Pythonマルチプロセッシング出力

Question

私はプログレスバー機能を持っています。他の機能が処理をしている間、このプログレスバーを実行したいと思います。私はマルチプロセッシングモジュールを使って簡単なテストコードを書いていますが、うまく機能しませんでした。私のコード：

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() 

fooが現在の時刻を最初に表示することを望んでいました。その後progress_barが10秒間カウントダウンします。最後に、fooは最後に別の時刻を出力します。

Tue Apr 5 11:49:47 2016 
Tue Apr 5 11:49:57 2016 
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Pythonでこれを修正する方法はあります：

Tue Apr 5 11:49:47 2016 
100% =================================================>| 
Tue Apr 5 11:49:57 2016 

は、しかし、このようなもの、私は出力からもらったのですか？ ありがとうございます！関数のパラメータが最初に評価されるため、そしてあなたのp1を定義するときとp2あなたが実際にp2のインスタンス化でp1、およびprogress_bar()の定義で機能を実行するfoo1('this')を呼んでいる

Answer 1

はこれを試してみてください印刷します

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() 

お知らせ異なるP1プロセス。

Answer 2

これが起こっています。これを示して簡単な例

、以下を参照：

def fn(): 
    print 'called' 
    return 1 

target1 = fn() 
target = fn 

print target1 
print target 

これが印刷されます。

>>> 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 

私は以下Thread S（編集での作業あなたの例を得た：いくつかを検討していた後、 Processexamplesのように、以下のコードと同じ構文で作業する必要があります（ちょうどインポートを変更し、の代わりにProcessを使用してください）。）、何らかの理由で、たとえ例をコピーしても、Processの印刷方法を取得できません。私のカスタムPythonのセットアップのために、しかし全くわからないかもしれません）：。

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() 

これは

Tue Apr 05 15:09:34 2016 
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