从列表字母中打印所有可能的长度为 10 的单词的组合，其中重复“A”两次

Question

I have a list of 5 letters ['A', 'B', 'N', 'M','E'].

I want to print all the words (word means a sequence of letters, it doesn't have to be a valid English word) of length 10 letters that have exactly two letters A. Order is important.

I have tried with itertools.product as it appeared to be the most promising solution:

from itertools import product

letters = ['A', 'B', 'N', 'M','E']

for word in product(letters, repeat=10):
    res = ''.join(str(x) for x in word)
    print(res)

The problem with this approach is that I can't really control the number of occurrences of the letter A as it returns the word composed of 10 letters of A.

Is there a solution for this? Thanks

EDIT 1 Example of possible words: BANAMEMNEB: it has only twice the letter A, we don't care about other letters.

Answer 1

The way to do it efficiently is to iterate over all combinations of the positions of the letters 'A', and for each such combination - iterate over all the ways other letters can be positioned, and just insert the 'A's there.

Be warned that with your inputs this will produce almost 3 million words!

On my machine it was printing the words for so long that I had to manually stop execution.

So here is the smaller example:

letters = ['A', 'b', 'c']
num_As = 2
num_total = 4

from itertools import combinations, product
for indices_A in combinations(range(num_total), num_As):
    for rest in product(letters[1:], repeat=num_total - num_As):
        rest = list(rest)
        for index_A in indices_A:
            rest.insert(index_A, 'A')
        print(''.join(rest))

AAbb
AAbc
AAcb
AAcc
AbAb
AbAc
AcAb
AcAc
AbbA
AbcA
AcbA
AccA
bAAb
bAAc
cAAb
cAAc
bAbA
bAcA
cAbA
cAcA
bbAA
bcAA
cbAA
ccAA

Answer 2

This is basically @Vladimir-Fokow's answer translated to your use case:

#!/usr/bin/env python
from itertools import combinations, product
letters =  ['A', 'B', 'N', 'M','E']
first, rest = letters[0], letters[1:]
total_length = 10
first_count = 2
for others in product(rest, repeat=total_length - first_count):
  for indices in combinations(range(total_length),first_count):
    result = list(others)
    for i in indices:
      result.insert(i,first)
    print(''.join(result))

On my iMac this only took about 3 seconds to print out all 2,949,120 words.

Answer 3

A solution would be to filter out the words that do not contain exactly two 'A', as in the following simple code:

from itertools import product
from collections import Counter

letters = ['A', 'B', 'N', 'M','E']

for word in product(letters, repeat=10):
    res = ''.join(word)  # equivalent to res = ''.join(str(x) for x in word) !
    ctr = Counter(res)
    if ctr['A'] == 2:
        print(res)

This can also be written with one-line style coding, as follows:

res = [''.join(w) for w in product(letters, repeat=10) if Counter(w)['A'] == 2]

Answer 4

First determine the two indices where the letter "A" will occur, then produce the letters at the other 8 indices:

from itertools import combinations, product

def solve():
    for i, j in combinations(range(10), 2):
        for b in map(list, product("BNME", repeat=8)):
            b.insert(i, "A")
            b.insert(j, "A")
            yield "".join(b)

To print:

for s in solve():
    print(s)

...but don't wait for it. There are 9*10/2 * 4 ⁸ possible combinations, which is almost 3 million.