How can we use tqdm in a parallel execution with joblib?

Question

I want to run a function in parallel, and wait until all parallel nodes are done, using joblib. Like in the example:

from math import sqrt
from joblib import Parallel, delayed
Parallel(n_jobs=2)(delayed(sqrt)(i ** 2) for i in range(10))

But, I want that the execution will be seen in a single progressbar like with tqdm, showing how many jobs has been completed.

How would you do that?

Answer 1

Just put range(10) inside tqdm(...)! It probably seemed too good to be true for you, but it really works (on my machine):

from math import sqrt
from joblib import Parallel, delayed  
from tqdm import tqdm  
result = Parallel(n_jobs=2)(delayed(sqrt)(i ** 2) for i in tqdm(range(100000)))

Answer 2

I've created pqdm a parallel tqdm wrapper with concurrent futures to comfortably get this done, give it a try!

To install

pip install pqdm

and use

from pqdm.processes import pqdm
# If you want threads instead:
# from pqdm.threads import pqdm

args = [1, 2, 3, 4, 5]
# args = range(1,6) would also work

def square(a):
    return a*a

result = pqdm(args, square, n_jobs=2)

Answer 3

As noted above, solutions that simply wrap the iterable passed to joblib.Parallel() do not truly monitor the progress of execution. Instead, I suggest subclassing Parallel and overriding the print_progress() method, as follows:

import joblib
from tqdm.auto import tqdm

class ProgressParallel(joblib.Parallel):
    def __call__(self, *args, **kwargs):
        with tqdm() as self._pbar:
            return joblib.Parallel.__call__(self, *args, **kwargs)

    def print_progress(self):
        self._pbar.total = self.n_dispatched_tasks
        self._pbar.n = self.n_completed_tasks
        self._pbar.refresh()

Answer 4

Modifying nth's great answer to permit a dynamic flag to use TQDM or not and to specify the total ahead of time so that the status bar fills in correctly.

from tqdm.auto import tqdm
from joblib import Parallel

class ProgressParallel(Parallel):
    def __init__(self, use_tqdm=True, total=None, *args, **kwargs):
        self._use_tqdm = use_tqdm
        self._total = total
        super().__init__(*args, **kwargs)

    def __call__(self, *args, **kwargs):
        with tqdm(disable=not self._use_tqdm, total=self._total) as self._pbar:
            return Parallel.__call__(self, *args, **kwargs)

    def print_progress(self):
        if self._total is None:
            self._pbar.total = self.n_dispatched_tasks
        self._pbar.n = self.n_completed_tasks
        self._pbar.refresh()

Answer 5

Here's possible workaround

def func(x):
    time.sleep(random.randint(1, 10))
    return x

def text_progessbar(seq, total=None):
    step = 1
    tick = time.time()
    while True:
        time_diff = time.time()-tick
        avg_speed = time_diff/step
        total_str = 'of %n' % total if total else ''
        print('step', step, '%.2f' % time_diff, 
              'avg: %.2f iter/sec' % avg_speed, total_str)
        step += 1
        yield next(seq)

all_bar_funcs = {
    'tqdm': lambda args: lambda x: tqdm(x, **args),
    'txt': lambda args: lambda x: text_progessbar(x, **args),
    'False': lambda args: iter,
    'None': lambda args: iter,
}

def ParallelExecutor(use_bar='tqdm', **joblib_args):
    def aprun(bar=use_bar, **tq_args):
        def tmp(op_iter):
            if str(bar) in all_bar_funcs.keys():
                bar_func = all_bar_funcs[str(bar)](tq_args)
            else:
                raise ValueError("Value %s not supported as bar type"%bar)
            return Parallel(**joblib_args)(bar_func(op_iter))
        return tmp
    return aprun

aprun = ParallelExecutor(n_jobs=5)

a1 = aprun(total=25)(delayed(func)(i ** 2 + j) for i in range(5) for j in range(5))
a2 = aprun(total=16)(delayed(func)(i ** 2 + j) for i in range(4) for j in range(4))
a2 = aprun(bar='txt')(delayed(func)(i ** 2 + j) for i in range(4) for j in range(4))
a2 = aprun(bar=None)(delayed(func)(i ** 2 + j) for i in range(4) for j in range(4))

Answer 6

If your problem consists of many parts, you could split the parts into k subgroups, run each subgroup in parallel and update the progressbar in between, resulting in k updates of the progress.

This is demonstrated in the following example from the documentation.

>>> with Parallel(n_jobs=2) as parallel:
...    accumulator = 0.
...    n_iter = 0
...    while accumulator < 1000:
...        results = parallel(delayed(sqrt)(accumulator + i ** 2)
...                           for i in range(5))
...        accumulator += sum(results)  # synchronization barrier
...        n_iter += 1

https://pythonhosted.org/joblib/parallel.html#reusing-a-pool-of-workers