多个条件有效

Question

如何在python中以更有效的方式执行以下代码？输入标志是二进制值。输出取决于标志的所有可能排列。

def f1():

    return 1

def f2():

    return 2

def f3():

    return 3    


def g(p1, p2, p3):

    if p1 ==  1 & p2 == 0 & p3 == 0:
        f1()

    elif: p1 == 0 & p2 == 1 & p3 == 0:
        f2()

    elif: p1 == 0 & p2 == 0 & p3 == 1:
        f3()

    elif: p1 == 1 & p2 == 1 & p3 == 1:
        f1()
        f2()

等等。

Answer 1

您可以将这三位组合成一个数字，然后像这样测试该数字的值：

def g(p1, p2, p3):

   v = (p1 << 2) + (p2 << 1) + p3

    if v == 4: # 100
        f1()
    elif v == 2: # 010
        f2()
    elif v == 1: # 001
        f3()
    elif v == 7: # 111
        f1()
        f2()

Answer 2

如果你想使用参数 (p1, p2, p3) 作为标志，你总是可以使用 *args 将这些参数打包为一个列表（参见 this, this and this）并将你的函数放在一个列表中（是的，Python 让你这样做）并得到类似的东西：

def f1():
    return 1

def f2():
    return 2

def f3():
    return 3 

def g(*ps):
    functions = [f1, f2, f3]
    for i, p in enumerate(ps):
        if p == 1:  # Could do just `if p:` (0 evaluates to False, anything else to True)
            print(functions[i])()  # Notice the () to actually call the function 

if __name__ == "__main__":
    print("Run 1 0 0")
    g(1, 0, 0)
    print("Run 1 1 0")
    g(1, 1, 0)
    print("Run 0 1 0")
    g(0, 1, 0)
    print("Run 1 1 1")
    g(1, 1, 1)

根据ShadowRanger's to this answer, you could even shorten the code a bit more. For instance, using zip：

def g(*ps):
    functions = [f1, f2, f3]
    for function, p in zip(functions, ps):
        if p:
            print(function())

或使用 itertools.compress（您需要在文件顶部 import itertools）：

def g(*ps):
    functions = [f1, f2, f3]
    for function in itertools.compress(functions, ps):
        print(function())

多个条件有效

multiple conditionals effciently

python

loops

combinatorics