Question

我收到了我需要检查内部档案的途径清单。当然,如果给我一个根基和一个子基点,就没有必要处理该子基。例如

c:	est  // process this
c:	estpics // do not process this
c:	est2 // process this

我怎么能告诉(跨平台),一条道路不是另一个路子。我最好希望这成为跨平台,只要它们不是周期性的联系,就不会担心这种联系(更糟糕的是,我最终将两次处理数据)。

Answer 1

我将保留你已经处理过的一套目录,然后对每条新途径进行检查,看看其母子是否已经在处理之前的表格中存在:

import os.path

visited = set()
for path in path_list:
    head, tail = os.path.split(path)
    while head and tail:
        if head in visited:
            break
        head, tail = os.path.split(head)
    else:
        process(path)
        visited.add(path)

注:path_list 应当进行分类,以便各子公司如果存在的话,总是在母系之后。

Answer 2

def is_subdir(path, directory):
    path = os.path.realpath(path)
    directory = os.path.realpath(directory)

    relative = os.path.relpath(path, directory)

    if relative.startswith(os.pardir):
        return False
    else:
        return True

Answer 3

跟踪已经处理过的目录(按正常格式处理),如果你已经看到这些目录,则不再处理。与此类似:

from os.path import realpath, normcase, sep

dirs = [r"C:	est", r"C:	estpics", r"C:	est2"]

processed = []

for dir in dirs:
    dir = normcase(realpath(dir)) + sep
    if not any(dir.startswith(p) for p in processed):
        processed.append(dir)
        process(dir)            # your code here

Answer 4

固定和简化jgoeders 原文:

def is_subdir(suspect_child, suspect_parent):
    suspect_child = os.path.realpath(suspect_child)
    suspect_parent = os.path.realpath(suspect_parent)

    relative = os.path.relpath(suspect_child, start=suspect_parent)

    return not relative.startswith(os.pardir)

Answer 5

这里是“is_subdir”的通用功能一号。

Python3.x compatible (works with bytes and str, matching os.path which also supports both).
Normalizes paths for comparison.
(parent hierarchy and case to work on ms-windows).
Avoids using os.path.relpath which will raise an exception on ms-windows if the paths are on different drives. (C:foo -> D:ar)

法典:

def is_subdir(path, directory):
    """
    Returns true if *path* in a subdirectory of *directory*.
    """
    import os
    from os.path import normpath, normcase, sep
    path = normpath(normcase(path))
    directory = normpath(normcase(directory))
    if len(path) > len(directory):
        sep = sep.encode( ascii ) if isinstance(directory, bytes) else sep
        if path.startswith(directory.rstrip(sep) + sep):
            return True
    return False

Answer 6

这里使用的解决方案一以Andrew Clarks的回答为基础,确保清单的分类,使儿童处于父母的状态,并使用<代码>normpath和>normcase,以固定连接c:users和c:/users的视窗。

  def unique_path_roots(paths):
    visited = set()
    paths = list(set(paths))

    for path in sorted(paths,key=cmp_to_key(locale.strcoll)):
        path = normcase(normpath(realpath(path)))

        head, tail = os.path.split(path)
        while head and tail:
            if head in visited:
                break
            head, tail = os.path.split(head)
        else:
            yield path
            visited.add(path)

友情链接