Finding index of values in a list dynamically

There are a few ways to do this.

I'd do it by using zip and groupby.

First:

>>> list(zip(list_1, list_2))
[('A-1', 'iPad'),
 ('A-1', 'iPod'),
 ('A-1', 'iPhone'),
 ('A-2', 'Windows'),
 ('A-2', 'X-box'),
 ('A-3', 'Kindle')]

Now:

>>> import itertools, operator
>>> [(key, list(group)) for key, group in 
...  itertools.groupby(zip(list_1, list_2), operator.itemgetter(0))]
[('A-1', [('A-1', 'iPad'), ('A-1', 'iPod'), ('A-1', 'iPhone')]),
 ('A-2', [('A-2', 'Windows'), ('A-2', 'X-box')]),
 ('A-3', [('A-3', 'Kindle')])]

So, you just want each group, ignoring the key, and you only want the second element of each element in the group. You can get the second element of each group with another comprehension, or just by unzipping:

>>> [list(zip(*group))[1] for key, group in
...  itertools.groupby(zip(list_1, list_2), operator.itemgetter(0))]
[('iPad', 'iPod', 'iPhone'), ('Windows', 'X-box'), ('Kindle',)]

I would personally find this more readable as a sequence of separate iterator transformations than as one long expression. Taken to the extreme:

>>> ziplists = zip(list_1, list_2)
>>> pairs = itertools.groupby(ziplists, operator.itemgetter(0))
>>> groups = (group for key, group in pairs)
>>> values = (zip(*group)[1] for group in groups)
>>> [list(value) for value in values]

… but a happy medium of maybe 2 or 3 lines is usually better than either extreme.

Usually I'm the one rushing to a groupby solution ;^) but here I'll go the other way and manually insert into an OrderedDict:

list_1 = ['A-1','A-1','A-1','A-2','A-2','A-3']
list_2 = ['iPad','iPod','iPhone','Windows','X-box','Kindle']

from collections import OrderedDict

d = OrderedDict()
for code, product in zip(list_1, list_2):
    d.setdefault(code, []).append(product)

produces a d looking like

>>> d
OrderedDict([('A-1', ['iPad', 'iPod', 'iPhone']), 
             ('A-2', ['Windows', 'X-box']), ('A-3', ['Kindle'])])

with easy access:

>>> d["A-2"]
['Windows', 'X-box']

and we can get the list-of-lists in list_1 order using .values():

>>> d.values()
[['iPad', 'iPod', 'iPhone'], ['Windows', 'X-box'], ['Kindle']]

If you've noticed that no one is telling you how to make a bunch of independent lists with names like list_a1 and so on-- that's because that's a bad idea. You want to keep the data together in something which you can (at a minimum) iterate over easily, and both dictionaries and list of lists qualify.

Maybe something like this?

#!/usr/local/cpython-3.3/bin/python

import pprint
import collections

def main():
    list_1 = ['A-1','A-1','A-1','A-2','A-2','A-3']
    list_2 = ['iPad','iPod','iPhone','Windows','X-box','Kindle']

    result = collections.defaultdict(list)
    for list_1_element, list_2_element in zip(list_1, list_2):
        result[list_1_element].append(list_2_element)

    pprint.pprint(result)


main()

Using itertools.izip_longest and itertools.groupby:

>>> from itertools import groupby, izip_longest
>>> inds = [next(g)[0] for k, g in groupby(enumerate(list_1), key=lambda x:x[1])]

First group items of list_1 and find the starting index of each group:

>>> inds
[0, 3, 5]

Now use slicing and izip_longest as we need pairs list_2[0:3], list_2[3:5], list_2[5:]:

>>> [list_2[x:y] for x, y in izip_longest(inds, inds[1:])]
[['iPad', 'iPod', 'iPhone'], ['Windows', 'X-box'], ['Kindle']]

To get a list of dicts you can something like:

>>> inds = [next(g) for k, g in groupby(enumerate(list_1), key=lambda x:x[1])]
>>> {k: list_2[ind1: ind2[0]] for (ind1, k), ind2 in
                                   zip_longest(inds, inds[1:], fillvalue=[None])}
{'A-1': ['iPad', 'iPod', 'iPhone'], 'A-3': ['Kindle'], 'A-2': ['Windows', 'X-box']}

You could do this if you want simple code, it's not pretty, but gets the job done.

list_1 = ['A-1','A-1','A-1','A-2','A-2','A-3']
list_2 = ['iPad','iPod','iPhone','Windows','X-box','Kindle']
list_1a = []
list_1b = []
list_1c = []
place = 0
for i in list_1[::1]:
    if list_1[place] == 'A-1':
        list_1a.append(list_2[place])
    elif list_1[place] == 'A-2':
        list_1b.append(list_2[place])
    else:
        list_1c.append(list_2[place])
    place += 1