How to get file path + file name into a list?

 

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  • How do I list all files of a directory? 31 answers

    • Since you have access to the directory path you could just do: 

    dir = "train_dataham"
output = map(lambda p: os.path.join(dir, p), os.listdir(dir))

    or simpler 

    output = [os.path.join(dir, p) for p in os.listdir(dir)]

    Where os.path.join will join your directory path with the filenames inside it.

    If you're on Python 3.5 or higher, skip os.listdir in favor of os.scandir , which is both more efficient and does the work for you ( path is an attribute of the result objects): 

    hamFileNames = [entry.path for entry in os.scandir(r"train_dataham")]

    This also lets you cheaply filter ( scandir includes some file info for free, without stat -ing the file), eg to keep only files (no directories or special file-system objects): 

    hamFileNames = [entry.path for entry in os.scandir(r"train_dataham") if entry.is_file()]

    If you're on 3.4 or below, you may want to look at the PyPI scandir module (which provides the same API on earlier Python).

    Also note: I used a raw string for the path; while h happens to work without it, you should always use raw strings for Windows path literals, or you'll get a nasty shock when you try to use "train_datafoo" (where f is the ASCII form feed character), while r"train_datafoo" works just fine (because the r prefix prevents backslash interpolation of anything but the quote character).

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