python - 列出函数的所有内部函数？

Question

在python中，您可以执行fname.__code__.co_names来检索函数引用的函数和全局事物的列表。如果我做fname.__code__.co_varnames，我相信这包括内部函数。

有没有办法做到inner.__code__.co_names？通过开始看起来像'inner'的字符串，由co_varnames返回？

Answer 1

我不你认为你可以检查代码对象，因为内部函数是懒惰的，它们的代码对象只有c按时完成。您可能想要看的是ast模块。这里有一个简单的例子：

import ast, inspect 

# this is the test scenario 
def function1(): 
    f1_var1 = 42 
    def function2(): 
     f2_var1 = 42 
     f2_var2 = 42 
     def function3(): 
      f3_var1 = 42 

# derive source code for top-level function 
src = inspect.getsource(function1) 

# derive abstract syntax tree rooted at top-level function 
node = ast.parse(src) 

# next, ast's walk method takes all the difficulty out of tree-traversal for us 
for x in ast.walk(node): 
    # functions have names whereas variables have ids, 
    # nested-classes may all use different terminology 
    # you'll have to look at the various node-types to 
    # get this part exactly right 
    name_or_id = getattr(x,'name', getattr(x,'id',None)) 
    if name_or_id: 
     print name_or_id 

的结果是：功能1，功能2，f1_var1，功能3，f2_var1，f2_var2，f3_var1。强制免责声明：可能没有一个很好的理由做这种事情..但有乐趣:)

哦，如果你只想要的内部功能的名称？

print dict([[x.name,x] for x in ast.walk(ast.parse(inspect.getsource(some_function))) if type(x).__name__=='FunctionDef']) 

Answer 2

在Python 3.4+中，您可以使用dis.get_instructions获取名称。为了支持嵌套函数，以及你需要递归遍历你遇到的每个代码对象：

import dis 
import types 

def get_names(f): 
    ins = dis.get_instructions(f) 
    for x in ins: 
     try: 
      if x.opcode == 100 and '<locals>' in next(ins).argval\ 
               and next(ins).opcode == 132: 
       yield next(ins).argrepr 
       yield from get_names(x.argval) 
     except Exception: 
      pass 

演示：

def func(): 
    x = 1 
    y = 2 
    print ('foo') 
    class A: 
     def method(self): 
      pass 
    def f1(): 
     z = 3 
     print ('bar') 
     def f2(): 
      a = 4 
      def f3(): 
       b = [1, 2, 3] 
    def f4(): 
     pass 

print(list(get_names(func))) 

输出：

['f1', 'f2', 'f3', 'f4']