字节数组到十六进制字符串

Question

我将数据存储在字节数组中。如何将这些数据转换为十六进制字符串？

我的字节数组示例：

array_alpha = [ 133, 53, 234, 241 ]

Answer 1

使用str.format：

>>> array_alpha = [ 133, 53, 234, 241 ]
>>> print ''.join('{:02x}'.format(x) for x in array_alpha)
8535eaf1

或使用 format

>>> print ''.join(format(x, '02x') for x in array_alpha)
8535eaf1

注意： 在format语句中，02表示0在必要时最多填充2个前导s。这很重要，因为[0x1, 0x1, 0x1] i.e. (0x010101)它将被格式化为"111"而不是"010101"

或bytearray与配合使用binascii.hexlify：

>>> import binascii
>>> binascii.hexlify(bytearray(array_alpha))
'8535eaf1'

这是Python 3.6.1中上述方法的基准：

from timeit import timeit
import binascii

number = 10000

def using_str_format() -> str:
    return "".join("{:02x}".format(x) for x in test_obj)

def using_format() -> str:
    return "".join(format(x, "02x") for x in test_obj)

def using_hexlify() -> str:
    return binascii.hexlify(bytearray(test_obj)).decode('ascii')

def do_test():
    print("Testing with {}-byte {}:".format(len(test_obj), test_obj.__class__.__name__))
    if using_str_format() != using_format() != using_hexlify():
        raise RuntimeError("Results are not the same")

    print("Using str.format       -> " + str(timeit(using_str_format, number=number)))
    print("Using format           -> " + str(timeit(using_format, number=number)))
    print("Using binascii.hexlify -> " + str(timeit(using_hexlify, number=number)))

test_obj = bytes([i for i in range(255)])
do_test()

test_obj = bytearray([i for i in range(255)])
do_test()

结果：

Testing with 255-byte bytes:
Using str.format       -> 1.459474583090427
Using format           -> 1.5809937679100738
Using binascii.hexlify -> 0.014521426401399307
Testing with 255-byte bytearray:
Using str.format       -> 1.443447684109402
Using format           -> 1.5608712609513171
Using binascii.hexlify -> 0.014114164661833684

使用的方法format的确提供了其他格式选项，例如用空格" ".join，逗号", ".join，大写字母"{:02X}".format(x)/format(x, "02X")等分隔数字，但是会影响性能。