Binary numbers of N digits

Question

for generating binary numbers in n digit
i did this to get upto 16 binary numbers.

n = 6                       # for 6 digits
for i in xrange(16):
    b = bin(i)[2:]
    l = len(b)
    b = str(0) * (n - l) + b
    print b

it results like this

000000
000001
000010
000011
000100
000101
000110
000111
001000
001001
001010
001011
001100
001101
001110
001111

but what i want is to get these values without adding a series of 0s in prefix.
can anyone help me for this.
thanks

Solution 1

Remove the line that pad 0s.

n = 6
for i in xrange(16):
    b = bin(i)[2:]
    l = len(b)
    #b = str(0) * (n - l) + b  # <--------
    print b

---

If you mean padding number without string oeprator, use str.format with b type format:

n = 6
for i in xrange(16):
    print '{:0{}b}'.format(i, n)
    # OR  print '{:06b}'.format(i)

    # OR  print '{:b}'.format(i)  if you want no leading 0s.

Solution 2

If you're asking for a different method:

n = 6
for i in xrange(16):
    b = bin(i)[2:].zfill(n)
    print b

str.zfill(n) pads the string with zeros on the left so that it is at least of length n.

---

If you just don't want the leading zeros:

for i in xrange(16):
    b = bin(i)[2:]
    print b

Solution 3

You can use list comprehension and bit conversions together like this: n = range

def binary_list(n):
    return ['{:0{}b}'.format(i, n) for i in range(n*n-1)]

print(binary_list(3)) -->
['000', '001', '010', '011', '100', '101', '110', '111']

or if you want to store each number in its own index do:

def binary_list(n):
    return [[int(j) for j in '{:0{}b}'.format(i, n)] for i in range(n*n-1)]

print(binary_list(3))--->
[[0, 0, 0], [0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0], [1, 0, 1], [1, 1, 0], [1, 1, 1]]