yield

invoking yield for a generator in another function

由 ≡放荡痞女 提交于 2019-12-12 08:33:33
问题 suppose I have some manager object. This object's API has a main_hook function, that gets another function f as it's argument, and runs the given f in a loop, doing some stuff in between each iteration: def main_hook(self,f): while (self.shouldContinue()): #do some preparations f(self) #do some tear down Now, I also have (more accurately, would like to have ) a function stop_and_do_stuff , that once called, stops main_hook dead in it's tracks, returns the control to whichever func called main

Generator recovery using decorator

由 泪湿孤枕 提交于 2019-12-12 08:33:27
问题 Let's have a class that has function that fails from time to time but after some actions it just works perfectly. Real life example would be Mysql Query that raises _mysql_exceptions.OperationalError: (2006, 'MySQL server has gone away') but after client reconnection it works fine. I've tried to write decorator for this: def _auto_reconnect_wrapper(func): ''' Tries to reconnects dead connection ''' def inner(self, *args, _retry=True, **kwargs): try: return func(self, *args, **kwargs) except

How does method yield work?

由 痞子三分冷 提交于 2019-12-12 07:56:25
问题 In javadoc there is said that yield method Causes the currently executing thread object to temporarily pause and allow other threads to execute. And Katherine Sierra and Bert Bates SCJP book says that yield() is supposed to do is make the currently running thread head back to runnable to allow other threads of the same priority to get their turn. So what actually method is doing? 回答1: Given a multi-threaded application, yield will cause the currently executing thread to pause execution and be

Range with floating point numbers and negative steps

由 半城伤御伤魂 提交于 2019-12-12 06:37:39
问题 I wrote the following for creating a range with negative floating point steps: def myRange(start, stop, step): s = start if step < 0: while s > stop: yield s s += step if step > 0: while s < stop: yield s s += step But the output of r = myRange(1,0,-0.1) looks rather strange >>> r = myRange(1,0,-0.1) >>> for n in r: print n ... 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 1.38777878078e-16 where does this last number come from? And why is it not 0? 回答1: Not all floating point numbers can be

Using “yield” in a function

由 試著忘記壹切 提交于 2019-12-12 03:34:37
问题 I want to generate something like that in a function that receives 1 argument n using yield to generate: 1 1+2 1+2+3 … … 1+2+3+⋯+n−1+n That is my last try: def suite(n): total = 0 for i in n: total+=i yield total and this is what I receive: Traceback (most recent call last): File "notebook", line 4, in suite TypeError: 'int' object is not iterable 回答1: Your error is here: for i in n: n is an integer and not an iterable. Perhaps you wanted to use xrange() (Python 2 only) or range()

Avoid Multiple Next () Statement in Python Generator

由 主宰稳场 提交于 2019-12-12 02:37:23
问题 I'm using a library that returns a generator. Is there a way to start at a particular iteration without using multiple next () statement? In a simple for loop, I could do the following. array = [2, 5, 1, 4, 3] for i in array [2:]: # do something In a generator, I couldn't do as shown above. Instead I'll have to use multiple next () statements to start at the 3rd index. When attempting to do the same as the for loop, I get an error that said, "generator is not scriptable." 回答1: itertools

生成器generator

由 ﹥＞﹥吖頭↗ 提交于 2019-12-11 16:15:25
在Python中， 一边循环一边计算后面元素的机制称为生成器 generator 一、列表生成器 语法： a = [i+1 for i in rang(10)] print(a) [1,2,3,4,5,6,7,8,9,10] 二、生成器 语法： 把列表生成器中的中括号 [] 换成 () 即可 例： (x*x for x in range(10)）生成的就是一个生成器。 如果要一个一个打印出来，可以通过 next() 函数获得generator的下一个返回值： generator保存的是算法，每次调用 next(g) 就计算出 g 的下一个元素的值，直到计算到最后一个元素，没有更多的元素时，抛出 StopIteration 的错误。 当然，上面这种不断调用 next(g) 实在是太变态了，正确的方法是使用 for 循环，因为generator也是可迭代(遍历)对象： >>> g = (x * x for x in range(10)) >>> for n in g: ... print(n) ... 0 1 4 9 16 25 36 49 64 81 三、函数生成器 def feib(n): a = 0 b = 1 count = 0 while count < n: tmp = a a = b b += tmp #print(b) yield b # 暂停，程序执行到这里

How to 'flatten' generators in python?

由 纵饮孤独 提交于 2019-12-11 09:09:55
问题 I have a problem with 'flattening' out some generators in python. Here is my code: import itertools as it test = [[1,2,3],[4,5],[6,7,8]] def comb(possible): if len(possible) != 1: for a in possible[0]: yield from it.product((a,), comb(possible[1:])) else: yield from possible[0] list(comb(test)) which gives me: [(1, (4, 6)), (1, (4, 7)), (1, (4, 8)), (1, (5, 6)), (1, (5, 7)), (1, (5, 8)), (2, (4, 6)), (2, (4, 7)), (2, (4, 8)), (2, (5, 6)), (2, (5, 7)), (2, (5, 8)), (3, (4, 6)), (3, (4, 7)), (3

Laravel - @yield() not allowed in if

由 偶尔善良 提交于 2019-12-11 07:43:30
问题 In my title I've been using @yield('title') for a while to create a dynamic title for every different page. I'd like to take it one step further and do something like this: @if(@yield('title') == 'post') <h1> This is the post page </h1> @endif But if I try that I get the error: "The "yield" expression can only be used inside a function" I've searched around and found things like Section::yield('title') but they all didn't work. How can I use the value of @yield('title') in an if statement? I

find_in_batches “NO BLOCK GIVEN (YIELD)”

由 感情迁移 提交于 2019-12-11 07:33:00
问题 I have a method where I take in a model and result_size . The first thing i try to do in this method is: array = model.logs.find_in_batches(:batch_size => result_size) But this doesn't work; instead it returns "No Block Given (Yield)". I guess I'm just unfamiliar with blocks and yields. If anyone could help me understand/fix this problem I would greatly appreciate it! Thanks in advance! 回答1: find_in_batches expects you to pass the values into a block, as so: model.logs.find_in_batches(:batch
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