How to divide a list in python?

Question

I have this list example list1=[['p1', 'p2', 'p3', 'p4'], ['p5', 'p6', 'p7']] and if any variable is the same as other variable them split it in a group. Lets say p1 and p4 are the same and p5 and p6. So I want a new list that looks like list2=[['p1', 'p4'], ['p2', 'p3'], ['p5', 'p6'], 'p7']. So how I need to divide them, pls help. I am using the newest version of python.

Ok to be more specific for "the same" in my program, I use p1 and p4 and if they give the same result for a certain character then I merge them in one group. Example

if dictionary.get(p1, character) is dictionary.get(p4, character)

if you have more questions just ask me.

Answer 1

Using the unique_everseen recipe and collections.Counters:

from collections import Counter

def solve(lst):
    counters = map(Counter, lst)
    return [ [uniq]*c[uniq] for seq, c in zip(lst, counters)
                                                for uniq in unique_everseen(seq)]

Demo:

>>> print(solve([[1, 2, 2, 1], [3, 4, 3]]))
[[1, 1], [2, 2], [3, 3], [4]]
>>> print(solve([['hello', 'hello', 'hello', 'what'], ['i', 'am', 'i']]))
[['hello', 'hello', 'hello'], ['what'], ['i', 'i'], ['am']]
>>> print(solve([[1,2,3,2,1],[9,8,7,8,9],[5,4,6,4,5]]))
[[1, 1], [2, 2], [3], [9, 9], [8, 8], [7], [5, 5], [4, 4], [6]]

As you can see this also preserves the order of the items.

---

Code for unique_everseen recipe:

from itertools import filterfalse

def unique_everseen(iterable, key=None):
    "List unique elements, preserving order. Remember all elements ever seen."
    # unique_everseen('AAAABBBCCDAABBB') --> A B C D
    # unique_everseen('ABBCcAD', str.lower) --> A B C D
    seen = set()
    seen_add = seen.add
    if key is None:
        for element in filterfalse(seen.__contains__, iterable):
            seen_add(element)
            yield element
    else:
        for element in iterable:
            k = key(element)
            if k not in seen:
                seen_add(k)
                yield element

Answer 2

The following will give you that result:

list1=[[1, 2, 2, 1], [3, 4, 3]]

print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]

[OUTPUT]
[[1, 1], [2, 2], [3, 3], [4]]

Example 1

list1=[['hello', 'hello', 'hello', 'what'], ['i', 'am', 'i']]

print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]

[OUTPUT]
[['what'], ['hello', 'hello', 'hello'], ['i', 'i'], ['am']]

Example 2

list1=[[1,2,3,2,1],[9,8,7,8,9],[5,4,6,4,5]]

print [[item]*lst.count(item) for lst in list1 for item in list(set(lst))]

[OUTPUT]
[[1, 1], [2, 2], [3], [8, 8], [9, 9], [7], [4, 4], [5, 5], [6]]