问题:如何将字节字符串转换为int?
如何在python中将字节字符串转换为int?
这样说: 'y\xcc\xa6\xbb'
我想出了一个聪明/愚蠢的方法:
sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))
我知道必须有内置的东西或在标准库中可以更简单地执行此操作…
这与转换可以使用int(xxx,16)的十六进制数字字符串不同,但是我想转换一个实际字节值的字符串。
更新:
我有点喜欢James的回答,因为它不需要导入另一个模块,但是Greg的方法更快:
>>> from timeit import Timer
>>> Timer('struct.unpack("<L", "y\xcc\xa6\xbb")[0]', 'import struct').timeit()
0.36242198944091797
>>> Timer("int('y\xcc\xa6\xbb'.encode('hex'), 16)").timeit()
1.1432669162750244
我的骇客方法:
>>> Timer("sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))").timeit()
2.8819329738616943
进一步更新:
有人在评论中问导入另一个模块有什么问题。好吧,导入模块不一定便宜,请看一下:
>>> Timer("""import struct\nstruct.unpack(">L", "y\xcc\xa6\xbb")[0]""").timeit()
0.98822188377380371
包括导入模块的成本,几乎抵消了此方法的所有优点。我认为,这仅包括在整个基准测试运行中一次导入一次的费用;看一下我每次强制重新加载时会发生什么:
>>> Timer("""reload(struct)\nstruct.unpack(">L", "y\xcc\xa6\xbb")[0]""", 'import struct').timeit()
68.474128007888794
不用说,如果您每次导入都执行此方法很多次,则成比例地减少了一个问题。也可能是I / O成本而不是CPU,因此它可能取决于特定计算机的容量和负载特性。
How can I convert a string of bytes into an int in python?
Say like this: 'y\xcc\xa6\xbb'
I came up with a clever/stupid way of doing it:
sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))
I know there has to be something builtin or in the standard library that does this more simply…
This is different from converting a string of hex digits for which you can use int(xxx, 16), but instead I want to convert a string of actual byte values.
UPDATE:
I kind of like James’ answer a little better because it doesn’t require importing another module, but Greg’s method is faster:
>>> from timeit import Timer
>>> Timer('struct.unpack("<L", "y\xcc\xa6\xbb")[0]', 'import struct').timeit()
0.36242198944091797
>>> Timer("int('y\xcc\xa6\xbb'.encode('hex'), 16)").timeit()
1.1432669162750244
My hacky method:
>>> Timer("sum(ord(c) << (i * 8) for i, c in enumerate('y\xcc\xa6\xbb'[::-1]))").timeit()
2.8819329738616943
FURTHER UPDATE:
Someone asked in comments what’s the problem with importing another module. Well, importing a module isn’t necessarily cheap, take a look:
>>> Timer("""import struct\nstruct.unpack(">L", "y\xcc\xa6\xbb")[0]""").timeit()
0.98822188377380371
Including the cost of importing the module negates almost all of the advantage that this method has. I believe that this will only include the expense of importing it once for the entire benchmark run; look what happens when I force it to reload every time:
>>> Timer("""reload(struct)\nstruct.unpack(">L", "y\xcc\xa6\xbb")[0]""", 'import struct').timeit()
68.474128007888794
Needless to say, if you’re doing a lot of executions of this method per one import than this becomes proportionally less of an issue. It’s also probably i/o cost rather than cpu so it may depend on the capacity and load characteristics of the particular machine.
回答 0
您还可以使用struct模块来执行此操作:
>>> struct.unpack("<L", "y\xcc\xa6\xbb")[0]
3148270713L
You can also use the struct module to do this:
>>> struct.unpack("<L", "y\xcc\xa6\xbb")[0]
3148270713L
回答 1
在Python 3.2和更高版本中,使用
>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='big')
2043455163
要么
>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='little')
3148270713
根据您的字节字符串的字节序。
这也适用于任意长度的字节字符串整数,并且通过指定,可用于以二进制补码的整数signed=True
。请参阅有关的文档from_bytes
。
In Python 3.2 and later, use
>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='big')
2043455163
or
>>> int.from_bytes(b'y\xcc\xa6\xbb', byteorder='little')
3148270713
according to the endianness of your byte-string.
