binaryToDecimal = lambda binary: sum([[int(2**index) for index in range(len(binary))][-index-1] for index, char in enumerate(binary) if int(char)])

I wrote a simple function that converts binary strings into decimal integers. How can this be improved/simplified?

I know about the int function, I wanted to implement the conversion myself.

  • \$\begingroup\$ what about using python: binaryToDecimal = lambda x: str(int(x,2)) \$\endgroup\$ Commented May 9, 2018 at 18:41
  • \$\begingroup\$ I'm aware that you can do that :) I just wanted to implement it myself \$\endgroup\$
    – Hum4n01d
    Commented May 9, 2018 at 18:41
  • \$\begingroup\$ Well, readability could be improved a lot by not using too much list comprehensions together :) \$\endgroup\$
    – IEatBagels
    Commented May 9, 2018 at 18:42
  • 1
    \$\begingroup\$ Definitely, I did that on purpose though to make it as compact as possible \$\endgroup\$
    – Hum4n01d
    Commented May 9, 2018 at 18:43
  • 2
    \$\begingroup\$ I added reinventing-the-wheel, because that's what you're doing here: inventing this even though a standard implementation exists. In the future, you may want to add this tag yourself, to signify to other users that you're knowingly not using the standard implementation. \$\endgroup\$
    – Daniel
    Commented May 9, 2018 at 19:28

1 Answer 1


Of course, the most efficient solution would be to let python parse the binary string as integer ;)

binaryToDecimal = lambda x: int(x,2)

But you already know that, so what about simplifying your code like this:

binaryToDecimal = lambda binary: sum(1<<(len(binary)-index-1) for index, char in enumerate(binary) if char=='1')
  • removing the creation of the inner list comprehension just to pick one value
  • not using power but bitshift (integer computation, faster, more accurate)
  • don't create a list to pass to sum, pass a generator comprehension (it's faster, saves memory)
  • in your condition, don't convert your number to integer, just test against the char value (it's faster)

Keep in mind that your parsing has no error checking!

  • \$\begingroup\$ Funny enough, I tested the character at first but moved to int to save characters (I was challenging myself to use the least characters) \$\endgroup\$
    – Hum4n01d
    Commented May 10, 2018 at 0:35
  • \$\begingroup\$ yes, it's good for code golfing, but not really for performance :) \$\endgroup\$ Commented May 10, 2018 at 15:12

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