I have a binary like this:
1101100110000110110110011000001011011000101001111101100010101000
and I want to convert it to utf-8. how can I do this in python?
4 Answers
Cleaner version:
>>> test_string = '1101100110000110110110011000001011011000101001111101100010101000'
>>> print ('%x' % int(test_string, 2)).decode('hex').decode('utf-8')
نقاب
Inverse (from @Robᵩ's comment):
>>> '{:b}'.format(int(u'نقاب'.encode('utf-8').encode('hex'), 16))
1: '1101100110000110110110011000001011011000101001111101100010101000'
15 Comments
Aidin.T
but it doesn't work properly. it shows something else, not the first text I just converted to binary
Aidin.T
worked, thank you. I think I should move the check to this answer. It's really simpler
Robᵩ
And the inverse would be:
s=u'نقاب'; print '{:b}'.format(int(s.encode('utf-8').encode('hex'), 16))
Igonato
@Robᵩ added to the answer with minor edit (I think in this case
.encode('utf-8') in unnecessary).Robᵩ
Note that
s = "سلام" and s = u"سلام" give different results. The former fails, the latter works. But let's stop solving the new problem. @Aidin.T, if you have a problem with encoding, please open a new question. |
Well, the idea I have is:
1. Split the string into octets
2. Convert the octet to hexadecimal using int and later chr
3. Join them and decode the utf-8 string into Unicode
This code works for me, but I'm not sure what does it print because I don't have utf-8 in my console (Windows :P ).
s = '1101100110000110110110011000001011011000101001111101100010101000'
u = "".join([chr(int(x,2)) for x in [s[i:i+8]
for i in range(0,len(s), 8)
]
])
d = u.decode('utf-8')
Hope this helps!
4 Comments
Paulo Bu
Hmmm, I'm somewhat suspicious of
unichr. Because OP says his binary is already utf-8. utf-8 has variable character length, so I just used chr to join the raw bytes in a string and decode them later into Unicode.Robᵩ
@JoranBeasley - I disagree, assuming Python2. In that step he is collecting bytes, not characters. Only after he has the utf-8-encoded byte string does he want to convert.
Paulo Bu
@Robᵩ: That's my point. Nice answer, love the
split('........'). I think is basically the same idea as mine. +1Robᵩ
+1 - This is the same technique as mine (so obviously I approve), plus you explained yours. Questioner should move the check to this better answer.
>>> s='1101100110000110110110011000001011011000101001111101100010101000'
>>> print (''.join([chr(int(x,2)) for x in re.split('(........)', s) if x ])).decode('utf-8')
نقاب
>>>
Or, the inverse:
>>> s=u'نقاب'
>>> ''.join(['{:b}'.format(ord(x)) for x in s.encode('utf-8')])
'1101100110000110110110011000001011011000101001111101100010101000'
>>>
1 Comment
Aidin.T
there is another question, how can I convert my text to binary by python? I mean the inverse form of my question
Use:
def bin2text(s): return "".join([chr(int(s[i:i+8],2)) for i in xrange(0,len(s),8)])
>>> print bin2text("01110100011001010111001101110100")
>>> test
2 Comments
Aidin.T
for my text it returns this: '\xd9\x86\xd9\x82\xd8\xa7\xd8\xa8', how can I get it in the right way of showing?
Christian Ternus
hdorodor a similar hex-dump utility and copy-paste the first few lines?