Question

我想编写一个函数，该函数返回具有文件完整路径的字符串（如果不在目录树中，则返回完整字符串）。

例如。

pc = ["home",
["Documents",
[ "Tools", "alex.txt", "sport.pdf",             
"row" ],
[ "Python", "flatten.py", "set.md" ],
],
["Downloads",
[ "Music",
[ "Movies", "Creed.mp4", "Grinch.avi" ],
"Raplh.avi", "22", "Reg.mp4"
],
],
"trec.txt", "doc.html"
]

finder（pc，'sport.pdf'）应该返回字符串： “主页/文档/工具/sport.pdf”

我尝试过：

path =""

def finder(pc, file_name):

global path

for i in range(len(pc)-1):
    if isinstance(pc[i], list):
        finder(pc[i], file_name)
    else:
        if pc[i]==file_name:       
            path="/"+file_name
return(path)        

print(finder(pc, 'sport.pdf'))

返回：

/sport.pdf

但是我如何获得完整路径：主页/Documents/Tools/sport.pdf

预先感谢

Answer 1

您可以将递归与生成器一起使用：

pc = ['home', ['Documents', ['Tools', 'alex.txt', 'sport.pdf', 'row'], ['Python', 'flatten.py', 'set.md']], ['Downloads', ['Music', ['Movies', 'Creed.mp4', 'Grinch.avi'], 'Raplh.avi', '22', 'Reg.mp4']], 'trec.txt', 'doc.html']
def finder(_tree, _filename, _current=''):
  if  _filename in _tree:
    yield f'{_current}/{_filename}'
  else:
    _dir, *_files = _tree
    for _row in _files:
      yield from finder(_row, _filename, f'{_current}/{_dir}' if _current else _dir)

print(list(finder(pc, 'sport.pdf'))[0])

输出：

'home/Documents/sport.pdf'

Python：使用递归返回文件的完整路径

1 个答案: