Python 字节数组到十六进制字符串
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Byte Array to Hex String
提问by Jamie Wright
I have data stored in a byte array. How can I convert this data into a hex string?
我将数据存储在字节数组中。如何将此数据转换为十六进制字符串？
Example of my byte array:
我的字节数组示例：
array_alpha = [ 133, 53, 234, 241 ]
采纳答案by falsetru
Using str.format:
使用str.format：
>>> array_alpha = [ 133, 53, 234, 241 ]
>>> print ''.join('{:02x}'.format(x) for x in array_alpha)
8535eaf1
or using format
或使用 format
>>> print ''.join(format(x, '02x') for x in array_alpha)
8535eaf1
Note:In the format statements, the
02means it will pad with up to 2 leading0s if necessary. This is important since[0x1, 0x1, 0x1] i.e. (0x010101)would be formatted to"111"instead of"010101"
注意：在格式语句中，这
02意味着0如果需要，它将最多填充 2 个前导。这很重要，因为[0x1, 0x1, 0x1] i.e. (0x010101)将被格式化为"111"而不是"010101"
or using bytearraywith binascii.hexlify:
或者使用bytearray具有binascii.hexlify：
>>> import binascii
>>> binascii.hexlify(bytearray(array_alpha))
'8535eaf1'
Here is a benchmark of above methods in Python 3.6.1:
这是 Python 3.6.1 中上述方法的基准测试：
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()
Result:
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
Methods using formatdo provide additional formatting options, as example separating numbers with spaces " ".join, commas ", ".join, upper-case printing "{:02X}".format(x)/format(x, "02X"), etc., but at a cost of great performance impact.
使用的方法format确实提供了额外的格式化选项，例如用空格" ".join、逗号", ".join、大写打印"{:02X}".format(x)/format(x, "02X")等分隔数字，但代价是对性能的影响很大。
回答by kindall
hex_string = "".join("%02x" % b for b in array_alpha)
回答by kindall
Or, if you are a fan of functional programming:
或者，如果您是函数式编程的粉丝：
>>> a = [133, 53, 234, 241]
>>> "".join(map(lambda b: format(b, "02x"), a))
8535eaf1
>>>
回答by ostrokach
If you have a numpy array, you can do the following:
如果您有一个 numpy 数组，则可以执行以下操作：
>>> import numpy as np
>>> a = np.array([133, 53, 234, 241])
>>> a.astype(np.uint8).data.hex()
'8535eaf1'
回答by orip
Consider the hex() methodof the bytestype on Python 3.5 and up:
考虑十六进制（）方法的的bytes关于Python 3.5和至多类型：
>>> array_alpha = [ 133, 53, 234, 241 ]
>>> print(bytes(array_alpha).hex())
8535eaf1
EDIT: it's also much faster than hexlify(modified @falsetru's benchmarks above)
编辑：它也比hexlify（上面修改了@falsetru的基准）快得多
from timeit import timeit
N = 10000
print("bytearray + hexlify ->", timeit(
'binascii.hexlify(data).decode("ascii")',
setup='import binascii; data = bytearray(range(255))',
number=N,
))
print("byte + hex ->", timeit(
'data.hex()',
setup='data = bytes(range(255))',
number=N,
))
Result:
结果：
bytearray + hexlify -> 0.011218150997592602
byte + hex -> 0.005952142993919551
