Python многопроцессорный выход

Question

У меня есть функция индикатора выполнения. Я хотел бы запустить этот индикатор выполнения, пока другая функция выполняет обработку. Я написал простой тестовый код с использованием многопроцессорного модуля, и он не работал хорошо. Мой код:

from multiprocessing import Process 
import time 

def foo(thread): 
    print time.ctime() 
    time.sleep(10) 
    print time.ctime() 

def progress_bar(timer = 10): 
    digits = 4 
    delete = '\b' * 6 
    time_slot = float(timer)/100 
    for i in range(1, 101): 
     delete_bar = '\b' * 52 
     if i == 1: 
      bar = '|' + ' ' * 50 + '|' 
     else: 
      bar = '|' + '=' * (i/2 - 1) + '>' + ' ' * (50 - i/2) + '|' 
     print "{0}{1:{2}}{3}{4}".format(delete, str(i) + '%', digits, bar, delete_bar), 
     time.sleep(time_slot) 
    print '' 

def main(): 
    p1 = Process(target = foo1('this')) 
    p1.start() 
    p2 = Process(target = progress_bar()) 
    p2.start() 
    p1.join() 
    p2.join() 

if __name__ == "__main__": 
    main() 

Я надеялся, что foo напечатает текущее время в первую очередь. Затем progress_bar приходит к обратному отсчету в течение 10 секунд. Наконец, foo выведет другое время в конце.

Tue Apr 5 11:49:47 2016 
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Tue Apr 5 11:49:57 2016 

Однако то, что я получил от выхода что-то вроде этого:

Tue Apr 5 11:49:47 2016 
Tue Apr 5 11:49:57 2016 
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Есть ли способ исправить это в Python? Большое спасибо!

Answer 1

Попробуйте это:

from multiprocessing import Process 
import time 

def foo1(thread): 
    print time.ctime() 
    time.sleep(10.5) 
    print time.ctime() 

def progress_bar(timer = 10): 
    digits = 4 
    delete = '\b' * 6 
    time_slot = float(timer)/100 
    for i in range(1, 101): 
     delete_bar = '\b' * 52 
     if i == 1: 
      bar = '|' + ' ' * 50 + '|' 
     else: 
      bar = '|' + '=' * (i/2 - 1) + '>' + ' ' * (50 - i/2) + '|' 
     print "{0}{1:{2}}{3}{4}".format(delete, str(i) + '%', digits, bar, delete_bar), 
     time.sleep(time_slot) 
    print '' 

def main(): 
    p1 = Process(target=foo1, args=('this',)) 
    # p1 = Process(target = foo1('this')) 
    p1.start() 
    time.sleep(0.1) 
    p2 = Process(target=progress_bar) 
    p2.start() 
    p1.join() 
    p2.join() 

if __name__ == "__main__": 
    main() 

Обратите внимание на разные p1 процесса.

Answer 2

Это происходит потому, что параметры функции вычисляются в первую очередь, и при определении вашей p1 и p2 вы на самом деле вызова foo1('this'), который выполняет функцию при определении p1 и progress_bar() в конкретизации p2.

Для простого примера, который демонстрирует это, смотрите ниже:

def fn(): 
    print 'called' 
    return 1 

target1 = fn() 
target = fn 

print target1 
print target 

Это печатает:

>>> called # Got called as soon as you called fn via fn() 
>>> 1 # Assigned the return value of fn to target1 
>>> <function fn at 0x12DA77F0> # Didn't get called, assigned the fn definition to target 

Я получил ваш пример работы с Thread с ниже (Edit: После просмотра в некоторых , кажется, что они должны работать с тем же синтаксисом, что и приведенный ниже код (просто измените импорт и используйте Process вместо Thread), но по какой-то причине я просто не могу получить подход Process к печати даже после копирования примера. Может быть из-за моей установки пользовательских питона, но не совсем уверен).

from threading import Thread 
import time 

def foo(thread): 
    print(time.ctime()) 
    time.sleep(10) 
    print(time.ctime()) 

def progress_bar(timer = 10): 
    digits = 4 
    delete = '\b' * 6 
    time_slot = float(timer)/100 
    for i in range(1, 101): 
     delete_bar = '\b' * 52 
     if i == 1: 
      bar = '|' + ' ' * 50 + '|' 
     else: 
      bar = '|' + '=' * (i/2 - 1) + '>' + ' ' * (50 - i/2) + '|' 
     print("{0}{1:{2}}{3}{4}".format(delete, str(i) + '%', digits, bar, delete_bar),) 
     time.sleep(time_slot) 
    print('') 

def main(): 
    t1 = Thread(target=foo, args=('this',)) # Notice, not foo('this') <- this executes foo('this') at definition 
    t1.start() 
    t2 = Thread(target=progress_bar) # Again, notice, no parens - target is just the function definition 
    t2.start() 
    t1.join() 
    t2.join() 

Печатается

Tue Apr 05 15:09:34 2016 
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