Q:

I want a program that writes every possible combination to a different line of a text file

I want to write a program that would print every combination of a set of variables to a text file, creating a word list. Each answer should be written on a separate line and write all of the results for 1 digit, 2 digits, and 3 digits to a single text file.
Is there a simple way I can write a python program that can accomplish this? Here is an example of the output I am expecting when printing all the binary number combinations possible for 1, 2, and 3 digits:
Output:
0  
1

00  
01  
10  
11

000  
001  
010  
011  
100  
101  
110  
111

A:

# Given two lists of strings, return a list of all ways to concatenate
# one from each.
def combos(xs, ys):
    return [x + y for x in xs for y in ys]

digits = ['0', '1']
for c in combos(digits, combos(digits, digits)):
    print c

#. 000
#. 001
#. 010
#. 011
#. 100
#. 101
#. 110
#. 111

A:

A naïve solution which solves the problem and is general enough for any application you might have is this:
def combinations(words, length):
    if length == 0:
        return []
    result = [[word] for word in words]
    while length > 1:
        new_result = []
        for combo in result:
            new_result.extend(combo + [word] for word in words)
        result = new_result[:]
        length -= 1
    return result

Basically, this gradually builds up a tree in memory of all the combinations, and then returns them. It is memory-intensive, however, and so is impractical for large-scale combinations.
Another solution for the problem is, indeed, to use counting, but then to transform the numbers generated into a list of words from the wordlist. To do so, we first need a function (called number_to_list()):
def number_to_list(number, words):
    list_out = []
    while number:
        list_out = [number % len(words)] + list_out
        number = number // len(words)
    return [words[n] for n in list_out]

This is, in fact, a system for converting decimal numbers to other bases. We then write the counting function; this is relatively simple, and will make up the core of the application:
def combinations(words, length):
    numbers = xrange(len(words)**length)
    for number in numbers:
        combo = number_to_list(number, words)
        if len(combo) < length:
            combo = [words[0]] * (length - len(combo)) + combo
        yield combo

This is a Python generator; making it a generator allows it to use up less RAM. There is a little work to be done after turning the number into a list of words; this is because these lists will need padding so that they are at the requested length. It would be used like this:
>>> list(combinations('01', 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']]

As you can see, you get back a list of lists. Each of these sub-lists contains a sequence of the original words; you might then do something like map(''.join, list(combinations('01', 3))) to retrieve the following result:
['000', '001', '010', '011', '100', '101', '110', '111']

You could then write this to disk; a better idea, however, would be to use the built-in optimizations that generators have and do something like this:
fileout = open('filename.txt', 'w')
fileout.writelines(
    ''.join(combo) for combo in combinations('01', 3))
fileout.close()

This will only use as much RAM as necessary (enough to store one combination). I hope this helps.

