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Adventures in Python with concurrent.futures

I use a lot of Python for scripting, and in particular to perform repetitive tasks. A lot of my scripts are a variant off the following:

for task in get_tasks_to_do():
    perform(task)

This is a sequential programming pattern – the tasks run one after another, with only one task running at a time. It keeps the code simple, and because it’s only ever doing one task at a time, it’s easy to follow the script as it’s running. It performs task 1, then task 2, then task 3, and so on.

This is fine when the tasks are few or fast, but if if I have lots of slow tasks, the overall script is very slow.

I’ve written a lot of scripts recently where the tasks are slow – for example, making HTTP requests – and I want to run more than one task at once. Each task spends a lot of time waiting for an external resource – say, a response from a distant server – and that time could be spent doing something more useful, like making the next request.

I’ve been thinking about ways to modify my scripts so they run more than one task at once. While a task is waiting for external resources, another task is doing work to take advantage of that idle time. Because there are lots of things happening at once, the script finishes faster. This is a concurrent programming pattern.

Concurrent programming trades simplicity for speed. The code gets more complicated, and there are more ways for it to go wrong, but it’s usually faster. It’s not better or worse than sequential programming; it’s just a different set of tradeoffs.

I’ve been experimenting with Python’s concurrent.futures module as a way to introduce concurrency into my scripts. In this post, I’m going to show off some of the patterns I’ve been using – both so you can use them, and to help me understand them better.

(This post was originally the solution to a single problem: listing all the rows in a DynamoDB table. The concurrency was a key part of the script, and I realised it was worth digging into that aspect on its own.)

Running a small, fixed number of tasks

If I have a small number of tasks, I schedule them all in one go, and wait for them all to complete. Here’s a simple example:

import concurrent.futures


with concurrent.futures.Executor() as executor:
    futures = {
        executor.submit(perform, task) for task in get_tasks_to_do()
    }

    for fut in concurrent.futures.as_completed(futures):
        print(f"The outcome is {fut.result()}")

We start by creating an Executor, which manages all the tasks that are running – either in separate processes or threads. Using the with statement creates a context manager, which ensures any stray threads or processes get cleaned up properly when we’re done.

In real code, this would be a ThreadPoolExecutor or a ProcessPoolExecutor – I’ve been using ThreadPoolExecutor without any arguments, because that’s been fast enough for my scripts, and I don’t understand the difference. If you want to squeeze the performance of your scripts, you can probably do some fine-tuning here.

Then I use a set comprehension to start all the tasks, using executor.submit() to schedule each task. This creates a Future object, which represents the task to be done.

Once all the tasks have been scheduled, I call concurrent.futures_as_completed, which yields the futures as they’re done – that is, as each task completes. The .result() method gives me the return value of perform(task), or throws an exception if it failed.

The Future object doesn’t hold any context, so if you want to match the results to the original task, you need to track that yourself:

import concurrent.futures


with concurrent.futures.Executor() as executor:
    futures = {
        executor.submit(perform, task): task
        for task in get_tasks_to_do()
    }

    for fut in concurrent.futures.as_completed(futures):
        original_task = futures[fut]
        print(f"The outcome of {original_task} is {fut.result()}")

Rather than creating a set, we’re creating a dict that maps each Future to its original task. When the Future completes, we look in the dict to find the task.

There’s a concrete example of this pattern in the Python docs.

This approach schedules all the tasks immediately, and creates a Future for each of them. That’s fine if you have a small number of tasks, but if you have lots of tasks it means you’re using lots of memory, and at some point your program might just crash. If we have a lot of tasks, we want to limit how many we’re working on at a time.

Running a large, fixed number of tasks

Often my tasks come from a lazy generator, because lazy generators are more memory efficient if you’re only processing one thing at a time. For example, I might have a list of tasks in a text file, and I read the tasks from the file line-by-line.

The pattern above loads all the tasks immediately, potentially using a lot of memory. It would be nicer if it was only keeping a small number of tasks in memory – the ones that were currently being worked on.

One approach to handle a very large number of tasks would be to break them into small chunks, and process each chunk in turn using the pattern above. Using my chunked_iterable() method from a previous post, we could do something like:

for task_set in chunked_iterable(get_tasks_to_do(), HOW_MANY_TASKS_AT_ONCE):
    with concurrent.futures.Executor() as executor:
        futures = {
            executor.submit(perform, task)
            for task in task_set
        }

        for fut in concurrent.futures.as_completed(futures):
            print(f"The outcome is {fut.result()}")

This code is fairly simple, but we’re losing some of the efficiency gains – it queues up N pieces of work, gets through them all, then loads another N pieces of work. When it’s near the end of a chunk, it’s only processing a few pieces of work concurrently.

This graph shows the number of tasks available to the executor against time – you can see when each chunk starts and ends. We’re getting some of the benefits of concurrency, but we could be doing better.

time tasks in the pool

It would be better if we could keep the pool “topped up” – as one task finishes, we schedule the next one.

