How to return a list of numbers of the power of 2?

Question

def problem(n):
myList = []
for j in range(0, n):
    number = 2 ** j
    myList.append(number)
return myList

I want this code to return the powers

Answer 1

Yet another simple and efficient solution (every step in O(1)) without the power operator:

def problem(n):
    return [1 << i for i in range(n)]

The 1 << i operations is a bitwise operation which translates to 2 ^ i for i integer positive.

https://en.wikipedia.org/wiki/Arithmetic_shift

Answer 2

As @JBernardo pointed out, I assume there is a typo in your question.

def squares(n):
    power = n
    square_list = []
    for i in range(1,n+1):
        square_list.append(2 ** i)
    return square_list


print squares(4)

will return

[2,4,8,16]

Answer 3

just use range(1,n+1) and it should all work out. range is a little confusing for some because it does not include the endpoint. So, range(3) returns the list [0,1,2] and range(1,3) returns [1,2].

---

As a side note, simple loops of the form:

out = []
for x in ...:
   out.append(...)

should typically be replaced by list comprehensions:

out = [ 2**j for j in range(1,n+1) ]

Answer 4

import math
n = input()
a = [i for i in xrange(2, n+1) if (math.log(i)/math.log(2)).is_integer()]
print a
>>> [2, 4, 8, 16 ...]

Returns list of powers of 2 less than or equal to n

Explanation:
A number can only be a power of 2 if its log divided by the log of 2 is an integer.

eg. log(32) = log(2^5) = 5 * log(2)
5 * log(2) when divided by log(2) gives 5 which is an integer.

Also there will be floor(math.log(n, 2)) elements in the list, as that is the formula for the number of powers of 2 below n if n itself is not a power of 2

Answer 5

If you want to do every step in O(1):

def gen(x):
   i = 2
   for n in range(x + 1):
       yield i
       i <<= 1

>>> list(gen(4))
[2, 4, 8, 16, 32]

PS: There's a typo in the question and if you want 4 numbers for gen(4), use range(x) instead