2

我正在使用以下函数从目标目录向下获取系统中的所有文件大小。

def get_files(target):
    # Get file size and modified time for all files from the target directory and down.
    # Initialize files list
    filelist = []
    # Walk the directory structure
    for root, dirs, files in os.walk(target):
        # Do not walk into directories that are mount points
        dirs[:] = filter(lambda dir: not os.path.ismount(os.path.join(root, dir)), dirs)
        for name in files:
            # Construct absolute path for files
            filename = os.path.join(root, name)
            # Test the path to account for broken symlinks
            if os.path.exists(filename):
                # File size information in bytes
                size = float(os.path.getsize(filename))
                # Get the modified time of the file
                mtime = os.path.getmtime(filename)
                # Create a tuple of filename, size, and modified time
                construct = filename, size, str(datetime.datetime.fromtimestamp(mtime))
                # Add the tuple to the master filelist
                filelist.append(construct)
    return(filelist)

如何修改它以包含包含目录和目录总大小的第二个列表？我试图将此操作包含在一个函数中，希望比在单独的函数中执行第二次遍历以获取目录信息和大小更有效。

这个想法是能够报告前 20 个最大文件的排序列表，以及前 10 个最大目录的第二个排序列表。

感谢你们的任何建议。

4

2 回答 2

1

我在字典而不是列表中输出目录，但看看你是否喜欢它：

def get_files(target):
    # Get file size and modified time for all files from the target directory and down.
    # Initialize files list
    filelist = []
    dirdict = {}
    # Walk the directory structure
    for root, dirs, files in os.walk(target):
        # Do not walk into directories that are mount points
        dirs[:] = filter(lambda dir: not os.path.ismount(os.path.join(root, dir)), dirs)
        for name in files:
            # Construct absolute path for files
            filename = os.path.join(root, name)
            # Test the path to account for broken symlinks
            if os.path.exists(filename):
                # File size information in bytes
                size = float(os.path.getsize(filename))
                # Get the modified time of the file
                mtime = os.path.getmtime(filename)
                # Create a tuple of filename, size, and modified time
                construct = filename, size, str(datetime.datetime.fromtimestamp(mtime))
                # Add the tuple to the master filelist
                filelist.append(construct)
                if root in dirdict.keys():
                    dirdict[root] += size
                else:
                    dirdict[root] = size
    return(filelist, dirdict)

如果您希望 dirdict 作为元组列表，只需执行以下操作：

dirdict.items()
于 2012-05-12T01:16:44.673 回答
0

我有很多脚本可以做这种事情，我刚刚将“bigfiles.py”上传到 github http://github.com/sente/sys-utils/blob/master/bigfiles.py

它不计算总累积目录大小，但可以轻松修改它。

我还有其他代码可以在给定深度求和总目录大小，例如：

In [7]: t = build_tree_from_directory('/scratch/stu/')

In [8]: pprint.pprint(walk_tree(t,depth=0))
{'name': 'ROOT', 'size': 6539880514}

In [9]: pprint.pprint(walk_tree(t,depth=0))
{'name': 'ROOT', 'size': 6539880514}

In [10]: pprint.pprint(walk_tree(t,depth=1))
{'children': [{'name': 'apache2-gzip', 'size': 112112512},
              {'name': 'gitnotes', 'size': 897104422},
              {'name': 'finder', 'size': 3810736368},
              {'name': 'apache2', 'size': 1719919406}],
 'name': 'ROOT'}

In [12]: pprint.pprint(walk_tree(t,depth=2))
{'children': [{'children': [{'name': 'vhost', 'size': 103489662}],
               'name': 'apache2-gzip'},
              {'children': [{'name': '2', 'size': 533145458},
                            {'name': 'notes.git', 'size': 363958964}],
               'name': 'gitnotes'},
              {'children': [{'name': 'gzipped', 'size': 3810736368},
                            {'name': 'output.txt', 'size': 0}],
               'name': 'finder'},
              {'children': [{'name': 'sente_combined.log', 'size': 0},
                            {'name': 'lisp_ssl.log', 'size': 0},
                            {'name': 'vhost', 'size': 1378778576},
                            {'name': 'other_vhosts_access.log', 'size': 0},
                            {'name': 'ssl_error.log', 'size': 0},
                            {'name': 'ssl_access.log', 'size': 0},
                            {'name': 'sente_test.log', 'size': 0}],
               'name': 'apache2'}],
 'name': 'ROOT'}

FS 只爬行一次，但树一旦创建就需要遍历以获得完整大小，如果您从叶节点开始并朝着根部工作，您可以最有效地计算每个 dir 的总大小。

于 2012-05-12T01:53:51.407 回答