按值列出组列表

Question

假设我有一个这样的列表:

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

我怎样才能最优雅地将其分组以在Python中获取此列表输出:

list = [["A", "C"], ["B"], ["D", "E"]]

所以这些值按secound值分组但订单仍然保留......

Answer 1

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

Answer 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

.

编辑

另一个不需要导入的解决方案,更具可读性,保留订单,并且比前一个解决方案长22%:

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

Answer 3

霍华德的答案简洁而优雅,但在最坏的情况下也是O(n ^ 2).对于具有大量分组键值的大型列表,您需要先对列表进行排序,然后使用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']]

编辑:

看到eyequem的答案之后,我改变了这个: itemgetter(1)比...更好lambda x: x[1].

Answer 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']]
>>>