CODEWITHSUNDEEP
pythonsorting
Netro
Netro

Reputation: 7297

Sorting list of dictionaries---what is the default behaviour (without key parameter)?

I m trying to sort a list of dict using sorted

>>> help(sorted)
Help on built-in function sorted in module __builtin__:

sorted(...)
    sorted(iterable, cmp=None, key=None, reverse=False) --> new sorted list

I have just given list to sorted and it sorts according to id.

>>>l = [{'id': 4, 'quantity': 40}, {'id': 1, 'quantity': 10}, {'id': 2, 'quantity': 20}, {'id': 3, 'quantity': 30}, {'id': 6, 'quantity': 60}, {'id': 7, 'quantity': -30}]
>>> sorted(l) # sorts by id
[{'id': -1, 'quantity': -10}, {'id': 1, 'quantity': 10}, {'id': 2, 'quantity': 20}, {'id': 3, 'quantity': 30}, {'id': 4, 'quantity': 40}, {'id': 6, 'quantity': 60}, {'id': 7, 'quantity': -30}]
>>> l.sort()
>>> l # sorts by id
[{'id': -1, 'quantity': -10}, {'id': 1, 'quantity': 10}, {'id': 2, 'quantity': 20}, {'id': 3, 'quantity': 30}, {'id': 4, 'quantity': 40}, {'id': 6, 'quantity': 60}, {'id': 7, 'quantity': -30}]

Many example of sorted says it requires key to sort the list of dict. But I didn't give any key. Why it didn't sort according to quantity? How did it choose to sort with id? I tried another example with name & age,

>>> a
[{'age': 1, 'name': 'john'}, {'age': 3, 'name': 'shyam'}, {'age': 30,'name': 'ram'}, {'age': 15, 'name': 'rita'}, {'age': 5, 'name': 'sita'}]
>>> sorted(a) # sorts by age
[{'age': 1, 'name': 'john'}, {'age': 3, 'name': 'shyam'}, {'age': 5, 'name':'sita'}, {'age': 15, 'name': 'rita'}, {'age': 30, 'name': 'ram'}]

>>> a.sort() # sorts by age
>>> a
[{'age': 1, 'name': 'john'}, {'age': 3, 'name': 'shyam'}, {'age': 5, 'name':'sita'}, {'age': 15, 'name': 'rita'}, {'age': 30, 'name': 'ram'}]

Here it sorts according to age but not name. What am I missing in default behavior of these method?

Upvotes: 4

Views: 106

Answers (2)

Mark Perryman
Mark Perryman

Reputation: 779

By default it will compare against the first difference it finds. If you are sorting dictionaries this is quite dangerous (consistent yet undefined).

Pass a function to key= parameter that takes a value from the list (in this case a dictionary) and returns the value to sort against.

>>> a
[{'age': 1, 'name': 'john'}, {'age': 3, 'name': 'shyam'}, {'age': 30,'name': 'ram'}, {'age': 15, 'name': 'rita'}, {'age': 5, 'name': 'sita'}]
>>> sorted(a, key=lambda d : d['name']) # sorts by name
[{'age': 1, 'name': 'john'}, {'age': 30, 'name': 'ram'}, {'age': 15, 'name': 'rita'}, {'age': 3, 'name': 'shyam'}, {'age': 5, 'name': 'sita'}]

See https://wiki.python.org/moin/HowTo/Sorting

The key parameter is quite powerful as it can cope with all sorts of data to be sorted, although maybe not very intuitive.

Upvotes: 0

Alex Hall
Alex Hall

Reputation: 36023

From some old Python docs:

Mappings (dictionaries) compare equal if and only if their sorted (key, value) lists compare equal. Outcomes other than equality are resolved consistently, but are not otherwise defined.

Earlier versions of Python used lexicographic comparison of the sorted (key, value) lists, but this was very expensive for the common case of comparing for equality. An even earlier version of Python compared dictionaries by identity only, but this caused surprises because people expected to be able to test a dictionary for emptiness by comparing it to {}.

Ignore the default behaviour and just provide a key.

Upvotes: 3

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