I assign a value to a variable x in the following way:
import wave
w = wave.open('/usr/share/sounds/ekiga/voicemail.wav', 'r')
x = w.readframes(1)
When I type x I get:
'\x1e\x00'
So x got a value. But what is that? Is it hexadecimal? type(x) and type(x[0]) tell me that x and x[0] a strings. Can anybody tell me how should I interpret this strings? Can I transform them into integer?
The interactive interpreter echoes unprintable characters like that. The string contains two bytes, 0x1E and 0x00. You can convert it to an integer with struct.unpack("<h", x) (little endian, 2 bytes, signed).
Yes, it is in hexadecimal, but what it means depends on the other outputs of the wav file e.g. the sample width and number of channels. Your data could be read in two ways, 2 channels and 1 byte sample width (stereo sound) or 1 channel and 2 byte sample width (mono sound). Use x.getparams(): the first number will be the number of channels and the second will be the sample width.
This Link explains it really well.
It's a two byte string:
>>> x='\x1e\x00'
>>> map(ord, list(x))
[30, 0]
>>> [ord(i) for i in x]
[30, 0]
This strings represent bytes. I guess you can turn them into an integer with struct package, which allows interpreting strings of bytes.
Related
I have a long Hex string that represents a series of values of different types. I need to convert this Hex String into bytes or bytearray so that I can extract each value from the raw data. How can I do this?
For example, the string "ab" should convert to the bytes b"\xab" or equivalent byte array. Longer example:
>>> # what to use in place of `convert` here?
>>> convert("8e71c61de6a2321336184f813379ec6bf4a3fb79e63cd12b")
b'\x8eq\xc6\x1d\xe6\xa22\x136\x18O\x813y\xeck\xf4\xa3\xfby\xe6<\xd1+'
Suppose your hex string is something like
>>> hex_string = "deadbeef"
Convert it to a bytearray (Python 3 and 2.7):
>>> bytearray.fromhex(hex_string)
bytearray(b'\xde\xad\xbe\xef')
Convert it to a bytes object (Python 3):
>>> bytes.fromhex(hex_string)
b'\xde\xad\xbe\xef'
Note that bytes is an immutable version of bytearray.
Convert it to a string (Python ≤ 2.7):
>>> hex_data = hex_string.decode("hex")
>>> hex_data
"\xde\xad\xbe\xef"
There is a built-in function in bytearray that does what you intend.
bytearray.fromhex("de ad be ef 00")
It returns a bytearray and it reads hex strings with or without space separator.
provided I understood correctly, you should look for binascii.unhexlify
import binascii
a='45222e'
s=binascii.unhexlify(a)
b=[ord(x) for x in s]
Assuming you have a byte string like so
"\x12\x45\x00\xAB"
and you know the amount of bytes and their type you can also use this approach
import struct
bytes = '\x12\x45\x00\xAB'
val = struct.unpack('<BBH', bytes)
#val = (18, 69, 43776)
As I specified little endian (using the '<' char) at the start of the format string the function returned the decimal equivalent.
0x12 = 18
0x45 = 69
0xAB00 = 43776
B is equal to one byte (8 bit) unsigned
H is equal to two bytes (16 bit) unsigned
More available characters and byte sizes can be found here
The advantages are..
You can specify more than one byte and the endian of the values
Disadvantages..
You really need to know the type and length of data your dealing with
You can use the Codecs module in the Python Standard Library, i.e.
import codecs
codecs.decode(hexstring, 'hex_codec')
You should be able to build a string holding the binary data using something like:
data = "fef0babe"
bits = ""
for x in xrange(0, len(data), 2)
bits += chr(int(data[x:x+2], 16))
This is probably not the fastest way (many string appends), but quite simple using only core Python.
A good one liner is:
byte_list = map(ord, hex_string)
This will iterate over each char in the string and run it through the ord() function. Only tested on python 2.6, not too sure about 3.0+.
-Josh
I'm using the following code to pack an integer into an unsigned short as follows,
raw_data = 40
# Pack into little endian
data_packed = struct.pack('<H', raw_data)
Now I'm trying to unpack the result as follows. I use utf-16-le since the data is encoded as little-endian.
def get_bin_str(data):
bin_asc = binascii.hexlify(data)
result = bin(int(bin_asc.decode("utf-16-le"), 16))
trimmed_res = result[2:]
return trimmed_res
print(get_bin_str(data_packed))
Unfortunately, it throws the following error,
result = bin(int(bin_asc.decode("utf-16-le"), 16)) ValueError: invalid
literal for int() with base 16: '㠲〰'
How do I properly decode the bytes in little-endian to binary data properly?
Use unpack to reverse what you packed. The data isn't UTF-encoded so there is no reason to use UTF encodings.
