[Tutor] for loop results into list

[Micheal Beatty]

  On 09/05/2010 02:22 PM, Evert Rol wrote:
>>>> I'm having a little problem figuring out how to accomplish this simple task. I'd like to take a list of 6 numbers and add every permutation of those numbers in groups of four. For example for 1, 2, 3, 4, 5, 6 add 1 + 1 + 1 +1 then 1 + 1 + 1 +2 etc. until reaching 6 + 6 + 6 + 6. Using a for loop, that was the easy part, now I'd like to take the results and count the number of times each number occurs.
>>>> My problem occurs when I try to create a list from the results of the for loop, it puts each individual number into its own list. I've looked everywhere for the solution to this and can find nothing to help.
>>>>
>>>> Any suggestions would be much appreciated
>>> If you had some code, that would be very helpful. Now it's a bit of guesswork what exactly you have (code tends to be clearer than a full paragraph or two of text).
>>> At least, I currently don't understand what your problem is (or what your for-loop involves).
>>> Eg, are you looping and calling a function recursively, do you have four nested loops (or nested list comprehensions)? Or some other convenient loop to step through all combinations?
>>>
>>> Anway, if you have a recent Python version (2.7 or 3.1), the itertools module provides a handy utiity: http://docs.python.org/py3k/library/itertools.html#itertools.combinations_with_replacement
>>> Eg,
>>>
>>>>>> map(sum, combinations_with_replacement(range(1,7), 4))
>>> [4, 5, 6, 7, 8, 9, 6, 7, 8, 9, 10, 8, 9, 10, 11, 10, 11, 12, 12, 13, 14, 7, 8, 9, 10, 11, 9, 10, 11, 12, 11, 12, 13, 13, 14, 15, 10, 11, 12, 13, 12, 13, 14, 14, 15, 16, 13, 14, 15, 15, 16, 17, 16, 17, 18, 19, 8, 9, 10, 11, 12, 10, 11, 12, 13, 12, 13, 14, 14, 15, 16, 11, 12, 13, 14, 13, 14, 15, 15, 16, 17, 14, 15, 16, 16, 17, 18, 17, 18, 19, 20, 12, 13, 14, 15, 14, 15, 16, 16, 17, 18, 15, 16, 17, 17, 18, 19, 18, 19, 20, 21, 16, 17, 18, 18, 19, 20, 19, 20, 21, 22, 20, 21, 22, 23, 24]
>>>
>>> seems to do what you want.
>>>
>>> But, I'd still say to adopt your own code first, and when you've learned from that, just use the one-liner above. You're most welcome to ask your question, best done in combination with code, actual output and expected output. Then we can point you in the right direction.
>>>
>>> Cheers,
>>>
>>>    Evert
>>>
>> Thanks Evert, here is the code.
>>
>>
>> fourdsix = [1, 2, 3, 4, 5, 6]
>> for i in fourdsix:
>>     for j in fourdsix:
>>         for k in fourdsix:
>>             for l in fourdsix:
>>                 fourdsix_result = [i, j, k, l]
>>                 attribs = sum(fourdsix_result) - min(fourdsix_result)
>>                 print attribs
>>
>> This gives me the proper results,
> I'm not sure I understand that, because now you have a subtraction here; not sure what that does.

It removes the lowest number in the group of 4. It's intended to 
replicate rolling four six sided dice, eliminating the lowest number and 
adding the remaining 3 dice roll results. I didn't mention it cause I 
felt it wasn't germane to the problem and it might needlessly complicate 
the situation.

>> now it's just a matter of getting it into a list so I can further work with the data.
>> I've tried the following
>> attrib_list = [attribs]
>>
>> and
>> attrib_list = []
>> attrib_list.append(attribs)
>> print attrib_list
> This one should work, unless you've put it incorrectly inside the code. But otherwise, it will create the list you're looking for.
> So then it's a matter of stepping through the numbers inside the list and counting their occurrences.

I've tried this in every place in the code, outside the loop, inside at 
each step of the loop and none of them get me what I want. Andre's 
suggestion that I put the empty list before the loop helped but it's 
still giving me multiple lists.
> In fact, that could be done straight inside the quadruple nested for loops. One way I can think of, is using a dictionary, where the sum of the four-element list is the key, and the value increases by one each time. Eg:
>
> try:
>     counter[attribs] += 1
> except KeyError:
>     counter[attribs] = 1
>
> or with dict.setdefault:
>
> counter2[attribs] = counter2.setdefault(attribs, 0) + 1
>
>
> (oh, and sorry: I missed the part where you wanted to count the number of times the sum appears in your initial email.)
>
>> but these both only create a list of the last number.
>>
>> I'm sure there is a pre-existing module that will do all of this for me but I'm trying to learn by doing.
> Definitely.
> It's often fun to be surprised later by the amount of utilities available in the Python standard library, that would have solved a problem in a one-liner (also a bit frustrating, but not too much).
>
> Cheers,
>
>    Evert
>