如何从简单函数中提取名称？

Question

我有这段代码:

import inspect
import ast

def func(foo):
    return foo.bar - foo.baz

s = inspect.getsource(func)
xx = ast.parse(s)

class VisitCalls(ast.NodeVisitor):
    def visit_Name(self, what):
        if what.id == 'foo':
            print ast.dump(what.ctx)

VisitCalls().visit(xx)

从函数'func'我想提取:

['foo.bar', 'foo.baz']

或类似的东西:

(('foo', 'bar'), ('foo', 'baz))

编辑

一些背景解释为什么我认为我需要这样做

我想将一个普通的python函数的代码转换为电子表格公式.

所以我需要转换:

foo.bar - foo.baz

至:

=A1-B1

示例电子表格http://img441.imageshack.us/img441/1451/84516405.png

**再次编辑*

到目前为止我得到了什么.

以下程序输出:

('A1', 5)
('B1', 3)
('C1', '= A1 - B1')

代码:

import ast, inspect
import codegen # by Armin Ronacher
from collections import OrderedDict

class SpreadSheetFormulaTransformer(ast.NodeTransformer):
    def __init__(self, sym):
        self.sym = sym
    def visit_Attribute(self, node):
        name = self.sym[id(eval(codegen.to_source(node)))]
        return ast.Name(id=name, ctx=ast.Load())

def create(**kwargs):
    class Foo(object): pass
    x = Foo()
    x.__dict__.update(kwargs)
    return x

def register(x,y):
    cell[y] = x
    sym[id(x)] = y

def func(foo):
    return foo.bar - foo.baz

foo = create(bar=5, baz=3)
cell = OrderedDict()
sym = {}

register(foo.bar, 'A1')
register(foo.baz, 'B1')

source = inspect.getsource(func)
tree = ast.parse(source)
guts = tree.body[0].body[0].value
SpreadSheetFormulaTransformer(sym).visit(guts)

code = '= ' + codegen.to_source(guts)
cell['C1'] = code

for x in cell.iteritems():
    print x

我在这里找到了一些资源:Python内部:使用Python AST 我在这里抓住了一个有效的codegen模块.

Answer 1

import ast, inspect
import codegen # by Armin Ronacher

def func(foo):
    return foo.bar - foo.baz

names = []

class CollectAttributes(ast.NodeVisitor):
    def visit_Attribute(self, node):
        names.append(codegen.to_source(node))

source = inspect.getsource(func)

tree = ast.parse(source)
guts = tree.body[0].body[0].value
CollectAttributes().visit(guts)
print names

输出：

['foo.bar', 'foo.baz']