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免责声明：不确定标题是否合适，这是我的第一篇文章，所以我正在学习如何组织这个

代码背景：

我使用 Apophysis7x 创建分形火焰（迭代图像）并在 Fractal Fr0st 中打开 .flame 文件（此文件未渲染，它只是保存要在另一个编辑器中打开的信息），我可以在其中自动化例如不断旋转的迭代从而创建动画。Fractal Fr0st 使用 python 进行此类自动化，但它们是不变的。我制作音乐并希望有一天能完成下面的代码，以根据我的歌曲的 BPM 进行轮换。在编辑与程序关联的实际 python 文件之前，我开始自己编写代码以生成滴答声序列（该代码现在并不完全重要，但在最后发布）

到目前为止有关问题的信息：

我马上就知道 Python 不是最快的语言，但我不能 100% 确定我的代码是否不够高效，无法跟上我想要输出的内容。请查看我的代码和评论，因为我试图尽可能彻底地解释发生了什么。

计时器.py：

import time

#Feel free to test as you'd like
#Suppose we have a song at 140 Beat Per Minute (BPM) -> There is approximately 2.33 Beats Per Second (BPS)
#So it takes approximately 0.43 (s) per beat
#So  if freq = 0.43, leng = *length of song*
#This program should alternate printing "tick" "tock" (or some action that I hope to code in the future)
#The Tick Tock output should come out every (in this case) .43 seconds

class timer(object):
    "takes in a frequency and duration of loop in seconds"

    def __init__(self,freq,leng):
        "input a frequency and duration in seconds"
        #rounded to 2 decimal places

        self.freq= round (freq,2)
        self.leng= round (leng,2)

    def ticker (self):
        "Starts the time and loop to be broken when max time is reached"

        counter = 0 #recorded out of curiosity to check just how many loops were made overall
        ticker = 0 #alternates the "tick" "tock"
        initTime = time.time() #records the initial upon starting the loop
        rnded = 0 #initial rounded time set at 0 *see the first if statement below*


        while True:

            stop = time.time() #The time recorded during the loop *stopwatch so-to-speak*
            passed = stop - initTime #The time passed from initial launch to time during the loop

            if rnded == self.leng:
                "When the time that has passed reaches time limit print stats and break"

                return print("Loops:".ljust(10).rjust(14) + "Initial Time:".ljust(17).rjust(21) + "Time Stopped:".ljust(17).rjust(21)+ "Passed:".ljust(11).rjust(15) + "Total Ticks:".ljust(16).rjust(20) + "\n" + (str (counter) ).ljust(10).rjust(14) +(str (initTime) ).ljust(17).rjust(21)+ (str (stop) ).ljust(17).rjust(21)+(str (rnded) ).ljust(11).rjust(15) + (str (ticker) ).ljust(11).rjust(15) )

            elif round(passed,2) == rnded:
                "Prevents duplicates, ie: previous stopped at .9999 (1.0) and current loop sopped at 1.001 (1.0) skip" 
                #If your current value happened to be rounded to the previous value don't continue

                #uncomment bellow for debugging
                #print ("*TEST* \t Initiated: ", initTime, "\t Time stopped: ",stop, "\nTime passed: ", rnded, "\t Tick Counter: ", ticker, "\n ---")
                pass

            else:
                "Tick tock code"

                rnded = round(passed,2) #now set current time differnce value after being checked for duplicates
                if rnded % self.freq == 0 and rnded != 0:

                    "if the time passed mod the frequency is 0 continue to execute tick tock"

                    if ticker % 2 == 0:
                        #gives stats for debugging purposes, delete, after "tick" / "tock" if desired
                        #prints in the meantime in the future I would like an action to be done with music in another program as precise as possible

