Python 如何在模块中以文本方式查找导入的名称
我编写了一个名为Python 如何在模块中以文本方式查找导入的名称,python,regex,python-2.7,Python,Regex,Python 2.7,我编写了一个名为buildRegex的方法,该方法给定一个名称(类型为str),返回一个regex对象,该对象从中查找。。。导入。。。Python模块中的name语句 例如,下面是buildRegex的预期行为: >>> regObj = buildRegex('foo') >>> regObj.search('from a import fool') is None True >>> regObj.search('from a import
buildRegex
的方法,该方法给定一个名称(类型为str
),返回一个regex
对象,该对象从中查找。。。导入。。。Python模块中的name
语句
例如,下面是buildRegex
的预期行为:
>>> regObj = buildRegex('foo')
>>> regObj.search('from a import fool') is None
True
>>> regObj.search('from a import foo') is not None
True
>>> regObj.search('from a.b.c import foo as food') is None
True
>>> regObj.search('from a.b.c import fool, bar as foo') is not None
True
到目前为止,我所做的工作适用于上述所有示例(以及更多示例):
buildRegex
假设搜索的模块没有SyntaxError
s,这是正常的
我的问题是,在查找导入的名称foo
时,我还需要知道它是否是其他名称的别名。即,如果模块具有以下语句:
from a.b.c import bar as foo
我想知道什么是别名,在本例中,是bar
。目前,由于正则表达式中的断言lookaheads
,这是不可能的。最后,我的问题是:
如何重构正则表达式以避免丢失这些信息,例如,如果给定的名称是别名,那么它的别名就在
regex
之一中?我建议不要编写复杂的正则表达式来解析导入,实际上可以使用ast.parse
将源代码解析为抽象语法树,并从中查找名称。asast.parse
保证正确解析Python。比如:
import ast
class ImportFinder(ast.NodeVisitor):
def __init__(self):
self.imports = []
def visit_Import(self, node):
names = []
for i in node.names:
names.append((i.name, i.asname))
self.imports.append(['import', names])
def visit_ImportFrom(self, node):
module = node.module
level = node.level # how many dots
names = []
for i in node.names:
names.append((i.name, i.asname))
self.imports.append(('from', level, module, names))
def parse_imports(source):
tree = ast.parse(source)
finder = ImportFinder()
finder.visit(tree)
return finder.imports
用法示例:
import pprint
pprint.pprint(parse_imports('''
from foo import bar, baz, frob
from .. import bar as spam, baz as ham, frob
import bar.baz
import bar.foo as baf
'''))
打印出:
[('from', 0, 'foo', [('bar', None), ('baz', None), ('frob', None)]),
('from', 2, None, [('bar', 'spam'), ('baz', 'ham'), ('frob', None)]),
['import', [('bar.baz', None)]],
['import', [('bar.foo', 'baf')]]]
from
行中的整数给出了模块名称前的
数。我建议不要编写复杂的正则表达式来解析导入,而是实际使用ast.parse
将源代码解析为抽象语法树,并从中查找名称。asast.parse
保证正确解析Python。比如:
import ast
class ImportFinder(ast.NodeVisitor):
def __init__(self):
self.imports = []
def visit_Import(self, node):
names = []
for i in node.names:
names.append((i.name, i.asname))
self.imports.append(['import', names])
def visit_ImportFrom(self, node):
module = node.module
level = node.level # how many dots
names = []
for i in node.names:
names.append((i.name, i.asname))
self.imports.append(('from', level, module, names))
def parse_imports(source):
tree = ast.parse(source)
finder = ImportFinder()
finder.visit(tree)
return finder.imports
import inspect
import importlib
import ast
class Imports(ast.NodeVisitor):
def visit_Import(self, node):
print("In Import")
for imp in node.names:
if imp.asname is not None:
print("module name = {}, alias = {}".format(imp.name, imp.asname))
else:
print("module name = {}".format(imp.name))
print()
def visit_ImportFrom(self, node):
print("In ImportFrom")
for imp in node.names:
if imp.asname is not None:
print("module = {}\nname = {}\nalias = {}\nlevel = {}\n".
format(node.module, imp.name, imp.asname, node.level))
else:
print("module = {}\nname = {}\nlevel = {}\n".
