使用字典选择要执行的函数

Question

我正在尝试使用函数式编程来创建包含键和要执行的函数的字典:

myDict={}
myItems=("P1","P2","P3",...."Pn")
def myMain(key):
    def ExecP1():
        pass
    def ExecP2():
        pass
    def ExecP3():
        pass
        ...
    def ExecPn():
        pass  

现在,我已经看到用于在模块中查找已定义函数的代码,我需要执行以下操作:

    for myitem in myItems:
        myDict[myitem] = ??? #to dynamically find the corresponding function

所以我的问题是,如何列出所有Exec函数,然后使用字典将它们分配给所需的项目？所以最后我会myDict["P1"]() #this will call ExecP1()

我真正的问题是我有大量的这些项目,我创建了一个库来处理它们,所以最终用户只需要调用 myMain("P1")

我认为使用检查模块,但我不确定如何做到这一点.

我有理由避免:

def ExecPn():
    pass
myDict["Pn"]=ExecPn

是我必须保护代码,因为我使用它来提供我的应用程序中的脚本功能.

Answer 1

简化,简化和简化:

def p1(args):
    whatever

def p2(more args):
    whatever

myDict = {
    "P1": p1,
    "P2": p2,
    ...
    "Pn": pn
}

def myMain(name):
    myDict[name]()

这就是你所需要的.

dict.get如果name引用无效函数,您可以考虑使用可调用默认值 -

def myMain(name):
    myDict.get(name, lambda: 'Invalid')()

(从Martijn Pieters那里得到了这个巧妙的技巧)

Answer 2

不为此感到自豪,但是:

def myMain(key):
    def ExecP1():
        pass
    def ExecP2():
        pass
    def ExecP3():
        pass
    def ExecPn():
        pass 
    locals()['Exec' + key]()

但是我建议你把它们放在模块/类中,这真的太可怕了.

Answer 3

简化,简化,简化 + DRY:

tasks = {}
task = lambda f: tasks.setdefault(f.__name__, f)

@task
def p1():
    whatever

@task
def p2():
    whatever

def my_main(key):
    tasks[key]()

Answer 4

# index dictionary by list of key names

def fn1():
    print "One"

def fn2():
    print "Two"

def fn3():
    print "Three"

fndict = {"A": fn1, "B": fn2, "C": fn3}

keynames = ["A", "B", "C"]

fndict[keynames[1]]()

# keynames[1] = "B", so output of this code is

# Two

Answer 5

你可以只使用

myDict = {
    "P1": (lambda x: function1()),
    "P2": (lambda x: function2()),
    ...,
    "Pn": (lambda x: functionn())}
myItems = ["P1", "P2", ..., "Pn"]

for item in myItems:
    myDict[item]()