Question

名单如下:

mylist = [( a , [(0,1), (1,2), (2,3)]),
         ( b , [(0,1), (1,2), (2,3)]),
         ( c , [(0,1), (1,2), (2,3)])]

我在守则中采用了不同的行动。

能否取得如下产出?

newlist = [( a , [0.5, 1.5, 2.5]),
          ( b , [0.5, 1.5, 2.5]),
          ( c , [0.5, 1.5, 2.5])]

投入清单的内容按范围分列,产出清单由从每一范围平均得出。谢谢。

Answer 1

>>> l = [( a , [(0,1), (1,2), (2,3)]),
... ( b , [(0,1), (1,2), (2,3)]),
... ( c , [(0,1), (1,2), (2,3)])]
>>> newl = [(i[0], [(j[0]+j[1])/2 for j in i[1]]) for i in l]
>>> newl
[( a , [0.5, 1.5, 2.5]), ( b , [0.5, 1.5, 2.5]), ( c , [0.5, 1.5, 2.5])]

这部建筑在3号山;在2号山,你需要

>>> newl = [(i[0], [(j[0]+j[1])/2.0 for j in i[1]]) for i in l]

这些清单核对表相当于以下更精确的编码:

>>> newl = []
>>> for i in l:
...     temp = (i[0], [])
...     for j in i[1]:
...         temp[1].append((j[0]+j[1])/2.0)
...     newl.append(temp)

Answer 2

清单缩略语是冰,但我喜欢的甲型六氯环己烷技术是map(),适用于结构的每一要素:

ls = (( a , [(0,1), (1,2), (2,3)]),
      ( b , [(0,1), (1,2), (2,3)]),
      ( c , [(0,1), (1,2), (2,3)]))

for li in ls:
    print(li[0], list(map(lambda x: sum(x)/len(x), li[1])))

正如Tim所说,这3台作品需要float(len(x),才能进入2台。

Answer 3

仅与您的名单(有效的字节部分)一起工作,您可使用一份清单:

numbers = [(0,1), (1,2), (2,3)]

print [sum(x) / float(len(x)) for x in numbers]

这应该使你走正确的方向,并给你:

[0.5, 1.5, 2.5]

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