Try this:

int(s, base=2)

for example,

for i in ['10101', '11100', '11010', '00101']:
    print(int(i, base=2))

@metatoaster also mentioned this.

Note on terminology: a binary literal is a piece of text, normally appearing in source code, so I've shown code that results in the same text, in a string object. A binary literal is not a number, but it evaluates to a number in Python code, so I've also shown code to calculate that number. 0b01101101 and 109 are different literals (one binary, one decimal), but when evaluated they give exactly the same result, an integer equal to one hundred and nine.

If you want to write the integer 109 as a single byte over a serial connection, then what you want to write is the single character with code 109, not a binary literal.

my_string = "01101101"

my_binary_literal = "0b" + my_string
my_string_converted_to_integer = int(my_string, 2)
my_single_character = chr(my_string_converted_to_integer)
assert my_single_character == 'm'

You could convert b'a' to its corresponding integer ord value, and then print the int in binary format using '{:b}'.format:

In [6]: '{:b}'.format(ord(b'a'))
Out[6]: '1100001'

• Reading a large file one-byte-at-a-time can be very slow. You'll get better performance by reading more bytes per call to f.read. You can iterate over the contents of the file in chunks of 1024 bytes using:

  with open(filename, 'rb') as f:
      for chunk in iter(lambda: f.read(1024), b''):
  

• Similarly, calling print once for each byte can be very slow. You'll get better performance by printing more bytes per call to print. So you could use a list comprehension to loop over the bytes in chunk, convert each to its string-binary format and then use ''.join to join the strings together:

  print(''.join(['{:b}'.format(ord(c)) for c in chunk]), end='')
  

• Use bare except is considered a bad practice. If you choose to use try..except here, list only those Exceptions you wish to handle:

  try:
      ...
  except IOError:          
  

def slurpInstructions(filename):
    with open(filename, 'rb') as f:
        for chunk in iter(lambda: f.read(1024), b''):
            print(''.join(['{:b}'.format(c) for c in chunk]), end='')

If you want a binary file,

>>> import struct
>>> myFile=open('binaryFoo','wb')
>>> myStr='10010101110010101'
>>> x=int(myStr,2)
>>> x
76693
>>> struct.pack('i',x)
'\x95+\x01\x00'
>>> myFile.write(struct.pack('i',x))
>>> myFile.close()
>>> quit()

$ cat binaryFoo
�+$

Is this what you are looking for?

Python doesn't have a separate type for binary numbers when you type 0b1 there it just gets interpreted as base 2 and is still stored as a base 10 integer at the end so for converting from a string '0b1' you can just convert it to int while telling it that the value is in base 2

y = '0b1'
print(int(y, 2))

the bin function also just gives you a binary representation of the integer as a string

This will convert each byte in your string to 8 binary digits and separate them with a space. You can easily change the separator.

data = " ".join(bin(ord(b))[2:].rjust(8, "0") for b in data)