FileNotFoundError：[Errno 2]没有这样的文件或目录：'corpus_or_AB_FMC.xlsx‘

Question

我有一个包含很多excel文件的目录。我的目标是读取所有这些excel文件，并从中提取一些信息。我使用下面的脚本读取目录，但仍然收到错误。文件被识别了，但是代码告诉它不是在创建它们，这很奇怪，因为有一行代码打印文件的名称。但是，当阅读熊猫的文件时，它是不成立的。

/home/geta/kelo/eXP/Test/corpus
-----File in processed : corpus_or_AB_FMC.xlsx
Traceback (most recent call last):
  File "test_vec.py", line 111, in <module>
    sentences = pd.read_excel(file, sheet_name= 0)
  File "/home/getalp/kelodjoe/anaconda3/lib/python3.7/site-packages/pandas/util/_decorators.py", line 208, in wrapper
    return func(*args, **kwargs)
  File "/home/geta/kelo/anaconda3/lib/python3.7/site-packages/pandas/io/excel/_base.py", line 310, in read_excel
    io = ExcelFile(io, engine=engine)
  File "/home/geta/kelo/anaconda3/lib/python3.7/site-packages/pandas/io/excel/_base.py", line 819, in __init__
    self._reader = self._engines[engine](self._io)
  File "/home/geta/kelo/anaconda3/lib/python3.7/site-packages/pandas/io/excel/_xlrd.py", line 21, in __init__
    super().__init__(filepath_or_buffer)
  File "/home/geta/kelo/anaconda3/lib/python3.7/site-packages/pandas/io/excel/_base.py", line 359, in __init__
    self.book = self.load_workbook(filepath_or_buffer)
  File "/home/geta/kelo/anaconda3/lib/python3.7/site-packages/pandas/io/excel/_xlrd.py", line 36, in load_workbook
    return open_workbook(filepath_or_buffer)
  File "/home/geta/kelo/anaconda3/lib/python3.7/site-packages/xlrd/__init__.py", line 111, in open_workbook
    with open(filename, "rb") as f:

代码如下：

dir = "/home/geta/kelo/eXP/Test/corpus"
for root, subdirs, files in os.walk(dir):
    print(root)
    for file in files:
        #print(files)
        print("-----File in processed :", file)
        # -----File in processed : corpus_or_AB_FMC.xlsx  # this file si located in the corpus directory
        sentences = pd.read_excel(file, sheet_name= 0)
        data_id = sentences.identifiant
        print("Total phrases: ", len(data_id))
        data = sentences.verbatim

        data_label = sentences.etiquette
        #print(type(data_id))
        #print(type(data))
        #number = LabelEncoder()
        # 0 = C; 1= F; 2= M
        #data_label = number.fit_transform(sentences.etiquette.astype('str'))
        #print(data_label)
        
        print("etiquette  :" , sentences['etiquette'].unique())
        classes = sentences['etiquette'].unique()
        len_classes = len(classes)

Answer 1

file是文件的名称，不包括该文件的路径。使用os.path.join

sentences = pd.read_excel(os.path.join(root, file), sheet_name=0)

以连接文件名及其绝对路径。