This also works for bytestring-integers of arbitrary length, and for two’s-complement signed integers by specifying signed=True
. See the docs for from_bytes
.
回答 2
正如Greg所说的,如果要处理二进制值,则可以使用struct,但是如果您只有一个“十六进制数”,但是以字节格式,则可能需要将其转换为:
s = 'y\xcc\xa6\xbb'
num = int(s.encode('hex'), 16)
…与以下内容相同:
num = struct.unpack(">L", s)[0]
…除了适用于任何数量的字节。
As Greg said, you can use struct if you are dealing with binary values, but if you just have a “hex number” but in byte format you might want to just convert it like:
s = 'y\xcc\xa6\xbb'
num = int(s.encode('hex'), 16)
…this is the same as:
num = struct.unpack(">L", s)[0]
…except it’ll work for any number of bytes.
回答 3
我使用以下函数在int,hex和字节之间转换数据。
def bytes2int(str):
return int(str.encode('hex'), 16)
def bytes2hex(str):
return '0x'+str.encode('hex')
def int2bytes(i):
h = int2hex(i)
return hex2bytes(h)
def int2hex(i):
return hex(i)
def hex2int(h):
if len(h) > 1 and h[0:2] == '0x':
h = h[2:]
if len(h) % 2:
h = "0" + h
return int(h, 16)
def hex2bytes(h):
if len(h) > 1 and h[0:2] == '0x':
h = h[2:]
if len(h) % 2:
h = "0" + h
return h.decode('hex')
资料来源:http : //opentechnotes.blogspot.com.au/2014/04/convert-values-to-from-integer-hex.html
I use the following function to convert data between int, hex and bytes.
def bytes2int(str):
return int(str.encode('hex'), 16)
def bytes2hex(str):
return '0x'+str.encode('hex')
def int2bytes(i):
h = int2hex(i)
return hex2bytes(h)
def int2hex(i):
return hex(i)
def hex2int(h):
if len(h) > 1 and h[0:2] == '0x':
h = h[2:]
if len(h) % 2:
h = "0" + h
return int(h, 16)
def hex2bytes(h):
if len(h) > 1 and h[0:2] == '0x':
h = h[2:]
if len(h) % 2:
h = "0" + h
return h.decode('hex')
Source: http://opentechnotes.blogspot.com.au/2014/04/convert-values-to-from-integer-hex.html
回答 4
import array
integerValue = array.array("I", 'y\xcc\xa6\xbb')[0]
警告:以上内容是特定于平台的。“ I”说明符和string-> int转换的字节序都取决于您的特定Python实现。但是,如果要一次转换许多整数/字符串,则数组模块可以快速完成转换。
import array
integerValue = array.array("I", 'y\xcc\xa6\xbb')[0]
Warning: the above is strongly platform-specific. Both the “I” specifier and the endianness of the string->int conversion are dependent on your particular Python implementation. But if you want to convert many integers/strings at once, then the array module does it quickly.
回答 5
在Python 2.x中,您可以将格式说明符<B
用于无符号字节,以及<b
用于带struct.unpack
/的有符号字节struct.pack
。
例如:
令x
='\xff\x10\x11'
data_ints = struct.unpack('<' + 'B'*len(x), x) # [255, 16, 17]
和:
data_bytes = struct.pack('<' + 'B'*len(data_ints), *data_ints) # '\xff\x10\x11'
那*
是必须的!
看到 https://docs.python.org/2/library/struct.html#format-characters获取格式说明符列表。
In Python 2.x, you could use the format specifiers <B
for unsigned bytes, and <b
for signed bytes with struct.unpack
/struct.pack
.
E.g:
Let x
= '\xff\x10\x11'
data_ints = struct.unpack('<' + 'B'*len(x), x) # [255, 16, 17]
And:
data_bytes = struct.pack('<' + 'B'*len(data_ints), *data_ints) # '\xff\x10\x11'
That *
is required!
