Python 组合多个os.walk运行的输出

我有多个目录dirs=[dir1，dir2，…]

这些目录的结构如下所示：

dir1
  subdir1
    folder1
    file1
    file2
  subdir2
dir2
  subdir1
    folder2
      file3
    file4
  subdir2

请注意，子曲面的名称是相同的。dir1和dir2都有相同的命名子目录。我需要的是打印一个html表，该表组合了来自dir1和dir2的文件和文件夹，如下所示：

subdir1
  folder1
  folder2
    file3
  file1
  file2
  file4
subdir2

还有一个警告是，我需要知道每个文件和文件夹的路径，以便链接到它

到目前为止，我使用os.walk为dir1创建了树，并从中创建了一个html表，其中的每一行都在一个列表中。然后我对所有其他目录执行os.walk，对于每个目录，遍历该列表，直到basename相同，然后插入文件和文件夹。但这是非常缓慢的。我相信有一个非常聪明的五线解决方案可以达到同样的效果

def get_table(self, teams=['test1', 'test2']):
    paths = []
    table = []
    for team in teams:
        paths.append(config.basepath + '/' + team)
    for path in paths:
        if not table:
            for root, dirs, files in os.walk(path):
                dirs = sorted(dirs)
                files = sorted(files)
                team = self.get_team(path) # extracts the 'dir' from path
                level = root.replace(path, '').count(os.sep)
                indent = ' ' * 4 * (level)
                subindent = ' ' * 4 * (level + 1)
                table.append('{0}<tr class="{2}"><td>{1}</td><td>{2}</td></tr>'.format(indent, os.path.basename(root), team))
                for f in files:
                    table.append('{0}<tr class="{2}"><td>{1}</td><td>{2}</td></tr>'.format(subindent, f, team))
        else:
            for root, dirs, files in os.walk(path):
                dirs = sorted(dirs)
                files = sorted(files)
                team = self.get_team(path)
                level = root.replace(path, '').count(os.sep)
                indent = ' ' * 4 * (level)
                subindent = ' ' * 4 * (level + 1)
                for idx, line in enumerate(table):
                    if os.path.basename(root) in line:
                        for f in files:
                            table.insert(idx+1, '{0}<tr class="{2}"><td>{1}</td><td>{2}</td></tr>'.format(subindent, f, team))

---

我已经用它运行了，但是，我相信有一个更好的解决方案：

for path in paths:
        for root, dirs, files in os.walk(path):
            dirs = sorted(dirs)
            files = sorted(files)
            team = self.get_team(path)
            level = root.replace(path, '').count(os.sep)
            indent = ' ' * 4 * (level)
            subindent = ' ' * 4 * (level + 1)
            basename = os.path.basename(root)
            if firstrun:
                table.append('{0}<tr class="{2}"><td>{1}</td><td>{2}</td></tr>'.format(indent, basename, team))
                coretasks[basename] = len(table) - 1
                for f in files:
                    table.append('{0}<tr class="{2}"><td>{1}</td><td>{2}</td></tr>'.format(subindent, f, team))
            else:
                parsed_folders = []
                if basename in coretasks:
                    inserted_files = 0
                    for f in files:
                        table.insert(coretasks[basename] + 1, '{0}<tr class="{2}"><td>{1}</td><td>{2}</td></tr>'.format(subindent, f, team))
                        inserted_files += 1
                    parsed_folders.append(basename)
                    for coretask in coretasks.keys():
                        if not coretask in parsed_folders: coretasks[coretask] += inserted_files
        firstrun = False
    print('\n'.join(table))

---

我会把提取部分和归还部分分开

提取：

def process(path, d={}):
    print('initial', d)
    for i in os.scandir(path):
        if i.is_file():
            if i.name in d: raise Exception(i.path +
                            "already present")
            d[i.name] = None
        elif i.is_dir():
            if not i.name in d: d[i.name] = {}
            process(i.path, d[i.name])
    print('final', d)
    return d

显示：

def process(path, d={}):
    print('initial', d)
    for i in os.scandir(path):
        if i.is_file():
            if i.name in d: raise Exception(i.path +
                            "already present")
            d[i.name] = None
        elif i.is_dir():
            if not i.name in d: d[i.name] = {}
            process(i.path, d[i.name])
    print('final', d)
    return d

根据您提出的结构，它提供了：

>>> process('dir1')
>>> d = process('dir2')
>>> print(d)
{'subdir1': {'folder2': {'file3': None}, 'file1': None, 'file2': None, 'folder1': {}, 'file4': None}, 'subdir2': {}}
>>> display(d)
subdir1
  file1
  file2
  file4
  folder1
  folder2
    file3
subdir2

---

这样，您只需更改HTML格式的显示部分…

我认为您意外地发布了两次处理函数，而没有显示函数。