python获取目录及子目录下（指定后缀）的文件名或文件路径

"""递归获取文件"""
import os


def getfiles(data_path, suffix=None, data_file_list=None, with_path=False):
    """
    获取目录及子目录下所有文件名或者文件路径
    :param data_path: 数据根目录
    :param suffix: 后缀名
    :param data_file_list: 返回的文件名或路径
    :param with_path: 是否返回路径
    :return: 文件名或文件路径
    """
    if data_file_list is None:
        data_file_list = []
    if suffix is not None:
        if not isinstance(suffix, list):
            suffix = [suffix]
    data_files = os.listdir(data_path)
    for data_file in data_files:
        data_file_path = os.path.join(data_path, data_file)
        if os.path.isdir(data_file_path):
            getfiles(data_file_path, suffix, data_file_list, with_path)
        else:
            if suffix is None:
                if with_path:
                    data_file_list.append(data_file_path)
                else:
                    data_file_list.append(data_file)
            else:
                _, data_suffix = os.path.splitext(data_file)
                data_suffix = data_suffix.split('.')[-1]
                if data_suffix in suffix:
                    if with_path:
                        data_file_list.append(data_file_path)
                    else:
                        data_file_list.append(data_file)
    return data_file_list