如何用Python编写一系列承诺?

是否可以仅使用Python 3.6.1 Standard Library编写promise(或任务)序列?

例如,JavaScript中的序列承诺写成:

const SLEEP_INTERVAL_IN_MILLISECONDS = 200;

const alpha = function alpha (number) {
    return new Promise(function (resolve, reject) {
        const fulfill = function() {
            return resolve(number + 1);
        };

        return setTimeout(fulfill, SLEEP_INTERVAL_IN_MILLISECONDS);
    });
};

const bravo = function bravo (number) {
    return new Promise(function (resolve, reject) {
        const fulfill = function() {
            return resolve(Math.ceil(1000*Math.random()) + number);
        };
        return setTimeout(fulfill, SLEEP_INTERVAL_IN_MILLISECONDS);
    });
};

const charlie = function charlie (number) {
    return new Promise(function (resolve, reject) {
        return (number%2 == 0) ? reject(number) : resolve(number);
    });
};

function run() {
    return Promise.resolve(42)
        .then(alpha)
        .then(bravo)
        .then(charlie)
        .then((number) => {
            console.log('success: ' + number)
        })
        .catch((error) => {
            console.log('error: ' + error);
        });
}

run();

具有异步处理结果的每个函数also returns a Promise,将由紧随其后的承诺解决/拒绝。

我知道promises-2.01basyncio 3.4.3这样的库,我正在寻找Python STL解决方案。因此,如果需要导入非STL库,我更喜欢使用RxPython。


解决方案

下面是一个使用Asyncio和async/await语法的类似程序:

import asyncio
import random

async def alpha(x):
    await asyncio.sleep(0.2)
    return x + 1 

async def bravo(x):
    await asyncio.sleep(0.2)
    return random.randint(0, 1000) + x

async def charlie(x):
    if x % 2 == 0:
        return x
    raise ValueError(x, 'is odd')

async def run():
    try:
        number = await charlie(await bravo(await alpha(42)))
    except ValueError as exc:
        print('error:', exc.args[0])
    else:
        print('success:', number)

if __name__ == '__main__':
    loop = asyncio.get_event_loop()
    loop.run_until_complete(run())
    loop.close()

编辑:如果您对反应流感兴趣,可以考虑使用aiostream。

这里有一个简单的示例:

import asyncio
from aiostream import stream, pipe

async def main():
    # This stream computes 11² + 13² in 1.5 second
    xs = (
        stream.count(interval=0.1)      # Count from zero every 0.1 s
        | pipe.skip(10)                 # Skip the first 10 numbers
        | pipe.take(5)                  # Take the following 5
        | pipe.filter(lambda x: x % 2)  # Keep odd numbers
        | pipe.map(lambda x: x ** 2)    # Square the results
        | pipe.accumulate()             # Add the numbers together
    )
    print('11² + 13² = ', await xs)

if __name__ == '__main__':
    loop = asyncio.get_event_loop()
    loop.run_until_complete(main())
    loop.close()

documentation中有更多示例。

免责声明:我是项目维护者。

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