Question

给定一个嵌套的元组列表（这是第0个元素上itertools.groupby操作的输出：

l = [[(95, 'studentD')], [(97, 'studentB')], [(98, 'studentA'), (98, 'studentC')]]

有一种简单的方法可以获得这个吗？

{ 95 : 'studentD', 97 : 'studentB', 98: ['studentA', 'studentC']}

注意单个元素不会放在列表中。此外，我可以保证内部列表中的所有元组具有相同的x[0]。

这是我的解决方案：

In [203]: d = {}

In [204]: for x in l:
     ...:     d.update({x[0][0] : ([y for _, y in x] if len(x) > 1 else x[0][1]) })
     ...:     

In [205]: d
Out[205]: {95: 'studentD', 97: 'studentB', 98: ['studentA', 'studentC']}

我缺少更优雅的选择吗？

Answer 1

最好直接从itertools.groupby的结果进行分组，而不是尝试将整个结果重新处理成dict。

从原始问题的副本开始：

from itertools import groupby

l = [(95, 'studentD'), (97, 'studentB'), (98, 'studentA'), (98, 'studentC')]

d = {}
for k, g in groupby(l, lambda x: x[0]):
    g = [x for _, x in g]
    d[k] = g[0] if len(g) == 1 else g

print d
# {97: 'studentB', 98: ['studentA', 'studentC'], 95: 'studentD'}

Answer 2

如果你必须重新加工，并且正在寻找优雅的，如果不是Pythonic，那该怎么样：

car

但如果我们回到Car步骤，我们可以做一些令人讨厌的事情，如：

>>> l = [[(95, 'studentD')], [(97, 'studentB')], [(98, 'studentA'), (98, 'studentC')]]
>>> 
>>> d = {}
>>> 
>>> for x in l:
...     grades, students = zip(*x)
...     d[grades[0]] = students if len(students) > 1 else students[0]
... 
>>> print(d)
{95: 'studentD', 97: 'studentB', 98: ('studentA', 'studentC')}
>>> ^D

使用赋值结构来简化触摸。

Answer 3

以下是使用groupby

的另一个想法

import collections
l = [(95, 'studentD'), (97, 'studentB'), (98, 'studentA'), (98, 'studentC')]
result_dict = collections.defaultdict(list)
for x, y in l:
    result_dict[x].append(y)

Answer 4

如果您接受所有项目都在列表中： Flatten + defaultdict

from collections import defaultdict

l = [[(95, 'studentD')], [(97, 'studentB')], [(98, 'studentA'), (98, 'studentC')]]
flat = tuple(item for sublist in l for item in sublist)

d = defaultdict(list)

for k,v in flat:
    d[k].append(v)

给出：

defaultdict(list,
            {95: ['studentD'], 97: ['studentB'], 98: ['studentA', 'studentC']})

可以删除那些：

for k,v in d.items():
    if len(v) == 1:
        d[k] = v[0]

Answer 5

L = [[(95, 'studentD')], [(97, 'studentB')], [(98, 'studentA'), (98, 'studentC')]]

answer = {}
for subl in L:
    for grade, student in subl:
        if grade not in answer:
            answer[grade] = []
        answer[grad].append(student)

将此嵌套的元组列表转换为字典？

5 个答案: