十六进制转储文件的pythonic方法 [英] pythonic way to hex dump files

问题描述

我的问题很简单:

有什么方法可以通过bash命令的pythonic方式进行编码吗?

Is there any way to code in a pythonic way that bash command?

hexdump -e '2/1 "%02x"' file.dat

很明显，没有使用os，popen或任何快捷方式;)

Obviously, without using os, popen, or any shortcut ;)

尽管我没有明确指定，但是如果代码在Python3.x中可以正常工作，那就太好了

although I've not explicitly specified, it would be great if the code was functional in Python3.x

谢谢！

推荐答案

如果您只关心Python 2.x，请

If you only care about Python 2.x, line.encode('hex') will encode a chunk of binary data into hex. So:

with open('file.dat', 'rb') as f:
    for chunk in iter(lambda: f.read(32), b''):
        print chunk.encode('hex')

_{(默认情况下，IIRC，hexdump每行打印32对十六进制；如果没有，只需将32更改为16或其他内容……)}

_{(IIRC, hexdump by default prints 32 pairs of hex per line; if not, just change that 32 to 16 or whatever it is…)}

_{如果两个参数} iter 看起来莫名其妙，请单击帮助链接；一旦您有了主意，它就不会太复杂.

如果您关心Python 3.x，请 encode 仅适用于将Unicode字符串转换为字节的编解码器；任何以其他方式转换的编解码器(或任何其他组合)，则必须使用

If you care about Python 3.x, encode only works for codecs that convert Unicode strings to bytes; any codecs that convert the other way around (or any other combination), you have to use codecs.encode to do it explicitly:

with open('file.dat', 'rb') as f:
    for chunk in iter(lambda: f.read(32), b''):
        print(codecs.encode(chunk, 'hex'))

或者最好使用 hexlify :

Or it may be better to use hexlify:

with open('file.dat', 'rb') as f:
    for chunk in iter(lambda: f.read(32), b''):
        print(binascii.hexlify(chunk))

如果您除了打印输出外还想做些其他事情，而不是将整个文件读入内存，您可能想要创建一个迭代器.您可以将其放在函数中，然后将print更改为yield，该函数将恰好返回您想要的迭代器.或使用genexpr或map呼叫:

If you want to do something besides print them out, rather than read the whole file into memory, you probably want to make an iterator. You could just put this in a function and change that print to a yield, and that function returns exactly the iterator you want. Or use a genexpr or map call:

with open('file.dat', 'rb') as f:
    chunks = iter(lambda: f.read(32), b'')
    hexlines = map(binascii.hexlify, chunks)

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