If we have an iterable of Futures, we can find ones that have completed with concurrent.futures.wait(). It returns a 2-tuple of finished and unfinished futures:

finished, unfinished = concurrent.futures.wait(
    futures, return_when=concurrent.futures.FIRST_COMPLETED
)

The return_when parameter lets us choose to wait for the first Future to complete, to throw an exception, or for everything to complete (which is equivalent to as_completed). In this case, I’m waiting for the first Future. When it’s done, we’ll go ahead and schedule the next one.

Here’s what it looks like:

import concurrent.futures
import itertools


tasks_to_do = get_tasks_to_do()

with concurrent.futures.ThreadPoolExecutor() as executor:

    # Schedule the first N futures.  We don't want to schedule them all
    # at once, to avoid consuming excessive amounts of memory.
    futures = {
        executor.submit(perform, task)
        for task in itertools.islice(tasks_to_do, HOW_MANY_TASKS_AT_ONCE)
    }

    while futures:
        # Wait for the next future to complete.
        done, futures = concurrent.futures.wait(
            futures, return_when=concurrent.futures.FIRST_COMPLETED
        )

        for fut in done:
            print(f"The outcome is {fut.result()}")

        # Schedule the next set of futures.  We don't want more than N futures
        # in the pool at a time, to keep memory consumption down.
        for task in itertools.islice(tasks_to_do, len(done)):
            futures.add(
                executor.submit(perform, task)
            )

A lot of heavy lifting is being done by itertools.islice(iterable, N), which generates the first N elements of an iterable (or the whole thing, if there are less than N elements left).

We start by scheduling the initial batch of futures, then loop as long as there are uncompleted futures. When a future completes, we split the set into done or futures (the new set of uncompleted futures).

We process the completed futures (for fut in done:), then schedule an equivalent number of new futures. This way, the executor always has plenty of tasks to do, and we’re making efficient use of the concurrency. This is what the number of tasks available looks like:

time tasks in the pool

The pool is kept topped up with new tasks to start, until we run out of tasks and it gradually runs down to zero. In practice, the exact shape will be less regular, but it gives the general idea.

This pattern makes it a little harder to track the task a future was associated with, because concurrenct.futures.wait() always returns a set, regardless of what iterable was passed in. But it’s still possible, like so:

with concurrent.futures.ThreadPoolExecutor() as executor:

    # Schedule the first N futures.  We don't want to schedule them all
    # at once, to avoid consuming excessive amounts of memory.
    futures = {
        executor.submit(perform, task): task
        for task in itertools.islice(tasks_to_do, HOW_MANY_TASKS_AT_ONCE)
    }

    while futures:
        # Wait for the next future to complete.
        done, _ = concurrent.futures.wait(
            futures, return_when=concurrent.futures.FIRST_COMPLETED
        )

        for fut in done:
            original_task = futures.pop(fut)
            print(f"The outcome of {original_task} is {fut.result()}")

        # Schedule the next set of futures.  We don't want more than N futures
        # in the pool at a time, to keep memory consumption down.
        for task in itertools.islice(tasks_to_do, len(done)):
            fut = executor.submit(perform, task)
            futures[fut] = task

As before, we’re storing the context in a dict futures. When a Future completes, we .pop() from the dict to delete the entry and retrieve the original task. When we schedule a new task, we store it in the dict. The futures dict still contains all the unfinished Futures, so we can continue to use it as a check for when the process is complete.

Running tasks that might have follow-up work

This is the problem I was originally trying to solve – I’d schedule some Futures based on initial tasks, but then the result of those Futures might uncover more tasks.

For example, the DynamoDB Scan API is paginated. You don’t know where the page boundaries are in advance – each response tells you how to request the next page. When you get page 1, that tells you how to request page 2. When you get page 2, that tells you how to request page 3, and so on. This means you can only schedule a task to request page 2 after you’ve received page 1.

When I first tried to do this, I kept having problems – everything I scheduled in the initial batch would complete, but I’d never see the result of any new Futures that I’d scheduled. I had to try a simpler problem (running large, fixed number of tasks, above) to learn about concurrent.futures.wait(), which can be used to solve this problem.

We can make a small modification to the for fut in done: loop in the example above:

for fut in done:
    print(f"The outcome is {fut.result()}")

    if has_follow_up_task(fut.result()):
        new_task = get_follow_up_task(fut.result())
        futures.add(
            executor.submit(perform, new_task)
        )

If there’s a follow-up task to run, it schedules that task as another Future. It’s now in the pool of running Futures, and it will get checked the next time we call concurrent.futures.wait(). (In theory you could schedule multiple new tasks here, but be careful – if you create too many, you could trigger a fork bomb.)

Eventually your process should stop scheduling follow-up tasks, or it’ll get stuck in an infinite loop and never terminate.

