coroutine

Python中 yield 关键字的用途是什么？ 它有什么作用？ 例如，我试图理解这段代码 1 ： def _get_child_candidates(self, distance, min_dist, max_dist): if self._leftchild and distance - max_dist < self._median: yield self._leftchild if self._rightchild and distance + max_dist >= self._median: yield self._rightchild 这是呼叫者： result, candidates = [], [self] while candidates: node = candidates.pop() distance = node._get_dist(obj) if distance <= max_dist and distance >= min_dist: result.extend(node._values) candidates.extend(node._get_child_candidates(distance, min_dist, max_dist)) return result 调用 _get_child_candidates 方法时会发生什么？ 是否返回列表？

I'm evaluating different patterns for periodic execution (actual sleep/delays ommited for brevity) using the Python 3 asyncio framework, and I have two pieces of code that behave diffrently and I can't explain why. The first version, which uses yield from to call itself recursively exhausts the stack in about 1000 iterations, as I expected. The second version calls the coroutine recursively, but delegates actual event loop execution to asyncio.async and doesn't exhaust the stack. Can you explain in detail why the stack isn't being used by the second version? What are the differences between

It is very convenient to use Tasks to express a lazy collection / a generator. Eg: function fib() Task() do prev_prev = 0 prev = 1 produce(prev) while true cur = prev_prev + prev produce(cur) prev_prev = prev prev = cur end end end collect(take(fib(), 10)) Output: 10-element Array{Int64,1}: 1 1 2 3 5 8 13 21 34 However, they do not follow good iterator conventions at all. They are as badly behaved as they can be They do not use the returned state state start(fib()) == nothing #It has no state So they are instead mutating the iterator object itself. An proper iterator uses its state, rather

I'm working on something that requires fast coroutines and I believe numba could speed up my code. Here's a silly example: a function that squares its input, and adds to it the number of times its been called. def make_square_plus_count(): i = 0 def square_plus_count(x): nonlocal i i += 1 return x**2 + i return square_plus_count You can't even nopython=False JIT this, presumably due to the nonlocal keyword. But you don't need nonlocal if you use a class instead: def make_square_plus_count(): @numba.jitclass({'i': numba.uint64}) class State: def __init__(self): self.i = 0 state = State() @numba

Closed . This question needs to be more focused . It is not currently accepting answers. Want to improve this question? Update the question so it focuses on one problem only by editing this post . Closed 2 years ago . I've heard about co-routines long ago, but never used them. As I know, co-routines are similar to generators. Why do we need co-routines in Python? Generator uses yield to return values. Python generator functions can also consume values using a (yield) statement. In addition two new methods on generator objects, send() and close() , create a framework for objects that consume

I'm trying to get an understanding of how I can use co-routines to "pause" a script and wait until some processing is done before resuming. Perhaps I'm looking at co-routines in the wrong way. But my attempt is structured similar to the example given in this answer . The loop in loop.lua never reaches a second iteration, and hence never reaches the i == 4 condition required to exit the running loop in the C code. If I do not yield in loop.lua , then this code performs as expected. main.cpp #include <lua/lua.hpp> bool running = true; int lua_finish(lua_State *) { running = false; printf("lua

I'm looking to be able to yield from a number of async coroutines. Asyncio's as_completed is kind of close to what I'm looking for (i.e. I want any of the coroutines to be able to yield at any time back to the caller and then continue), but that only seems to allow regular coroutines with a single return. Here's what I have so far: import asyncio async def test(id_): print(f'{id_} sleeping') await asyncio.sleep(id_) return id_ async def test_gen(id_): count = 0 while True: print(f'{id_} sleeping') await asyncio.sleep(id_) yield id_ count += 1 if count > 5: return async def main(): runs = [test

using HttpServer http = HttpHandler() do request::Request, response::Response show(request) Response("Hello there") end http.events["error"] = (client, error) -> println(error) http.events["listen"] = (port) -> println("Listening on $port") server = Server(http) t = @async run(server, 3000) This starts a simple little web server asynchronously. The problem is I have no idea how to stop it. I've been going through the Julia documentation and trying to find some function that will remove this task from the queue ( kill , interrupt , etc.) but nothing seems to work. How can I kill this task? I