如何在Python中逐字节读取文件以及如何将bytelist打印为二进制文件？ [英] How to read a file byte by byte in Python and how to print a bytelist as a binary?

问题描述

我试图逐字节读取一个文件，但我不知道该怎么做。我想这样做：

$ $ $ $ $ $ $ $ $ $ $ $ $' ：
byte = file.read（8）
＃做某事...

<那么这是否使变量字节在每个循环的开头都包含8个下一位？这些字节实际上并不重要。唯一重要的是我需要读取一个8位堆栈文件。

编辑：

我也收集这些字节在列表中，我想要打印它们，以便他们不打印出ASCII字符，但作为原始字节，即当我打印bytelist它给出的结果为

  ['10010101'，'00011100'，....] 
  

$为了回答你的问题的第二部分，要转换为二进制文件，你可以使用 format string 和

     >>>> byte ='a'
>>>格式（ord（byte））
'01100001'
  

请注意格式垫与正确数量的前导零，这似乎是您的要求。此方法需要Python 2.6或更高版本。

I'm trying to read a file byte by byte, but I'm not sure how to do that. I'm trying to do it like that:

file = open(filename, 'rb')
while 1:
   byte = file.read(8)
   # Do something...

So does that make the variable byte to contain 8 next bits at the beginning of every loop? It doesn't matter what those bytes really are. The only thing that matters is that I need to read a file in 8-bit stacks.

EDIT:

Also I collect those bytes in a list and I would like to print them so that they don't print out as ASCII characters, but as raw bytes i.e. when I print that bytelist it gives the result as

['10010101', '00011100', .... ]

解决方案

To answer the second part of your question, to convert to binary you can use a format string and the ord function:

>>> byte = 'a'
>>> '{0:08b}'.format(ord(byte))
'01100001'

Note that the format pads with the right number of leading zeros, which seems to be your requirement. This method needs Python 2.6 or later.

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