full-speed-python习题解答(八)-- Coroutines

Python协同(coroutines)类似于生成器,会使用yield关键字,但它不是生成数据,而是通常消费(consume)数据。


Exercise with coroutines

1. Create a coroutine named “square” that prints the square of any sent value.
def square():
    print('square')
    while True:
        val = yield
        print(val*val)

s=square()
next(s)
s.send(2)
s.send(3)
s.send(4)
s.close()
s.send(5)

2. Implement the “minimize” coroutine that keeps and prints the minimum value that

is sent to the function.

def minimize():
    print("minimize")
    var = yield
    min = var
    print('min',min)
    while True:
        curr = yield
        if min>=curr:
            min=curr
        print("min",min)

m = minimize()
next(m)
m.send(3)
m.send(2)
m.send(5)


Exercise with coroutines

1. Implement a producer-consumer pipeline where the values squared by the producer
are sent to two consumers. One should store and print the minimum value sent so
far and the other the maximum value.
def producer(consumers):
    print('producer ready')
    try:
        while True:
            var = yield
            for consumer in consumers:
                consumer.send(var*var)
    except GeneratorExit:
        for consumer in consumers:
            consumer.close()

def consumer(name,func):
    print(f"{name} ready")
    var = yield
    max = var
    min = var
    print(f"current {func} value is {var}")
    
    try:
        while True:
            curr = yield
            if func == "max":
                if max<=curr:
                    max=curr
                print(f"current max value is {max}")
            else:
                if min>=curr:
                    min=curr
                print(f"current min value is {min}")
    except GeneratorExit:
        print(f"{name} closed")

conmax = consumer("conmax","max")
conmin = consumer("conmin","min")
proc = producer([conmax,conmin])

next(conmax)
next(conmin)
next(proc)

proc.send(1)
proc.send(2)
proc.send(5)
proc.send(0)
2. Implement a producer-consumer pipeline where the values squared by the producer
are dispatched to two consumers, one at a time. The first value should be sent to
consumer 1, the second value to consumer 2, third value to consumer 1 again, and
so on. Closing the producer should force the consumers to print a list with the
numbers that each one obtained.
def turn_on(label):
    if label==1:
        return 0
    elif label==0:
        return 1
def producer(consumers):
    print("producer ready")
    try:
        index = 0
        while True:
            var = yield
            consumers[index].send(var*var)
            index = turn_on(index)
    except GeneratorExit:
        for consumer in consumers:
            consumer.close()
list=[[] for i in range(2)]
def consumer(name,label):
    print(f"{name} ready")
    
    try:
        while True:
                var = yield
                list[label].append(var)
                print(f"consumer{label+1}",var)
                
    except GeneratorExit:
        print(f"consumer{label+1}",list[label])
        

con1 = consumer("consumer1",0)
con2 = consumer("consumer2",1)
prod = producer([con1,con2])

next(con1)
next(con2)
next(prod)

prod.send(1)
prod.send(2)
prod.send(3)
prod.send(4)
prod.close()


版权声明

本文仅代表作者观点,不代表本站立场。
本文系作者授权发表,未经许可,不得转载。

评论