I'm trying to do something like this:
mylist.sort(key=lambda x: await somefunction(x)) But I get this error:
SyntaxError: 'await' outside async function Which makes sense because the lambda is not async.
I tried to use async lambda x: ... but that throws a SyntaxError: invalid syntax.
Pep 492 states:
Syntax for asynchronous lambda functions could be provided, but this construct is outside of the scope of this PEP.
But I could not find out if that syntax was implemented in CPython.
Is there a way to declare an async lambda, or to use an async function for sorting a list?
5 Answers
You can't. There is no async lambda, and even if there were, you coudln't pass it in as key function to list.sort(), since a key function will be called as a synchronous function and not awaited. An easy work-around is to annotate your list yourself:
mylist_annotated = [(await some_function(x), x) for x in mylist] mylist_annotated.sort() mylist = [x for key, x in mylist_annotated] Note that await expressions in list comprehensions are only supported in Python 3.6+. If you're using 3.5, you can do the following:
mylist_annotated = [] for x in mylist: mylist_annotated.append((await some_function(x), x)) mylist_annotated.sort() mylist = [x for key, x in mylist_annotated] An "async lambda" can be emulated by combining a lambda with an async generator:
key=lambda x: (await somefunction(x) for _ in '_').__anext__() It is possible to move the ( ).__anext__() to a helper, which likely makes the pattern clearer as well:
def head(async_iterator): return async_iterator.__anext__() key=lambda x: head(await somefunction(x) for _ in '_') Note that the sort method/function in the standard library are not async. One needs an async version, such as asyncstdlib.sorted (disclaimer: I maintain this library):
import asyncstdlib as a mylist = await a.sorted(mylist, key=lambda x: head(await somefunction(x) for _ in '_')) Understanding the lambda ...: (...).__anext__() pattern
An "async lambda" would be an anonymous asynchronous function, or in other words an anonymous function evaluating to an awaitable. This is in parallel to how async def defines a named function evaluating to an awaitable.
The task can be split into two parts: An anonymous function expression and a nested awaitable expression.
An anonymous function expression is exactly what a
lambda ...: ...is.An awaitable expression is only allowed inside a coroutine function; however:
- An (asynchronous) generator expression implicitly creates a (coroutine) function. As an async generator only needs async to run, it can be defined in a sync function (since Python 3.7).
- An asynchronous iterable can be used as an awaitable via its
__anext__method.
These three parts are directly used in the "async lambda" pattern:
# | regular lambda for the callable and scope # | | async generator expression for an async scope # v v v first item as an awaitable key=lambda x: (await somefunction(x) for _ in '_').__anext__() The for _ in '_' in the async generator is only to have exactly one iteration. Any variant with at least one iteration will do.
await cannot be included in a lambda function.
The solutions here can be shortened to:
from asyncio import coroutine, run my_list = [. . .] async def some_function(x) -> coroutine: . . . my_list.sort(key=lambda x: await some_function(x)) # raises a SyntaxError my_list.sort(key=lambda x: run(some_function(x)) # works If you already defined a separate async function, you can simplify MisterMiyagi's answer even a bit more:
mylist = await a.sorted( mylist, key=somefunction) If you want to change the key after awaiting it, you can use asyncstdlib.apply:
mylist = await a.sorted( mylist, key=lambda x: a.apply(lambda after: 1 / after, some_function(x))) Here is a complete example program:
import asyncio import asyncstdlib as a async def some_function(x): return x async def testme(): mylist=[2, 1, 3] mylist = await a.sorted( mylist, key=lambda x: a.apply(lambda after: 1 / after, some_function(x))) print(f'mylist is: {mylist}') if __name__ == "__main__": asyncio.run(testme()) The answer from Sven Marnach has an Edge case.
If you try and sort a list that has 2 items that produce the same search key but are different and are not directly sortable, it will crash.
mylist = [{'score':50,'name':'bob'},{'score':50,'name':'linda'}] mylist_annotated = [(x['score'], x) for x in mylist] mylist_annotated.sort() print( [x for key, x in mylist_annotated] ) Will give:
TypeError: '<' not supported between instances of 'dict' and 'dict' Fortunately I had an easy solution - my data had a unique key in that was sortable, so I could put that as the second key:
mylist = [{'score':50,'name':'bob','unique_id':1},{'score':50,'name':'linda','unique_id':2}] mylist_annotated = [(x['score'], x['unique_id'], x) for x in mylist] mylist_annotated.sort() print( [x for key, unique, x in mylist_annotated] ) I guess if your data doesn't have a naturally unique value in, you can insert one before trying to sort? A uuid maybe?
EDIT: As suggested in comment (Thanks!), you can also use operator.itemgetter:
import operator mylist = [{'score':50,'name':'bob'},{'score':50,'name':'linda'}] mylist_annotated = [(x['score'], x) for x in mylist] mylist_annotated.sort(key=operator.itemgetter(0)) print( [x for key, x in mylist_annotated] ) 
