CODEWITHSUNDEEP
pythonbit-manipulation
user4881368
user4881368

Reputation:

Multiply list with binary

I am trying to create a powerset by multiplying binary counter with array to get 

Value of Counter            Subset
    000                    -> Empty set
    001                    -> a
    010                    -> b
    011                    -> ab
    100                    -> c
    101                    -> ac
    110                    -> bc
    111                    -> abc

My code:

A = ['a','b','c']
n = len(A)
for i in range(2**n):
    print format(i,'b') and A

I get all the outputs as ['a','b','c'] . Is there a different way to do the print statement to get the required result.

Upvotes: 1

Views: 192

Answers (3)

Abhijith Asokan
Abhijith Asokan

Reputation: 1875

You cant use bitwise operation between a string and a list.

I tried the following way.

A = ['a','b','c']
n = len(A)
for i in range(2**n):
    s = format(i,'b')[::-1]
    print(''.join( (A[j] for j in range(len(s)) if s[j]=='1' )))

Upvotes: 0

FHTMitchell
FHTMitchell

Reputation: 12157

Try this

d = {0b1: 'a', 0b10: 'b', 0b100: 'c'}
for i in range(0b1000):
    s = ''
    for k, v in d.items():
        if i & k:
             s += v
    print s

Edit: Some explination

So you're trying to associate 'a', 'b', 'c' to 1, 2 and 4 respectively. To do this I used the dictionary above (using binary literals for clarity).

Then, looping through all 0 <= i < 8, you can see if there is a 1 at the relevant position using the & operator.

e.g. is there a 1 at the second position from the right: 0b00 & 0b10 == 0b01 & 0b10 == 0 is false, whereas 0b10 & 0b10 == 0b11 & 0b10 != 0is true.

Upvotes: 0

tobias_k
tobias_k

Reputation: 82899

The expression format(i,'b') and A is not the same as "bitwise-and of the binary number i with the elements of list A". X and Y is logical-and and is interpreted as Y if X else X, thus you always get A as a result.

You can use a list comprehension, checking the bitwise & of 2**i (i being the current index) and the running number. Also, no explicit conversion to binary is needed.

>>> [[e for i, e in enumerate(A) if 2**i & x] for x in range(2**n)]
[[], ['a'], ['b'], ['a', 'b'], ['c'], ['a', 'c'], ['b', 'c'], ['a', 'b', 'c']]

Upvotes: 1

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