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This is python3 code:
>>> bytes(json.dumps({'Ä':0}), "utf-8")
b'{"\\u00c4": 0}'
json.dumps() returns unicode string and bytes() returns its' bytes representation - string encoded into utf-8.
How do I achieve the same result in Lua? I need a bytes representation of a json object which contains non-ascii chars.
You have to do it manually.
local function utf8_to_unicode(utf8str, pos)
local code, size = utf8str:byte(pos), 1
if code >= 0xC0 and code < 0xFE then
local mask = 64
code = code - 128
repeat
local next_byte = utf8str:byte(pos + size) or 0
if next_byte >= 0x80 and next_byte < 0xC0 then
code, size = (code - mask - 2) * 64 + next_byte, size + 1
else
code, size = utf8str:byte(pos), 1
end
mask = mask * 32
until code < mask
end
-- returns code, number of bytes in this utf8 char
return code, size
end
function utf8_to_python(utf8str)
local pos = 1
local z = ''
while pos <= #utf8str do
local unicode, size = utf8_to_unicode(utf8str, pos)
pos = pos + size
if unicode < 0x80 then
z = z..string.char(unicode)
elseif unicode < 0x10000 then
z = z..string.format('\\\\u%04x', unicode)
else
z = z..string.format('\\\\U%08x', unicode)
end
end
return z
end
Usage:
local json = require('json')
local x = {['Ä'] = 0}
local y = json.encode(x)
print(y) --> {"Ä":0}
local z = utf8_to_python(y)
print(z) --> {"\\u00c4":0}
A simpler version using string.gsub:
local function python_escape(str)
return (string.gsub(
str,
-- leading byte followed by one or more continuation bytes;
-- decimal version for Lua 5.1: "[\194-\244][\128-\191]+",
"[\xC2-\xF4][\x80-\xBF]+"
function (non_ASCII)
local codepoint = utf8.codepoint(non_ASCII)
if codepoint <= 0xFFFF then
return ("\\u%04x"):format(codepoint)
else
return ("\\U%08x"):format(codepoint)
end
end))
end
I put parentheses around the return value (string.gsub(--[[...]])) to strip away the second return value of string.gsub (the number of replacements).
I'm reading a binary file in python and the documentation for the file format says:
Flag (in binary)Meaning
1 nnn nnnn Indicates that there is one data byte to follow
that is to be duplicated nnn nnnn (127 maximum)
times.
0 nnn nnnn Indicates that there are nnn nnnn bytes of image
data to follow (127 bytes maximum) and that
there are no duplications.
n 000 0000 End of line field. Indicates the end of a line
record. The value of n may be either zero or one.
Note that the end of line field is required and
that it is reflected in the length of line record
field mentioned above.
When reading the file I'm expecting the byte I'm at to return 1 nnn nnnn where the nnn nnnn part should be 50.
I've been able to do this using the following:
flag = byte >> 7
numbytes = int(bin(byte)[3:], 2)
But the numbytes calculation feels like a cheap workaround.
Can I do more bit math to accomplish the calculation of numbytes?
How would you approach this?
The classic approach of checking whether a bit is set, is to use binary "and" operator, i.e.
x = 10 # 1010 in binary
if x & 0b10: # explicitly: x & 0b0010 != 0
print('First bit is set')
To check, whether n^th bit is set, use the power of two, or better bit shifting
def is_set(x, n):
return x & 2 ** n != 0
# a more bitwise- and performance-friendly version:
return x & 1 << n != 0
is_set(10, 1) # 1 i.e. first bit - as the count starts at 0-th bit
>>> True
You can strip off the leading bit using a mask ANDed with a byte from file. That will leave you with the value of the remaining bits:
mask = 0b01111111
byte_from_file = 0b10101010
value = mask & byte_from_file
print bin(value)
>> 0b101010
print value
>> 42
I find the binary numbers easier to understand than hex when doing bit-masking.
EDIT: Slightly more complete example for your use case:
LEADING_BIT_MASK = 0b10000000
VALUE_MASK = 0b01111111
values = [0b10101010, 0b01010101, 0b0000000, 0b10000000]
for v in values:
value = v & VALUE_MASK
has_leading_bit = v & LEADING_BIT_MASK
if value == 0:
print "EOL"
elif has_leading_bit:
print "leading one", value
elif not has_leading_bit:
print "leading zero", value
If I read your description correctly:
if (byte & 0x80) != 0:
num_bytes = byte & 0x7F
there you go:
class ControlWord(object):
"""Helper class to deal with control words.
Bit setting and checking methods are implemented.