See https://docs.python.org/2/library/struct.html#format-characters for a list of the format specifiers.
回答 6
>>> reduce(lambda s, x: s*256 + x, bytearray("y\xcc\xa6\xbb"))
2043455163
测试1:逆:
>>> hex(2043455163)
'0x79cca6bb'
测试2:字节数> 8:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAAA"))
338822822454978555838225329091068225L
测试3:加1:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAAB"))
338822822454978555838225329091068226L
测试4:附加一个字节,说“ A”:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAABA"))
86738642548474510294585684247313465921L
测试5:除以256:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAABA"))/256
338822822454978555838225329091068226L
结果等于预期的测试4的结果。
>>> reduce(lambda s, x: s*256 + x, bytearray("y\xcc\xa6\xbb"))
2043455163
Test 1: inverse:
>>> hex(2043455163)
'0x79cca6bb'
Test 2: Number of bytes > 8:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAAA"))
338822822454978555838225329091068225L
Test 3: Increment by one:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAAB"))
338822822454978555838225329091068226L
Test 4: Append one byte, say ‘A’:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAABA"))
86738642548474510294585684247313465921L
Test 5: Divide by 256:
>>> reduce(lambda s, x: s*256 + x, bytearray("AAAAAAAAAAAAAABA"))/256
338822822454978555838225329091068226L
Result equals the result of Test 4, as expected.
回答 7
我一直在努力寻找适用于Python 2.x的任意长度字节序列的解决方案。最后,我写了这个,有点麻烦,因为它执行字符串转换,但是可以用。
Python 2.x的函数,任意长度
def signedbytes(data):
"""Convert a bytearray into an integer, considering the first bit as
sign. The data must be big-endian."""
negative = data[0] & 0x80 > 0
if negative:
inverted = bytearray(~d % 256 for d in data)
return -signedbytes(inverted) - 1
encoded = str(data).encode('hex')
return int(encoded, 16)
此功能有两个要求:
输入data
必须为bytearray
。您可以这样调用函数:
s = 'y\xcc\xa6\xbb'
n = signedbytes(s)
数据必须是大端的。如果您有一个小端值,则应首先将其取反:
n = signedbytes(s[::-1])
当然,仅在需要任意长度时才应使用此选项。否则,请遵循更多标准方法(例如struct
)。
I was struggling to find a solution for arbitrary length byte sequences that would work under Python 2.x. Finally I wrote this one, it’s a bit hacky because it performs a string conversion, but it works.
Function for Python 2.x, arbitrary length
def signedbytes(data):
"""Convert a bytearray into an integer, considering the first bit as
sign. The data must be big-endian."""
negative = data[0] & 0x80 > 0
if negative:
inverted = bytearray(~d % 256 for d in data)
return -signedbytes(inverted) - 1
encoded = str(data).encode('hex')
return int(encoded, 16)
This function has two requirements:
The input data
needs to be a bytearray
. You may call the function like this:
s = 'y\xcc\xa6\xbb'
n = signedbytes(s)
The data needs to be big-endian. In case you have a little-endian value, you should reverse it first:
n = signedbytes(s[::-1])
Of course, this should be used only if arbitrary length is needed. Otherwise, stick with more standard ways (e.g. struct
).
回答 8
如果版本> = 3.2,则int.from_bytes是最佳解决方案。“ struct.unpack”解决方案需要一个字符串,因此它不适用于字节数组。这是另一种解决方案:
def bytes2int( tb, order='big'):
if order == 'big': seq=[0,1,2,3]
elif order == 'little': seq=[3,2,1,0]
i = 0
for j in seq: i = (i<<8)+tb[j]
return i
hex(bytes2int([0x87,0x65,0x43,0x21]))返回’0x87654321’。
它处理大小字节序,很容易修改为8个字节
int.from_bytes is the best solution if you are at version >=3.2. The “struct.unpack” solution requires a string so it will not apply to arrays of bytes. Here is another solution:
def bytes2int( tb, order='big'):
if order == 'big': seq=[0,1,2,3]
elif order == 'little': seq=[3,2,1,0]
i = 0
for j in seq: i = (i<<8)+tb[j]
return i
hex( bytes2int( [0x87, 0x65, 0x43, 0x21])) returns ‘0x87654321’.