If there’s a follow-up task, we also need to skip scheduling a task from the initial batch of tasks – otherwise futures, the pool of currently running Futures, will keep growing. Here’s how that works:

import concurrent.futures
import itertools


with concurrent.futures.ThreadPoolExecutor() as executor:

    # Schedule the first N futures.  We don't want to schedule them all
    # at once, to avoid consuming excessive amounts of memory.
    futures = {
        executor.submit(perform, task)
        for task in itertools.islice(initial_tasks, HOW_MANY_TASKS_AT_ONCE)
    }

    while futures:
        # Wait for the next future to complete.
        done, futures = concurrent.futures.wait(
            futures, return_when=concurrent.futures.FIRST_COMPLETED
        )

        # Process the results of any completed futures, then schedule any
        # follow-up tasks.  If there's a follow-up task, we don't want
        # to schedule a replacement task from the initial batch.
        new_tasks_to_schedule = 0

        for fut in done:
            print(f"The outcome is {fut.result()}")

            if has_follow_up_task(fut.result()):
                new_task = get_follow_up_task(fut.result())
                futures.add(
                    executor.submit(perform, new_task)
                )
            else:
                new_tasks_to_schedule += 1

        # Schedule the next set of futures.  We don't want more than N futures
        # in the pool at a time, to keep memory consumption down.
        for task in itertools.islice(initial_tasks, new_tasks_to_schedule):
            futures.add(
                executor.submit(perform, task)
            )

And as before, we can use a dict to track the task a future was started from:

import concurrent.futures
import itertools


with concurrent.futures.ThreadPoolExecutor() as executor:

    # Schedule the first N futures.  We don't want to schedule them all
    # at once, to avoid consuming excessive amounts of memory.
    futures = {
        executor.submit(perform, task): task
        for task in itertools.islice(initial_tasks, HOW_MANY_TASKS_AT_ONCE)
    }

    while futures:
        # Wait for the next future to complete.
        done, _ = concurrent.futures.wait(
            futures, return_when=concurrent.futures.FIRST_COMPLETED
        )

        # Process the results of any completed futures, then schedule any
        # follow-up tasks.  If there's a follow-up task, we don't want
        # to schedule a replacement task from the initial batch.
        new_tasks_to_schedule = 0

        for fut in done:
            original_task = futures.pop(fut)
            print(f"The outcome of {original_task} is {fut.result()}")

            if has_follow_up_task(fut.result()):
                new_task = get_follow_up_task(fut.result())
                fut = executor.submit(perform, new_task)
                futures[fut] = new_task
            else:
                new_tasks_to_schedule += 1

        # Schedule the next set of futures.  We don't want more than N futures
        # in the pool at a time, to keep memory consumption down.
        for task in itertools.islice(initial_tasks, new_tasks_to_schedule):
            fut = executor.submit(perform, task)
            futures[fut] = task

Using this code

When I want to use this code, I copy/paste it into my newest script, and fill in the parts that have changed. If you’d like to try the code, you can do the same!

Most of my scripts are only ever intended to run on one machine, so I optimise for speed on that computer. I experiment with ThreadPoolExecutor and ProcessPoolExecutor and the max_workers variable until I get something that’s fast enough. It isn’t necessarily optimal – it’s just fast enough that it’s not worth fiddling any further. Getting a script from 5 minutes to 30 seconds is pretty good. Spending time to save another 5–10 seconds is harder to justify.

Now I have these templates and a write-up, I hope I’ll be able to use this concurrency code much more. As well as being easier to find, I’m more confident that I understand how the code works, and where the rough edges are. Once again, a detailed walkthrough of code is as useful as me as for the reader.

Update, 2 May 2020: A common problem with this code is when your list of tasks is a list. If tasks_to_do is a list, you’ll see it run the same tasks repeatedly.

That’s because on each call to itertools.islice(), you’re passing the complete list each time. It only reads the list; it doesn’t modify it. That means that when you call it more than once, it doesn’t know you’ve already gone through some elements of the list, so it starts from the first element again. You can see this in the interactive console:

  >>> import itertools
  >>> numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9]
  >>> list(itertools.islice(numbers, 3))
  [1, 2, 3]
  >>> list(itertools.islice(numbers, 3))
  [1, 2, 3]
  >>> list(itertools.islice(numbers, 3))
  [1, 2, 3]

The fix is to create an iterator for the list, and pass that around – it holds the state of “how many elements have I already been through”. Observe:

  >>> iterator = iter(numbers)
  >>> list(itertools.islice(iterator, 3))
  [1, 2, 3]
  >>> list(itertools.islice(iterator, 3))
  [4, 5, 6]
  >>> list(itertools.islice(iterator, 3))
  [7, 8, 9]

Update, 4 January 2022: I’ve created a GitHub repo which has a nicer version of this code, which allows you to call a single function, rather than interspersing your function/inputs with the concurrency code:

from concurrently import concurrently

for (input, output) in concurrently(fn=perform, inputs=get_tasks_to_do()):
    print(input, output)

I’d recommend using this instead of the code above (although the above still serves as a valuable explanation of how it works). See alexwlchan/concurrently.