并行运行函数并查看其进度

Question

我正在使用 joblib 在四个核心上并行运行四个进程。我想在不同的线路上分别查看四个进程的进度。 然而，我看到的是进度被写在同一行上，直到第一个进程完成。

from math import factorial
from decimal import Decimal, getcontext
from joblib import Parallel, delayed
from tqdm import trange
import time

def calc(n_digits):
    # number of iterations
    n = int(n_digits+1/14.181647462725477)
    n = n if n >= 1 else 1

    # set the number of digits for our numbers
    getcontext().prec = n_digits+1

    t    = Decimal(0)
    pi   = Decimal(0)
    deno = Decimal(0)

    for k in trange(n):
        t = ((-1)**k)*(factorial(6*k))*(13591409+545140134*k)
        deno = factorial(3*k)*(factorial(k)**3)*(640320**(3*k))
        pi += Decimal(t)/Decimal(deno)

    pi = pi * Decimal(12) / Decimal(640320 ** Decimal(1.5))
    pi = 1/pi
    
    # no need to round
    return pi


def parallel_with_joblib():
    # Define the number of cores to use
    n_cores = 4

    # Define the tasks (e.g., compute first 100, 200, 300, 400 digits of pi)
    tasks = [1000, 1500, 700, 1200]


    # Run tasks in parallel
    results = Parallel(n_jobs=n_cores)(delayed(calc)(n) for n in tasks)


if __name__ == "__main__":
    parallel_with_joblib()

我还希望将这四行标记为“Job 1 of 4”、“Job 2 of 4”等。

Answer 1

您可以使用

position

参数并行显示条形，并使用

desc

参数标记它们。 我已向您的

calc

函数添加了足够的参数。

from math import factorial
from decimal import Decimal, getcontext
from joblib import Parallel, delayed
from tqdm import trange
import time


def calc(n_digits, pos, total):
    # number of iterations
    n = int(n_digits + 1 / 14.181647462725477)
    n = n if n >= 1 else 1

    # set the number of digits for our numbers
    getcontext().prec = n_digits + 1

    t = Decimal(0)
    pi = Decimal(0)
    deno = Decimal(0)

    for k in trange(n, position=pos, desc=f"Job {pos + 1} of {total}", leave=False):
        t = ((-1) ** k) * (factorial(6 * k)) * (13591409 + 545140134 * k)
        deno = factorial(3 * k) * (factorial(k) ** 3) * (640320 ** (3 * k))
        pi += Decimal(t) / Decimal(deno)

    pi = pi * Decimal(12) / Decimal(640320 ** Decimal(1.5))
    pi = 1 / pi

    # no need to round
    return pi


def parallel_with_joblib():
    # Define the number of cores to use
    n_cores = 4

    # Define the tasks (e.g., compute first 100, 200, 300, 400 digits of pi)
    tasks = [1000, 1500, 700, 1200]

    # Run tasks in parallel
    results = Parallel(n_jobs=n_cores)(delayed(calc)(n, pos, len(tasks)) for (pos, n) in enumerate(tasks))


if __name__ == "__main__":
    parallel_with_joblib()