I want to convert a hex string to utf-8
a = '0xb3d9'
to
동 (http://www.unicodemap.org/details/0xB3D9/index.html)
First, obtain the integer value from the string of a, noting that a is expressed in hexadecimal:
a_int = int(a, 16)
Next, convert this int to a character. In python 2 you need to use the unichr method to do this, because the chr method can only deal with ASCII characters:
a_chr = unichr(a_int)
Whereas in python 3 you can just use the chr method for any character:
a_chr = chr(a_int)
So, in python 3, the full command is:
a_chr = chr(int(a, 16))
Related
I have an info_address that I want to convert to delimited hex
info_address_original = b'002dd748'
What i want is
info_address_coded = b'\x00\x2d\xd7\x48'
I tried this solution
info_address_original = b'002dd748'
info_address_intermediary = info_address_original.decode("utf-8") # '002dd748'
info_address_coded = bytes.fromhex( info_address_intermediary ) # b'\x00-\xd7H'
and i get
info_address_coded = b'\x00-\xd7H'
What my debugger shows
How would one go about correctly turning a bytes string like that to delimited hex? It worked implicitly in Python 2 but it doesn't work the way i would want in Python 3.
This is only a representation of the bytes. '-' is the same as '\x2d'.
>>> b'\x00\x2d\xd7\x48' == b'\x00-\xd7H'
True
The default representation of a byte string is to display the character value for all ascii printable characters and the encoded \xhh representation where hh is the hexadecimal value of the byte.
That means that b'\x00\x2d\xd7\x48' and `b'\x00-\xd7H' are the exact same string containing 4 bytes.
I have encounter a case where I need to convert a string of character into a character string in python.
s = "\\x80\\x78\\x07\\x00\\x75\\xb3"
print s #gives: \x80\x78\x07\x00\x75\xb3
what I want is that, given the string s, I can get the real character store in s. which in this case is "\x80, \x78, \x07, \x00, \x75, and \xb3"(something like this)�xu�.
You can use string-escape encoding (Python 2.x):
>>> s = "\\x80\\x78\\x07\\x00\\x75\\xb3"
>>> s.decode('string-escape')
'\x80x\x07\x00u\xb3'
Use unicode-escape encoding (in Python 3.x, need to convert to bytes first):
>>> s.encode().decode('unicode-escape')
'\x80x\x07\x00u³'
you can simply write a function, taking the string and returning the converted form!
something like this:
def str_to_chr(s):
res = ""
s = s.split("\\")[1:] #"\\x33\\x45" -> ["x33","x45"]
for(i in s):
res += chr(int('0'+i, 16)) # converting to decimal then taking the chr
return res
remember to print the return of the function.
to find out what does each line do, run that line, if still have questions comment it... i'll answer
or you can build a string from the byte values, but that might not all be "printable" depending on your encoding, example:
# -*- coding: utf-8 -*-
s = "\\x80\\x78\\x07\\x00\\x75\\xb3"
r = ''
for byte in s.split('\\x'):
if byte: # to get rid of empties
r += chr(int(byte,16)) # convert to int from hex string first
print (r) # given the example, not all bytes are printable char's in utf-8
HTH, Edwin
This question already has answers here:
What's the correct way to convert bytes to a hex string in Python 3?
(9 answers)
Closed 7 years ago.
I have an ANSI string Ď–ór˙rXüď\ő‡íQl7 and I need to convert it to hexadecimal like this:
06cf96f30a7258fcef5cf587ed51156c37 (converted with XVI32).
The problem is that Python cannot encode all characters correctly (some of them are incorrectly displayed even here, on Stack Overflow) so I have to deal with them with a byte string.
So the above string is in bytes this: b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
And that's what I need to convert to hexadecimal.
So far I tried binascii with no success, I've tried this:
h = ""
for i in b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7':
h += hex(i)
print(h)
It prints:
0x60xcf0x960xf30xa0x720x830xff0x720x580xfc0xef0x5c0xf50x870xed0x510x150x6c0x37
Okay. It looks like I'm getting somewhere... but what's up with the 0x thing?
When I remove 0x from the string like this:
h.replace("0x", "")
I get 6cf96f3a7283ff7258fcef5cf587ed51156c37 which looks like it's correct.
But sometimes the byte string has a 0 next to a x and it gets removed from the string resulting in a incorrect hexadecimal string. (the string above is missing the 0 at the beginning).
Any ideas?
If you're running python 3.5+, bytes type has an new bytes.hex() method that returns string representation.
