How to use await in a python lambda

Question

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?

Solution 1

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]

Solution 2

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.

Solution 3

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

Solution 4

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())

Solution 5

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] )

View more solutions

Comments

[iCart]

I was getting a SyntaxError: 'await' expressions in comprehensions are not supported, so i had to do this (for future reference): 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] And now it works, thanks!

[Sven Marnach]

@iCart Right, that's a restriction in Python 3.5, which is lifted in the upcoming Python 3.6.

[James]

Found a very edge case - see my answer :-)

[Sven Marnach]

I think the best solution for this edge case would be to pass operator.itemgetter(0) as key function to sort(). Tuples are sorted lexicographically, so equal keys will result in a comparison of the second item. By explicitly selecting only the first item as the sort key we prevent that second comparison.