如何避免每个命令的如果？[关闭]

Question

我正在为自己建立一个虚拟助手，如果像这里举个例子，要求个人执行每一个命令是错误和无效的：

WAKE = 'hello'


while True:
    print("Mic ready")
    text = voice.get_audio()
    if WAKE in text:
        if "note" in text:
            voice.speak("What do you want to note?")
            print("What do you want to note?")
            text = voice.get_audio()
            Vcm.note(text)
        else if "timer" in text:
            voice.speak("How long is that timer supposed to run?")
            print("How long is that timer supposed to run?")
            text = voice.get_audio()
            Vcm.timer(text)
        else:
            voice.speak("At your service")
            print("At your service")
            text = voice.get_audio()

有没有更有效的方法来检查要执行的命令？

Answer 1

while True:
    print("Mic ready")
    text = voice.get_audio()
    if WAKE in text:
        words = text.split(" ")
        for word in words:
            command = commands.get(word)
            if command:
                command(text)
                break
        else:
            vcm.default()
    ```

commands = {
'note': lambda text: note(text)

}

Answer 2

您可以将命令分成函数：

def note():
    #note body

def timer():
    #timer body

def default():
    #default body

然后将命令包装到字典中：

commands = {
    'note': lambda: note(),
    'timer': lambda: timer()
}

并分别定义默认值：

default_command = lambda: default()

然后您可以检查文本中的内容：

for word in text:
   command = commands.get(word, None)
   if command not None:
       command()
       break
else:
    default_command()

不确定这是否是您想要的，但如果您有许多命令，它可能会使您的代码更加稀疏。

编辑：

根据lambda函数的bool值是否为false？，lambda表达式表现为true。所以你也可以简化最后一点！

for word in text:
    command = commands.get(word, None)
    if command:
        command()
        break
else:
    default_command()