在 Python 3 中将 int 转换为字节

Question

I was trying to build this bytes object in Python 3:

b'3\r\n'

so I tried the obvious (for me), and found a weird behaviour:

>>> bytes(3) + b'\r\n'
b'\x00\x00\x00\r\n'

Apparently:

>>> bytes(10)
b'\x00\x00\x00\x00\x00\x00\x00\x00\x00\x00'

I've been unable to see any pointers on why the bytes conversion works this way reading the documentation. However, I did find some surprise messages in this Python issue about adding format to bytes (see also Python 3 bytes formatting ):

http://bugs.python.org/issue3982

This interacts even more poorly with oddities like bytes(int) returning zeroes now

and:

It would be much more convenient for me if bytes(int) returned the ASCIIfication of that int; but honestly, even an error would be better than this behavior. (If I wanted this behavior - which I never have - I'd rather it be a classmethod, invoked like "bytes.zeroes(n)".)

Can someone explain me where this behaviour comes from?

Answer 1

From python 3.2 you can do

>>> (1024).to_bytes(2, byteorder='big')
b'\x04\x00'

https://docs.python.org/3/library/stdtypes.html#int.to_bytes

def int_to_bytes(x: int) -> bytes:
    return x.to_bytes((x.bit_length() + 7) // 8, 'big')
    
def int_from_bytes(xbytes: bytes) -> int:
    return int.from_bytes(xbytes, 'big')

Accordingly, x == int_from_bytes(int_to_bytes(x)) . Note that the above encoding works only for unsigned (non-negative) integers.

For signed integers, the bit length is a bit more tricky to calculate:

def int_to_bytes(number: int) -> bytes:
    return number.to_bytes(length=(8 + (number + (number < 0)).bit_length()) // 8, byteorder='big', signed=True)

def int_from_bytes(binary_data: bytes) -> Optional[int]:
    return int.from_bytes(binary_data, byteorder='big', signed=True)

Answer 2

That's the way it was designed - and it makes sense because usually, you would call bytes on an iterable instead of a single integer:

>>> bytes([3])
b'\x03'

The docs state this , as well as the docstring for bytes :

 >>> help(bytes)
 ...
 bytes(int) -> bytes object of size given by the parameter initialized with null bytes

Answer 3

You can use the struct's pack :

In [11]: struct.pack(">I", 1)
Out[11]: '\x00\x00\x00\x01'

The ">" is the byte-order (big-endian) and the "I" is theformat character . So you can be specific if you want to do something else:

In [12]: struct.pack("<H", 1)
Out[12]: '\x01\x00'

In [13]: struct.pack("B", 1)
Out[13]: '\x01'

This works the same on both python 2 and python 3 .

Note: the inverse operation (bytes to int) can be done with unpack .

Answer 4

Python 3.5+ introduces %-interpolation ( printf -style formatting) for bytes :

>>> b'%d\r\n' % 3
b'3\r\n'

See PEP 0461 -- Adding % formatting to bytes and bytearray .

On earlier versions, you could use str and .encode('ascii') the result:

>>> s = '%d\r\n' % 3
>>> s.encode('ascii')
b'3\r\n'

Note: It is different from what int.to_bytes produces :

>>> n = 3
>>> n.to_bytes((n.bit_length() + 7) // 8, 'big') or b'\0'
b'\x03'
>>> b'3' == b'\x33' != '\x03'
True

Answer 5

The documentation says:

bytes(int) -> bytes object of size given by the parameter
              initialized with null bytes

The sequence:

b'3\r\n'

It is the character '3' (decimal 51) the character '\\r' (13) and '\\n' (10).

Therefore, the way would treat it as such, for example:

>>> bytes([51, 13, 10])
b'3\r\n'

>>> bytes('3', 'utf8') + b'\r\n'
b'3\r\n'

>>> n = 3
>>> bytes(str(n), 'ascii') + b'\r\n'
b'3\r\n'

Tested on IPython 1.1.0 & Python 3.2.3

Answer 6

The ASCIIfication of 3 is "\\x33" not "\\x03" !

That is what python does for str(3) but it would be totally wrong for bytes, as they should be considered arrays of binary data and not be abused as strings.

The most easy way to achieve what you want is bytes((3,)) , which is better than bytes([3]) because initializing a list is much more expensive, so never use lists when you can use tuples. You can convert bigger integers by using int.to_bytes(3, "little") .

Initializing bytes with a given length makes sense and is the most useful, as they are often used to create some type of buffer for which you need some memory of given size allocated. I often use this when initializing arrays or expanding some file by writing zeros to it.

Answer 7

int (including Python2's long ) can be converted to bytes using following function:

import codecs

def int2bytes(i):
    hex_value = '{0:x}'.format(i)
    # make length of hex_value a multiple of two
    hex_value = '0' * (len(hex_value) % 2) + hex_value
    return codecs.decode(hex_value, 'hex_codec')

The reverse conversion can be done by another one:

import codecs
import six  # should be installed via 'pip install six'

long = six.integer_types[-1]

def bytes2int(b):
    return long(codecs.encode(b, 'hex_codec'), 16)

Both functions work on both Python2 and Python3.

Answer 8

I was curious about performance of various methods for a single int in the range [0, 255] , so I decided to do some timing tests.

Based on the timings below, and from the general trend I observed from trying many different values and configurations, struct.pack seems to be the fastest, followed by int.to_bytes , bytes , and with str.encode (unsurprisingly) being the slowest. Note that the results show some more variation than is represented, and int.to_bytes and bytes sometimes switched speed ranking during testing, but struct.pack is clearly the fastest.

