How to generate 5-character strings combinations (1 digit, two equal letters and two different equal letters) without duplication

问题

I am trying to generate combinations of a 5-character strings consisting of four letters (exactly two are equal and another repeated two are equal) and one digit.

Example for CORRECT combinations:

1aabb  
b1aab  
ca3ac  

Example for INCORRECT combinations:

1aaaa  -> incorrect because there are more than 2 equal letters
1aaab  -> Same as the previous
1abcd  -> No 2 equal letters + 2 equal different letters  

This is the code I am using:

from itertools import combinations, permutations, product

LETTERS = 'bcdfghjklmnpqrstvwxz'
DIGITS = '2456789'

def aabb1(letters=LETTERS, digits=DIGITS):
    """Generate the distinct 5-character strings consisting of four
    letters (exactly two are equal and another repeat two are equal) and one digit.

    """
    combs = []
    for (a, b), (i, j), (x, y), d, k in product(
            permutations(letters, 2),   # Two letters (a repeated).
            combinations(range(4), 2),  # Positions for the repeated letter.
            combinations(range(2), 2),  # Positions for the second repeated letter.
            digits,                     # One digit.
            range(5)):                  # Positions for the digit.
        result = []
        result[i:i] = a,
        result[j:j] = a,
        result[x:x] = b,
        result[y:y] = b,
        result[k:k] = d,
        combs.append(''.join(result))

    print(len(combs))
    return combs

It prints that I have 79,800 combinations but this is incorrect because I am counting duplicated combinations:

The problem is because it chooses some letter, for example a to appear twice and then repeated letter, like f, to appear twice so we will get something like: a3faf but later it chooses the first letter as f and the second as a and get again a3faf.
In math I can solve it with dividing by 2:

But not sure how to do it properly in my code.

Can you suggest how I can prevent it in my code ? Meaning, get the combinations without duplication.

回答1:

Change permutations(letters, 2) to combinations(letters, 2). permutations() will deliver ('a', 'b') and ('b', 'a'), but combinations() will deliver just ('a', 'b'). Your combinations for letter positions takes care of all orderings of those letters so you don't need to see them twice.

Edit: In addition to the previous fix, calculating the positions of the second letter based on the first letter finally fixes it. So if 'a' is at index 0 and 2 then 'b' must be at index 1 and 4.

def aabb1(letters=LETTERS, digits=DIGITS):
    """Generate the distinct 5-character strings consisting of four
    letters (exactly two are equal and another repeat two are equal) and one digit.

    """
    letterdxs = set(range(4))
    combs = []
    for (a, b), (i, j), d, k in product(
            combinations(letters, 2),   # Two letters (a repeated).
            combinations(range(4), 2),  # Positions for the 1st repeated letter.
            digits,                     # One digit.
            range(5)):                  # Positions for the digit.
        x, y = letterdxs.difference((i, j))
        result = []
        result[i:i] = a,
        result[j:j] = a,
        result[x:x] = b,
        result[y:y] = b,
        result[k:k] = d,
        combs.append(''.join(result))
    print(len(combs))
    return combs

回答2:

You can write a recursive function:

#1aabb  
#b1aab  
#ca3ac  
from collections import Counter
LETTERS = 'bcdfghjklmnpqrstvwxz'
DIGITS = '2456789'
def combinations(d, current = []):
   if len(current) == 5:
      yield ''.join(current)
   else:
      for i in d:
        _d = Counter(current)
        if i.isdigit() and not any(c.isdigit() for c in current):
          yield from combinations(d, current+[i])
        elif (not current or _d.get(i, 0) == 1 or sum(c.isalpha() for c in current) < 2) and i.isalpha():
          yield from combinations(d, current+[i])

result = list(combinations(LETTERS+DIGITS))

Output (first 100 results):

['bcbc2', 'bcbc4', 'bcbc5', 'bcbc6', 'bcbc7', 'bcbc8', 'bcbc9', 'bcb2c', 'bcb4c', 'bcb5c', 'bcb6c', 'bcb7c', 'bcb8c', 'bcb9c', 'bccb2', 'bccb4', 'bccb5', 'bccb6', 'bccb7', 'bccb8', 'bccb9', 'bcc2b', 'bcc4b', 'bcc5b', 'bcc6b', 'bcc7b', 'bcc8b', 'bcc9b', 'bc2bc', 'bc2cb', 'bc4bc', 'bc4cb', 'bc5bc', 'bc5cb', 'bc6bc', 'bc6cb', 'bc7bc', 'bc7cb', 'bc8bc', 'bc8cb', 'bc9bc', 'bc9cb', 'bdbd2', 'bdbd4', 'bdbd5', 'bdbd6', 'bdbd7', 'bdbd8', 'bdbd9', 'bdb2d', 'bdb4d', 'bdb5d', 'bdb6d', 'bdb7d', 'bdb8d', 'bdb9d', 'bddb2', 'bddb4', 'bddb5', 'bddb6', 'bddb7', 'bddb8', 'bddb9', 'bdd2b', 'bdd4b', 'bdd5b', 'bdd6b', 'bdd7b', 'bdd8b', 'bdd9b', 'bd2bd', 'bd2db', 'bd4bd', 'bd4db', 'bd5bd', 'bd5db', 'bd6bd', 'bd6db', 'bd7bd', 'bd7db', 'bd8bd', 'bd8db', 'bd9bd', 'bd9db', 'bfbf2', 'bfbf4', 'bfbf5', 'bfbf6', 'bfbf7', 'bfbf8', 'bfbf9', 'bfb2f', 'bfb4f', 'bfb5f', 'bfb6f', 'bfb7f', 'bfb8f', 'bfb9f', 'bffb2', 'bffb4']

回答3:

For fixed length and format this straightforward code generates 39900 combinations:

LETTERS = 'bcdfghjklmnpqrstvwxz'
DIGITS = '2456789'

def insdig(s, d):
    for i in range(5):
        ss = s[:i] + d + s[i:]
        print(ss)

def aabb1():
    for dig in DIGITS:
        for i in range(len(LETTERS)-1):
            for j in range(i+1, len(LETTERS)):
                a = LETTERS[i]
                b = LETTERS[j]
                insdig(a+a+b+b, dig)
                insdig(a+b+a+b, dig)
                insdig(b+a+a+b, dig)
                insdig(a+b+b+a, dig)
                insdig(b+a+b+a, dig)
                insdig(b+b+a+a, dig)

aabb1()

来源：https://stackoverflow.com/questions/52224466/how-to-generate-5-character-strings-combinations-1-digit-two-equal-letters-and