基於python中的子列表項將子列表合並到列表中

Question

輸入：

我有這個有序列表。

[[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

期望輸出

[[1,['A','B','D']],[2, ['A','D']], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

由於這兩個子列表的第一項是相同的。

我也可以轉換成帶有鍵和那些值對的字典。 喜歡

{1:['A','B','D'],2:['A','D'],3:['C']}

執行此操作的最簡單和最簡單的方法是什么？

Answer 1

如果數據是有序的，那么itertools.groupby是一個很好的方法：

>>> from itertools import groupby
>>> from operator import itemgetter
>>> data = [[1, 'A'], [1, 'B'], [2, 'A'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]
>>> final_data = []
>>> final_data = []
>>> for k, g in groupby(data, itemgetter(0)):
...     group = list(g)
...     if len(group) == 1:
...         final_data.append(group[0])
...     else:
...         final_data.append([k, [sub[1] for sub in group]])
...
>>> final_data
[[1, ['A', 'B']], [2, 'A'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]
>>>

如果您想要字典中的結果，那就更容易了：

>>> grouped_dict = {}
>>> for num, letter in data:
...     grouped_dict.setdefault(num, []).append(letter)
...
>>> grouped_dict
{1: ['A', 'B'], 2: ['A'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']}
>>>

Answer 2

您可以使用itertools模塊中的groupby ，如下例所示：

a = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

final = []
for k, v in groupby(sorted(a, key=lambda x: x[0]), lambda x: x[0]):
        bb = list(v)
        if len(bb) > 1:
            final.append([k, [j for _, j in bb]])
        else:
            final.append([k, bb[0][1]])

# OR:
# Within a list comprehension
# final = [[k, [j[1] for j in list(v)]] for k, v in groupby(sorted(a, key=lambda x: x[0]), lambda x: x[0])]

print(final)

輸出：

[[1, ['A', 'B', 'D']], 
[2, ['A', 'D']],
[3, 'C'], 
[4, 'D'], 
[5, 'B'], 
[6, 'D']]

然后要將最終列表轉換為字典，您可以執行以下操作：

final_dict = {k:v if isinstance(v, list) else [v] for k, v in final}
print(final_dict)

輸出：

{1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']}}

Answer 3

您可以直接從輸入創建字典。

from collections import defaultdict

input = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

d = defaultdict(list)
for el in input: d[el[0]].append(el[1])

d的輸出將是：

{1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']}

Answer 4

如果順序不重要，並且您仍然需要字典：

import collections

your_list = [[1,'A'], [1,'B'], [1,'D'], [2,'A'], [2,'D'], [3,'C'], [4,'D'], [5,'B'], [6,'D']]

result = collections.defaultdict(list)
for k, v in your_list:
    result[k].append(v)

# {1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']}

您也可以在沒有collections.defaultdict情況下執行此操作（可能會受到一些性能損失，取決於鍵頻率）：

your_list = [[1,'A'], [1,'B'], [1,'D'], [2,'A'], [2,'D'], [3,'C'], [4,'D'], [5,'B'], [6,'D']]

result = {}
for k, v in your_list:
    result[k] = result.get(k, []) + [v]

# {1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: ['C'], 4: ['D'], 5: ['B'], 6: ['D']}

Answer 5

我發現最好做相反的事情，而不是先制作一個列表，然后再制作一個字典，我先制作了字典，然后再制作了一個列表。

輸入：

in_list = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

代碼：

mydict = {}
for sublist in in_list:
    if sublist[0] in mydict.keys():
        mydict[sublist[0]] = [*mydict[sublist[0]],sublist[1]]
    else:
        mydict[sublist[0]] = sublist[1]

輸出：

>>> mydict
{1: ['A', 'B', 'D'], 2: ['A', 'D'], 3: 'C', 4: 'D', 5: 'B', 6: 'D'}

用字典做一個簡單的列表：

mylist = list(mydict.items())

輸出：

>>> mylist
[(1, ['A', 'B', 'D']), (2, ['A', 'D']), (3, 'C'), (4, 'D'), (5, 'B'), (6, 'D')]

用字典制作另一個列表：

mylist = = [[k,v] for k,v in mydict.items()]

與...一樣：

mylist = []
for key, value in mydict.items():

輸出：

>>> mylist
[[1, ['A', 'B', 'D']], [2, ['A', 'D']], [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

Answer 6

in_list
out_list = []
sublist = []
i = 0
for l in in_list:
    if l[0] != i:
        i = l[0]
        sublist = []
        out_list.append([i, sublist])
    sublist.append(l[1])
dico = dict( out_list)

Answer 7

在 python 文檔示例中，他們使用的https://docs.python.org/2/library/collections.html#defaultdict-examples解決了您帖子中的相同問題。

>>> s = [('yellow', 1), ('blue', 2), ('yellow', 3), ('blue', 4), ('red', 1)]
>>> d = defaultdict(list)
>>> for k, v in s:
...     d[k].append(v)
...
>>> d.items()
[('blue', [2, 4]), ('red', [1]), ('yellow', [1, 3])]

然后我用 defaultdict 投票了一個答案。

Answer 8

dic = {} res_lst = [] lst = [[1, 'A'], [1, 'B'],[1, 'D'], [2, 'A'],[2,'D'] , [3, 'C'], [4, 'D'], [5, 'B'], [6, 'D']]

轉換為列表對

for lst_item in lst:
    if lst_item[0] in dic:
        for item in lst_item[1:]:
            dic[lst_item[0]].append(item)
    else:
        dic[lst_item[0]] = lst_item[1:]

轉換成鍵值對

for item in dic:
    lst1 = []
    lst1.append(item)
    if(len(dic[item]) == 1):
        lst1.append(dic[item][0])
    else:
        lst1.append(dic[item])
    res_lst.append(lst1)
print(res_lst)