Python-将字符串转换为m位数字

Question

Suppose I have a string str="hello", and every char can be converted to ascii code, which is:

"104, 101, 108, 108, 111" = "0b01101000, 0b01100101, 0b01101100, 0b01101100, 0b01101111"

and each char is 8-bit long, each number is smaller than 2^8.

Now I wonder how can I convert arbitrary m-bit long number. For example, if m=10, then it becomes

"0b0110100001, 0b1001010110, 0b1100011011, 0b0001101111" = "417, 598, 795, 111"

and each number is smaller than 2^m.

My current idea is:

str = "hello"
tmp = ""
for i in str:
    tmp = tmp + bin(ord(i))[2:].zfill(8)

print tmp

m=10
for i in xrange(0,len(tmp),m):
    print int(tmp[i:i+m],2)

But I wonder whether there is some more efficient way to do that?

Thanks!

Answer 1

If you are looking for standard conversions like binary, octal, hex; you could use format as:

for binary:

numbers = "417, 598, 795, 111"
print ','.join(["{0:b}".format(int(n,10)) for n in numbers.split(',')])

output:

110100001,1001010110,1100011011,1101111

You can also use int for conversion (it will not work for non valid digits):

print ','.join([str(int(n,16)) for n in numbers.split(',')])

output:

1047,1432,1941,273

Answer 2

You could do something general like this:

s = "hello"
print ['0b{:0{bits}b}'.format(ord(c), bits=10) for c in s]

Output:

['0b0001101000', '0b0001100101', '0b0001101100', '0b0001101100', '0b0001101111']