It handles big and little endianness and is easily modifiable for 8 bytes
回答 9
如上文使用所提unpack
的功能结构是一个很好的方式。如果要实现自己的功能,则还有另一种解决方案:
def bytes_to_int(bytes):
result = 0
for b in bytes:
result = result * 256 + int(b)
return result
As mentioned above using unpack
function of struct is a good way. If you want to implement your own function there is an another solution:
def bytes_to_int(bytes):
result = 0
for b in bytes:
result = result * 256 + int(b)
return result
回答 10
在python 3中,您可以通过以下方式轻松地将字节字符串转换为整数列表(0..255)
>>> list(b'y\xcc\xa6\xbb')
[121, 204, 166, 187]
In python 3 you can easily convert a byte string into a list of integers (0..255) by
>>> list(b'y\xcc\xa6\xbb')
[121, 204, 166, 187]
回答 11
一种使用array.array的快速方法,我已经使用了一段时间:
预定义变量:
offset = 0
size = 4
big = True # endian
arr = array('B')
arr.fromstring("\x00\x00\xff\x00") # 5 bytes (encoding issues) [0, 0, 195, 191, 0]
诠释为:(阅读)
val = 0
for v in arr[offset:offset+size][::pow(-1,not big)]: val = (val<<8)|v
来自int:(写)
val = 16384
arr[offset:offset+size] = \
array('B',((val>>(i<<3))&255 for i in range(size)))[::pow(-1,not big)]
这些可能会更快一些。
编辑:
对于某些数字,这是一项性能测试(Anaconda 2.3.0),与以下各项相比,显示出稳定的平均读数reduce()
:
========================= byte array to int.py =========================
5000 iterations; threshold of min + 5000ns:
______________________________________code___|_______min______|_______max______|_______avg______|_efficiency
⣿⠀⠀⠀⠀⡇⢀⡀⠀⠀⠀⠀⠀⠀⡇⠀⠀⠀⡀⠀⢰⠀⠀⠀⢰⠀⠀⠀⢸⠀⠀⢀⡇⠀⢀⠀⠀⠀⠀⢠⠀⠀⠀⠀⢰⠀⠀⠀⢸⡀⠀⠀⠀⢸⠀⡇⠀⠀⢠⠀⢰⠀⢸⠀
⣿⣦⣴⣰⣦⣿⣾⣧⣤⣷⣦⣤⣶⣾⣿⣦⣼⣶⣷⣶⣸⣴⣤⣀⣾⣾⣄⣤⣾⡆⣾⣿⣿⣶⣾⣾⣶⣿⣤⣾⣤⣤⣴⣼⣾⣼⣴⣤⣼⣷⣆⣴⣴⣿⣾⣷⣧⣶⣼⣴⣿⣶⣿⣶
val = 0 \nfor v in arr: val = (val<<8)|v | 5373.848ns | 850009.965ns | ~8649.64ns | 62.128%
⡇⠀⠀⢀⠀⠀⠀⡇⠀⡇⠀⠀⣠⠀⣿⠀⠀⠀⠀⡀⠀⠀⡆⠀⡆⢰⠀⠀⡆⠀⡄⠀⠀⠀⢠⢀⣼⠀⠀⡇⣠⣸⣤⡇⠀⡆⢸⠀⠀⠀⠀⢠⠀⢠⣿⠀⠀⢠⠀⠀⢸⢠⠀⡀
⣧⣶⣶⣾⣶⣷⣴⣿⣾⡇⣤⣶⣿⣸⣿⣶⣶⣶⣶⣧⣷⣼⣷⣷⣷⣿⣦⣴⣧⣄⣷⣠⣷⣶⣾⣸⣿⣶⣶⣷⣿⣿⣿⣷⣧⣷⣼⣦⣶⣾⣿⣾⣼⣿⣿⣶⣶⣼⣦⣼⣾⣿⣶⣷
val = reduce( shift, arr ) | 6489.921ns | 5094212.014ns | ~12040.269ns | 53.902%
