I have the following sequence with missing 1s and 0s:

"01010xx101xxx001"

And I need to print all possible sequences instead of the missing places marked by "x"

How can I do it if I might also need to change the number of "x"s?

CodePudding user response：

You can do the following, using itertools.product:

from itertools import product

def combs(string):
    for p in map(iter, product("01", repeat=string.count("x"))):
        yield "".join(c if c in "01" else next(p) for c in string)
        # maybe more consistent:
        # yield "".join(next(p) if c == "x" else c for c in string)

for c in combs("01010xx101xxx001"):
    print(c)

0101000101000001
0101000101001001
0101000101010001
0101000101011001
0101000101100001
0101000101101001
# ...

Some documentation on the utils used here:

• iter
• map
• next
• str.join
• str.count
• itertools.product
• generators

CodePudding user response：

There's already a pretty good answer, and I suggest you study it. Here's another solution that I think walks you a bit more through each step:

l = '1010xx101xxx001'

# count x's
count = len([c for c in l if c == 'x'])

# There will be 2 ^ {count} combinations, so iterate from 0 to 2 ^ {count} - 1
for i in range(2 ** count):
    # Get the binary representation of i
    b = str(bin(i)).replace('0b', '')

    # Pad with zero's so we get {count} digits ('1' becomes '00001')
    b_p = '0' * (count - len(b))   b

    # Replace x's one at a time
    out = l
    for digit in b_p:
        out = out.replace('x', digit, 1)
    print(out)