[FIXED] Python how to use await in a lambda

Issue

I want to use 'await' for the lambda function as shown below.

self.voice_channel.play(discord.FFmpegPCMAudio('music.mp3'), after=lambda e: await self.test(x))

SyntaxError: 'await' outside async function

How can I fix it to make it feel like this?

Solution

As the other answer said, you can't put await in a lambda. But if await is the outermost operator in the expression, you can omit both the async and await and have the lambda directly return the awaitable:

self.voice_channel.play(discord.FFmpegPCMAudio('music.mp3'),
                        after=lambda e: self.test(x))

Without the await, the lambda simply returns self.test(x). As that value was to be awaited, self.test(x) must obviously return an awaitable object. This awaitable object will be received by code that at some point invokes o.after(e) and awaits the result. In other words, await o.after(e) will be equivalent to await self.test(x), which is what you wanted to accomplish.

In general, an async def function:

async def foo():
    return await bar()

can be replaced with:

def foo():
    return bar()

Although foo() is not async in the latter case, the end result is the same: await foo() will return the result of awaiting bar(). See this answer for another example.

Answered By - user4815162342