>>> import struct
>>> data_packed = struct.pack('<H', 40)
>>> data_packed.hex() # the two little-endian bytes are 0x28 (40) and 0x00 (0)
2800
>>> data = struct.unpack('<H',data_packed)
>>> data
(40,)
unpack returns a tuple, so index it to get the single value
>>> data = struct.unpack('<H',data_packed)[0]
>>> data
40
To print in binary use string formatting. Either of these work work best. bin() doesn't let you specify the number of binary digits to display and the 0b needs to be removed if not desired.
>>> format(data,'016b')
'0000000000101000'
>>> f'{data:016b}'
'0000000000101000'
You have not said what you are trying to do, so let's assume your goal is to educate yourself. (If you are trying to pack data that will be passed to another program, the only reliable test is to check if the program reads your output correctly.)
Python does not have an "unsigned short" type, so the output of struct.pack() is a byte array. To see what's in it, just print it:
>>> data_packed = struct.pack('<H', 40)
>>> print(data_packed)
b'(\x00'
What's that? Well, the character (, which is decimal 40 in the ascii table, followed by a null byte. If you had used a number that does not map to a printable ascii character, you'd see something less surprising:
>>> struct.pack("<H", 11)
b'\x0b\x00'
Where 0b is 11 in hex, of course. Wait, I specified "little-endian", so why is my number on the left? The answer is, it's not. Python prints the byte string left to right because that's how English is written, but that's irrelevant. If it helps, think of strings as growing upwards: From low memory locations to high memory. The least significant byte comes first, which makes this little-endian.
Anyway, you can also look at the bytes directly:
>>> print(data_packed[0])
40
Yup, it's still there. But what about the bits, you say? For this, use bin() on each of the bytes separately:
>>> bin(data_packed[0])
'0b101000'
>>> bin(data_packed[1])
'0b0'
The two high bits you see are worth 32 and 8. Your number was less than 256, so it fits entirely in the low byte of the short you constructed.
What's wrong with your unpacking code?
Just for fun let's see what your sequence of transformations in get_bin_str was doing.
>>> binascii.hexlify(data_packed)
b'2800'
Um, all right. Not sure why you converted to hex digits, but now you have 4 bytes, not two. (28 is the number 40 written in hex, the 00 is for the null byte.) In the next step, you call decode and tell it that these 4 bytes are actually UTF-16; there's just enough for two unicode characters, let's take a look:
>>> b'2800'.decode("utf-16-le")
'㠲〰'
In the next step Python finally notices that something is wrong, but by then it does not make much difference because you are pretty far away from the number 40 you started with.
To correctly read your data as a UTF-16 character, call decode directly on the byte string you packed.
>>> data_packed.decode("utf-16-le")
'('
>>> ord('(')
40
I have a long Hex string that represents a series of values of different types. I need to convert this Hex String into bytes or bytearray so that I can extract each value from the raw data. How can I do this?
For example, the string "ab" should convert to the bytes b"\xab" or equivalent byte array. Longer example:
>>> # what to use in place of `convert` here?
>>> convert("8e71c61de6a2321336184f813379ec6bf4a3fb79e63cd12b")
b'\x8eq\xc6\x1d\xe6\xa22\x136\x18O\x813y\xeck\xf4\xa3\xfby\xe6<\xd1+'
Suppose your hex string is something like
>>> hex_string = "deadbeef"
Convert it to a bytearray (Python 3 and 2.7):
>>> bytearray.fromhex(hex_string)
bytearray(b'\xde\xad\xbe\xef')
Convert it to a bytes object (Python 3):
>>> bytes.fromhex(hex_string)
b'\xde\xad\xbe\xef'
Note that bytes is an immutable version of bytearray.
Convert it to a string (Python ≤ 2.7):
>>> hex_data = hex_string.decode("hex")
>>> hex_data
"\xde\xad\xbe\xef"
There is a built-in function in bytearray that does what you intend.
bytearray.fromhex("de ad be ef 00")
It returns a bytearray and it reads hex strings with or without space separator.
provided I understood correctly, you should look for binascii.unhexlify
import binascii
a='45222e'
s=binascii.unhexlify(a)
b=[ord(x) for x in s]
Assuming you have a byte string like so
"\x12\x45\x00\xAB"
and you know the amount of bytes and their type you can also use this approach
import struct
bytes = '\x12\x45\x00\xAB'
val = struct.unpack('<BBH', bytes)
#val = (18, 69, 43776)
As I specified little endian (using the '<' char) at the start of the format string the function returned the decimal equivalent.
0x12 = 18
0x45 = 69
0xAB00 = 43776
B is equal to one byte (8 bit) unsigned
H is equal to two bytes (16 bit) unsigned
More available characters and byte sizes can be found here
The advantages are..
You can specify more than one byte and the endian of the values
Disadvantages..
You really need to know the type and length of data your dealing with
You can use the Codecs module in the Python Standard Library, i.e.
import codecs
codecs.decode(hexstring, 'hex_codec')
You should be able to build a string holding the binary data using something like:
data = "fef0babe"
bits = ""
for x in xrange(0, len(data), 2)
bits += chr(int(data[x:x+2], 16))
This is probably not the fastest way (many string appends), but quite simple using only core Python.
A good one liner is:
byte_list = map(ord, hex_string)
This will iterate over each char in the string and run it through the ord() function. Only tested on python 2.6, not too sure about 3.0+.
-Josh
I have a long Hex string that represents a series of values of different types. I need to convert this Hex String into bytes or bytearray so that I can extract each value from the raw data. How can I do this?
For example, the string "ab" should convert to the bytes b"\xab" or equivalent byte array. Longer example:
>>> # what to use in place of `convert` here?
>>> convert("8e71c61de6a2321336184f813379ec6bf4a3fb79e63cd12b")
b'\x8eq\xc6\x1d\xe6\xa22\x136\x18O\x813y\xeck\xf4\xa3\xfby\xe6<\xd1+'
Suppose your hex string is something like
>>> hex_string = "deadbeef"
Convert it to a bytearray (Python 3 and 2.7):
>>> bytearray.fromhex(hex_string)
bytearray(b'\xde\xad\xbe\xef')
Convert it to a bytes object (Python 3):
>>> bytes.fromhex(hex_string)
b'\xde\xad\xbe\xef'
Note that bytes is an immutable version of bytearray.
Convert it to a string (Python ≤ 2.7):
>>> hex_data = hex_string.decode("hex")
>>> hex_data
"\xde\xad\xbe\xef"
There is a built-in function in bytearray that does what you intend.
bytearray.fromhex("de ad be ef 00")
It returns a bytearray and it reads hex strings with or without space separator.
provided I understood correctly, you should look for binascii.unhexlify
import binascii
a='45222e'
s=binascii.unhexlify(a)
b=[ord(x) for x in s]
Assuming you have a byte string like so
"\x12\x45\x00\xAB"
and you know the amount of bytes and their type you can also use this approach
import struct
bytes = '\x12\x45\x00\xAB'
val = struct.unpack('<BBH', bytes)
#val = (18, 69, 43776)
As I specified little endian (using the '<' char) at the start of the format string the function returned the decimal equivalent.
0x12 = 18
0x45 = 69
0xAB00 = 43776
B is equal to one byte (8 bit) unsigned
H is equal to two bytes (16 bit) unsigned
More available characters and byte sizes can be found here
The advantages are..
You can specify more than one byte and the endian of the values
Disadvantages..
You really need to know the type and length of data your dealing with
You can use the Codecs module in the Python Standard Library, i.e.
import codecs
codecs.decode(hexstring, 'hex_codec')
You should be able to build a string holding the binary data using something like:
data = "fef0babe"
bits = ""
for x in xrange(0, len(data), 2)
bits += chr(int(data[x:x+2], 16))
This is probably not the fastest way (many string appends), but quite simple using only core Python.
A good one liner is:
byte_list = map(ord, hex_string)
This will iterate over each char in the string and run it through the ord() function. Only tested on python 2.6, not too sure about 3.0+.
-Josh
I'm implementing PKCS#7 padding right now in Python and need to pad chunks of my file in order to amount to a number divisible by sixteen. I've been recommended to use the following method to append these bytes:
input_chunk += '\x00'*(-len(input_chunk)%16)
What I need to do is the following:
input_chunk_remainder = len(input_chunk) % 16
input_chunk += input_chunk_remainder * input_chunk_remainder
Obviously, the second line above is wrong; I need to convert the first input_chunk_remainder to a single byte string. How can I do this in Python?
In Python 3, you can create bytes of a given numeric value with the bytes() type; you can pass in a list of integers (between 0 and 255):
>>> bytes([5])
b'\x05'
bytes([5] * 5)
b'\x05\x05\x05\x05\x05'
An alternative method is to use an array.array() with the right number of integers:
>>> import array
>>> array.array('B', 5*[5]).tobytes()
b'\x05\x05\x05\x05\x05'
or use the struct.pack() function to pack your integers into bytes:
>>> import struct
>>> struct.pack('{}B'.format(5), *(5 * [5]))
b'\x05\x05\x05\x05\x05'
There may be more ways.. :-)
In Python 2 (ancient now), you can do the same by using the chr() function:
>>> chr(5)
'\x05'
>>> chr(5) * 5
'\x05\x05\x05\x05\x05'
In Python3, the bytes built-in accepts a sequence of integers. So for just one integer:
>>> bytes([5])
b'\x05'
Of course, thats bytes, not a string. But in Python3 world, OP would probably use bytes for the app he described, anyway.