How to split an integer into a list of 2^n?-CodePudding

I want to split a number into a sequence of 2^n values.
For example, 80 = 64 16, 123 = 64 32 16 8 2 1

I came up with a naive way of doing this like the following:

nums = []
for d in np.binary_repr(123):
    nums = [nn << 1 for nn in nums]
    nums.append(int(d))

print(nums)
>>> [64, 32, 16, 8, 0, 2, 1]

I'll probably have to write this in C too. So, I'd like to know more C ways of achieving this (in Python, for the moment)

CodePudding user response：

for (int p= 1; p <= n; p<<= 1) if (p & n) cout << p;

or if you want zeroes,

for (int p= 1; p <= n; p<<= 1) cout << p & n;

The formatting is up to you.

CodePudding user response：

You could write a one-line lambda function to do what you want:

list2n = lambda n: [2**p for p, d in enumerate(bin(n)[:1:-1]) if d == '1']
list2n(123)
list2n(0)
list2n(64)

Output:

[1, 2, 8, 16, 32, 64]
[]
[64]

Note python has a builtin bin function that does (roughly) the same thing as np.binary_repr.

CodePudding user response：

you could try something like the following :

l = 123
r= reversed(list(format(l, "b")))
k = list(reversed([int(x)*2**num for num, x in enumerate(r)]))
print(k)

->[64, 32, 16, 8, 0, 2, 1]

which could be also done on one line, but this is for sure not the most efficient way to do it, if this would be the point.

CodePudding user response：

You certainly don't need numpy for something this trivial and you definitely don't want to be converting to/from strings.

def bits(n):
    rv = []
    b = 1
    while b <= n:
        rv.insert(0, n & b)
        b <<= 1
    return rv

print(bits(123))

Output:

[64, 32, 16, 8, 0, 2, 1]

Note:

Might be better to append rather than insert and then reverse the list before returning

CodePudding user response：

One possible way to do it in C :

int p{1};        
while (n)
{
    if (n & 0x1)
    {
        std::cout << p;
    }
    else
    {
        std::cout << 0;
    }
    n >>= 1;
    p <<= 1;
}