Início / coding / Perguntas / 79570287
Accepted
Rhys
Asked: 2025-04-12 18:03:03 +0800 CST

python 3 - concurrent.futures - obter número de thread sem adicionar uma função

  • 772

Atualmente este código imprime,

MainThread done at sec 1
MainThread done at sec 1
MainThread done at sec 2
MainThread done at sec 1
MainThread done at sec 0
MainThread done at sec 3
MainThread done at sec 2
MainThread done at sec 5
MainThread done at sec 4

Preciso imprimir

MainThread 1 done at sec 1
MainThread 3 done at sec 1
MainThread 2 done at sec 2
MainThread 3 done at sec 1
MainThread 1 done at sec 0
MainThread 2 done at sec 3
MainThread 2 done at sec 2
MainThread 3 done at sec 5
MainThread 1 done at sec 4

Como posso fazer isso, modificando esta linha de código?

print(threading.current_thread().name + ' done at sec ' + str(future.result()))

Aqui está o código completo

import concurrent.futures
import threading
import pdb
import time
import random


def Threads2(time_sleep):
    time.sleep(time_sleep)
    return time_sleep


all_sections = [random.randint(0, 5) for iii in range(10)]

test1 = []
with concurrent.futures.ThreadPoolExecutor(max_workers=3, thread_name_prefix="Ok") as executor:
    working_threads = {executor.submit(Threads2, time_sleep): time_sleep for index, time_sleep in enumerate(all_sections)}

    for future in concurrent.futures.as_completed(working_threads):
        print(threading.current_thread().name + ' done at sec ' + str(future.result()))
python

1 respostas

  1. Best Answer

    futurenão sabe qual thread executou a função, você precisa armazenar isso sozinho.

    você pode encapsular o functor em um wrapper que retornará o que você quiser do thread em execução.

    import concurrent.futures
import threading
import pdb
import time
import random
from functools import wraps

def wrap_function(func):
    @wraps(func)
    def wrapper_functor(*args, **kwargs):
        return (threading.current_thread().name, 
                threading.current_thread().ident, 
                func(*args, **kwargs)) # calls the wrapped function
    return wrapper_functor  # return a functor that wraps the function

def Threads2(time_sleep):
    time.sleep(time_sleep)
    return time_sleep

all_sections = [random.randint(0, 5) for iii in range(10)]

test1 = []
with concurrent.futures.ThreadPoolExecutor(
        max_workers=3, thread_name_prefix="Ok") as executor:
    working_threads = {
        executor.submit(wrap_function(Threads2), time_sleep): time_sleep for
        index, time_sleep in enumerate(all_sections)}

    for future in concurrent.futures.as_completed(working_threads):
        thread_name, thread_id, result = future.result()
        print(thread_name, thread_id, 'done at sec', str(result))

    Ok_0 18532 done at sec 0
Ok_2 12424 done at sec 1
Ok_0 18532 done at sec 3
Ok_2 12424 done at sec 2
Ok_1 19356 done at sec 4
Ok_0 18532  done at sec 4
Ok_1 19356  done at sec 3
Ok_2 12424  done at sec 4
Ok_0 18532  done at sec 3
Ok_1 19356  done at sec 4

    se você quiser, 0,1,2então você pode usarthread_name.split('_')[-1]

    uma forma supercomprimida de fazer isso com um lambda (que não deve passar por uma revisão de código) é

    working_threads = {
    executor.submit(lambda *args, **kwargs: (
        threading.current_thread().name,
        threading.current_thread().ident,
        Threads2(*args, **kwargs)), time_sleep): time_sleep for
    index, time_sleep in enumerate(all_sections)}
    • 1

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