Pythonic way to sorting list of namedtuples by field name

Question

I want to sort a list of named tuples without having to remember the index of the fieldname. My solution seems rather awkward and was hoping someone would have a more elegant solution.

from operator import itemgetter
from collections import namedtuple

Person = namedtuple('Person', 'name age score')
seq = [
    Person(name='nick', age=23, score=100),
    Person(name='bob', age=25, score=200),
]

# sort list by name
print(sorted(seq, key=itemgetter(Person._fields.index('name'))))
# sort list by age
print(sorted(seq, key=itemgetter(Person._fields.index('age'))))

Thanks, Nick

Is the field name always given as a string or does the solution by @clyfish also work? — jamylak, Aug 23 '12 at 08:51
I wasn't trying to do anything dynamic, so both solutions work perfectly. — Nick, Aug 23 '12 at 09:01

jamylak · Accepted Answer · 2016-07-06T12:58:33.570

108

from operator import attrgetter
from collections import namedtuple

Person = namedtuple('Person', 'name age score')
seq = [Person(name='nick', age=23, score=100),
       Person(name='bob', age=25, score=200)]

Sort list by name

sorted(seq, key=attrgetter('name'))

Sort list by age

sorted(seq, key=attrgetter('age'))

edited Jul 06 '16 at 12:58

answered Aug 23 '12 at 08:48

jamylak

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score 69 · Answer 2 · answered Aug 23 '12 at 08:50

69

sorted(seq, key=lambda x: x.name)
sorted(seq, key=lambda x: x.age)

answered Aug 23 '12 at 08:50

clyfish

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I think this is more elegant than using `attrgetter` – zenpoy Aug 23 '12 at 08:55
3

I prefer the attrgetter, but that is just taste. An advantage is also if I were to get the fields to sort on dynamically. Then I could just pass the string. – Nick Aug 23 '12 at 09:02
3

@zenpoy Keep in mind `attrgetter` performs much better and `lambda`s arent usually considered elegant – jamylak Dec 10 '14 at 22:45
2

and `sorted(seq, key=lambda x: [x.age, x.name])` sorts by multiple attributes – SpeedCoder5 May 24 '19 at 00:00

André C. Andersen · Answer 3 · 2017-05-27T17:40:37.623

I tested the two alternatives given here for speed, since @zenpoy was concerned about performance.

Testing script:

import random
from collections import namedtuple
from timeit import timeit
from operator import attrgetter

runs = 10000
size = 10000
random.seed = 42
Person = namedtuple('Person', 'name,age')
seq = [Person(str(random.randint(0, 10 ** 10)), random.randint(0, 100)) for _ in range(size)]

def attrgetter_test_name():
    return sorted(seq.copy(), key=attrgetter('name'))

def attrgetter_test_age():
    return sorted(seq.copy(), key=attrgetter('age'))

def lambda_test_name():
    return sorted(seq.copy(), key=lambda x: x.name)

def lambda_test_age():
    return sorted(seq.copy(), key=lambda x: x.age)

print('attrgetter_test_name', timeit(stmt=attrgetter_test_name, number=runs))
print('attrgetter_test_age', timeit(stmt=attrgetter_test_age, number=runs))
print('lambda_test_name', timeit(stmt=lambda_test_name, number=runs))
print('lambda_test_age', timeit(stmt=lambda_test_age, number=runs))

Results:

attrgetter_test_name 44.26793992166096
attrgetter_test_age 31.98247099677627
lambda_test_name 47.97959511074551
lambda_test_age 35.69356267603864

Using lambda was indeed slower. Up to 10% slower.

EDIT:

Further testing shows the results when sorting using multiple attributes. Added the following two test cases with the same setup:

def attrgetter_test_both():
    return sorted(seq.copy(), key=attrgetter('age', 'name'))

def lambda_test_both():
    return sorted(seq.copy(), key=lambda x: (x.age, x.name))

print('attrgetter_test_both', timeit(stmt=attrgetter_test_both, number=runs))
print('lambda_test_both', timeit(stmt=lambda_test_both, number=runs))

Results:

attrgetter_test_both 92.80101586919373
lambda_test_both 96.85089983147456

Lambda still underperforms, but less so. Now about 5% slower.

Testing is done on Python 3.6.0.

Thanks for the multi attribute sort case :) – Black Stallion Sep 12 '18 at 09:50 — Black Stallion, Sep 12 '18 at 09:50

score 4 · Answer 4 · answered Mar 10 '18 at 19:18

since nobody mentioned using itemgetter(), here how you do using itemgetter().

from operator import itemgetter
from collections import namedtuple

Person = namedtuple('Person', 'name age score')
seq = [
    Person(name='nick', age=23, score=100),
    Person(name='bob', age=25, score=200),
]

# sort list by name
print(sorted(seq, key=itemgetter(0)))

# sort list by age
print(sorted(seq, key=itemgetter(1)))

Sumudu Fernando · Answer 5 · 2021-10-13T09:46:58.340

4

This might be a bit too 'magical' for some, but I'm partial to:

# sort list by name
print(sorted(seq, key=Person.name.fget))

Edit: this assumes namedtuple uses the property() built-in to implement the accessors, because it leverages the fget attribute on such a property (see documentation). This may still be true in some implementations but it seems CPython no longer does that, which I think is related to optimization work referenced in https://bugs.python.org/issue32492 (so, since 3.8). Such fragility is the cost of the "magic" I mentioned; namedtuple certainly doesn't promise to use property().

Writing Person.name.__get__ is better (works before & after the implementation change) but is maybe not worth the arcaneness vs. just writing it more plainly as lambda p: p.name

edited Oct 13 '21 at 09:46

answered Jan 10 '19 at 07:39

Sumudu Fernando

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Could you please add some context to your answer? I tried it but am getting *AttributeError: '_collections._tuplegetter' object has no attribute 'fget'*. – simlev Oct 12 '21 at 08:52
Sure. This approach used to work but it depended on a particular implementation detail of namedtuple. That seems to have been changed in https://bugs.python.org/issue32492. Will add an edit to the answer shortly. – Sumudu Fernando Oct 13 '21 at 09:16

Pythonic way to sorting list of namedtuples by field name

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