[Python] 动态函数调用(通过函数名)
2018-04-09 update
利用python中的内置函数 eval() ,函数说明:
def eval(*args, **kwargs): # real signature unknown """ Evaluate the given source in the context of globals and locals. The source may be a string representing a Python expression or a code object as returned by compile(). The globals must be a dictionary and locals can be any mapping, defaulting to the current globals and locals. If only globals is given, locals defaults to it. """ pass
样例1:
def function2(name, age): print("name: %s, age: %s" % (name, age)) if __name__ == '__main__': eval("function2")("Alice", 11) # 或者: args = ["Alice", 11] kwargs = {} eval("function2")(*args, **kwargs) """ 输出结果都是: name: Alice, age: 11 """
样例2:
class Test(object): states = [u"大于等于零", u"大于等于二"] state2function = {u"大于等于零": "check_gt0", u"大于等于二": "check_gt2"} @staticmethod def check_gt0(x): return x >= 0 @staticmethod def check_gt2(x): return x >= 2 def predict(self, x): for state in Test.states: check_ans = eval("Test." + Test.state2function[state])(x) # 调用Test类中的方法 print(state, Test.state2function[state], x, check_ans) if __name__ == '__main__': test = Test() test.predict(x=-1) test.predict(x=1) test.predict(x=2) """ 输出: 大于等于零 check_gt0 -1 False 大于等于二 check_gt2 -1 False 大于等于零 check_gt0 1 True 大于等于二 check_gt2 1 False 大于等于零 check_gt0 2 True 大于等于二 check_gt2 2 True """
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2017-08-09
由字符串函数名得到对应的函数
把函数作为参数的用法比较直观:
def func(a, b): return a + b def test(f, a, b): print f(a, b) test(func, 3, 5)
但有些情况下,‘要传递哪个函数’这个问题事先还不确定,例如函数名与某变量有关。可以利用 func = globals().get(func_name) 来得到函数:
def func_year(s): print 'func_year:', s def func_month(s): print 'func_month:', s strs = ['year', 'month'] for s in strs: globals().get('func_%s' % s)(s) """ 输出: func_year: year func_month: month """
参考:https://blog.csdn.net/lifestxx/article/details/78757345