将嵌套列表转换为字典

Question

我试图将其转换为字典的嵌套列表如下所示：

my_dict = {}
book_ratings = [["Ben"],["5", "0", "1", "4"], ["Sally"],["0", "7", "3", "3"]]

我正在尝试 return 命名 ["Ben"]、["Sally"] 作为键和评级 ["5","0","1","4"] , ["0","7","3","3"] 作为值。

希望输出：

 {"Ben": ["5," "0", "1", "4"], "Sally": ["0", "7", "3", "3"]}

Answer 1

如果 book_ratings 的结构是 Name, List, Name, List, ... 你可以用这个例子来构造字典：

book_ratings = [["Ben"],["5", "0", "1", "4"], ["Sally"],["0", "7", "3", "3"]]

i = iter(book_ratings)
my_dict = dict((a[0], b) for a, b in zip(i, i))

print(my_dict)

打印：

{'Ben': ['5', '0', '1', '4'], 'Sally': ['0', '7', '3', '3']}

或者：

my_dict = dict((a, b) for (a,), b in zip(book_ratings[::2], book_ratings[1::2]))

Answer 2

你可以用字典理解来完成它，而不需要定义一个空的字典：

book_ratings = [["Ben"],["5", "0", "1", "4"], ["Sally"],["0", "7", "3", "3"]]
new_dict = {book_ratings[i][0]:book_ratings[i+1] for i in range(0,len(book_ratings),2)}
new_dict

输出：

{'Ben': ['5', '0', '1', '4'], 'Sally': ['0', '7', '3', '3']}

Answer 3

您可以使用 iter 和一些 zip 魔法来获取所有其他密钥。但是因为你的键在列表中，你只想要它们的单个值，你需要使用字典理解：

book_ratings = [["Ben"],["5", "0", "1", "4"], ["Sally"],["0", "7", "3", "3"]]
my_dict = {k[0]: v for k, v in zip(*([iter(book_ratings)]*2))}

{'Ben': ['5', '0', '1', '4'], 'Sally': ['0', '7', '3', '3']}

Answer 4

也许是这样的：

my_dict = {}
book_ratings = [['Ben'],['5', '0', '1', '4'], ['Sally'],['0', '7', '3', '3']]

i=0
while i<len(book_ratings):
  if not book_ratings[i][0].isnumeric():
    my_dict[book_ratings[i][0]] = book_ratings[i+1]
    i+=2
  else:
    i+=1

print(my_dict)

输出：

{'Ben': ['5', '0', '1', '4'], 'Sally': ['0', '7', '3', '3']}

Answer 5

多一个方案，对新手来说可能更简单

my_dict =  {}
book_ratings = [['Ben'],[5, 0, 1, 4], ['Sally'],[0, 7, 3, 3]]
for i, book in enumerate(book_ratings):
    if (i==0) or (i%2==0):
        try:
            my_dict[book[0]] = book_ratings[i+1]
        except:
            pass
        
print(my_dict)

打印

  {'Ben': [5, 0, 1, 4], 'Sally': [0, 7, 3, 3]}

Answer 6

简单的字典组合：

>>> it = iter(book_ratings)
>>> {k: next(it) for k, in it}
{'Ben': ['5', '0', '1', '4'], 'Sally': ['0', '7', '3', '3']}

采用已接受答案的解决方案（f1 和 f2）和我的解决方案（f3）进行基准测试，三轮，数字是以秒为单位的时间，因此更低=更快：

2.31 f1
2.08 f2
1.39 f3

2.30 f1
2.03 f2
1.34 f3

2.30 f1
2.08 f2
1.31 f3

基准代码：

from timeit import repeat

book_ratings = []
for i in range(1000):
    book_ratings += [["Ben" + str(i)],["5", "0", "1", "4"]]    

def f1():
    i = iter(book_ratings)
    return dict((a[0], b) for a, b in zip(i, i))

def f2():
    return dict((a, b) for (a,), b in zip(book_ratings[::2], book_ratings[1::2]))

def f3():
    it = iter(book_ratings)
    return {k: next(it) for k, in it}

for _ in range(3):
    for f in f1, f2, f3:
        t = min(repeat(f, number=10000))
        print('%.2f' % t, f.__name__)
    print()

将嵌套列表转换为字典

Convert nested list to dictionary

python

dictionary

nested-lists