嵌套列表到dict [英] Nested list to dict
本文介绍了嵌套列表到dict的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我尝试创建 dict
通过嵌套列表
:
groups = [['Group1','A','B'],['Group2','C','D']]
对于x中的x,如果y = x [0]},对于x中的y,则L = [{y:x [0])对于x中的x,
d = {k:v对于k的L中的d,d中的v, )}
print(d)
{'B':'Group1','C':'Group2','D':'Group2','A'
但似乎有点复杂。
有更好的解决方案吗?
解决方案
如何:
<$对于行[1:]中的k的行,p = p>
d = {k:row [0]}
这给出:
>>> {k:row [0] for row in group for k in row [1:]}
{'D':'Group2','B':'Group1','C':'Group2',' A':'Group1'}
所以你遍历每个
。该行的第一个元素被视为值(组
row [0]
),并且迭代 row [1:]
获取所有密钥 k
。
很奇怪,因为它可能会似乎,这个表达式也可以当你给它一个空行(如 groups = [[],['A','B']]
)。那是因为 row [1:]
将是空的,因此行[0]
部分是从不评估:
>>> groups = [[],['A','B']]
>>> {k:row [0] for row in group for k in row [1:]}
{'B':'A'}
I try create dict
by nested list
:
groups = [['Group1', 'A', 'B'], ['Group2', 'C', 'D']]
L = [{y:x[0] for y in x if y != x[0]} for x in groups]
d = { k: v for d in L for k, v in d.items()}
print (d)
{'B': 'Group1', 'C': 'Group2', 'D': 'Group2', 'A': 'Group1'}
But it seems a bit complicated.
Is there a better solution?
解决方案
What about:
d = {k:row[0] for row in groups for k in row[1:]}
This gives:
>>> {k:row[0] for row in groups for k in row[1:]}
{'D': 'Group2', 'B': 'Group1', 'C': 'Group2', 'A': 'Group1'}
So you iterate over every row
in the groups
. The first element of the row is taken as value (row[0]
) and you iterate over row[1:]
to obtain all the keys k
.
Weird as it might seem, this expression also works when you give it an empty row (like groups = [[],['A','B']]
). That is because row[1:]
will be empty and thus the row[0]
part is never evaluated:
>>> groups = [[],['A','B']]
>>> {k:row[0] for row in groups for k in row[1:]}
{'B': 'A'}
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