                        print ("Tick \t Initiated: ", initTime, "\t Time stopped: ",stop, "\nTime passed: ", rnded, "\t Tick Counter: ", ticker, "\n ---")

                    elif ticker % 2 == 1:
                        print ("Tock \t Initiated: ", initTime, "\t Time stopped: ",stop, "\nTime passed: ", rnded, "\t Tick Counter: ", ticker, "\n ---")

                    ticker +=1 #only increases if tick/tock was processed        

        counter += 1 #counts how many times loop was accessed, use this as you'd like

    def chngFreq(self, freq):
        'Changes current frequency'
        self.freq = round (freq,2)

    def chngLeng(self,leng):
        'Changes current duration'
        self.leng = round(leng,2)

示例 1：

随着频率变快，输出会跳过多个“滴答”/“滴答”（我只是在测试，例如，当频率为 0.01（s）且总时间为 4（s）时，总共应该有 399 个滴答声但它只达到了 5 个滴答声。我的理论是，由于它确实需要时间来完成循环（但是时间非常短），它可能会跳过一个滴答声，因为在从 . 01 到 .02 在四舍五入之前可能是一个勾号，但代码处理速度不够快。我四舍五入到最后尽可能准确，并制作了“elif”部分以防止重复（请参阅上面的代码）

输出：

>>>s = timer(.01,4)
>>>s.ticker()
Tick     Initiated:  1502409793.6775877      Time stopped:  1502409793.6937597 
Time passed:  0.02   Tick Counter:  0 
---
Tock     Initiated:  1502409793.6775877      Time stopped:  1502409793.7562232 
Time passed:  0.08   Tick Counter:  1 
---
Tick     Initiated:  1502409793.6775877      Time stopped:  1502409793.8409605 
Time passed:  0.16   Tick Counter:  2 
 ---
Tock     Initiated:  1502409793.6775877      Time stopped:  1502409793.9940765 
Time passed:  0.32   Tick Counter:  3 
---
Tick     Initiated:  1502409793.6775877      Time stopped:  1502409794.9579115 
Time passed:  1.28   Tick Counter:  4 
---
Loops:        Initial Time:        Time Stopped:        Passed:        Total Ticks:    
5473896      1502409793.6775877    1502409797.680171    4.0            5

请注意，有时只计算 2^x 的刻度，即： .01、.02、.04、.08、.16、.32... 而不是： .01、.02、.03、.04、 .05, .06...

我知道这不是我在代码中给出的示例，但是每当我尝试使用更近似的频率（例如 0.23 或 .01）时都是如此，但是无论精度如何，频率越大，即：4.29 它更可能准确返回“刻度”4.29、8.58、12.87 等...

示例 2

在以下示例中，您可以说它工作得非常好，因为它在刻度之间有更长的时间来处理

>>>s = timer(1,4)
>>>s.ticker()
Tick     Initiated:  1502411198.8088021      Time stopped:  1502411199.8091898 
Time passed:  1.0    Tick Counter:  0 
---
Tock     Initiated:  1502411198.8088021      Time stopped:  1502411200.8089292 
Time passed:  2.0    Tick Counter:  1 
---
Tick     Initiated:  1502411198.8088021      Time stopped:  1502411201.81057 
Time passed:  3.0    Tick Counter:  2 
---
Tock     Initiated:  1502411198.8088021      Time stopped:  1502411202.8121786 
Time passed:  4.0    Tick Counter:  3 
---
Loops:        Initial Time:        Time Stopped:        Passed:        Total Ticks:    
5586557      1502411198.8088021   1502411202.8328226    4.0            4

如果有任何代码不清楚，请告诉我，但请务必先阅读每条评论，并在需要时运行代码/修改。

一旦我之前的问题得到解决，我希望修改 Fractal Fr0st 中使用的以下代码的“if x.animate:”部分。

默认.py：

update_flame = False

while True:
    for x in flame.xform:
        if x.animate:
            x.rotate(-3)
    preview()

