Binary to Decimal For Loop

Question

Hi I am a new programmer and I trying to complete one my assignments which requires to me convert a binary number to a denary one. I am not getting any errors however i dont get the correct denary equivalent, please help. This is what i've done so far.

binary = "10111"
denary = 0
length=len(binary)
for i in range(length-1,-1,-1):
 if binary[i] == "1":
 denary += (2**i)
else:
 denary += 0
print(denary)

and the output is:

29

• juanpa.arrivillaga: The problem is that i will corresponding to the decimal digit in the reversed order...
• dan04: I suppose that your assignment doesn't let you use int(binary, 2)?
• armallah: @dan04 yes, we were told not use that
• armallah: @usr2564301 no, i just needed it for python 3, sorry for tagging python 2.7

Answer 1

You're coming from the wrong direction. You can use binary[::-1] or reversed(binary) to reverse the array.

binary = "10111"
denary = 0
for i, d in enumerate(reversed(binary)):
 if d == "1":
 denary += (2**i)
print(denary)

---

Also note that you can do this:

denary = int(binary, 2) # Parses string on base 2 to integer base 10
print(denary)

• Nathan Vērzemnieks: Nice and simple, although enumerate(reversed(binary)) is, in my opinion, more readable. (It also looks to be about 20% faster.)
• Stefan Falk: @NathanVērzemnieks Thank you for the hint. Going to include it in my answer :)

Answer 2

You can use a reverse list like this:

binary = "10111" # needs to be reversed so the lowest bit is in front for ease of computing
denary = 0
# ind = index, bit = the bitvalue as string of the reversed string 
for ind, bit in enumerate(binary[::-1]): # reversed copy of string
 denary += int(bit)*2**ind # if bit is 0 this evaluates to 0, else to the power of 2
print(denary)