Python:在满足特定条件时跳过代码块(计算)而不使用 "if"语句

Question

我有一小段代码，我正在尝试以更好的方式编写它，因为它有一堆“if”语句。这是一些大项目的小代码。问题是这样的:当代码运行时，函数“f”、“g”或/和“k”可以返回 None 或数值数据。每当返回 None 值时，就必须跳过其余的计算，因为无法完成数学运算(发生在这些函数中)。我尝试使用 TRY/CATCH 方法重写代码，但无法使其工作。我试图避免“if”语句并重写简洁的方式。我很感激你的帮助。

<小时/>

def f(output):
    #some code which computes output which be None or numerical
    return [output*1,2]
def g(Y):
    #some code which computes Y which be None or numerical
    return Y*3
def k(output):
  #some code which computes output which be None or numerical
  return output*4
def foutput():
  #some code which computes "value" which be None or numerical 
  value=2.0
  return 1.0*value


#####START
#some code
output=foutput()

if output is not None:
    print 'S1'
    [output,A]=f(output)
    if output is not None:
        print 'S2'
        [a,b,c,Y]=[1,2,3,k(output)]
        if Y is not None:
            print 'S3'
            A=g(Y)
        else:
            [Q,A,output]=[None,None,None]
    else:
        [Q,A,output]=[None,None,None]
else:
    [Q,A,output]=[None,None,None]

Answer 1

确定每个步骤中将引发的错误，然后将这些异常添加到 try..except 中。在这个玩具示例中，它们都是 TypeError，但我将添加 ValueError 作为演示:

def f(output):
    #some code which computes output which be None or numerical
    return [output*1,2]
def g(Y):
    #some code which computes Y which be None or numerical
    return Y*3
def k(output):
  #some code which computes output which be None or numerical
  return output*4
def foutput():
  #some code which computes "value" which be None or numerical 
  value=2.0
  return 1.0*value


output=foutput()

try:
    print 'S1'
    output, A = f(output)
    print 'S2'
    a, b, c, Y = 1, 2, 3, k(output)
    print 'S3'
    A = g(Y)
except (ValueError, TypeError):
    Q = A = output = None
else:
    Q = 'success' # if none of this fails, you might want a default value for Q