Itertools - Merge two list to get all possible combinations

Question

I have two lists: a and b.

a is a list which contains three or more string, while b is a list of separators.

I need to generate all the possible combinations of a and then "merge" the result with all possible combination of b (See the example for better understanding).

I ended up using this code:

from itertools import permutations, combinations, product

a = ["filename", "timestamp", "custom"]
b = ["_", "-", ".", ""]

output = []

for com in combinations(b, len(a) - 1):
    for per in product(com, repeat=len(a) - 1):
        for ear_per in permutations(a):
            out = ''.join(map(''.join, zip(list(ear_per[:-1]), per))) + list(ear_per)[-1]
            output.append(out)

# For some reason the algorithm is generating duplicates
output = list(dict.fromkeys(output))

for o in output:
    print o

This is a sample of the output (which is correct, it is what I need in this case):

timestamp.customfilename
filenamecustom.timestamp
custom_filenametimestamp
timestamp_custom_filename
timestamp-filename.custom
custom_filename-timestamp
filename.timestamp-custom
. . .
filename.custom.timestamp
filename-customtimestamp
custom-timestamp_filename
filename_custom-timestamp
filename.timestampcustom
timestampcustom-filename
custom-timestamp.filename
filenamecustom_timestamp
timestamp.custom_filename
custom.timestampfilename
timestampfilename.custom
customfilename_timestamp
filenametimestamp-custom
custom-filenametimestamp
timestampfilename-custom
timestamp-custom-filename
custom.filenametimestamp
customfilenametimestamp
timestampfilename_custom
custom_filename.timestamp
custom-timestamp-filename
custom-timestampfilename
filename_timestamp.custom
. . .
filename.custom-timestamp
timestamp_filenamecustom
custom_timestampfilename
timestamp.custom.filename
timestamp.filename-custom
filename-custom-timestamp
customfilename.timestamp
filename_timestamp_custom
timestamp_filename.custom
customtimestampfilename
filenamecustomtimestamp
custom.timestamp_filename
filename_customtimestamp
. . .
timestamp-customfilename
filename_custom.timestamp

There are two main problems with this algorithm:

1. It generates some duplicated lines, so I always need to delete them (slow on bigger sets of data)

2. if len(a) > len(b) + 2 the script won't start. I that case I would need to repeat the separator to cover len(a) - 1 available spaces between words contained in a.

Answer 1

You may be looking for this:

a = ["filename", "timestamp", "custom"]
b = ["_", "-", ".", ""]
count = 0

def print_sequence(sol_words, sol_seps):
  global count 
  print("".join([sol_words[i] + sep for (i, sep) in enumerate(sol_seps)] + [sol_words[-1]]))
  count += 1

def backtrack_seps(sol_words, seps, sol_seps, i):
  for (si, sep) in enumerate(seps):
    sol_seps[i] = sep

    if i == len(sol_words) - 2:
      print_sequence(sol_words, sol_seps)
    else:
      backtrack_seps(sol_words, seps, sol_seps, i + 1)

def bt_for_sep(sol_words, seps):
  backtrack_seps(sol_words, seps, [''] * (len(sol_words) - 1), 0)

def backtrack_words(active, words, seps, sol_words, i):
  for (wi, word) in enumerate(words):
    if active[wi]:
      sol_words[i] = word
      active[wi] = False

      if i == len(words) - 1:
        bt_for_sep(sol_words, seps)
      else:
        backtrack_words(active, words, seps, sol_words, i + 1)

      active[wi] = True

backtrack_words([True] * len(a), set(a), set(b), [''] * len(a), 0)
print(count) #96

Usually when you need to enumerate all the possibilities of a certain set of values, backtracking is the way. The scheme for backtracking is always the same, and it is repeated for separators after using it to permute words.

---

EDIT

The second part of the problem, described as finding the combinations of separators, is actually the problem of finding all the dispositions with repetitions. Doing this was simpler than I thought: after you choose a separator from seps, instead of removing it (or disable it in this case), you just keep it in.

Answer 2

This may be a possible solution. It takes the permutations of a, (3*2 = 6), interleaved with the product of b (2 at a time here, 4*4 == 16), to get a total of 6 * 16 == 96 results.

from itertools import permutations, chain, zip_longest, product

a = ["filename", "timestamp", "custom"]
b = ["_", "-", ".", ""]

i=0
for perm in permutations(a):
    for prod in product(b,repeat=len(a)-1):
        tpls = list(chain.from_iterable(zip_longest(perm, prod, fillvalue='')))
        print(''.join(tpls))
        i += 1
print(i)