0

例如,我有列表列表

res = [[None,'A','B'],[19980228, 'd1', 't1'],[19980302, 'd2', 't2'],[19980303, 'd3', 't3']]

现在我需要将上述结果框定为元组列表的字典,例如

{'A': [(19980228, 'd1'), (19980302, 'd2'), (19980303, 'd3')], 
 'B': [(19980228, 't1'), (19980302, 't2'), (19980303, 't3')]}

请让我知道这个概念,在此先感谢。

4

3 回答 3

2

这应该这样做:

x, A, B = res[0]
output = {A:[], B:[]}
for a,b,c in res[1:]:
  output[A].append((a, b))
  output[B].append((a, c))
于 2012-04-20T07:42:26.127 回答
2
>>> res = [[None,'A','B'],[19980228, 'd1', 't1'],[19980302, 'd2', 't2'],[19980303, 'd3', 't3']]
>>> {a:[(r[0], r[i]) for r in res[1:]] for i, a in enumerate(res[0]) if a}
{'A': [(19980228, 'd1'), (19980302, 'd2'), (19980303, 'd3')], 'B': [(19980228, 't1'), (19980302, 't2'), (19980303, 't3')]}

注意:正如@shuzOMGchen 指出的那样,这需要在 python 2.7 和 3.0 中添加的“字典理解”,因此如果您使用的是较早版本,则必须稍微更改代码。这里没有使用字典理解(这很丑,我只是试图从上面复制我的逻辑)

>>> res = [[None,'A','B'],[19980228, 'd1', 't1'],[19980302, 'd2', 't2'],[19980303, 'd3', 't3']]
>>> d = {}
>>> for i, a in enumerate(res[0]):
...     if a:
...         d[a] = [(r[0], r[i]) for r in res[1:]]
... 
>>> d
{'A': [(19980228, 'd1'), (19980302, 'd2'), (19980303, 'd3')], 'B': [(19980228, 't1'), (19980302, 't2'), (19980303, 't3')]}
于 2012-04-20T07:46:37.013 回答
0
>>> res = [[None,'A','B'],[19980228, 'd1', 't1'],[19980302, 'd2', 't2'],[19980303, 'd3', 't3']]
>>> d = dict((x[0],x[1:]) for x in zip(*res))
>>> nums = d.pop(None)
>>> for key in d:
        d[key] = zip(nums,d[key])


>>> d
{'A': [(19980228, 'd1'), (19980302, 'd2'), (19980303, 'd3')], 'B': [(19980228, 't1'), (19980302, 't2'), (19980303, 't3')]}
于 2012-04-20T11:15:28.593 回答