在 Python 中将 ASCII 码列表转换为字符串（字节数组）

Question

I have a list of integer ASCII values that I need to transform into a string (binary) to use as the key for a crypto operation. (I am re-implementing java crypto code in python)

This works (assuming an 8-byte key):

key = struct.pack('BBBBBBBB', 17, 24, 121, 1, 12, 222, 34, 76)

However, I would prefer to not have the key length and unpack() parameter list hardcoded.

How might I implement this correctly, given an initial list of integers?

Thanks!

Answer 1

For Python 2.6 and later if you are dealing with bytes then a bytearray is the most obvious choice:

>>> str(bytearray([17, 24, 121, 1, 12, 222, 34, 76]))
'\x11\x18y\x01\x0c\xde"L'

To me this is even more direct than Alex Martelli's answer - still no string manipulation or len call but now you don't even need to import anything!

Answer 2

I much prefer the array module to the struct module for this kind of tasks (ones involving sequences of homogeneous values):

>>> import array
>>> array.array('B', [17, 24, 121, 1, 12, 222, 34, 76]).tostring()
'\x11\x18y\x01\x0c\xde"L'

no len call, no string manipulation needed, etc -- fast, simple, direct, why prefer any other approach?!

Answer 3

This is reviving an old question, but in Python 3, you can just use bytes directly:

>>> bytes([17, 24, 121, 1, 12, 222, 34, 76])
b'\x11\x18y\x01\x0c\xde"L'

Answer 4

struct.pack('B' * len(integers), *integers)

*sequence means "unpack sequence" - or rather, "when calling f(..., *args,...) , let args = sequence ".

Answer 5

key = "".join( chr( val ) for val in myList )

Answer 6

Shorter version of previous using map() function (works for python 2.7):

"".join(map(chr, myList))

Answer 7

另一种方法（Python 2.7+）：

"".join(str(x) for x in int_array)