I want to run a function in parallel, and wait until all parallel nodes are done, using joblib. Like in the example:
我想要并行运行一个函数,并使用joblib等待所有并行节点完成。的例子:
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.
但是,我希望在一个像tqdm这样的单个progressbar中可以看到执行,说明已经完成了多少工作。
How would you do that?
你会怎么做?
3 个解决方案
#1
5
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.
如果您的问题包含许多部分,您可以将这些部分分成k个子组,并行运行每个子组,并在其中更新progressbar,从而获得进度的k个更新。
This is demonstrated in the following example from the documentation.
下面的示例从文档中演示了这一点。
>>> 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
https://pythonhosted.org/joblib/parallel.html reusing-a-pool-of-workers
#2
7
Here's possible workaround
这是可能的解决方案
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))
#3
1
Just put range(10)
inside tqdm(...)
! It probably seemed too good to be true for you, but it really works (on my machine):
只需要在tqdm(…)中设置范围(10)!它可能看起来太好了,对你来说可能不真实,但它确实有效(在我的机器上):
from math import sqrt
from joblib import Parallel, delayed
import multiprocessing
from tqdm import tqdm
result = Parallel(n_jobs=2)(delayed(sqrt)(i ** 2) for i in tqdm(range(100000)))
#1
5
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.
如果您的问题包含许多部分,您可以将这些部分分成k个子组,并行运行每个子组,并在其中更新progressbar,从而获得进度的k个更新。
This is demonstrated in the following example from the documentation.
下面的示例从文档中演示了这一点。
>>> 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
https://pythonhosted.org/joblib/parallel.html reusing-a-pool-of-workers
#2
7
Here's possible workaround
这是可能的解决方案
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))
#3
1
Just put range(10)
inside tqdm(...)
! It probably seemed too good to be true for you, but it really works (on my machine):
只需要在tqdm(…)中设置范围(10)!它可能看起来太好了,对你来说可能不真实,但它确实有效(在我的机器上):
from math import sqrt
from joblib import Parallel, delayed
import multiprocessing
from tqdm import tqdm
result = Parallel(n_jobs=2)(delayed(sqrt)(i ** 2) for i in tqdm(range(100000)))