Question

我有一个列表：

loc_list=[['1,', "'Houston'"], 
         ['4,', "'Stafford'"], 
         ['5,', "'Bellaire'"], 
         ['5,', "'Sugarland'"], 
         ['5,', "'Houston'"], 
         ['6,', "'Atlanta'"], 
         ['6,', "'Sacramento'"], 
         ['7,', "'Milwaukee'"], 
         ['8,', "'Chicago'"], 
         ['8,', "'Dallas'"], 
         ['8,', "'Philadephia'"], 
         ['8,', "'Seattle'"], 
         ['8,', "'Miami'"], 
         ['9,', "'Arlington'"], 
         ['11,', "'Austin'"]]

现在我想在第一个位置合并具有相同元素的列表。

Desired_list=[['1,', "'Houston'"], 
             ['4,', "'Stafford'"], 
             ['5,', "'Bellaire','Sugarland','Houston'"],
             ['6,', "'Atlanta','Sacramento'"],
             ['7,', "'Milwaukee'"], 
             ['8,', "'Chicago','Dallas','Philadephia','Seattle','Miami'"], 
             ['9,', "'Arlington'"], 
             ['11,', "'Austin'"]]

我很无能。

Answer 1

你可以这样做：

desired_list = {}
for elem in loc_list:
  if not elem[0] in desired_list:
    desired_list[elem[0]] = []
  desired_list[elem[0]].append(elem[1])


print desired_list

这不是列表而是字典输出是：

{'9,': ["'Arlington'"], '8,': ["'Chicago'", "'Dallas'", "'Philadephia'", "'Seattle'", "'Miami'"], '11,': ["'Austin'"], '7,': ["'Milwaukee'"], '6,': ["'Atlanta'", "'Sacramento'"], '5,': ["'Bellaire'", "'Sugarland'", "'Houston'"], '4,': ["'Stafford'"], '1,': ["'Houston'"]}

Answer 2

分开您的疑虑。根据您的索引范围，只需使用数组或哈希表。

然后对于每个项目，你可以拥有一套，并根据你的愿望，你可以在你的集合中具有多样性。

对于初始构造，您需要遍历列表。

Answer 3

使用类似dict的结构会更好，defaultdict set似乎是最佳选择。

以下是一个例子：

>>> from collections import defaultdict
>>> d = defaultdict(set)
>>> for index,city in loc_list:
...   d[index].add(city)
... 
>>> for k,v in d.items():
...   print(k,v)
... 
4, {"'Stafford'"}
6, {"'Atlanta'", "'Sacramento'"}
1, {"'Houston'"}
5, {"'Houston'", "'Bellaire'", "'Sugarland'"}
9, {"'Arlington'"}
8, {"'Philadephia'", "'Dallas'", "'Seattle'", "'Chicago'", "'Miami'"}
7, {"'Milwaukee'"}
11, {"'Austin'"}

BTW：您的输入列表看起来有点奇怪，键是带有尾随昏迷的字符串（而不是数字）：'4,'; 值有双引号："'Chicago'"。

Answer 4

我已经写过这个解决方案，我觉得它非常清晰有效。使用defaultdict（集合包中的数据类型），您可以轻松获得相同的结果：

import collections as co

loc_list=[['1,', "'Houston'"], 
         ['4,', "'Stafford'"], 
         ['5,', "'Bellaire'"], 
         ['5,', "'Sugarland'"], 
         ['5,', "'Houston'"], 
         ['6,', "'Atlanta'"], 
         ['6,', "'Sacramento'"], 
         ['7,', "'Milwaukee'"], 
         ['8,', "'Chicago'"], 
         ['8,', "'Dallas'"], 
         ['8,', "'Philadephia'"], 
         ['8,', "'Seattle'"], 
         ['8,', "'Miami'"], 
         ['9,', "'Arlington'"], 
         ['11,', "'Austin'"]]


def transform_list():
    count = co.defaultdict(list)

    for sublist in loc_list:
        count[sublist[0]].append(sublist[1])

    return count

if __name__ == '__main__':
    count = transform_list()
    print(count)

将列表传递给defaultdict构造函数，您可以在指定新密钥时自动创建空列表。我建议你考虑这个例子：defaultdict examples

合并共享公共索引元素的列表中的子列表

4 个答案: