How to convert string to byte array in Python

Question

Say that I have a 4 character string, and I want to convert this string into a byte array where each character in the string is translated into its hex equivalent. e.g.

str = "ABCD"

I'm trying to get my output to be

array('B', [41, 42, 43, 44])

Is there a straightforward way to accomplish this?

• Tim Pietzcker: What you want is not possible, at least not in this exact form. A bytearray of type B contains 1-byte integers, and they are always represented in decimal.

Answer 1

Just use a bytearray() which is a list of bytes.

Python2:

s = "ABCD"
b = bytearray()
b.extend(s)

Python3:

s = "ABCD"
b = bytearray()
b.extend(map(ord, s))

By the way, don't use str as a variable name since that is builtin.

• Pithikos: @KevanAhlquist my bad. Fixed it now.
• Diego Herranz: For Python 3 this looks cleaner to me: s = "ABCD", b = bytearray(), b.extend(s.encode())
• Diego Herranz: Regarding encode(), it returns a bytes object which naturally extends a bytearray.
• 9000: map(ord, s) will return values > 255 unless your strings are strictly ASCII. Please update your answer to include something like s.encode('utf-8'). (Note that UTF-8 is a strict superset of ASCII, so it does not alter ASCII strings in any way.)
• user70960: @9000 it is incorrect to use .encode() as well as .encode('utf-8'). Use map(ord, ...) if you don't want you bytes to be transformed. repl.it/repls/MistySubtleVisitors just press run and see the result.

Answer 2

encode function can help you here, encode returns an encoded version of the string

In [44]: str = "ABCD"
In [45]: [elem.encode("hex") for elem in str]
Out[45]: ['41', '42', '43', '44']

or you can use array module

In [49]: import array
In [50]: print array.array('B', "ABCD")
array('B', [65, 66, 67, 68])

• avasal: however as you can see,, array module gives a ascii value of string elements, which doesn't match with your expected output
• Fabio Picchi: This is the accepted answer and does not work in Python3. Could you please add the python3 version as pointed in other answers?

Answer 3

An alternative to get a byte array is to encode the string in ascii: b=s.encode('ascii').

• FuriousGeorge: Assuming that the string is ASCII to begin with. If you have s = '\x80', that's not going to work.
• mbrc: what if s = '\x80' what then to do?

Answer 4

Depending on your needs, this can be one step or two steps

1. use encode() to convert string to bytes, immutable
2. use bytearray() to convert bytes to bytearray, mutable

s="ABCD"
encoded=s.encode('utf-8')
array=bytearray(encoded)

The following validation is done in Python 3.7

>>> s="ABCD"
>>> encoded=s.encode('utf-8')
>>> encoded
b'ABCD'
>>> array=bytearray(encoded)
>>> array
bytearray(b'ABCD')

Answer 5

This work in both Python 2 and 3:

>>> bytearray(b'ABCD')
bytearray(b'ABCD')

Note string started with b.

To get individual chars:

>>> print("DEC HEX ASC")
... for b in bytearray(b'ABCD'):
... print(b, hex(b), chr(b))
DEC HEX ASC
65 0x41 A
66 0x42 B
67 0x43 C
68 0x44 D

Hope this helps

Answer 6

This works for me (Python 2)

s = "ABCD"
b = bytearray(s)
# if you print whole b, it still displays it as if its original string
print b
# but print first item from the array to see byte value
print b[0]

Reference: http://www.dotnetperls.com/bytes-python

Answer 7

s = "ABCD"
from array import array
a = array("B", s)

If you want hex:

print map(hex, a)

• MikeB2019x: Does not work in repl.it. Returns: "TypeError: cannot use a str to initialize an array with typecode 'B' "

Answer 8

Since none of the answers is producing exactly array('B', [41, 42, 43, 44]) and the answer by avasal fails in Python 3, I post here my alternative:

import array
s = 'ABCD'
a = array.array('B', [ord(c) for c in s])
print(a)

which prints

array('B', [65, 66, 67, 68])

Note that 65-68 is the correct ASCII for "ABCD".

Answer 9

for python 3 it worked for what @HYRY posted. I needed it for a returned data in a dbus.array. This is the only way it worked

s = "ABCD"

from array import array

a = array("B", s)