Results in CPython 3.7 on Windows:

Testing with 63:
bytes_: 100000 loops, best of 5: 3.3 usec per loop
to_bytes: 100000 loops, best of 5: 2.72 usec per loop
struct_pack: 100000 loops, best of 5: 2.32 usec per loop
chr_encode: 50000 loops, best of 5: 3.66 usec per loop

Test module (named int_to_byte.py ):

"""Functions for converting a single int to a bytes object with that int's value."""

import random
import shlex
import struct
import timeit

def bytes_(i):
    """From Tim Pietzcker's answer:
    https://stackoverflow.com/a/21017834/8117067
    """
    return bytes([i])

def to_bytes(i):
    """From brunsgaard's answer:
    https://stackoverflow.com/a/30375198/8117067
    """
    return i.to_bytes(1, byteorder='big')

def struct_pack(i):
    """From Andy Hayden's answer:
    https://stackoverflow.com/a/26920966/8117067
    """
    return struct.pack('B', i)

# Originally, jfs's answer was considered for testing,
# but the result is not identical to the other methods
# https://stackoverflow.com/a/31761722/8117067

def chr_encode(i):
    """Another method, from Quuxplusone's answer here:
    https://codereview.stackexchange.com/a/210789/140921

    Similar to g10guang's answer:
    https://stackoverflow.com/a/51558790/8117067
    """
    return chr(i).encode('latin1')

converters = [bytes_, to_bytes, struct_pack, chr_encode]

def one_byte_equality_test():
    """Test that results are identical for ints in the range [0, 255]."""
    for i in range(256):
        results = [c(i) for c in converters]
        # Test that all results are equal
        start = results[0]
        if any(start != b for b in results):
            raise ValueError(results)

def timing_tests(value=None):
    """Test each of the functions with a random int."""
    if value is None:
        # random.randint takes more time than int to byte conversion
        # so it can't be a part of the timeit call
        value = random.randint(0, 255)
    print(f'Testing with {value}:')
    for c in converters:
        print(f'{c.__name__}: ', end='')
        # Uses technique borrowed from https://stackoverflow.com/q/19062202/8117067
        timeit.main(args=shlex.split(
            f"-s 'from int_to_byte import {c.__name__}; value = {value}' " +
            f"'{c.__name__}(value)'"
        ))

Answer 9

Although the prior answer by brunsgaard is an efficient encoding, it works only for unsigned integers. This one builds upon it to work for both signed and unsigned integers.

def int_to_bytes(i: int, *, signed: bool = False) -> bytes:
    length = ((i + ((i * signed) < 0)).bit_length() + 7 + signed) // 8
    return i.to_bytes(length, byteorder='big', signed=signed)

def bytes_to_int(b: bytes, *, signed: bool = False) -> int:
    return int.from_bytes(b, byteorder='big', signed=signed)

# Test unsigned:
for i in range(1025):
    assert i == bytes_to_int(int_to_bytes(i))

# Test signed:
for i in range(-1024, 1025):
    assert i == bytes_to_int(int_to_bytes(i, signed=True), signed=True)

For the encoder, (i + ((i * signed) < 0)).bit_length() is used instead of just i.bit_length() because the latter leads to an inefficient encoding of -128, -32768, etc.

---

Credit: CervEd for fixing a minor inefficiency.

Answer 10

From bytes docs :

Accordingly, constructor arguments are interpreted as for bytearray().

Then, from bytearray docs :

The optional source parameter can be used to initialize the array in a few different ways:

• If it is an integer, the array will have that size and will be initialized with null bytes.

Note, that differs from 2.x (where x >= 6) behavior, where bytes is simply str :

>>> bytes is str
True

PEP 3112 :

The 2.6 str differs from 3.0's bytes type in various ways; most notably, the constructor is completely different.

Answer 11

The behaviour comes from the fact that in Python prior to version 3 bytes was just an alias for str . In Python3.x bytes is an immutable version of bytearray - completely new type, not backwards compatible.

Answer 12

Some answers don't work with large numbers.

Convert integer to the hex representation, then convert it to bytes:

def int_to_bytes(number):
    hrepr = hex(number).replace('0x', '')
    if len(hrepr) % 2 == 1:
        hrepr = '0' + hrepr
    return bytes.fromhex(hrepr)

Result:

>>> int_to_bytes(2**256 - 1)
b'\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff\xff'

Answer 13

As you want to deal with binary representation, the best is to use ctypes .

import ctypes
x = ctypes.c_int(1234)
bytes(x)

You must use the specific integer representation (signed/unsigned and the number of bits: c_uint8 , c_int8 , c_unit16 ,...).

Answer 14

I think you can convert the int to str first, before you convert to byte. That should produce the format you want.

bytes(str(your_number),'UTF-8') + b'\r\n'

It works for me in py3.8.

Answer 15

>>> chr(116).encode()
b't'

Answer 16

If you don't care about the performance, you can convert an int to str first.

number = 1024
str(number).encode()

Answer 17

If the question is how to convert an integer itself (not its string equivalent) into bytes, I think the robust answer is:

>>> i = 5
>>> i.to_bytes(2, 'big')
b'\x00\x05'
>>> int.from_bytes(i.to_bytes(2, 'big'), byteorder='big')
5

More information on these methods here:

1. https://docs.python.org/3.8/library/stdtypes.html#int.to_bytes
2. https://docs.python.org/3.8/library/stdtypes.html#int.from_bytes