按值分组列表

Question

假设我有一个这样的列表：

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

如何最优雅地将其分组以在Python中获得此列表输出：

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

所以这些值按次值分组，但顺序保留下来...

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

不需要导入的另一种解决方案，它更具可读性，可以保留订单，并且比前一个短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

霍华德的答案简洁明了，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：

我在看到眼影的答案后改变了这一点：itemgetter(1)比lambda x: x[1]更好。

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']]
>>> 

Answer 5

我不了解优雅，但这当然是可行的：

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

这将保留每个键第一次出现的顺序，以及每个键的项目顺序。它要求密钥是可散列的，但不以其他方式为其赋予含义。

Answer 6

len = max(key for (item, key) in list)
newlist = [[] for i in range(len+1)]
for item,key in list:
  newlist[key].append(item)

您可以通过单个列表理解来完成，也许更优雅，但是O（n ** 2）：

[[item for (item,key) in list if key==i] for i in range(max(key for (item,key) in list)+1)]