字符串（二进制）到int in python？

Question

I want to make an int from a string in python. I thought to translate the string first to binary like this:

st = 'string'
binSt = ' '.join(format(ord(x), '08b') for x in st)

returns this:

01110011 01110100 01110010 01101001 01101110 01100111

And then I want to translate the binary( in groups of 8) to integers which should to return this:

115 116 114 105 110 103

How can I do this? Is there maybe a special function in python or something?

Answer 1

You can use the int() function:

result = ''
st = 'string'
binSt = ' '.join(format(ord(x), '08b') for x in st)
binSt_split = binSt.split()
for split in binSt_split:
    result = result + str(int(split,2) + ' '
print result

Answer 2

You can simply do

r = [int(numbs, 2) for numbs in binSt.split(' ')]

int(str, baseNumber) will read the string str and convert it to int using baseNumber

so int("0xFF", 16) = 255 and int("11", 2) = 3

Answer 3

Why not use a bytearray ?

>>> barr = bytearray('string')
>>> barr[0]
115

Bytearray does exactly what you want -- It interprets each character in the string as an integer in the range from 0 -> 255 .

Answer 4

Using the solution for binary to int here: Convert base-2 binary number string to int

 binSt = ' '.join([str(int(format(ord(x), '08b'), 2)) for x in st])

And if you just want an array of ints

 int_array = [int(format(ord(x), '08b'), 2) for x in st]

Addressing SpoonMeiser comments. you can avoid intermediate conversions with ord(x)

 int_array = [ord(x) for x in st]

Answer 5

String to binary and then to decimal :

st = 'string'
binSt = ' '.join(format(ord(x), '08b') for x in st)
binSt
##'01110011 01110100 01110010 01101001 01101110 01100111'
bin=binSt.split()
bin
##['01110011', '01110100', '01110010', '01101001', '01101110', '01100111']
print(map(lambda x: int(x,2), bin))
##[115, 116, 114, 105, 110, 103]

## is used for outputs