[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
"""
*************************************************************************************************************************************************************
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
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