Group list by values

values = set(map(lambda x:x[1], mylist))
newlist = [[y[0] for y in mylist if y[1]==x] for x in values]

from operator import itemgetter
from itertools import groupby

lki = [["A",0], ["B",1], ["C",0], ["D",2], ["E",2]]
lki.sort(key=itemgetter(1))

glo = [[x for x,y in g]
       for k,g in  groupby(lki,key=itemgetter(1))]

print glo

.

EDIT

Another solution that needs no import , is more readable, keeps the orders, and is 22 % shorter than the preceding one:

oldlist = [["A",0], ["B",1], ["C",0], ["D",2], ["E",2]]

newlist, dicpos = [],{}
for val,k in oldlist:
    if k in dicpos:
        newlist[dicpos[k]].extend(val)
    else:
        newlist.append([val])
        dicpos[k] = len(dicpos)

print newlist

Howard's answer is concise and elegant, but it's also O(n^2) in the worst case. For large lists with large numbers of grouping key values, you'll want to sort the list first and then use itertools.groupby:

>>> from itertools import groupby
>>> from operator import itemgetter
>>> seq = [["A",0], ["B",1], ["C",0], ["D",2], ["E",2]]
>>> seq.sort(key = itemgetter(1))
>>> groups = groupby(seq, itemgetter(1))
>>> [[item[0] for item in data] for (key, data) in groups]
[['A', 'C'], ['B'], ['D', 'E']]

Edit:

I changed this after seeing eyequem's answer: itemgetter(1) is nicer than lambda x: x[1].