这是原始性能测试,因此省略了endian pow-flip。
所shift
显示的函数与for循环应用相同的移位或运算,并且该函数的迭代性能arr
仅次于array.array('B',[0,0,255,0])
dict
。
我可能还应该注意到,效率是通过对平均时间的准确性来衡量的。
A decently speedy method utilizing array.array I’ve been using for some time:
predefined variables:
offset = 0
size = 4
big = True # endian
arr = array('B')
arr.fromstring("\x00\x00\xff\x00") # 5 bytes (encoding issues) [0, 0, 195, 191, 0]
to int: (read)
val = 0
for v in arr[offset:offset+size][::pow(-1,not big)]: val = (val<<8)|v
from int: (write)
val = 16384
arr[offset:offset+size] = \
array('B',((val>>(i<<3))&255 for i in range(size)))[::pow(-1,not big)]
It’s possible these could be faster though.
EDIT:
For some numbers, here’s a performance test (Anaconda 2.3.0) showing stable averages on read in comparison to reduce()
:
========================= byte array to int.py =========================
5000 iterations; threshold of min + 5000ns:
______________________________________code___|_______min______|_______max______|_______avg______|_efficiency
⣿⠀⠀⠀⠀⡇⢀⡀⠀⠀⠀⠀⠀⠀⡇⠀⠀⠀⡀⠀⢰⠀⠀⠀⢰⠀⠀⠀⢸⠀⠀⢀⡇⠀⢀⠀⠀⠀⠀⢠⠀⠀⠀⠀⢰⠀⠀⠀⢸⡀⠀⠀⠀⢸⠀⡇⠀⠀⢠⠀⢰⠀⢸⠀
⣿⣦⣴⣰⣦⣿⣾⣧⣤⣷⣦⣤⣶⣾⣿⣦⣼⣶⣷⣶⣸⣴⣤⣀⣾⣾⣄⣤⣾⡆⣾⣿⣿⣶⣾⣾⣶⣿⣤⣾⣤⣤⣴⣼⣾⣼⣴⣤⣼⣷⣆⣴⣴⣿⣾⣷⣧⣶⣼⣴⣿⣶⣿⣶
val = 0 \nfor v in arr: val = (val<<8)|v | 5373.848ns | 850009.965ns | ~8649.64ns | 62.128%
⡇⠀⠀⢀⠀⠀⠀⡇⠀⡇⠀⠀⣠⠀⣿⠀⠀⠀⠀⡀⠀⠀⡆⠀⡆⢰⠀⠀⡆⠀⡄⠀⠀⠀⢠⢀⣼⠀⠀⡇⣠⣸⣤⡇⠀⡆⢸⠀⠀⠀⠀⢠⠀⢠⣿⠀⠀⢠⠀⠀⢸⢠⠀⡀
⣧⣶⣶⣾⣶⣷⣴⣿⣾⡇⣤⣶⣿⣸⣿⣶⣶⣶⣶⣧⣷⣼⣷⣷⣷⣿⣦⣴⣧⣄⣷⣠⣷⣶⣾⣸⣿⣶⣶⣷⣿⣿⣿⣷⣧⣷⣼⣦⣶⣾⣿⣾⣼⣿⣿⣶⣶⣼⣦⣼⣾⣿⣶⣷
val = reduce( shift, arr ) | 6489.921ns | 5094212.014ns | ~12040.269ns | 53.902%
This is a raw performance test, so the endian pow-flip is left out.
The shift
function shown applies the same shift-oring operation as the for loop, and arr
is just array.array('B',[0,0,255,0])
as it has the fastest iterative performance next to dict
.
I should probably also note efficiency is measured by accuracy to the average time.
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