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我想知道是否有更快的方法来实现在 python 中返回区分大小写路径的函数。我想出的解决方案之一适用于 linux 和 windows，但需要我迭代 os.listdir，这可能很慢。

此解决方案适用于不需要足够速度的应用程序和上下文：

def correctPath(start, path):
    'Returns a unix-type case-sensitive path, works in windows and linux'
    start = unicode(start);
    path = unicode(path);
    b = '';
    if path[-1] == '/':
        path = path[:-1];
    parts = path.split('\\');
    d = start;
    c = 0;
    for p in parts:
        listing = os.listdir(d);
        _ = None;
        for l in listing:
            if p.lower() == l.lower():
                if p != l:
                    c += 1;
                d = os.path.join(d, l);
                _ = os.path.join(b, l);
                break;
        if not _:
            return None;
        b = _;

    return b, c; #(corrected path, number of corrections)

>>> correctPath('C:\\Windows', 'SYSTEM32\\CmD.EXe')
(u'System32\\cmd.exe', 2)

但是，当上下文从 50,000 多个条目的大型数据库中收集文件名时，这将不会那么快。

一种方法是为每个目录创建一个字典树。将字典树与路径的目录部分匹配，如果发生键未命中，执行 os.listdir 为新目录查找并创建一个字典条目，并删除未使用的部分或保留变量计数器作为一种方法为每个目录分配一个“生命周期”。

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2 回答 2

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以下是对您自己的代码的轻微重写，并进行了三处修改：在匹配之前检查文件名是否已经正确，在测试之前将列表处理为小写，使用索引查找相关的“真实大小写”文件。

def corrected_path(start, path):
    '''Returns a unix-type case-sensitive path, works in windows and linux'''
    start = unicode(start)
    path = unicode(path)
    corrected_path = ''
    if path[-1] == '/':
        path = path[:-1]
    parts = path.split('\\')
    cd = start
    corrections_count = 0

    for p in parts:
        if not os.path.exists(os.path.join(cd,p)): # Check it's not correct already
            listing = os.listdir(cd)

            cip = p.lower()
            cilisting = [l.lower() for l in listing]

            if cip in cilisting:
                l = listing[ cilisting.index(cip) ] # Get our real folder name
                cd = os.path.join(cd, l)
                corrected_path = os.path.join(corrected_path, l)
                corrections_count += 1
            else:
                return False # Error, this path element isn't found
        else:
            cd = os.path.join(cd, p)
            corrected_path = os.path.join(corrected_path, p)

    return corrected_path, corrections_count

我不确定这是否会更快，尽管正在进行的测试少了一点，而且一开始的“已经正确”的捕获可能会有所帮助。

于 2013-04-13T23:45:26.213 回答
0

带有不区分大小写缓存的扩展版本，用于提取更正的路径：

import os,re

def corrected_paths(start, pathlist):
    ''' This wrapper function takes a list of paths to correct vs. to allow caching '''

    start = unicode(start)
    pathlist = [unicode(path[:-1]) if path[-1] == '/' else unicode(path) for path in pathlist ]

    # Use a dict as a cache, storing oldpath > newpath for first-pass replacement
    # with path keys from incorrect to corrected paths
    cache = dict() 
    corrected_path_list = []
    corrections_count = 0
    path_split = re.compile('(/+|\+)')

    for path in pathlist:
        cd = start
        corrected_path = ''
        parts = path_split.split(path)

        # Pre-process against the cache
        for n,p in enumerate(parts):
            # We pass *parts to send through the contents of the list as a series of strings
            uncorrected_path= os.path.join( cd, *parts[0:len(parts)-n] ).lower() # Walk backwards
            if uncorrected_path in cache:
                # Move up the basepath to the latest matched position
                cd = os.path.join(cd, cache[uncorrected_path])
                parts = parts[len(parts)-n:] # Retrieve the unmatched segment
                break; # First hit, we exit since we're going backwards

        # Fallback to walking, from the base path cd point
        for n,p in enumerate(parts):

            if not os.path.exists(os.path.join(cd,p)): # Check it's not correct already
            #if p not in os.listdir(cd): # Alternative: The above does not work on Mac Os, returns case-insensitive path test

                listing = os.listdir(cd)

                cip = p.lower()
                cilisting = [l.lower() for l in listing]

                if cip in cilisting:

                    l = listing[ cilisting.index(cip) ] # Get our real folder name
                    # Store the path correction in the cache for next iteration
                    cache[ os.path.join(cd,p).lower() ] = os.path.join(cd, l)
                    cd = os.path.join(cd, l)
                    corrections_count += 1

                else:
                    print "Error %s not in folder %s" % (cip, cilisting)
                    return False # Error, this path element isn't found

            else:
                cd = os.path.join(cd, p)

        corrected_path_list.append(cd)

    return corrected_path_list, corrections_count

在一组路径的示例运行中，这大大减少了 listdirs 的数量（这显然取决于您的路径有多相似）：

corrected_paths('/Users/', ['mxF793/ScRiPtS/meTApaTH','mxF793/ScRiPtS/meTApaTH/metapAth/html','mxF793/ScRiPtS/meTApaTH/metapAth/html/css','mxF793/ScRiPts/PuBfig'])
([u'/Users/mxf793/Scripts/metapath', u'/Users/mxf793/Scripts/metapath/metapath/html', u'/Users/mxf793/Scripts/metapath/metapath/html/css', u'/Users/mxf793/Scripts/pubfig'], 14)
([u'/Users/mxf793/Scripts/metapath', u'/Users/mxf793/Scripts/metapath/metapath/html', u'/Users/mxf793/Scripts/metapath/metapath/html/css', u'/Users/mxf793/Scripts/pubfig'], 5)

在此过程中，我意识到 Mac OSX Python 上的路径匹配返回路径匹配，就好像它们不区分大小写一样，因此存在性测试总是成功。在这种情况下，listdir 可以向上移动以替换它。

于 2013-04-14T16:04:46.273 回答