>>> h = b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
>>> h.hex()
'06cf96f30a7283ff7258fcef5cf587ed51156c37'
Otherwise you can use binascii.hexlify() to do the same thing
>>> import binascii
>>> binascii.hexlify(h).decode('utf8')
'06cf96f30a7283ff7258fcef5cf587ed51156c37'
As per the documentation, hex() converts “an integer number to a lowercase hexadecimal string prefixed with ‘0x’.” So when using hex() you always get a 0x prefix. You will always have to remove that if you want to concatenate multiple hex representations.
But sometimes the byte string has a 0 next to a x and it gets removed from the string resulting in a incorrect hexadecimal string. (the string above is missing the 0 at the beginning).
That does not make any sense. x is not a valid hexadecimal character, so in your solution it can only be generated by the hex() call. And that, as said above, will always create a 0x. So the sequence 0x can never appear in a different way in your resulting string, so replacing 0x by nothing should work just fine.
The actual problem in your solution is that hex() does not enforce a two-digit result, as simply shown by this example:
>>> hex(10)
'0xa'
>>> hex(2)
'0x2'
So in your case, since the string starts with b\x06 which represents the number 6, hex(6) only returns 0x6, so you only get a single digit here which is the real cause of your problem.
What you can do is use format strings to perform the conversion to hexadecimal. That way you can both leave out the prefix and enforce a length of two digits. You can then use str.join to combine it all into a single hexadecimal string:
>>> value = b'\x06\xcf\x96\xf3\nr\x83\xffrX\xfc\xef\\\xf5\x87\xedQ\x15l7'
>>> ''.join(['{:02x}'.format(x) for x in value])
'06cf96f30a7283ff7258fcef5cf587ed51156c37'
This solution does not only work with a bytes string but with really anything that can be formatted as a hexadecimal string (e.g. an integer list):
>>> value = [1, 2, 3, 4]
>>> ''.join(['{:02x}'.format(x) for x in value])
'01020304'
I'm reading a wav audio file in Python using wave module. The readframe() function in this library returns frames as hex string. I want to remove \x of this string, but translate() function doesn't work as I want:
>>> input = wave.open(r"G:\Workspace\wav\1.wav",'r')
>>> input.readframes (1)
'\xff\x1f\x00\xe8'
>>> '\xff\x1f\x00\xe8'.translate(None,'\\x')
'\xff\x1f\x00\xe8'
>>> '\xff\x1f\x00\xe8'.translate(None,'\x')
ValueError: invalid \x escape
>>> '\xff\x1f\x00\xe8'.translate(None,r'\x')
'\xff\x1f\x00\xe8'
>>>
Any way I want divide the result values by 2 and then add \x again and generate a new wav file containing these new values. Does any one have any better idea?
What's wrong?
Indeed, you don't have backslashes in your string. So, that's why you can't remove them.
If you try to play with each hex character from this string (using ord() and len() functions - you'll see their real values. Besides, the length of your string is just 4, not 16.
You can play with several solutions to achieve your result:
'hex' encode:
'\xff\x1f\x00\xe8'.encode('hex')
'ff1f00e8'
Or use repr() function:
repr('\xff\x1f\x00\xe8').translate(None,r'\\x')
One way to do what you want is:
>>> s = '\xff\x1f\x00\xe8'
>>> ''.join('%02x' % ord(c) for c in s)
'ff1f00e8'
The reason why translate is not working is that what you are seeing is not the string itself, but its representation. In other words, \x is not contained in the string:
>>> '\\x' in '\xff\x1f\x00\xe8'
False
\xff, \x1f, \x00 and \xe8 are the hexadecimal representation of for characters (in fact, len(s) == 4, not 24).
Use the encode method:
>>> s = '\xff\x1f\x00\xe8'
>>> print s.encode("hex")
'ff1f00e8'
As this is a hexadecimal representation, encode with hex
>>> '\xff\x1f\x00\xe8'.encode('hex')
'ff1f00e8'
In Python, how can I represent an integer value (<256) as a string? For example:
i = 10
How can I create a string "s" that is one-byte long, and the byte has the value 10?
to clarify, I do not want a string "10". I want a string that its 1st (and only) byte has the value of 10.
by the way, I cannot create the string statically:
s = '\x0A'
because the value is not pre-defined. It is a dynamic number value.
You can use chr() function as:
>>> chr(60)
'<'
>>> chr(97)
'a'
>>> chr(67)
'C'
To convert back use ord() funtion as:
>>> ord('C')
67
In Python 2.x, you want:
s = chr(10)
In Python 3.x, strings are Unicode, so you want:
s = bytes([10])
why don't you just use chr?
chr(10)
Out[41]: '\n'
chr(255)
Out[42]: '\xff'
Found another answer using struct module working for Python 2:
import struct
struct.pack('B', i)