进度指标使PI计算太慢

Question

我编写了代码来计算PI到用户指定的小数位数。效果很好。我想添加一个进度指示，这样用户就不必连续数小时盯着闪烁的光标，而不显示进度。我得到了一些有用的东西，但它使代码变得非常、太慢。在下面的代码中，我在“冒犯”代码之前和之后用*‘S的双行线注释了进度指标部分。我怀疑是log10计算使事情变得太慢了。任何建议：

# A program to calculate PI to a user specified number of decimal digits
# The program uses the Chudnovsky algorithm.  
# Further details on the algorithm are availabler here:
# https://en.wikipedia.org/wiki/Chudnovsky_algorithm

# A generator which divides the user requested number of decimal places into 10% increments

def ten_percent(places):

    for i in range(int(places/10),int(places+1),int(places/10)):
        yield i

def pi_calc(places):

    # places is user specified number of decimal places that should be calculated.

    import decimal
    import time
    import numpy as np
    start_time = time.time()

    decimal.getcontext().prec = places + 3

    # The function makes succesive calculations of PI using the algorithm.
    # The current calculated PI value is subtracted from the succesive PI value
    # If the difference is less than the 1 x 10**(-places) the result is returned

    # Initialise some variables

    current_pi = 3
    next_pi = 3
    precision = 1 * 10**(-places-3)
    counter = 1
    precision_step = ten_percent(places)
    precision_check = next(precision_step)

    # Initialise terms used in the itteration

    L = 13591409
    X = 1
    M = 1
    K = 6
    S = decimal.Decimal(M*L)/X

    next_pi = 426880*decimal.Decimal(10005).sqrt()/S

    # Perform the itterative calculation until the required precision is achieved

    while abs(next_pi - current_pi) > precision:
        current_pi = next_pi

        # Calculate the next set of components for the PI approximation

        L += 545140134
        X *= -262537412640768000
        M = M*(K**3-16*K)//(counter)**3
        S += decimal.Decimal(M*L)/X
        K += 12
        counter += 1

        # Calculate the next approximation of PI

        next_pi = 426880*decimal.Decimal(10005).sqrt()/S

#******************************************************************************************************************************
#******************************************************************************************************************************
# This progress indication slows the code down too much to be practical
#
#
#       # Give the user some feedback on progress of the calculation
#
#       # The try statement is required because the error between successive pi calculations can become
#       # "infintly" small and then a string is returned instead of a number.
#
#       try:
#           test_num = abs(round(np.log10(abs(decimal.Decimal(next_pi - current_pi)))))
#       except:
#           pass
#       
#       if test_num >= precision_check:
#           print('Calculation steps: ' + str(counter-1) + ' Approximate decimal places: ' + str(precision_check))
#           if precision_check < places:
#               precision_check = next(precision_step)
#*******************************************************************************************************************************
#*******************************************************************************************************************************

    return decimal.Decimal(str(next_pi)[:places+2])

# Get the required number of decimal places from the user and call the function to perform
# the calculation

while True:

    try:
        places = int(input('To how many decimal places would you like to calculate PI? '))
    except:
        print('Please provide a valid integer value.')
        continue
    else:
        break

calculated_pi = pi_calc(places) 
print('The value of PI to '+ str(places) + ' decimal places is:\n%s' % (calculated_pi))

Answer 1

这一答案提出了各种修复性能的策略。它不考虑代码的其他方面。它不考虑数值分析。(如果有一种更便宜的方法来进行实际计算来衡量进度，那不是我能告诉你的。)

首先，这段代码经历了很多迭代。您知道，大多数情况下，您将没有达到下一个进度标记。如果您只通过循环每100次检查一次进度，您已经将花费在这一昂贵操作上的时间减少到了1%。(对于迭代次数较少的事情，您将失去准确性，但对于快速情况，进度条不太重要。)

另一种选择是在一个单独的过程中计算进度。主进程可以将next_pi和current_pi之间的差异推到堆栈上，当进程完成最后一次计算时，这个堆栈就会弹出。

最后，你可以估计/撒谎。图运行时间(或迭代次数)与数字数之比。这是个不错的功能吗？然后，你可以猜出这样的计算要花费多长时间，并以此为基础进行计算。我不是一个数学人，所以我不知道这是否实用。如果不是，也许它能让你朝着更好的方向思考。