python -优化分支逻辑

Question

我目前有一种令人讨厌的代码块，我正在为一个模拟螃蟹成长的程序编写代码(令人兴奋的东西……)。这个程序后来被更大的模拟所吸收，所以代码的速度很重要。我最慢的代码块之一包含了这个讨厌的分支逻辑。我希望有人能想出一种更有效的方法.

就上下文而言，这段代码基本上是在说：“我是在再生一只爪子吗?如果是哪一只。如果左/右爪子正在再生，那么它是占主导地位的还是不占主导地位的一只手?有鉴于此，应用这个数字xyz。”

if left_or_right_growing == 'left':
    if crab.rightclawCrusher == True:
        crab.rightclaw_size = new_crushersize
        if crab.moltnumber_for_claw_removal_left < crab.numberofmolts: #so it doesn't overwrite budding claw growth
            crab.leftclaw_size = max(new_pincersize * crab.proportion_of_new_claw_thats_grownback()+adj, crab.leftclaw_size)
    elif crab.rightclawCrusher == False:
        crab.rightclaw_size = new_pincersize + adj
        if crab.moltnumber_for_claw_removal_left < crab.numberofmolts: #so it doesn't overwrite budding claw growth
            crab.leftclaw_size = max(new_crushersize * crab.proportion_of_new_claw_thats_grownback(), crab.leftclaw_size)

elif left_or_right_growing == 'right':
    if crab.rightclawCrusher == True:
        crab.leftclaw_size = new_pincersize + adj
        if crab.moltnumber_for_claw_removal_right < crab.numberofmolts: #so it doesn't overwrite budding claw growth
            crab.rightclaw_size = max(new_crushersize  * crab.proportion_of_new_claw_thats_grownback(), crab.rightclaw_size)
    elif crab.rightclawCrusher == False:
        crab.leftclaw_size = new_crushersize
        if crab.moltnumber_for_claw_removal_right < crab.numberofmolts: #so it doesn't overwrite budding claw growth
            crab.rightclaw_size = max(new_pincersize  * crab.proportion_of_new_claw_thats_grownback() +adj, crab.rightclaw_size)

elif left_or_right_growing == 'both':
    pro_left, pro_right = crab.proportion_of_new_claw_thats_grownback()
    if pro_left > 1. or pro_right > 1.:
        print('ERROR IN TRANFORM:  pro_left: ' + str(pro_left) +'   pro_right:  ' + str(pro_right))
    if crab.rightclawCrusher == True:
        if crab.moltnumber_for_claw_removal_left < crab.numberofmolts: #so it doesn't overwrite budding claw growth
            crab.leftclaw_size = max(new_pincersize * pro_left+ adj, crab.leftclaw_size) 

        if crab.moltnumber_for_claw_removal_right < crab.numberofmolts:
            crab.rightclaw_size = max(new_crushersize  * pro_right, crab.rightclaw_size)                    

    else:  
        if crab.moltnumber_for_claw_removal_right < crab.numberofmolts: #so it doesn't overwrite budding claw growth
            crab.rightclaw_size = max(new_pincersize  * pro_right+ adj, crab.rightclaw_size)

        if crab.moltnumber_for_claw_removal_left < crab.numberofmolts:    
            crab.leftclaw_size = max(new_crushersize * pro_left, crab.leftclaw_size)

Answer 1

根据前面的人所说的，您可以为在三个条件之后出现的所有嵌套的if语句创建一个函数。然后你会把它们放在字典里，然后按这样的方式调用它们。下面是一个演示此方法的示例：

def fizz(x, y):
    return x*y

def foo(k, v):
    return k - v

def buzz(a, b):
    return a + b

然后你就这样编了一本字典：

opts = {'left':fizz, 'right':foo, 'both':buzz}

那么您的代码将如下所示：

if left_or_right_growing == 'left':
    opts['left'](1,2)
    ## output is 2

elif left_or_right_growing == 'right':
    opts['right'](3,4)
    ## output is -1

elif left_or_right_growing == 'both':
    opts['both'](5,6)
    ## output is 11

为了回答你向另一个人提出的问题，是的，这个方法最终应该比一堆嵌套的if语句更快。对代码进行基准测试，看看会发生什么。

编辑:使用我的方法的一些示例基准测试：

real    0m0.020s
user    0m0.010s
sys     0m0.007s

然后，当我使用您的方法并将函数替换为它自己的if循环时：

real    0m0.037s
user    0m0.012s
sys     0m0.009s

下面是用于测试函数方法的脚本：

import sys

def fizz(crab, x, y):
    if crab == 2:
        print 'Hello'
    else:
        return x*y

def foo(crab, k, v):
    if crab == 2:
        print 'Hello'
    else:
        return k - v

def buzz(crab, a, b):
    if crab == 2:
        print 'Hello'
    else:
        return a + b


opts = {'left':fizz, 'right':foo, 'both':buzz}

def main():
    left_or_right_growing = sys.argv[1]
    crab = int(sys.argv[2])
    if left_or_right_growing == 'left':
        opts['left'](crab,1,2)

    elif left_or_right_growing == 'right':
        opts['right'](crab,3,4)

    elif left_or_right_growing == 'both':
        opts['both'](crab,5,6)

if __name__ == '__main__':
    main()

为了测试if循环，我只需用一个简单的if循环替换函数。正如您所看到的，即使对于这样一个简单的任务，函数方法也更快。

Answer 2

您可以将这段代码放入一个函数中，以使其看起来更干净：

if crab.rightclawCrusher == True:
            crab.rightclaw_size = new_crushersize
            if crab.moltnumber_for_claw_removal_left < crab.numberofmolts: #so it doesn't overwrite budding claw growth
                crab.leftclaw_size = max(new_pincersize * crab.proportion_of_new_claw_thats_grownback()+adj, crab.leftclaw_size)
        elif crab.rightclawCrusher == False:
            crab.rightclaw_size = new_pincersize + adj
            if crab.moltnumber_for_claw_removal_left < crab.numberofmolts: #so it doesn't overwrite budding claw growth
                crab.leftclaw_size = max(new_crushersize * crab.proportion_of_new_claw_thats_grownback(), crab.leftclaw_size)