最后感谢一百万我知道这真的很长，但我真的想了解 python 发生了什么/它是如何运作的，不管我的最终目标是什么。

此外，如果任何其他方法更快/更有效，请告诉我。谢谢！！！

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1 回答 1

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经过一些研究和一些简单的混乱，我不仅找到了一些解决方案（我发现了真正的问题并且仍在发生），而且它也使我的代码更短。看一看。

新的和改进的代码：

import time

#Feel free to test as you'd like
#Suppose we have a song at 140 Beat Per Minute (BPM) -> There is approximately 2.33 Beats Per Second (BPS)
#So it takes approximately 0.43 (s) per beat
#So  if freq = 0.43, leng = *length of song*
#This program should alternate printing "tick" "tock" (or some action that I hope to code in the future)
#The Tick Tock output should come out every (in this case) .43 seconds

class timer2(object):
    "takes in a frequency and duration of loop in seconds"

    def __init__(self,freq,leng):
        "input a frequency and duration in seconds"
        #rounded to 2 decimal places

        self.freq= round (freq,2)
        self.leng= round (leng,2)

    def ticker (self):
        "Starts the time and loop to be broken by freq"

        self.ticker = 0 #alternates the "tick" "tock"
        self.passed = 0 

        while round (self.passed,2) != self.leng:

            time.sleep(self.freq)

            if self.ticker % 2 == 0:
            #prints in the meantime in the future I would like an action to be done with music in another program as precise as possible

                print ("***TICK*** \t Tick Counter: ", self.ticker)

            elif self.ticker % 2 == 1:

                print ("***TOCK*** \t Tick Counter: ", self.ticker)

            self.ticker +=1 #only increases if tick/tock was processed

            self.passed += self.freq
            round (self.passed,2)

            print ("Time passed: ", self.passed, "(s)","\n ---")

        print ("Ticks: \t Time Passed: \n {} \t {}(s)".format(self.ticker,self.passed))

    def chngFreq(self, freq):
        'Changes current frequency'
        self.freq = round (freq,2)

    def chngLeng(self,leng):
        'Changes current duration'
        self.leng = round(leng,2)

请注意许多 round() 方法，我将稍后解释。

我的基本输入：

>>>s = timer2(1,4)
>>>s.ticker()

这按预期工作：1,2,3,4。然而，回到我最初想要的，如果我的频率很小，比如 0.43，它就不能准确地工作。我包括 0.01 的统一频率的第一个和最后 4 个输出。

输出：

前四名：

>>>s = timer2(.01,4)
>>>s.ticker()
***TICK***   Tick Counter:  0
Time passed:  0.01 (s) 
---
***TOCK***   Tick Counter:  1
Time passed:  0.02 (s) 
---
***TICK***   Tick Counter:  2
Time passed:  0.03 (s) 
---
***TOCK***   Tick Counter:  3
Time passed:  0.04 (s) 
---

前四名：

***TICK***   Tick Counter:  396
Times passed:  3.9699999999999593 (s) 
---
***TOCK***   Tick Counter:  397
Times passed:  3.979999999999959 (s) 
---
***TICK***   Tick Counter:  398
Times passed:  3.989999999999959 (s) 
---
***TOCK***   Tick Counter:  399
Times passed:  3.9999999999999587 (s) 
---
Ticks:   Time Passed: 
 400     3.9999999999999587(s)

请注意，最后经过的时间应该是 4 (s)，就像频率为 1 时一样。另外，为什么经过的时间是一个非常难看的浮点数？前四个很好，之后它给出了长小数。我将打开另一个线程，给出一个简单的循环（我仍然希望对此进行修改）。

Python 没有正确添加，这很烦人...... .31 + .01 应该是：.32 及以后，.34、.35、.36、.37 等。即使将时间四舍五入到小数点后 2 位，它的不工作。IE。说即使我不能修复丑陋的浮点数（.310000015684668），四舍五入应该是（.31），但Python不会执行这个。

***如果我消除了所有的舍入，我仍然会得到很长的浮点数，所以这不是 ROUND () 方法开始***

于 2017-08-16T01:32:57.333 回答