我有一个递归调用的函数，我想知道递归的当前级别。下面的代码显示了我用来计算它的方法，但它没有给出预期的结果。
例如：要查找系统路径的递归级别：
import os
funccount = 0
def reccount(src):
global funccount
print "Function level of %s is %d" %(src, funccount)
def runrec(src):
global funccount
funccount = funccount + 1
lists = os.listdir(src)
if((len(lists) == 0)):
funccount = funccount - 1
reccount(src)
for x in lists:
srcname = os.path.join(src, x)
if((len(lists) - 1) == lists.index(x)):
if (not(os.path.isdir(srcname))):
funccount = funccount - 1
if os.path.isdir(srcname):
runrec(srcname)
runrec(C:test)
问题：给定目录路径，打印目录的递归级别
目录结构为：在我的目录结构中，我将调用函数"reccount（Test）"（函数将使用MainFolder的路径进行调用）。我想知道每个文件夹的递归调用级别。（仅限目录）
Test:
|----------doc
|----------share
|----------doc
|----------file1
|----------bin
|----------common
|----------doc
|----------extras
|----------file2
当我调用该过程时，我得到以下结果：
Function level of C:test is 1
Function level of C:testbin is 2
Function level of C:testbincommon is 3
Function level of C:testbincommondoc is 3
Function level of C:testdoc is 3
Function level of C:testextras is 3
Function level of C:testshare is 4
Function level of C:testsharedoc is 5
正如您所看到的，当它打印bin/common/doc的结果时，它打印的是3而不是4，并且所有后续结果都是错误的
def some_method(data, level=0):
some_method(..., level=level+1)
if __name__ == '__main__':
some_method(my_data)
from inspect import getouterframes, currentframe
import os
def runrec(src):
level = len(getouterframes(currentframe(1)))
print("Function level of {} is {}".format(src, level))
for x in os.listdir(src):
srcname = os.path.join(src, x)
if os.path.isdir(srcname):
runrec(srcname)
runrec('C:\test')
Function level of C:test is 1
Function level of C:testbin is 2
Function level of C:testbincommon is 3
Function level of C:testbincommondoc is 4
Function level of C:testdoc is 2
Function level of C:testextras is 2
Function level of C:testshare is 2
Function level of C:testsharedoc is 3
为什么不将递归级别存储在参数中？
def runrec(src, level=1):
# ...
runrec(new_src, level + 1)
这样，您就不需要全局变量：
def reccount(src, level):
print "Function count of {} is {}".format(src, level)