I once read the following function in a given program, but I am not very clear about what is this function used for? According SciPy.org,
dtype.newbyteorder(new_order='S')
Return a new dtype with a different byte order.
I do not quite understand what does it mean?
def _read32(bytestream):
dt = numpy.dtype(numpy.uint32).newbyteorder('>')
return numpy.frombuffer(bytestream.read(4), dtype=dt)[0]
Here's a simple example:
>>> dt = np.dtype(np.uint32)
>>> val = np.frombuffer(b'\x01\x02\x03\x04', dtype=dt)
>>> hex(val)
'0x4030201'
>>> val2 = np.frombuffer(b'\x01\x02\x03\x04', dtype=dt.newbyteorder('>'))
>>> hex(val2)
'0x1020304'
Byte order describes which order to pack a series of bytes (in this case, a bytes object from a b"..." literal) into the data type you asked for.
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?
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)
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
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.
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
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
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.
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)'"
))
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.
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.
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.
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.
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,...).
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'
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.
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:
https://docs.python.org/3.8/library/stdtypes.html#int.to_bytes
https://docs.python.org/3.8/library/stdtypes.html#int.from_bytes
>>> chr(116).encode()
b't'
I have the following code written in Matlab:
>>> fid = fopen('filename.bin', 'r', 'b')
>>> %separated r and b
>>> dim = fread(dim, 2, 'uint32');
if I use a "equivalent" code in Python
>>> fid = open('filename.bin', 'rb')
>>> dim = np.fromfile(fid, dtype=np.uint32)
I got a different value of dim when I use Python.
Someone knows how to open this file with permission like Matlab ('r' and 'b' separated) in Python?
Thanks in advance,
Rhenan
From Matlab docs I learn that your third parameter 'b' stands for Big-Endian ordering, is not a permission.
Most probably Numpy uses the little-endian order on your machine. To fix the problem try to specify explicitly the ordering in Numpy (as you do in Matlab):
>>> fid = open('filename.bin', 'rb')
>>> dim = np.fromfile(fid, dtype='>u4')
the dtype string stands for Big-Endian ('>'), unsigned integer ('u'), 4-bytes number.
See also Data type objects (dtype) in Numpy reference.
When iterating over a bytes object in Python 3, one gets the individual bytes as ints:
>>> [b for b in b'123']
[49, 50, 51]
How to get 1-length bytes objects instead?
The following is possible, but not very obvious for the reader and most likely performs bad:
>>> [bytes([b]) for b in b'123']
[b'1', b'2', b'3']
If you are concerned about performance of this code and an int as a byte is not suitable interface in your case then you should probably reconsider data structures that you use e.g., use str objects instead.
You could slice the bytes object to get 1-length bytes objects:
L = [bytes_obj[i:i+1] for i in range(len(bytes_obj))]
There is PEP 0467 -- Minor API improvements for binary sequences that proposes bytes.iterbytes() method:
>>> list(b'123'.iterbytes())
[b'1', b'2', b'3']
int.to_bytes
int objects have a to_bytes method which can be used to convert an int to its corresponding byte:
>>> import sys
>>> [i.to_bytes(1, sys.byteorder) for i in b'123']
[b'1', b'2', b'3']
As with some other other answers, it's not clear that this is more readable than the OP's original solution: the length and byteorder arguments make it noisier I think.
struct.unpack
Another approach would be to use struct.unpack, though this might also be considered difficult to read, unless you are familiar with the struct module:
>>> import struct
>>> struct.unpack('3c', b'123')
(b'1', b'2', b'3')
(As jfs observes in the comments, the format string for struct.unpack can be constructed dynamically; in this case we know the number of individual bytes in the result must equal the number of bytes in the original bytestring, so struct.unpack(str(len(bytestring)) + 'c', bytestring) is possible.)
Performance
>>> import random, timeit
>>> bs = bytes(random.randint(0, 255) for i in range(100))
>>> # OP's solution
>>> timeit.timeit(setup="from __main__ import bs",
stmt="[bytes([b]) for b in bs]")
46.49886950897053
>>> # Accepted answer from jfs
>>> timeit.timeit(setup="from __main__ import bs",
stmt="[bs[i:i+1] for i in range(len(bs))]")
20.91463226894848
>>> # Leon's answer
>>> timeit.timeit(setup="from __main__ import bs",
stmt="list(map(bytes, zip(bs)))")
27.476876026019454
>>> # guettli's answer
>>> timeit.timeit(setup="from __main__ import iter_bytes, bs",
stmt="list(iter_bytes(bs))")
24.107485140906647
>>> # user38's answer (with Leon's suggested fix)
>>> timeit.timeit(setup="from __main__ import bs",
stmt="[chr(i).encode('latin-1') for i in bs]")
45.937552741961554
>>> # Using int.to_bytes
>>> timeit.timeit(setup="from __main__ import bs;from sys import byteorder",
stmt="[x.to_bytes(1, byteorder) for x in bs]")
32.197659170022234
>>> # Using struct.unpack, converting the resulting tuple to list
>>> # to be fair to other methods
>>> timeit.timeit(setup="from __main__ import bs;from struct import unpack",
stmt="list(unpack('100c', bs))")
1.902243083808571
struct.unpack seems to be at least an order of magnitude faster than other methods, presumably because it operates at the byte level. int.to_bytes, on the other hand, performs worse than most of the "obvious" approaches.
