Question

假设我在单个.py文件中包含以下内容：

class Graph( object ):
    def ReadGraph( file_name ):

def ProcessGraph(file_name, verbose):
    g=ReadGraph(file_name)

其中ProcessGraph是驱动程序类。当我输入

ProcessGraph('testcase.txt', verbose=True)

我收到此错误

NameError: global name 'ReadGraph' is not defined

有人可以解释如何修复此错误吗？

Answer 1

试试这个：

class Graph( object ):
    def ReadGraph( file_name ):
        # do something
        pass

def ProcessGraph(file_name, verbose):
    g = Graph()
    return g.ReadGraph(file_name)

Answer 2

ReadGraph位于Graph类的命名空间中，这就是为什么你不能把它称为高级函数。试试这个：

class Graph(object):
     @classmethod
     def ReadGraph(cls, file_name):
         # Something

def ProcessGraph(file_name, verbose):
     g=Graph.ReadGraph(file_name)

@classmethod装饰器可让您在不创建类实例的情况下调用类ReadGraph。

Answer 3

用@staticmethod

装饰它们

class Graph( object ):
    @staticmethod
    def ReadGraph( file_name ):
         print 'read graph'

    @staticmethod
    def ProcessGraph(file_name, verbose):
         g=ReadGraph(file_name)

if __name__=="__main__":
    Graph.ProcessGraph('f', 't')

输出＆＃39;你好＆＃39;。

staticmethod vs classmethod

Answer 4

创建Graph类的实例。

class Graph(object):
    def ReadGraph(file_name):
        pass

def ProcessGraph(file_name, verbose):
    g = Graph()
    out = g.ReaGraph(file_name)
    print out

在Python中调用类中的函数

4 个答案: