In Python, I am trying to sort by date with lambda. I can't understand my error message. The message is:
<lambda>() takes exactly 1 argument (2 given) The line I have is
a = sorted(a, lambda x: x.modified, reverse=True) 7 Answers
Use
a = sorted(a, key=lambda x: x.modified, reverse=True) # ^^^^ On Python 2.x, the sorted function takes its arguments in this order:
sorted(iterable, cmp=None, key=None, reverse=False) so without the key=, the function you pass in will be considered a cmp function which takes 2 arguments.
lst = [('candy','30','100'), ('apple','10','200'), ('baby','20','300')] lst.sort(key=lambda x:x[1]) print(lst) It will print as following:
[('apple', '10', '200'), ('baby', '20', '300'), ('candy', '30', '100')] You're trying to use key functions with lambda functions.
Python and other languages like C# or F# use lambda functions.
Also, when it comes to key functions and according to the documentation
Both list.sort() and sorted() have a key parameter to specify a function to be called on each list element prior to making comparisons.
...
The value of the key parameter should be a function that takes a single argument and returns a key to use for sorting purposes. This technique is fast because the key function is called exactly once for each input record.
So, key functions have a parameter key and it can indeed receive a lambda function.
In Real Python there's a nice example of its usage. Let's say you have the following list
ids = ['id1', 'id100', 'id2', 'id22', 'id3', 'id30'] and want to sort through its "integers". Then, you'd do something like
sorted_ids = sorted(ids, key=lambda x: int(x[2:])) # Integer sort and printing it would give
['id1', 'id2', 'id3', 'id22', 'id30', 'id100'] In your particular case, you're only missing to write key= before lambda. So, you'd want to use the following
a = sorted(a, key=lambda x: x.modified, reverse=True) In Python3:
from functools import cmp_to_key def compare(i1,i2): return i1-i2 events.sort(key=cmp_to_key(compare)) Take a look at this Example, you will understand:
Example 1:
a = input() a = sorted(a, key = lambda x:(len(x),x)) print(a) input: ["tim", "bob", "anna", "steve", "john","aaaa"]
output: ['bob', 'tim', 'aaaa', 'anna', 'john', 'steve']
input: ["tim", "bob", "anna", "steve", "john","aaaaa"]
output: ['bob', 'tim', 'anna', 'john', 'aaaaa', 'steve']
Example 2 (advanced):
a = ["tim", "bob", "anna", "steve", "john","aaaaa","zzza"] a = sorted(a, key = lambda x:(x[-1],len(x),x)) print(a) output: ['anna', 'zzza', 'aaaaa', 'bob', 'steve', 'tim', 'john']
Conclusion:
key = lambda x:(p1,p2,p3,p4,...,pn),
x is one element at a time from the stream of input.
p1,p2,p3...pn being properties based on which the stream of elements needs to be sorted.
based on priority order of p1>p2>p3>...>pn.
We can also add reverse=True, after the sorting condition, to sort the elements in reverse order.
Except for other complete answers, I know a great page for you to dive into.
You may find why there are key and cmp between python2.x and python3.
Another excellent example of a sorted function with lambda on Python3;
# list of sport teams with wins and losses sportTeams = [("Royals", (18, 12)), ("Rockets", (24, 6)), ("Cardinals", (20, 10)), ("Dragons", (22, 8)), ("Kings", (15, 15)), ("Chargers", (20, 10)), ("Jets", (16, 14)), ("Warriors", (25, 5))] # sort the teams by number of wins sortedTeams = sorted(sportTeams, key=lambda t: t[1][0], reverse=True) print(sortedTeams) Output:
[('Warriors', (25, 5)), ('Rockets', (24, 6)), ('Dragons', (22, 8)), ('Cardinals', (20, 10)), ('Chargers', (20, 10)), ('Royals', (18, 12)), ('Jets', (16, 14)), ('Kings', (15, 15))] 
