Question

我在目录中有很多excel文件，它们都有相同的标题行。其中一些excel文件具有多个工作表，这些工作表又具有相同的标题。我正在尝试遍历目录中的excel文件，并为每一个检查是否有多个工作表来连接它们以及其余的excel文件。

这是我尝试过的：

import pandas as pd
import os
import ntpath
import glob

dir_path = os.path.dirname(os.path.realpath(__file__))
os.chdir(dir_path)

for excel_names in glob.glob('*.xlsx'):
    # read them in
    i=0
    df = pd.read_excel(excel_names[i], sheet_name=None, ignore_index=True)
    cdf = pd.concat(df.values())
    cdf.to_excel("c.xlsx", header=False, index=False)
    excels = [pd.ExcelFile(name) for name in excel_names]

    # turn them into dataframes
    frames = [x.parse(x.sheet_names[0], header=None,index_col=None) for x in excels]

    # delete the first row for all frames except the first
    # i.e. remove the header row -- assumes it's the first
    frames[1:] = [df[1:] for df in frames[1:]]

    # concatenate them..
    combined = pd.concat(frames)

    # write it out
    combined.to_excel("c.xlsx", header=False, index=False)
    i+=1

但是随后我得到以下错误任何建议吗？

"concat excel.py", line 12, in <module>
    df = pd.read_excel(excel_names[i], sheet_name=None, ignore_index=True)
  File "/usr/local/lib/python2.7/site-packages/pandas/util/_decorators.py", line 188, in wrapper
    return func(*args, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/pandas/util/_decorators.py", line 188, in wrapper
    return func(*args, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/pandas/io/excel.py", line 350, in read_excel
    io = ExcelFile(io, engine=engine)
  File "/usr/local/lib/python2.7/site-packages/pandas/io/excel.py", line 653, in __init__
    self._reader = self._engines[engine](self._io)
  File "/usr/local/lib/python2.7/site-packages/pandas/io/excel.py", line 424, in __init__
    self.book = xlrd.open_workbook(filepath_or_buffer)
  File "/usr/local/lib/python2.7/site-packages/xlrd/__init__.py", line 111, in open_workbook
    with open(filename, "rb") as f:
IOError: [Errno 2] No such file or directory: 'G'

Answer 1

您的for语句将excel_names依次设置为每个文件名（因此，更好的变量名为excel_name）：

for excel_names in glob.glob('*.xlsx'):

但是在循环内您的代码确实如此

df = pd.read_excel(excel_names[i], sheet_name=None, ignore_index=True)

您显然希望excel_names是要从中提取一个元素的列表。但这不是一个列表，而是一个字符串。因此，您将获得第一个文件名的第一个字符。

使用python将

1 个答案: