I have a byte array b'string\x01' that i need to format to string1. I need to do this for any "string", followed by a byte e.g, b'string\t' to string9. Why is my way not correctly working?
I have tried to get the x = b'string\x01', i am trying to turn into "string1".
So i need to remove the '\x01', s = str(x).split("g",1) and then byte_part = s[1].rstrip('\'') so i get "\x01" on its own, but the next problem is:
I am trying to convert this string to a byte, so i can use int.from_bytes(byte_part,'little') and get the correct integer result. e.g. \x01 = 1.
What is happening is i am converting the string to a bytearray bytearray(string, 'utf-8') which then gives me bytearray(b'\\x01') then using int.from_bytes() gives me the result for b'\\x01' is 825260124 instead of b'\x01' being 1 i am after.
The method you are looking for is ord().
ord('\x01') # the result is 1
Also, following would convert your string and return the last number.
ord(a.decode().split('string')[1])
hope this helps.
Related
Similar to this other question on decoding a hex string, I have some code in a Python 2.7 script which has worked for years. I'm now trying to convert that script to Python 3.
OK, I apologize for not posting a complete question initially. I hope this clarifies the situation.
The issue is that I'm trying to convert an older Python 2.7 script to Python 3.8. For the most part the conversion has gone ok, but I am having issues converting the following code:
# get Register Stings
RegString = ""
for i in range(length):
if regs[start+i]!=0:
RegString = RegString + str(format(regs[start+i],'x').decode('hex'))
Here are some suppodrting data:
regs[start+0] = 20341
regs[start+1] = 29762
I think that my Python 2.7 code is converting these to HEX as "4f75" and "7442", respectively. And then to the characters "Ou" and "tB", respectively.
In Python 3 I get this error:
'str' object has no attribute 'decode'
My goal is to modify my Python 3 code so that the script will generate the same results.
str(format(regs[start+i],'x').decode('hex')) is a very verbose and round-about way of turning the non-zero integer values in regs[start:start + length] into individual characters of a bytestring (str in Python 2 should really be seen as a sequence of bytes). It first converts an integer value into a hexadecimal representation (a string), decodes that hexadecimal string to a (series) of string characters, then calls str() on the result (redundantly, the value is already a string). Assuming that the values in regs are integers in the range 0-255 (or even 0-127), in Python 2 this should really have been using the chr() function.
If you want to preserve the loop use chr() (to get a str string value) or if you need a binary value, use bytes([...]). So:
RegString = ""
for codepoint in regs[start:start + length]:
RegString += chr(codepoint)
or
RegString = b""
for codepoint in regs[start:start + length]:
RegString += bytes([codepoint])
Since this is actually converting a sequence of integers, you can just pass the whole lot to bytes() and filter out the zeros as you go:
# only take non-zero values
RegString = bytes(b for b in regs[start:start + length] if b)
or remove the nulls afterwards:
RegString = bytes(regs[start:start + length]).replace(b"\x00", b"")
If that's still supposed to be a string and not a bytes value, you can then decode it, with whatever encoding is appropriate (ASCII if the integers are in the range 0-127, or a more specific codec otherwise, in Python 2 this code produced a bytestring so look for other hints in the code as to what encoding they might have been using).
I am converting a string into integer using int function and it is working fine but i want to keep save zero digit that are at the start of the string.
string_value = '0123'
print(int(string_value))
result is 123
How can i format output 0123 as in integer type value not in string.
You can't, but if you want to put 0's (zero padding) at the beginning of your number, this is the way to do it.
"{:04}".format(123)
# '0123'
"{:05}".format(123)
# '00123'
Like every one said you can try above answers or the following :
string_value = '0123'
int_no = int(string_value)
print("%04d" % int_no)
print(string_value.zfill(4))
Both will give same answer
Impossible, you cannot get an integer value of 0123.
You should change your mind, you do not actually need 0123 in integer, but you need to keep zero when displaying it. So the question should change to how to format output.
I want to get the value of 99997 in big endian which is (2642804992) and then return the answer as a long value
here is my code in python:
v = 99997
ttm = pack('>i', v) # change the integer to big endian form
print ("%x"), {ttm}
r = long(ttm, 16) # convert to long (ERROR)
return r
Output: %x set(['\x00\x01\x86\x9d'])
Error: invalid literal for long() with base 16: '\x00\x01\x86\x9d'
As the string is already in hex form why isn't it converting to a long? How would I remove this error and what is the solution to this problem.
pack will return a string representation of the data you provide.
The string representation is different than a base 16 of a long number. Notice the \x before each number.
Edit:
try this
ttm = pack('>I',v)
final, = unpack('<I',ttm)
print ttm
Notice the use of I, this so the number is treated as an unsigned value
You have to use struct.unpack as a reverse operation to struct.pack.
r, = unpack('<i', ttm)
this will r set to -1652162304.
You just converted the integer value to big endian binary bytes.
