Group list by values

Solution 1:

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

Solution 2:

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

Solution 3:

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].

Solution 4:

>>> import collections
>>> D1 = collections.defaultdict(list)
>>> for element in L1:
...     D1[element[1]].append(element[0])
... 
>>> L2 = D1.values()
>>> print L2
[['A', 'C'], ['B'], ['D', 'E']]
>>>

Solution 5:

I don't know about elegant, but it's certainly doable:

oldlist = [["A",0], ["B",1], ["C",0], ["D",2], ["E",2]]
# change into: list = [["A", "C"], ["B"], ["D", "E"]]

order=[]
dic=dict()
for value,key in oldlist:
  try:
    dic[key].append(value)
  except KeyError:
    order.append(key)
    dic[key]=[value]
newlist=map(dic.get, order)

print newlist

This preserves the order of the first occurence of each key, as well as the order of items for each key. It requires the key to be hashable, but does not otherwise assign meaning to it.