将 dict 转换为嵌套列表，将键插入每个子列表

Question

我有解决这个问题的方法，但我很好奇是否有更好的方法。我有一个这样的字典：

dict1 = {
'key1':[['value1','value2','value3'],['value4','value5','value6']],
'key2':[['value7','value8','value9'],['value10','value11','value12']],
'key3':[['value13','value14','value15'],['value16','value17','value18']]}

我想将其转换为嵌套列表，并将键插入新的子列表，如下所示：

nestedlist = [
['value1','value2','key1','value3'],['value4','value5','key1','value6'],
['value7','value8','key1','value9'],['value10','value11','key2','value12'],
['value13','value14','key2','value15'],['value16','value17','key2','value18'],
['value10','value11','key3','value12'],['value13','value14','key3','value15'],
['value16','value17','key3','value18']]

我通过以下方式解决这个问题：

keys = [*dict1]
newlist = []
for item in keys:
    for item2 in dict1[item]:
        item2.insert(2,item)
        newlist.append(item2)

那么，我该如何改进这段代码？

Answer 1

我会做的非常相似。此处显示的唯一差异：

result = []
for k, l in dict1.items():
   for ll in l:
      ll.insert(2, k)
      result.append(ll)

无需对 dict1 进行列表解包或进行 [] 访问。 items() returns 包含字典的每个键和值的元组列表。

Answer 2

这是通过列表理解的一种方式：

res = [[w[0], w[1], k, w[2]] for k, v in dict1.items() for w in v]

# [['value1', 'value2', 'key1', 'value3'],
#  ['value4', 'value5', 'key1', 'value6'],
#  ['value7', 'value8', 'key2', 'value9'],
#  ['value10', 'value11', 'key2', 'value12'],
#  ['value13', 'value14', 'key3', 'value15'],
#  ['value16', 'value17', 'key3', 'value18']]

将 dict 转换为嵌套列表，将键插入每个子列表

Convert dict to nested list, inserting keys into each sublist

python

dictionary

python-3.5