format(node.module, imp.name, node.level))
print()
mod = "temp_test"
mod = importlib.import_module(mod)
p = ast.parse(inspect.getsource(mod))
Imports().visit(p)
用法示例:
import pprint
pprint.pprint(parse_imports('''
from foo import bar, baz, frob
from .. import bar as spam, baz as ham, frob
import bar.baz
import bar.foo as baf
'''))
打印出:
[('from', 0, 'foo', [('bar', None), ('baz', None), ('frob', None)]),
('from', 2, None, [('bar', 'spam'), ('baz', 'ham'), ('frob', None)]),
['import', [('bar.baz', None)]],
['import', [('bar.foo', 'baf')]]]
from
行中的整数表示模块名称前的
数
import inspect
import importlib
import ast
class Imports(ast.NodeVisitor):
def visit_Import(self, node):
print("In Import")
for imp in node.names:
if imp.asname is not None:
print("module name = {}, alias = {}".format(imp.name, imp.asname))
else:
print("module name = {}".format(imp.name))
print()
def visit_ImportFrom(self, node):
print("In ImportFrom")
for imp in node.names:
if imp.asname is not None:
print("module = {}\nname = {}\nalias = {}\nlevel = {}\n".
format(node.module, imp.name, imp.asname, node.level))
else:
print("module = {}\nname = {}\nlevel = {}\n".
format(node.module, imp.name, node.level))
print()
mod = "temp_test"
mod = importlib.import_module(mod)
p = ast.parse(inspect.getsource(mod))
Imports().visit(p)
输入:
from bisect import bisect_left as bs
import datetime
import time
import numpy as np
def foo():
from re import findall
class Foo():
def test(self):
from re import compile as cp, finditer as ft
输出:
In ImportFrom
module = bisect
name = bisect_left
alias = bs
level = 0
In Import
module name = datetime
In Import
module name = time
In Import
module name = numpy, alias = np
In ImportFrom
module = re
name = findall
level = 0
In ImportFrom
module = re
name = compile
alias = cp
level = 0
module = re
name = finditer
alias = ft
level = 0
类导入(名称)
导入语句。名称是别名节点的列表
类导入自(模块、名称、级别)
表示从x导入y。模块是一个由“from”名称组成的原始字符串,不带任何前导点,对于诸如from之类的语句也不带前导点。输入foo。level是一个整数,表示相对导入的级别(0表示绝对导入)
对于我来说,这个文档至少比实际的ast文档本身对所有节点做什么以及如何使用ast模块有更好的解释
您还可以使用直接传递模块或打开py文件并将内容传递给ast.parse:
with open("temp_test.py") as f:
p = ast.parse(f.read(), filename="<ast>", mode="exec")
Imports().visit(p)
输入:
from bisect import bisect_left as bs
import datetime
import time
import numpy as np
def foo():
from re import findall
class Foo():
def test(self):
from re import compile as cp, finditer as ft
输出:
In ImportFrom
module = bisect
name = bisect_left
alias = bs
level = 0
In Import
module name = datetime
In Import
module name = time
In Import
module name = numpy, alias = np
In ImportFrom
module = re
name = findall
level = 0
In ImportFrom
module = re
name = compile
alias = cp
level = 0
module = re
name = finditer
alias = ft
level = 0
类导入(名称)
导入语句。名称是别名节点的列表
类导入自(模块、名称、级别)
表示从x导入y。模块是一个由“from”名称组成的原始字符串,不带任何前导点,对于诸如from之类的语句也不带前导点。输入foo。level是一个整数,表示相对导入的级别(0表示绝对导入)
对于我来说,这个文档至少比实际的ast文档本身对所有节点做什么以及如何使用ast模块有更好的解释
您还可以使用直接传递模块或打开py文件并将内容传递给ast.parse:
with open("temp_test.py") as f:
p = ast.parse(f.read(), filename="<ast>", mode="exec")
Imports().visit(p)
表单当然可以使用regex进行解析,但考虑到可能有多行、多个导入逗号分隔等,您确定不想使用
ast.parse
?表单当然可以使用regex进行解析,但考虑到可能有多行、多个导入逗号分隔等,你确定你不想使用ast.parse
?这个答案和@Antii's都是好的和正确的,但是由于链接非常有用,这个答案和@Antii's都是好的和正确的,但是由于链接非常有用,这个答案和@Antii's都是好的和正确的。