如何将列表列表转换为词典列表?
更具体地说:我如何从中走出来:
[['a1', 'b1', 'c1', 'd1', 'e1', 'f1', 'g1', 'h1', 'i1'], ['a2', 'b2', 'c2', 'd2', 'e2', 'f2', 'g2', 'h2', 'i2'], ['a3', 'b3', 'c3', 'd3', 'e3', 'f3', 'g3', 'h3', 'i3'], ['a4', 'b4', 'c4', 'd4', 'e4', 'f4', 'g4', 'h4', 'i4'], ['a5', 'b5', 'c5', 'd5', 'e5', 'f5', 'g5', 'h5', 'i5'], ['a6', 'b6', 'c6', 'd6', 'e6', 'f6', 'g6', 'h6', 'i6'], ['a7', 'b7', 'c7', 'd7', 'e7', 'f7', 'g7', 'h7', 'i7'], ['a8', 'b8', 'c8', 'd8', 'e8', 'f8', 'g8', 'h8', 'i8'], ['a9', 'b9', 'c9', 'd9', 'e9', 'f9', 'g9', 'h9', 'i9']]
到此:
[{'a1': None, 'b1': None, 'c1': None, 'd1': None, 'e1': None, 'f1': None, 'g1': None, 'h1': None, 'i1': None}, #etc
答案 0 :(得分:22)
In [20]: l = [['a1', 'b1', 'c1', 'd1', 'e1', 'f1', 'g1', 'h1', 'i1'], ['a2', 'b2', 'c2', 'd2', 'e2', 'f2', 'g2', 'h2', 'i2'], ['a3', 'b3', 'c3', 'd3', 'e3', 'f3', 'g3', 'h3', 'i3'], ['a4', 'b4', 'c4', 'd4', 'e4', 'f4', 'g4', 'h4', 'i4'], ['a5', 'b5', 'c5', 'd5', 'e5', 'f5', 'g5', 'h5', 'i5'], ['a6', 'b6', 'c6', 'd6', 'e6', 'f6', 'g6', 'h6', 'i6'], ['a7', 'b7', 'c7', 'd7', 'e7', 'f7', 'g7', 'h7', 'i7'], ['a8', 'b8', 'c8', 'd8', 'e8', 'f8', 'g8', 'h8', 'i8'], ['a9', 'b9', 'c9', 'd9', 'e9', 'f9', 'g9', 'h9', 'i9']]
In [21]: map(dict.fromkeys, l)
Out[21]:
[{'a1': None,
'b1': None,
'c1': None,
'd1': None,
'e1': None,
'f1': None,
'g1': None,
'h1': None,
'i1': None},
{'a2': None,
'b2': None,
'c2': None,
'd2': None,
...
这适用于任何可迭代的迭代,而不仅仅是列表列表(当然,前提是第二级元素为hashable)。
在Python 2中,上面的代码返回一个列表。
在Python 3中,它返回一个可迭代的。如果您需要列表,可以使用list(map(dict.fromkeys, l))
。
答案 1 :(得分:3)
试试这个:
l = [['a1', 'b1', 'c1', 'd1', 'e1', 'f1', 'g1', 'h1', 'i1'], ['a2', 'b2', 'c2', 'd2', 'e2', 'f2', 'g2', 'h2', 'i2'], ['a3', 'b3', 'c3', 'd3', 'e3', 'f3', 'g3', 'h3', 'i3'], ['a4', 'b4', 'c4', 'd4', 'e4', 'f4', 'g4', 'h4', 'i4'], ['a5', 'b5', 'c5', 'd5', 'e5', 'f5', 'g5', 'h5', 'i5'], ['a6', 'b6', 'c6', 'd6', 'e6', 'f6', 'g6', 'h6', 'i6'], ['a7', 'b7', 'c7', 'd7', 'e7', 'f7', 'g7', 'h7', 'i7'], ['a8', 'b8', 'c8', 'd8', 'e8', 'f8', 'g8', 'h8', 'i8'], ['a9', 'b9', 'c9', 'd9', 'e9', 'f9', 'g9', 'h9', 'i9']]
res = []
for line in l:
res.append(dict((k, None) for k in line))
OR:
res = [dict((k, None) for k in line) for line in l]