Python 文件名不区分大小写？

Question

我需要根据文件名来加载一个文件，但我得到的文件名是不区分大小写的。也就是说，“A.txt”其实可能是“a.txt”。我该怎么快速做到这一点呢？（不是生成所有可能的名字然后一个个尝试）

Answer 1

这是一个简单的递归函数，用来执行Eli上面提到的搜索：

def find_sensitive_path(dir, insensitive_path):

    insensitive_path = insensitive_path.strip(os.path.sep)

    parts = insensitive_path.split(os.path.sep)
    next_name = parts[0]
    for name in os.listdir(dir):
        if next_name.lower() == name.lower():
            improved_path = os.path.join(dir, name)
            if len(parts) == 1:
                return improved_path
            else:
                return find_sensitive_path(improved_path, os.path.sep.join(parts[1:]))
    return None

Answer 2

你不能不先获取目录列表，然后把你要找的文件和目录里的每个文件都转换成相同的大小写来进行比较。因为文件系统是区分大小写的，这就是事实。

这里有一个我写的函数（其实是两个），可以完全实现这个功能，以不区分大小写的方式递归匹配文件名：http://portableapps.hg.sourceforge.net/hgweb/portableapps/development-toolkit/file/775197d56e86/utils.py#l78。

def path_insensitive(path):
    """
    Get a case-insensitive path for use on a case sensitive system.

    >>> path_insensitive('/Home')
    '/home'
    >>> path_insensitive('/Home/chris')
    '/home/chris'
    >>> path_insensitive('/HoME/CHris/')
    '/home/chris/'
    >>> path_insensitive('/home/CHRIS')
    '/home/chris'
    >>> path_insensitive('/Home/CHRIS/.gtk-bookmarks')
    '/home/chris/.gtk-bookmarks'
    >>> path_insensitive('/home/chris/.GTK-bookmarks')
    '/home/chris/.gtk-bookmarks'
    >>> path_insensitive('/HOME/Chris/.GTK-bookmarks')
    '/home/chris/.gtk-bookmarks'
    >>> path_insensitive("/HOME/Chris/I HOPE this doesn't exist")
    "/HOME/Chris/I HOPE this doesn't exist"
    """

    return _path_insensitive(path) or path


def _path_insensitive(path):
    """
    Recursive part of path_insensitive to do the work.
    """

    if path == '' or os.path.exists(path):
        return path

    base = os.path.basename(path)  # may be a directory or a file
    dirname = os.path.dirname(path)

    suffix = ''
    if not base:  # dir ends with a slash?
        if len(dirname) < len(path):
            suffix = path[:len(path) - len(dirname)]

        base = os.path.basename(dirname)
        dirname = os.path.dirname(dirname)

    if not os.path.exists(dirname):
        dirname = _path_insensitive(dirname)
        if not dirname:
            return

    # at this point, the directory exists but not the file

    try:  # we are expecting dirname to be a directory, but it could be a file
        files = os.listdir(dirname)
    except OSError:
        return

    baselow = base.lower()
    try:
        basefinal = next(fl for fl in files if fl.lower() == baselow)
    except StopIteration:
        return

    if basefinal:
        return os.path.join(dirname, basefinal) + suffix
    else:
        return

Answer 3

你可以列出文件所在的文件夹里的所有文件（用os.listdir），然后看看有没有和你要找的文件名相匹配的文件。匹配的时候可以把两个文件名都变成小写，然后进行比较。