在python中查找递归调用的级别

Question

我有一个递归调用的函数,我想知道当前的递归级别.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

目录结构是:在我的目录结构中,我将调用函数"reccount(Test)"(函数将通过MainFolder的路径调用).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:\test\bin is 2
    Function level of C:\test\bin\common is 3
    Function level of C:\test\bin\common\doc is 3
    Function level of C:\test\doc is 3
    Function level of C:\test\extras is 3
    Function level of C:\test\share is 4
    Function level of C:\test\share\doc is 5

正如您所看到的,当它打印bin/common/doc的结果时,它会打印3而不是4,并且所有后续结果都是错误的

Answer 1

def some_method(data, level=0):


    some_method(..., level=level+1)


if __name__ == '__main__':
    some_method(my_data)

Answer 2

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:\test\bin is 2
Function level of C:\test\bin\common is 3
Function level of C:\test\bin\common\doc is 4
Function level of C:\test\doc is 2
Function level of C:\test\extras is 2
Function level of C:\test\share is 2
Function level of C:\test\share\doc is 3

Answer 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)