How to sort objects by multiple keys in Python?
This answer works for any kind of column in the dictionary -- the negated column need not be a number.
def multikeysort(items, columns):
from operator import itemgetter
comparers = [((itemgetter(col[1:].strip()), -1) if col.startswith('-') else
(itemgetter(col.strip()), 1)) for col in columns]
def comparer(left, right):
for fn, mult in comparers:
result = cmp(fn(left), fn(right))
if result:
return mult * result
else:
return 0
return sorted(items, cmp=comparer)
You can call it like this:
b = [{u'TOT_PTS_Misc': u'Utley, Alex', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Russo, Brandon', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Chappell, Justin', u'Total_Points': 96.0},
{u'TOT_PTS_Misc': u'Foster, Toney', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Lawson, Roman', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Lempke, Sam', u'Total_Points': 80.0},
{u'TOT_PTS_Misc': u'Gnezda, Alex', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Kirks, Damien', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Worden, Tom', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Korecz, Mike', u'Total_Points': 78.0},
{u'TOT_PTS_Misc': u'Swartz, Brian', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Burgess, Randy', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Smugala, Ryan', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Harmon, Gary', u'Total_Points': 66.0},
{u'TOT_PTS_Misc': u'Blasinsky, Scott', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Carter III, Laymon', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Coleman, Johnathan', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Venditti, Nick', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Blackwell, Devon', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Kovach, Alex', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Bolden, Antonio', u'Total_Points': 60.0},
{u'TOT_PTS_Misc': u'Smith, Ryan', u'Total_Points': 60.0}]
a = multikeysort(b, ['-Total_Points', 'TOT_PTS_Misc'])
for item in a:
print item
Try it with either column negated. You will see the sort order reverse.
Next: change it so it does not use extra class....
2016-01-17
Taking my inspiration from this answer What is the best way to get the first item from an iterable matching a condition?, I shortened the code:
from operator import itemgetter as i
def multikeysort(items, columns):
comparers = [
((i(col[1:].strip()), -1) if col.startswith('-') else (i(col.strip()), 1))
for col in columns
]
def comparer(left, right):
comparer_iter = (
cmp(fn(left), fn(right)) * mult
for fn, mult in comparers
)
return next((result for result in comparer_iter if result), 0)
return sorted(items, cmp=comparer)
In case you like your code terse.
Later 2016-01-17
This works with python3 (which eliminated the cmp
argument to sort
):
from operator import itemgetter as i
from functools import cmp_to_key
def cmp(x, y):
"""
Replacement for built-in function cmp that was removed in Python 3
Compare the two objects x and y and return an integer according to
the outcome. The return value is negative if x < y, zero if x == y
and strictly positive if x > y.
https://portingguide.readthedocs.io/en/latest/comparisons.html#the-cmp-function
"""
return (x > y) - (x < y)
def multikeysort(items, columns):
comparers = [
((i(col[1:].strip()), -1) if col.startswith('-') else (i(col.strip()), 1))
for col in columns
]
def comparer(left, right):
comparer_iter = (
cmp(fn(left), fn(right)) * mult
for fn, mult in comparers
)
return next((result for result in comparer_iter if result), 0)
return sorted(items, key=cmp_to_key(comparer))
Inspired by this answer How should I do custom sort in Python 3?
This article has a nice rundown on various techniques for doing this. If your requirements are simpler than "full bidirectional multikey", take a look. It's clear the accepted answer and the blog post I just referenced influenced each other in some way, though I don't know which order.
In case the link dies here's a very quick synopsis of examples not covered above:
mylist = sorted(mylist, key=itemgetter('name', 'age'))
mylist = sorted(mylist, key=lambda k: (k['name'].lower(), k['age']))
mylist = sorted(mylist, key=lambda k: (k['name'].lower(), -k['age']))
I know this is a rather old question, but none of the answers mention that Python guarantees a stable sort order for its sorting routines such as list.sort()
and sorted()
, which means items that compare equal retain their original order.
This means that the equivalent of ORDER BY name ASC, age DESC
(using SQL notation) for a list of dictionaries can be done like this:
items.sort(key=operator.itemgetter('age'), reverse=True)
items.sort(key=operator.itemgetter('name'))
Note how the items are first sorted by the "lesser" attribute age
(descending), then by the "major" attribute name
, leading to the correct final order.
The reversing/inverting works for all orderable types, not just numbers which you can negate by putting a minus sign in front.
And because of the Timsort algorithm used in (at least) CPython, this is actually rather fast in practice.