This is useful mostly to embed in messages addressed to big-endian machines (PowerPC, M68K,...)
Converting to long like this means parsing the ttm string which should be 0x1869D as ASCII.
(and the print statement does not work either BTW)
If I just follow your question title: "Convert hexadecimal string to long":
just use long("0x1869D",16). No need to serialize it.
(BTW long only works in python 2. In python 3, you would have to use int since all numbers are represented in the long form)
Well, I'm answering to explain why it's bound to fail, but I'll edit my answer when I really know what you want to do.
This is a nice question.
Here is what you are looking for.
s = str(ttm)
for ch in r"\bx'":
s = s.replace(ch, '')
print(int(s, 16))
The problem is that ttm is similar to a string in some aspects. This is what is looks like: b'\x00\x01\x86\x9d'. Supplying it to int (or long) keeps all the non-hex characters. I removed them and then it worked.
After removing the non-hex-digit chars, you are left with 0001869d which is indeed 99997
Comment I tried it on Python 3. But on Python 2 it will be almost the same, you won't have the b attached to the string, but otherwise it's the same thing.
I am new to python so please bear with if it sounds a novice question. I looked it up but different sources tell me different things so I decided to ask it here. And doc is not very clear.
I have an int that I want to convert to bytes.
#1024 should give me 0000010000000000
print bytes([1024])
[1024]
If I use bytes() I assume I get a list since when I print I end up getting [1024]. However if I do
print bytes([1024])[0]
[
I get [ back. Why is this? So it does not return a list?
Is there a way to get the byte streams back given an integer? Ideally I want something as follows:
x = tobyte(1024)
print x[0] #should give me 0 00000000
print x[1] #should give me 8 00000100
I need to be able to use x[0] elsewhere in the code i.e. to be passed to base64 encode where I expect my data to be 64 bits
To get the individual bytes from an integer, use struct:
individual_bytes = struct.unpack("BB", struct.pack("<H", 1024))
First, "<I" packs the integer as a 16-bit value, using little-endian ordering. Then "BB" unpacks the byte string into two separate 8-bit values.
I am quite new in python and I need to solve this simple problem. Already there are several similar questions but still I cannot solve it.
I need to read a binary file, which is composed by several blocks of bytes. For example the header is composed by 6 bytes and I would like to extract those 6 bytes and transform ins sequence of binary characters like 000100110 011001 for example.
navatt_dir='C:/PROCESSING/navatt_read/'
navatt_filename='OSPS_FRMT_NAVATT____20130621T100954_00296_caseB.bin'
navatt_path=navatt_dir+navatt_filename
navatt_file=open(navatt_path, 'rb')
header=list(navatt_file.read(6))
print header
As result of the list i have the following
%run C:/PROCESSING/navatt_read/navat_read.py
['\t', 'i', '\xc0', '\x00', '\x00', 't']
which is not what i want.
I would like also to read a particular value in the binary file knowing the position and the length, without reading all the file. IS it possible
thanks
ByteArray
A bytearray is a mutable sequence of bytes (Integers where 0 ≤ x ≤ 255). You can construct a bytearray from a string (If it is not a byte-string, you will have to provide encoding), an iterable of byte-sized integers, or an object with a buffer interface. You can of course just build it manually as well.
An example using a byte-string:
string = b'DFH'
b = bytearray(string)
# Print it as a string
print b
# Prints the individual bytes, showing you that it's just a list of ints
print [i for i in b]
# Lets add one to the D
b[0] += 1
# And print the string again to see the result!
print b
The result:
DFH
[68, 70, 72]
EFH
This is the type you want if you want raw byte manipulation. If what you want is to read 4 bytes as a 32bit int, one would use the struct module, with the unpack method, but I usually just shift them together myself from a bytearray.
Printing the header in binary
What you seem to want is to take the string you have, convert it to a bytearray, and print them as a string in base 2/binary.
So here is a short example for how to write the header out (I read random data from a file named "dump"):
with open('dump', 'rb') as f:
header = f.read(6)
b = bytearray(header)
print ' '.join([bin(i)[2:].zfill(8) for i in b])
After converting it to a bytearray, I call bin() on every single one, which gives back a string with the binary representation we need, in the format of "0b1010". I don't want the "0b", so I slice it off with [2:]. Then, I use the string method zfill, which allows me to have the required amount of 0's prepended for the string to be 8 long (which is the amount of bits we need), as bin will not show any unneeded zeroes.
If you're new to the language, the last line might look quite mean. It uses list comprehension to make a list of all the binary strings we want to print, and then join them into the final string with spaces between the elements.
A less pythonic/convoluted variant of the last line would be:
result = []
for byte in b:
string = bin(i)[2:] # Make a binary string and slice the first two bytes
result.append(string.zfill(8)) # Append a 0-padded version to the results list
# Join the array to a space separated string and print it!
print ' '.join(result)
I hope this helps!