Python 3.x pycharm并打开文件
我有一个从目录中打开文件的def,在其中我有文件,并希望迭代每个文件以生成一些东西Python 3.x pycharm并打开文件,python-3.x,file,pycharm,with-statement,Python 3.x,File,Pycharm,With Statement,我有一个从目录中打开文件的def,在其中我有文件,并希望迭代每个文件以生成一些东西 def read_decks_from_disk(): deck_list = [] basepath = Path("Decks\\") for filename in basepath.iterdir(): if filename.is_file(): with open(filename) as file: deck_in
def read_decks_from_disk():
deck_list = []
basepath = Path("Decks\\")
for filename in basepath.iterdir():
if filename.is_file():
with open(filename) as file:
deck_info = file.read().splitlines()
d = {
"nome": deck_info[0],
"formato": deck_info[1],
"prezzo": deck_info[2]
}
deck_list.append(d)
return deck_list
当我使用:
with open(filename) as file:
我的建议如下:
Unexpected type(s):
(Path)
Possible types:
(Union[str, bytes, int])
(Union[str, bytes, int, PathLike])
我如何解决这个问题,为什么pycharm建议我这样做,即使代码仍然有效?请投票(在问题标题附近竖起大拇指)我认为截图会更好。添加了截图