Reputation:
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?
Upvotes: 288
Views: 687518
Reputation: 3843
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)'"
))
Upvotes: 4
Reputation: 943
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
Upvotes: 12
Reputation: 336158
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
Upvotes: 258
Reputation: 414265
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' != b'\x03'
True
Upvotes: 31
Reputation: 63312
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.
Upvotes: 6
Reputation: 5168
From python 3.2 you can use to_bytes
:
>>> (1024).to_bytes(2, byteorder='big')
b'\x04\x00'
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)
Upvotes: 354
Reputation: 248
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.
Upvotes: 0
Reputation: 49
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
,...).
Upvotes: 4
Reputation: 31555
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'
Upvotes: 0
Reputation: 1400
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.
Upvotes: 4
Reputation: 1325
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.
Upvotes: 8
Reputation: 375535
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 the format 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.
Upvotes: 52
Reputation: 48317
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
The 2.6 str differs from 3.0’s bytes type in various ways; most notably, the constructor is completely different.
Upvotes: 3
Reputation: 56477
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.
Upvotes: 4