Question

我正在提取电子表格并意识到有些会受密码保护，因此下面的代码无法读取某些电子表格，这很好。我正试图通过except行传递这些电子表格，我从zipfile导入了BadZipfile，试图将它们识别为错误。

但是，except函数不会将BadZipfile识别为错误，并且仍然会在Traceback中引发错误（如下所示）。我希望通过将except行留空，它会传递任何错误并再次循环以在剩余文件上运行try语句，但是它似乎不会将BadZipfile识别为错误。

在except语句中也需要KeyError，因为有些电子表格没有我想要提取的索引标签。也许我应该通过任何例外，因为我没有循环的文件可能会出现更多错误。我更愿意处理在此时工作的任何电子表格，并在稍后协调错误。

因此，如何确保BadZipfile被识别为错误，然后使用except语句传递任何错误？

CODE

import itertools
import glob
from openpyxl import load_workbook
from pandas import DataFrame
import pandas as pd
import os
from zipfile import BadZipfile

def get_data(ws):
        for row in ws.values:
            row_it = iter(row)
            for cell in row_it:
                if cell is not None:
                    yield itertools.chain((cell,), row_it)
                    break

def read_workbook(file_):
        wb = load_workbook(file_, data_only=True)
        for sheet in wb.worksheets:
            ws = sheet
        return DataFrame(get_data(ws))

path =r'dir'
allFiles = glob.glob(path + "/*.xlsx")
frame = pd.DataFrame()
list_ = []
for file_ in allFiles:
        parsed_file = read_workbook(file_)
        parsed_file['filename'] = os.path.basename(file_)
        parsed_file.set_index(parsed_file.columns[0], inplace = True)
        parsed_file.index.str.strip()

        try: 
                parsed_file.loc["Staff" : "Total  Staff"].copy()
                list_.append(parsed_file)
        except KeyError or BadZipfile:
            pass
frame = pd.concat(list_)
print(frame.dropna(axis='columns', thresh=2, inplace = True))

错误

Traceback (most recent call last):

  File "<ipython-input-47-2d5508ddf805>", line 1, in <module>
    runfile('C:dir.py', wdir='C:dir')

  File "C:\ProgramData\Anaconda2\lib\site-packages\spyder\utils\site\sitecustomize.py", line 880, in runfile
    execfile(filename, namespace)

  File "C:\ProgramData\Anaconda2\lib\site-packages\spyder\utils\site\sitecustomize.py", line 87, in execfile
    exec(compile(scripttext, filename, 'exec'), glob, loc)

  File "C:dir.py", line 35, in <module>
    parsed_file = read_workbook(file_)

  File "C:dir.py", line 25, in read_workbook
    wb = load_workbook(file_, data_only=True)

  File "C:\ProgramData\Anaconda2\lib\site-packages\openpyxl\reader\excel.py", line 164, in load_workbook
    archive = _validate_archive(filename)

  File "C:\ProgramData\Anaconda2\lib\site-packages\openpyxl\reader\excel.py", line 121, in _validate_archive
    archive = ZipFile(f, 'r', ZIP_DEFLATED)

  File "C:\ProgramData\Anaconda2\lib\zipfile.py", line 770, in __init__
    self._RealGetContents()

  File "C:\ProgramData\Anaconda2\lib\zipfile.py", line 813, in _RealGetContents
    raise BadZipfile, "File is not a zip file"

BadZipfile: File is not a zip file

Answer 1

首先，检查堆栈跟踪。你错误被抛出的地方没有except (KeyError, BadZipFile):。

第二，为了捕获多种类型的错误，您需要except KeyError or BadZipFile:而不是except KeyError，它只评估为or（了解Name,Channel0,Channel1,Channel2,Channel3,Channel4 CSBA10,125833,147883,184269,162270,151366 CSBA20,125533,145883,154269,152270,155365如何工作）

Answer 2

来自docs page on error handling：

try:
    # your code

except (KeyError, BadZipfile):
    pass

如果您想对错误做些什么，可以使用

try:
    # your code

except (KeyError, BadZipfile) as err:
    # do something

当您调用语句read_workbook时，您的load_workbook函数实际上发生了错误，因此您在发生错误时无法捕获错误。

Python：如何将引发的异常识别为错误并将其传递给except语句

2 个答案: