Decode base64 data as array in Python

I'm using this handy Javascript function to decode a base64 string and get an array in return.

This is the string:

base64_decode_array('6gAAAOsAAADsAAAACAEAAAkBAAAKAQAAJgEAACcBAAAoAQAA')

This is what's returned:

234,0,0,0,235,0,0,0,236,0,0,0,8,1,0,0,9,1,0,0,10,1,0,0,38,1,0,0,39,1,0,0,40,1,0,0

The problem is I don't really understand the javascript function:

var base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'.split("");
var base64inv = {}; 
for (var i = 0; i < base64chars.length; i++) { 
  base64inv[base64chars[i]] = i; 
}
function base64_decode_array (s)
{
  // remove/ignore any characters not in the base64 characters list
  //  or the pad character -- particularly newlines
  s = s.replace(new RegExp('[^'+base64chars.join("")+'=]', 'g'), "");

  // replace any incoming padding with a zero pad (the 'A' character is zero)
  var p = (s.charAt(s.length-1) == '=' ? 
          (s.charAt(s.leng开发者_StackOverflow中文版th-2) == '=' ? 'AA' : 'A') : ""); 

  var r = [];

  s = s.substr(0, s.length - p.length) + p;

  // increment over the length of this encrypted string, four characters at a time
  for (var c = 0; c < s.length; c += 4) {

    // each of these four characters represents a 6-bit index in the base64 characters list
    //  which, when concatenated, will give the 24-bit number for the original 3 characters
    var n = (base64inv[s.charAt(c)] << 18) + (base64inv[s.charAt(c+1)] << 12) +
            (base64inv[s.charAt(c+2)] << 6) + base64inv[s.charAt(c+3)];


    // split the 24-bit number into the original three 8-bit (ASCII) characters
    r.push((n >>> 16) & 255);
    r.push((n >>> 8) & 255);
    r.push(n & 255);


  }
   // remove any zero pad that was added to make this a multiple of 24 bits
  return r;
}

What's the function of those "<<<" and ">>>" characters. Or is there a function like this for Python?

---

Who cares. Python has easier ways of doing the same.

[ord(c) for c in '6gAAAOsAAADsAAAACAEAAAkBAAAKAQAAJgEAACcBAAAoAQAA'.decode('base64')]

---

In Python I expected you'd just use the base64 module...

... but in response to your question about << and >>>:

• << is the left-shift operator; the result is the first operand shifted left by the second operand number of bits; for example 5 << 2 is 20, as 5 is 101 in binary, and 20 is 10100.
• >>> is the non-sign-extended right-shift operator; the result is the first operand shifted right by the second operand number of bits... with the leftmost bit always being filled with a 0.

---

Why not just:

from binascii import a2b_base64, b2a_base64
encoded_data = b2a_base64(some_string)
decoded_string = a2b_base64(encoded_data)

def base64_decode_array(string):
    return [ord(c) for c in a2b_base64(string)]

---

Just for fun/completeness, I'll translate the javascript more literally: :)

# No particular reason to make a list of chars here instead of a string.
base64chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/'
lookup = dict((c, i) for (i, c) in enumerate(base64chars))

def base64_decode_array(s):
  # Filter out meaningless chars, especially newlines. No need for a regex.
  s = ''.join(c for c in s if c in base64chars + '=')

  # replace any incoming padding with a zero pad (the 'A' character is zero)
  # Their way:
  # p = ('AA' if s[-2] == '=' else 'A') if s[-1] == '=' else ''
  # s = s[:len(s) - len(p)] + p
  # My way (allows for more padding than that; 
  # '=' will only appear at the end anyway
  s = s.replace('=', 'A')

  r = []

  # Iterate over the string in blocks of 4 chars - an ugly hack
  # though we are preserving the original code's assumption that the text length
  # is a multiple of 4 (that's what the '=' padding is for) ;)
  for a, b, c, d in zip(*([iter(s)] * 4)):
    # Translate each letter in the quad into a 6-bit value and bit-shift them
    # together into a 24-bit value
    n = (lookup[a] << 18) + (lookup[b] << 12) + (lookup[c] << 6) + lookup[d]

    # split the 24-bit number into the original three 8-bit (ASCII) characters
    r += [(n >> 16) & 0xFF), (n >> 8) & 0xFF), (n & 0xFF)]

  return r