Python CSV阅读器TypeError:字节对象上的字符串模式

我想第一次使用python CSV阅读器.我有一个方法要求用户选择他们想要解析的文件,然后它将该文件路径传递给parse方法:

def parse(filename):
        parsedFile = []
        with open(filename, 'rb') as csvfile:
                dialect = csv.Sniffer().sniff(csvfile.read(), delimiters=';,|')
                csvfile.seek(0)
                reader = csv.reader(csvfile, dialect)

                for line in reader:
                    parsedFile.append(line)
                return(parsedFile)

def selectFile():
        print('start selectFile method')
        localPath = os.getcwd() + '\Files'
        print(localPath)
        for fileA in os.listdir(localPath):
                print (fileA)

        test = False
        while test == False:
                fileB = input('which file would you like to DeID? \n')
                conjoinedPath = os.path.join(localPath, fileB)
                test = os.path.isfile(conjoinedPath)


        userInput = input('Please enter the number corresponding to which client ' + fileB + ' belongs to. \n\nAcceptable options are: \n1.A \n2.B \n3.C \n4.D \n5.E \n')
        client = ''
        if (userInput == '1'):
                client = 'A'
        elif (userInput == '2'):
                client = 'B'
        elif (userInput == '3'):
                client = 'CServices'
        elif (userInput == '4'):
                client = 'D'
        elif (userInput == '5'):
                client = 'E'
        return(client, conjoinedPath)



def main():
       x, y = selectFile() 
       parse(y)


if __name__ == '__main__':
        main()

所有这些似乎按预期工作,但我得到一个:

TypeError: can't use a string pattern on a bytes-like object 

尝试打开文件名时(代码中的第3行).我试图将文件名转换为字符串类型和字节类型,似乎都不起作用.

这是输出:

>>> 
start selectFile method
C:\PythonScripts\DeID\Files
89308570_201601040630verifyppn.txt
89339985_201601042316verifyppn.txt
which file would you like to DeID? 
89339985_201601042316verifyppn.txt
Please enter the number corresponding to which client 89339985_201601042316verifyppn.txt belongs to. 

Acceptable options are: 
1.Client A
2.Client B
3.Client C
4.Client D
5.Client E
3
Traceback (most recent call last):
  File "C:\PythonScripts\DeID\DeIDvA1.py", line 107, in 
    main()
  File "C:\PythonScripts\DeID\DeIDvA1.py", line 103, in main
    parse(y)
  File "C:\PythonScripts\DeID\DeIDvA1.py", line 63, in parse
    dialect = csv.Sniffer().sniff(csvfile.read(), delimiters=';,|')
  File "C:\Python34\lib\csv.py", line 183, in sniff
    self._guess_quote_and_delimiter(sample, delimiters)
  File "C:\Python34\lib\csv.py", line 224, in _guess_quote_and_delimiter
    matches = regexp.findall(data)
TypeError: can't use a string pattern on a bytes-like object
>>> 

我不确定我做错了什么.

1> Jim Fasaraki..：

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这里不是要归咎的文件名,而是你打开文件的事实:

with open(filename, 'rb') as csvfile:

当'rb'模式指定的文件将在二进制模式打开,也就是说,该文件的内容被视为byte对象.文件:

'b'附加到模式后以二进制模式打开文件:现在以bytes对象的形式读取和写入数据.此模式应该用于所有不包含文本的文件.

然后,您尝试csv.Sniff().sniff()使用字符串模式在其中进行搜索,并且正如TypeError优雅地指出的那样,这是不允许的.

b从模式中删除并简单地使用r将可以解决问题.

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注意:Python 2.x在Unix机器上不会出现此行为.这是将对象bytes和str对象分离为不同类型的结果3.x.

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注意:您还应该在Py3上传递`newline =''`到`open`,以防止行结束转换(`csv`处理它本身,因为行结尾是CSV方言的一部分)并允许换行出现在引用字段正确.在Py2中,出于同样的原因,您以二进制模式打开.