這可能幫助:
import gc
def checker(checkee, maxdepth = 3):
def onlyDict(ls):
return filter(lambda x: isinstance(x, dict), ls)
collection = []
toBeInspected = {}
tBI = toBeInspected
gc.collect()
for dic in onlyDict(gc.get_referrers(checkee)):
for item, value in dic.iteritems():
if value is checkee:
collection.append(item)
elif item != "checker":
tBI[item] = value
def _auxChecker(checkee, path, collection, checked, current, depth):
if current in checked: return
checked.append(current)
gc.collect()
for dic in onlyDict(gc.get_referents(current)):
for item, value in dic.iteritems():
currentPath = path + "." + item
if value is checkee:
collection.append(currentPath)
else:
try:
_auxChecker(checkee, currentPath, collection,
checked, value, depth + 1)
if depth < maxdepth else None
except TypeError:
continue
checked = []
for item, value in tBI.iteritems():
_auxChecker(checkee, item, collection, checked, value, 1)
return collection
使用方法:
referrer = []
class Foo:
pass
noo = Foo()
bar = noo
import xml
import libxml2
import sys
import os
op = os.path
xml.foo = bar
foobar = noo
for x in checker(foobar, 5):
try:
y= eval(x)
referrer.append(x)
except:
continue
del x, y
PS:在checkee的屬性將不會進一步檢查,對於給checkee本身的遞歸或嵌套引用。
看來,遍歷'全局()項目()'會更容易些,但還是不漂亮......。 – mgilson
這似乎是一個可怕的情況。 – delnan
爲什麼你會需要這樣做?對於您正在查找的函數,它們是否始終是在同一個模塊中定義的類的類方法? –