"""
def __init__(self, value = 0):
self.value = int(value)
def set_bit(self, bit):
self.value |= bit
def check_bit(self, bit):
return self.value & bit != 0
def clear_bit(self, bit):
self.value &= ~bit
Instead of int(bin(byte)[3:], 2), you could simply use: int(bin(byte>>1),2)
not sure I got you correctly, but if I did, this should do the trick:
>>> x = 154 #just an example
>>> flag = x >> 1
>>> flag
1
>>> nb = x & 127
>>> nb
26
You can do it like this:
def GetVal(b):
# mask off the most significant bit, see if it's set
flag = b & 0x80 == 0x80
# then look at the lower 7 bits in the byte.
count = b & 0x7f
# return a tuple indicating the state of the high bit, and the
# remaining integer value without the high bit.
return (flag, count)
>>> testVal = 50 + 0x80
>>> GetVal(testVal)
(True, 50)
I have a file where the first byte contains encoded information. In Matlab I can read the byte bit by bit with var = fread(file, 8, 'ubit1'), and then retrieve each bit by var(1), var(2), etc.
Is there any equivalent bit reader in python?
Read the bits from a file, low bits first.
def bits(f):
bytes = (ord(b) for b in f.read())
for b in bytes:
for i in xrange(8):
yield (b >> i) & 1
for b in bits(open('binary-file.bin', 'r')):
print b
The smallest unit you'll be able to work with is a byte. To work at the bit level you need to use bitwise operators.
x = 3
#Check if the 1st bit is set:
x&1 != 0
#Returns True
#Check if the 2nd bit is set:
x&2 != 0
#Returns True
#Check if the 3rd bit is set:
x&4 != 0
#Returns False
With numpy it is easy like this:
Bytes = numpy.fromfile(filename, dtype = "uint8")
Bits = numpy.unpackbits(Bytes)
More info here:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.fromfile.html
You won't be able to read each bit one by one - you have to read it byte by byte. You can easily extract the bits out, though:
f = open("myfile", 'rb')
# read one byte
byte = f.read(1)
# convert the byte to an integer representation
byte = ord(byte)
# now convert to string of 1s and 0s
byte = bin(byte)[2:].rjust(8, '0')
# now byte contains a string with 0s and 1s
for bit in byte:
print bit
Joining some of the previous answers I would use:
[int(i) for i in "{0:08b}".format(byte)]
For each byte read from the file. The results for an 0x88 byte example is:
>>> [int(i) for i in "{0:08b}".format(0x88)]
[1, 0, 0, 0, 1, 0, 0, 0]
You can assign it to a variable and work as per your initial request.
The "{0.08}" is to guarantee the full byte length
To read a byte from a file: bytestring = open(filename, 'rb').read(1). Note: the file is opened in the binary mode.
To get bits, convert the bytestring into an integer: byte = bytestring[0] (Python 3) or byte = ord(bytestring[0]) (Python 2) and extract the desired bit: (byte >> i) & 1:
>>> for i in range(8): (b'a'[0] >> i) & 1
...
1
0
0
0
0
1
1
0
>>> bin(b'a'[0])
'0b1100001'
There are two possible ways to return the i-th bit of a byte. The "first bit" could refer to the high-order bit or it could refer to the lower order bit.
Here is a function that takes a string and index as parameters and returns the value of the bit at that location. As written, it treats the low-order bit as the first bit. If you want the high order bit first, just uncomment the indicated line.
def bit_from_string(string, index):
i, j = divmod(index, 8)
# Uncomment this if you want the high-order bit first
# j = 8 - j
if ord(string[i]) & (1 << j):
return 1
else:
return 0
The indexing starts at 0. If you want the indexing to start at 1, you can adjust index in the function before calling divmod.
Example usage:
>>> for i in range(8):
>>> print i, bit_from_string('\x04', i)
0 0
1 0
2 1
3 0
4 0
5 0
6 0
7 0
Now, for how it works:
A string is composed of 8-bit bytes, so first we use divmod() to break the index into to parts:
i: the index of the correct byte within the string
j: the index of the correct bit within that byte
We use the ord() function to convert the character at string[i] into an integer type. Then, (1 << j) computes the value of the j-th bit by left-shifting 1 by j. Finally, we use bitwise-and to test if that bit is set. If so return 1, otherwise return 0.
Supposing you have a file called bloom_filter.bin which contains an array of bits and you want to read the entire file and use those bits in an array.