I thought it might be useful to compare the runtimes of the different approaches so I made a benchmark (using my library simple_benchmark):
Probably unsurprisingly the NumPy solution is by far the fastest solution for large bytes object.
But if a resulting list is desired then both the NumPy solution (with the tolist()) and the struct solution are much faster than the other alternatives.
I didn't include guettlis answer because it's almost identical to jfs solution just instead of a comprehension a generator function is used.
import numpy as np
import struct
import sys
from simple_benchmark import BenchmarkBuilder
b = BenchmarkBuilder()
#b.add_function()
def jfs(bytes_obj):
return [bytes_obj[i:i+1] for i in range(len(bytes_obj))]
#b.add_function()
def snakecharmerb_tobytes(bytes_obj):
return [i.to_bytes(1, sys.byteorder) for i in bytes_obj]
#b.add_function()
def snakecharmerb_struct(bytes_obj):
return struct.unpack(str(len(bytes_obj)) + 'c', bytes_obj)
#b.add_function()
def Leon(bytes_obj):
return list(map(bytes, zip(bytes_obj)))
#b.add_function()
def rusu_ro1_format(bytes_obj):
return [b'%c' % i for i in bytes_obj]
#b.add_function()
def rusu_ro1_numpy(bytes_obj):
return np.frombuffer(bytes_obj, dtype='S1')
#b.add_function()
def rusu_ro1_numpy_tolist(bytes_obj):
return np.frombuffer(bytes_obj, dtype='S1').tolist()
#b.add_function()
def User38(bytes_obj):
return [chr(i).encode() for i in bytes_obj]
#b.add_arguments('byte object length')
def argument_provider():
for exp in range(2, 18):
size = 2**exp
yield size, b'a' * size
r = b.run()
r.plot()
since python 3.5 you can use % formatting to bytes and bytearray:
[b'%c' % i for i in b'123']
output:
[b'1', b'2', b'3']
the above solution is 2-3 times faster than your initial approach, if you want a more fast solution I will suggest to use numpy.frombuffer:
import numpy as np
np.frombuffer(b'123', dtype='S1')
output:
array([b'1', b'2', b'3'],
dtype='|S1')
The second solution is ~10% faster than struct.unpack (I have used the same performance test as #snakecharmerb, against 100 random bytes)
A trio of map(), bytes() and zip() does the trick:
>>> list(map(bytes, zip(b'123')))
[b'1', b'2', b'3']
However I don't think that it is any more readable than [bytes([b]) for b in b'123'] or performs better.
I use this helper method:
def iter_bytes(my_bytes):
for i in range(len(my_bytes)):
yield my_bytes[i:i+1]
Works for Python2 and Python3.
A short way to do this:
[bytes([i]) for i in b'123\xaa\xbb\xcc\xff']
The pickle module seems to use string escape characters when pickling; this becomes inefficient e.g. on numpy arrays. Consider the following
z = numpy.zeros(1000, numpy.uint8)
len(z.dumps())
len(cPickle.dumps(z.dumps()))
The lengths are 1133 characters and 4249 characters respectively.
z.dumps() reveals something like "\x00\x00" (actual zeros in string), but pickle seems to be using the string's repr() function, yielding "'\x00\x00'" (zeros being ascii zeros).
i.e. ("0" in z.dumps() == False) and ("0" in cPickle.dumps(z.dumps()) == True)
Try using a later version of the pickle protocol with the protocol parameter to pickle.dumps(). The default is 0 and is an ASCII text format. Ones greater than 1 (I suggest you use pickle.HIGHEST_PROTOCOL). Protocol formats 1 and 2 (and 3 but that's for py3k) are binary and should be more space conservative.
Solution:
import zlib, cPickle
def zdumps(obj):
return zlib.compress(cPickle.dumps(obj,cPickle.HIGHEST_PROTOCOL),9)
def zloads(zstr):
return cPickle.loads(zlib.decompress(zstr))
>>> len(zdumps(z))
128
z.dumps() is already pickled string i.e., it can be unpickled using pickle.loads():
>>> z = numpy.zeros(1000, numpy.uint8)
>>> s = z.dumps()
>>> a = pickle.loads(s)
>>> all(a == z)
True
An improvement to vartec's answer, that seems a bit more memory efficient (since it doesn't force everything into a string):
def pickle(fname, obj):
import cPickle, gzip
cPickle.dump(obj=obj, file=gzip.open(fname, "wb", compresslevel=3), protocol=2)
def unpickle(fname):
import cPickle, gzip
return cPickle.load(gzip.open(fname, "rb"))