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我正在尝试在 Python 中使用一个简单的异步示例，主要遵循这里的出色答案。

我的目标是设置一个上下文变量，并通过不断附加来跟踪一系列调用。我知道可以使用该方法访问上下文变量，并且可以使用该.get()方法更改它们的值.set()。sum()然而，在以下情况下，尽管从控制台可以明显看出对函数的一系列调用，但变量不会被修改。

编辑： 根据下面 Michael Butscher 的评论，我用 list: 替换了原始的上下文变量（它是一个字符串），output_list并使用.append(). 现在，这确实使我能够查看最终输出，但不能查看各个sum()方法中的中间输出。

完整代码：

import asyncio
import contextvars
import time

output_list = contextvars.ContextVar('output_list', default=list())

async def sleep():
    print(f'Time: {time.time() - start:.2f}')
    await asyncio.sleep(1)

async def sum(name, numbers):
    total = 0
    for number in numbers:
        print(f'Task {name}: Computing {total}+{number}')
        await sleep()
        total += number
    output_list.set(output_list.get().append(f"{name}"))
    print(f'Task {name}: Sum = {total}\n')
    print(f'Partial output from task {name}:', output_list.get())

start = time.time()

loop = asyncio.get_event_loop()
tasks = [
    loop.create_task(sum("A", [1, 2])),
    loop.create_task(sum("B", [1, 2, 3])),
]
loop.run_until_complete(asyncio.wait(tasks))
loop.close()

end = time.time()
print(f'Time: {end-start:.2f} sec')
print("Final output_str =", output_list.get())

如何迭代地跟随上下文变量列表的扩展output_list？

我想要的控制台输出是：

Task A: Computing 0+1
Time: 0.00
Task B: Computing 0+1
Time: 0.00
Task A: Computing 1+2
Time: 1.02
Task B: Computing 1+2
Time: 1.02
Task A: Sum = 3

Partial output from task A: ['A']
Task B: Computing 3+3
Time: 2.02
Task B: Sum = 6

Partial output from task B: ['A', 'B']
Time: 3.03 sec
Final output_str = ['A', 'B']

相反，我得到：

Task A: Computing 0+1
Time: 0.00
Task B: Computing 0+1
Time: 0.00
Task A: Computing 1+2
Time: 1.02
Task B: Computing 1+2
Time: 1.02
Task A: Sum = 3

Partial output from task A: None
Task B: Computing 3+3
Time: 2.02
Task B: Sum = 6

Partial output from task B: None
Time: 3.03 sec
Final output_str = ['A', 'B']
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1 回答 1

1

根据asyncio文档：

任务支持 contextvars 模块。创建任务时，它会复制当前上下文，然后在复制的上下文中运行其协程。

所以，如果你在程序的顶部声明cvar = contextvars.ContextVar('cvar', default='x')，当你创建一个任务时，这将复制当前上下文，如果你修改cvar它只会影响副本，但不会影响原始上下文。''这就是您在最终输出中得到 (empty string) 的主要原因。

要实现您想要的“跟踪”，您必须使用全局变量才能在任何地方修改它。但是，如果您想尝试asyncio并contextvars了解它是如何工作的，请参见下面的示例：

import asyncio
import contextvars
import time

output = contextvars.ContextVar('output', default='No changes at all') 

async def sleep():
    print(f'Time: {time.time() - start:.2f}')
    await asyncio.sleep(1)

async def sum(name, numbers):
    total = 0
    for number in numbers:
        print(f'Task {name}: Computing {total}+{number}')
        await sleep()
        total += number
        output.set(output.get()+name) #Here we modify the respective context
    print(f'Task {name}: Sum = {total}\n')
    print(f'Partial output from task {name}:', output.get())
    return output.get() #Here we return the variable modified
start = time.time()

# main() will have its own copy of the context
async def main():
    output.set('Changed - ') # Change output var in this function context
    # task1 and task2 will copy this context (In this contect output=='Changed - ')
    task1 = asyncio.create_task(sum("A", [1, 2])) #This task has its own copy of the context of main()
    task2 = asyncio.create_task(sum("B", [1, 2, 3])) #This task also has its own copy of the context of main()
    done, pending = await asyncio.wait({task1,task2})
    resultTask1 = task1.result() # get the value of return of task1
    resultTask2 = task2.result() # get the value of return of task1
    print('Result1: ', resultTask1)
    print('Result2: ', resultTask2)
    print('Variable output in main(): ',output.get()) # However, output in main() is sitill 'Changed - '
    output.set(output.get()+'/'+resultTask1+'/'+resultTask2) #Modify the var in this context
    print('Variable modified in main(): ', output.get())
    return output.get() #Return modified value

x = asyncio.run(main()) # Assign the return value to x

end = time.time()
print(f'Time: {end-start:.2f} sec')
print("Final output (without changes) =", output.get())
output.set(x)
print("Final output (changed) =", output.get())

##### OUTPUT #####
# Time: 0.00
# Task B: Computing 0+1
# Time: 0.00
# Task A: Computing 1+2
# Time: 1.01
# Task B: Computing 1+2
# Time: 1.01
# Task A: Sum = 3

# Partial output from task A: Changed - AA
# Task B: Computing 3+3
# Time: 2.02
# Task B: Sum = 6

# Partial output from task B: Changed - BBB
# Result1:  Changed - AA
# Result2:  Changed - BBB
# Variable output in main():  Changed -
# Variable modified in main():  Changed - /Changed - AA/Changed - BBB
# Time: 3.03 sec
# Final output (without changes) = No changes at all
# Final output (changed) = Changed - /Changed - AA/Changed - BBB

如您所见，不可能同时修改同一个变量。在task1修改它的副本的同时，task2也在修改它的副本。

于 2022-01-19T00:17:18.843 回答