Ato*_*nal 3 python abstract-syntax-tree
考虑下面的代码:
1 | x = 20
2 |
3 | def f():
4 | x = 0
5 | for x in range(10):
6 | x += 10
7 | return x
8 | f()
9 |
10| for x in range(10):
11| pass
12| x += 1
13| print(x)
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的值x的代码的执行之后以上10。现在,我怎么能得到所有一流的节点Name,其ids为x并指x这是一个在线路1,10,12和13被使用?
换句话说,的x内部f与xs 的其余部分不同。是否可以获取仅具有脚本和脚本的AST而不执行脚本的AST节点?
在AST树上行走时,请跟踪上下文;与全球范围内启动,那么当你遇到FunctionDef或ClassDef或Lambda节点,记录这方面作为一个堆栈(退出相关节点时再弹出堆栈)。
然后,您可以仅查看Name全局上下文中的节点。您也可以跟踪global标识符(我会在每个堆栈级别使用一组)。
import ast
class GlobalUseCollector(ast.NodeVisitor):
def __init__(self, name):
self.name = name
# track context name and set of names marked as `global`
self.context = [('global', ())]
def visit_FunctionDef(self, node):
self.context.append(('function', set()))
self.generic_visit(node)
self.context.pop()
# treat coroutines the same way
visit_AsyncFunctionDef = visit_FunctionDef
def visit_ClassDef(self, node):
self.context.append(('class', ()))
self.generic_visit(node)
self.context.pop()
def visit_Lambda(self, node):
# lambdas are just functions, albeit with no statements, so no assignments
self.context.append(('function', ()))
self.generic_visit(node)
self.context.pop()
def visit_Global(self, node):
assert self.context[-1][0] == 'function'
self.context[-1][1].update(node.names)
def visit_Name(self, node):
ctx, g = self.context[-1]
if node.id == self.name and (ctx == 'global' or node.id in g):
print('{} used at line {}'.format(node.id, node.lineno))
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演示(为中的示例代码提供了AST树t):
>>> GlobalUseCollector('x').visit(t)
x used at line 1
x used at line 10
x used at line 12
x used at line 13
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并global x在函数中使用:
>>> u = ast.parse('''\
... x = 20
...
... def g():
... global x
... x = 0
... for x in range(10):
... x += 10
... return x
...
... g()
... for x in range(10):
... pass
... x += 1
... print(x)
... ''')
>>> GlobalUseCollector('x').visit(u)
x used at line 1
x used at line 5
x used at line 6
x used at line 7
x used at line 8
x used at line 11
x used at line 13
x used at line 14
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