3

我有一个排序列表，其中包含重复的第一个元素。目前我正在迭代它以获得解决方案。

[['5th ave', 111, -30.00, 38.00],
['5th ave', 222, -30.00, 33.00],
['6th ave', 2224, -32.00, 34.90]]

我想要一个优雅的列表理解，将其转换为基于第一个元素的列表列表：

['5th ave', [[111, -30.00, 38.00] , [222, -30.00, 33.00]]

谢谢

4

3 回答 3

8

看起来像一份工作collections.defaultdict：

>>> from collections import defaultdict
>>> L = [['5th ave', 111, -30.00, 38.00],
... ['5th ave', 222, -30.00, 33.00],
... ['6th ave', 2224, -32.00, 34.90]]
>>> d = defaultdict(list)
>>> for sublist in L:
...     d[sublist[0]].append(sublist[1:])
... 
>>> print d.items()
[('5th ave', [[111, -30.0, 38.0], [222, -30.0, 33.0]]), ('6th ave', [[2224, -32.0, 34.9]])]

绝对没有理由进行列表理解。仅仅因为它的行数较少并不意味着它更 Pythonic。

于 2013-09-02T05:52:51.860 回答
1 
data = [['5th ave', 111, -30.00, 38.00],
['5th ave', 222, -30.00, 33.00],
['6th ave', 2224, -32.00, 34.90]]

previous   = ""
listOfData = []
result     = []
for currentItem in data:
    if currentItem[0] != previous:
        if listOfData:
            result.append([previous, listOfData])
            listOfData = []
        previous = currentItem[0]
    listOfData.append(currentItem[1:])

if listOfData:
    result.append([previous, listOfData])

print result

输出

[['5th ave', [[111, -30.0, 38.0], [222, -30.0, 33.0]]], ['6th ave', [[2224, -32.0, 34.9]]]]

这也维持了秩序。

编辑：

我defaultdict可以减少几行

from collections import defaultdict

data = [['5th ave', 111, -30.00, 38.00],
['5th ave', 222, -30.00, 33.00],
['6th ave', 2224, -32.00, 34.90]]

unique, Map = [], defaultdict(list)
for item in data:
    if item[0] not in unique: unique.append(item[0])
    Map[item[0]].append(item[1:])
print [(item, Map[item]) for item in unique]

这仍然维持秩序。

于 2013-09-02T06:12:02.290 回答
1

collections.defaultdict确实是要走的路，但我觉得它可能会更慢，这就是我想出这个的原因：

from itertools import imap

def RemDup(L):
    ListComp = {}
    for sublist in L:
        try: ListComp[sublist[0]].append(sublist[1:])
        except KeyError: ListComp[sublist[0]] = [sublist[1:]]
    return imap( list, ListComp.items() )

DupList = [['5th ave', 111, -30.00, 38.00],
['5th ave', 222, -30.00, 33.00],
['6th ave', 2224, -32.00, 34.90]]

print [ uniq for uniq in RemDup(DupList) ]
于 2013-09-02T06:16:29.737 回答