Custom Python List Sorting
I Was Refactoring Some Old Code of Mine and Came Across of This: Alist. Sort(Cmp_Items) Def Cmp_Items(A, B): If A. Foo > B. Foo: Return 1 Elif A. Foo == B...
I was refactoring some old code of mine and came across of this:
alist.sort(cmp_items)
def cmp_items(a, b):
if a.foo > b.foo:
return 1
elif a.foo == b.foo:
return 0
else:
return -1
The code works (and I wrote it some 3 years ago!) but I cannot find this thing documented anywhere in the Python docs and everybody uses sorted() to implement custom sorting. Can someone explain why this works?
6 Answers
As a side note, here is a better alternative to implement the same sorting:
alist.sort(key=lambda x: x.foo)
Or alternatively:
import operator
alist.sort(key=operator.attrgetter('foo'))
Check out the Sorting How To, it is very useful.
It's documented here.
The sort() method takes optional arguments for controlling the comparisons.
cmp specifies a custom comparison function of two arguments (list items) which should return a negative, zero or positive number depending on whether the first argument is considered smaller than, equal to, or larger than the second argument: cmp=lambda x,y: cmp(x.lower(), y.lower()). The default value is None.
Just like this example. You want sort this list.
[('c', 2), ('b', 2), ('a', 3)]
output:
[('a', 3), ('b', 2), ('c', 2)]
you should sort the tuples by the second item, then the first:
def letter_cmp(a, b):
if a[1] > b[1]:
return -1
elif a[1] == b[1]:
if a[0] > b[0]:
return 1
else:
return -1
else:
return 1
Then convert it to a key function:
from functools import cmp_to_key
letter_cmp_key = cmp_to_key(letter_cmp))
Now you can use your custom sort order:
[('c', 2), ('b', 2), ('a', 3)].sort(key=letter_cmp_key)
This does not work in Python 3.
You can use functools cmp_to_key to have old-style comparison functions work though.
from functools import cmp_to_key
def cmp_items(a, b):
if a.foo > b.foo:
return 1
elif a.foo == b.foo:
return 0
else:
return -1
cmp_items_py3 = cmp_to_key(cmp_items)
alist.sort(cmp_items_py3)
I know many have already posted some good answers. However I want to suggest one nice and easy method without importing any library.
l = [(2, 3), (3, 4), (2, 4)]
l.sort(key = lambda x: (-x[0], -x[1]) )
print(l)
l.sort(key = lambda x: (x[0], -x[1]) )
print(l)
Output will be
[(3, 4), (2, 4), (2, 3)]
[(2, 4), (2, 3), (3, 4)]
The output will be sorted based on the order of the parameters we provided in the tuple format
Even better:
student_tuples = [
('john', 'A', 15),
('jane', 'B', 12),
('dave', 'B', 10),
]
sorted(student_tuples, key=lambda student: student[2]) # sort by age
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
Taken from:
Elena Rostova holds a Master's degree in Public Health Journalism. She covers groundbreaking medical research, holistic wellness trends, mental health awareness, and nutritional science.
