How can I find same values in a list and group together a new list?

Question

From this list:

N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5] 

I'm trying to create:

L = [[1],[2,2],[3,3,3],[4,4,4,4],[5,5,5,5,5]] 

Any value which is found to be the same is grouped into it's own sublist. Here is my attempt so far, I'm thinking I should use a while loop?

global n  n = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5] #Sorted list l = [] #Empty list to append values to  def compare(val):    """ This function receives index values    from the n list (n[0] etc) """        global valin    valin = val     global count    count = 0      for i in xrange(len(n)):         if valin == n[count]: # If the input value i.e. n[x] == n[iteration]             temp = valin, n[count]              l.append(temp) #append the values to a new list              count +=1         else:           count +=1       for x in xrange (len(n)):     compare(n[x]) #pass the n[x] to compare function 

Answer 1

Use itertools.groupby:

from itertools import groupby  N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5]  print([list(j) for i, j in groupby(N)]) 

Output:

[[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5]] 

Side note: Prevent from using global variable when you don't need to.

Answer 2

Someone mentions for N=[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 1] it will get [[1], [2, 2], [3, 3, 3], [4, 4, 4, 4], [5, 5, 5, 5, 5], [1]]

In other words, when numbers of the list isn't in order or it is a mess list, it's not available.

So I have better answer to solve this problem.

from collections import Counter  N = [1,2,2,3,3,3,4,4,4,4,5,5,5,5,5] C = Counter(N)  print [ [k,]*v for k,v in C.items()]