first element, then by the second element. You can also sort by one field ascending and another descending for example:

sorted_list = sorted(list, key=lambda x: (x[0], -x[1]))

Answer 2

like this:

import operator
list1 = sorted(csv1, key=operator.itemgetter(1, 2))

Answer 3

Python has a stable sort, so provided that performance isn't an issue the simplest way is to sort it by field 2 and then sort it again by field 1.

That will give you the result you want, the only catch is that if it is a big list (or you want to sort it often) calling sort twice might be an unacceptable overhead.

list1 = sorted(csv1, key=operator.itemgetter(2))
list1 = sorted(list1, key=operator.itemgetter(1))

Doing it this way also makes it easy to handle the situation where you want some of the columns reverse sorted, just include the 'reverse=True' parameter when necessary.

Otherwise you can pass multiple parameters to itemgetter or manually build a tuple. That is probably going to be faster, but has the problem that it doesn't generalise well if some of the columns want to be reverse sorted (numeric columns can still be reversed by negating them but that stops the sort being stable).

So if you don't need any columns reverse sorted, go for multiple arguments to itemgetter, if you might, and the columns aren't numeric or you want to keep the sort stable go for multiple consecutive sorts.

Edit: For the commenters who have problems understanding how this answers the original question, here is an example that shows exactly how the stable nature of the sorting ensures we can do separate sorts on each key and end up with data sorted on multiple criteria:

DATA = [
    ('Jones', 'Jane', 58),
    ('Smith', 'Anne', 30),
    ('Jones', 'Fred', 30),
    ('Smith', 'John', 60),
    ('Smith', 'Fred', 30),
    ('Jones', 'Anne', 30),
    ('Smith', 'Jane', 58),
    ('Smith', 'Twin2', 3),
    ('Jones', 'John', 60),
    ('Smith', 'Twin1', 3),
    ('Jones', 'Twin1', 3),
    ('Jones', 'Twin2', 3)
]

# Sort by Surname, Age DESCENDING, Firstname
print("Initial data in random order")
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
First we sort by first name, after this pass all
Twin1 come before Twin2 and Anne comes before Fred''')
DATA.sort(key=lambda row: row[1])

for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
Second pass: sort by age in descending order.
Note that after this pass rows are sorted by age but
Twin1/Twin2 and Anne/Fred pairs are still in correct
firstname order.''')
DATA.sort(key=lambda row: row[2], reverse=True)
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

print('''
Final pass sorts the Jones from the Smiths.
Within each family members are sorted by age but equal
age members are sorted by first name.
''')
DATA.sort(key=lambda row: row[0])
for d in DATA:
    print("{:10s} {:10s} {}".format(*d))

This is a runnable example, but to save people running it the output is:

Initial data in random order
Jones      Jane       58
Smith      Anne       30
Jones      Fred       30
Smith      John       60
Smith      Fred       30
Jones      Anne       30
Smith      Jane       58
Smith      Twin2      3
Jones      John       60
Smith      Twin1      3
Jones      Twin1      3
Jones      Twin2      3

First we sort by first name, after this pass all
Twin1 come before Twin2 and Anne comes before Fred
Smith      Anne       30
Jones      Anne       30
Jones      Fred       30
Smith      Fred       30
Jones      Jane       58
Smith      Jane       58
Smith      John       60
Jones      John       60
Smith      Twin1      3
Jones      Twin1      3
Smith      Twin2      3
Jones      Twin2      3

Second pass: sort by age in descending order.
Note that after this pass rows are sorted by age but
Twin1/Twin2 and Anne/Fred pairs are still in correct
firstname order.
Smith      John       60
Jones      John       60
Jones      Jane       58
Smith      Jane       58
Smith      Anne       30
Jones      Anne       30
Jones      Fred       30
Smith      Fred       30
Smith      Twin1      3
Jones      Twin1      3
Smith      Twin2      3
Jones      Twin2      3

Final pass sorts the Jones from the Smiths.
Within each family members are sorted by age but equal
age members are sorted by first name.

Jones      John       60
Jones      Jane       58
Jones      Anne       30
Jones      Fred       30
Jones      Twin1      3
Jones      Twin2      3
Smith      John       60
Smith      Jane       58
Smith      Anne       30
Smith      Fred       30
Smith      Twin1      3
Smith      Twin2      3

Note in particular how in the second step the reverse=True parameter keeps the firstnames in order whereas simply sorting then reversing the list would lose the desired order for the third sort key.

Answer 4

list1 = sorted(csv1, key=lambda x: (x[1], x[2]) )

Answer 5

employees.sort(key = lambda x:x[1])
employees.sort(key = lambda x:x[0])

We can also use .sort with lambda 2 times because python sort is in place and stable. This will first sort the list according to the second element, x[1]. Then, it will sort the first element, x[0] (highest priority).

employees[0] = "Employee's Name"
employees[1] = "Employee's Salary"

This is equivalent to doing the following:

employees.sort(key = lambda x:(x[0], x[1]))

Answer 6

Sorting list of dicts using below will sort list in descending order on first column as salary and second column as age

d=[{'salary':123,'age':23},{'salary':123,'age':25}]
d=sorted(d, key=lambda i: (i['salary'], i['age']),reverse=True)

Output: [{'salary': 123, 'age': 25}, {'salary': 123, 'age': 23}]

Answer 7

If you want to sort the array based on, based on both ascending and then descending order follow the method mentioned below. For that, you can use the lambda function.
let us consider below example,
input: [[1,2],[3,3],[2,1],[1,1],[4,1],[3,1]]
expected output: [[4, 1], [3, 1], [3, 3], [2, 1], [1, 1], [1, 2]]
code used:

    arr = [[1,2],[3,3],[2,1],[1,1],[4,1],[3,1]]
    arr.sort(key=lambda ele: (ele[0], -ele[1]), reverse=True)
    # output [[4, 1], [3, 1], [3, 3], [2, 1], [1, 1], [1, 2]]

The negative sign is the reason the procedure the result expected.

Answer 8

In ascending order you can use:

sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]))

or in descending order you can use:

sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]),reverse=True)

Answer 9

After reading the answers in this thread, I wrote a general solution that will work for an arbitrary number of columns:

def sort_array(array, *columns):
    for col in columns:
        array.sort(key = lambda x:x[col])

The OP would call it like this:

sort_array(list1, 2, 1)

Which sorts first by column 2, then by column 1.
(Most important column goes last)

Answer 10

python 3 https://docs.python.org/3.5/howto/sorting.html#the-old-way-using-the-cmp-parameter

from functools import cmp_to_key

def custom_compare(x, y):
    # custom comparsion of x[0], x[1] with y[0], y[1]
    return 0

sorted(entries, key=lambda e: (cmp_to_key(custom_compare)(e[0]), e[1]))