First create the array where the bits will be stored after reading,
from bitarray import bitarray
a=bitarray(size) #same as the number of bits in the file
Open the file,
using open or with, anything is fine...I am sticking with open here,
f=open('bloom_filter.bin','rb')
Now load all the bits into the array 'a' at one shot using,
f.readinto(a)
'a' is now a bitarray containing all the bits
This is pretty fast I would think:
import itertools
data = range(10)
format = "{:0>8b}".format
newdata = (False if n == '0' else True for n in itertools.chain.from_iterable(map(format, data)))
print(newdata) # prints tons of True and False
I think this is a more pythonic way:
a = 140
binary = format(a, 'b')
The result of this block is:
'10001100'
I was to get bit planes of the image and this function helped me to write this block:
def img2bitmap(img: np.ndarray) -> list:
if img.dtype != np.uint8 or img.ndim > 2:
raise ValueError("Image is not uint8 or gray")
bit_mat = [np.zeros(img.shape, dtype=np.uint8) for _ in range(8)]
for row_number in range(img.shape[0]):
for column_number in range(img.shape[1]):
binary = format(img[row_number][column_number], 'b')
for idx, bit in enumerate("".join(reversed(binary))[:]):
bit_mat[idx][row_number, column_number] = 2 ** idx if int(bit) == 1 else 0
return bit_mat
Also by this block, I was able to make primitives image from extracted bit planes
img = cv2.imread('test.jpg', cv2.IMREAD_GRAYSCALE)
out = img2bitmap(img)
original_image = np.zeros(img.shape, dtype=np.uint8)
for i in range(original_image.shape[0]):
for j in range(original_image.shape[1]):
for data in range(8):
x = np.array([original_image[i, j]], dtype=np.uint8)
data = np.array([data], dtype=np.uint8)
flag = np.array([0 if out[data[0]][i, j] == 0 else 1], dtype=np.uint8)
mask = flag << data[0]
x[0] = (x[0] & ~mask) | ((flag[0] << data[0]) & mask)
original_image[i, j] = x[0]
I'm trying to slowly knock out all of the intricacies of python. Basically, I'm looking for some way, in python, to take a string of characters and push them all over by 'x' characters.
For example, inputing abcdefg will give me cdefghi (if x is 2).
My first version:
>>> key = 2
>>> msg = "abcdefg"
>>> ''.join( map(lambda c: chr(ord('a') + (ord(c) - ord('a') + key)%26), msg) )
'cdefghi'
>>> msg = "uvwxyz"
>>> ''.join( map(lambda c: chr(ord('a') + (ord(c) - ord('a') + key)%26), msg) )
'wxyzab'
(Of course it works as expected only if msg is lowercase...)
edit: I definitely second David Raznick's answer:
>>> import string
>>> alphabet = "abcdefghijklmnopqrstuvwxyz"
>>> key = 2
>>> tr = string.maketrans(alphabet, alphabet[key:] + alphabet[:key])
>>> "abcdefg".translate(tr)
'cdefghi'
I think your best bet is to look at string.translate. You may have to use make_trans to make the mapping you like.
I would do it this way (for conceptual simplicity):
def encode(s):
l = [ord(i) for i in s]
return ''.join([chr(i + 2) for i in l])
Point being that you convert the letter to ASCII, add 2 to that code, convert it back, and "cast" it into a string (create a new string object). This also makes no conversions based on "case" (upper vs. lower).
Potential optimizations/research areas:
Use of StringIO module for large strings
Apply this to Unicode (not sure how)
This solution works for both lowercase and uppercase:
from string import lowercase, uppercase
def caesar(text, key):
result = []
for c in text:
if c in lowercase:
idx = lowercase.index(c)
idx = (idx + key) % 26
result.append(lowercase[idx])
elif c in uppercase:
idx = uppercase.index(c)
idx = (idx + key) % 26
result.append(uppercase[idx])
else:
result.append(c)
return "".join(result)
Here is a test:
>>> caesar("abcdefg", 2)
'cdefghi'
>>> caesar("z", 1)
'a'
Another version. Allows for definition of your own alphabet, and doesn't translate any other characters (such as punctuation). The ugly part here is the loop, which might cause performance problems. I'm not sure about python but appending strings like this is a big no in other languages like Java and C#.
def rotate(data, n):
alphabet = list("abcdefghijklmopqrstuvwxyz")
n = n % len(alphabet)
target = alphabet[n:] + alphabet[:n]
translation = dict(zip(alphabet, target))
result = ""
for c in data:
if translation.has_key(c):
result += translation[c]
else:
result += c
return result
print rotate("foobar", 1)
print rotate("foobar", 2)
print rotate("foobar", -1)
print rotate("foobar", -2)
Result:
gppcbs
hqqdct
emmazq
dllzyp
The make_trans() solution suggested by others is the way to go here.
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.
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.
You can also use the struct module to do this:
>>> struct.unpack("<L", "y\xcc\xa6\xbb")[0]
3148270713L
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.
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
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.
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.
>>> 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.
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).
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
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
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]
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.