Question

我有一个数字列表如下：

select avg(c),count(c) from (select generate_series(1,10) union select null) as a(c);

我必须按如下方式更改数字：

select
  avg(c),
  count(c) count_column,
  count(*) count_star,
  sum(c),
  array_agg(c)
from (
  select generate_series(1,10) union select null order by 1
) as a(c);

        avg         | count_column | count_star | sum |          array_agg          
--------------------+--------------+------------+-----+-----------------------------
 5.5000000000000000 |           10 |         11 |  55 | {1,2,3,4,5,6,7,8,9,10,NULL}
(1 row)

或从l1到l2。

given = [[0,5,4,9,0],
         [2,1,4,9,9]]

预期答案是：

0>>0
1>>1
2>>2
3>>3
4>>3
5>>4
9>>5

这样做有多简单？

我的审判很脏：

l1 = [0,1,2,3,4,5,9]
l2 = [0,1,2,3,3,4,5]

我更喜欢numpy方法。

Answer 1

使用字典作为翻译：

>>> given = [[0,5,4,9,0],
...          [2,1,4,9,9]]
>>> translation = {4:3, 5:4, 9:5}
>>> [[translation.get(x, x) for x in sub] for sub in given]
[[0, 4, 3, 5, 0], [2, 1, 3, 5, 5]]

您只需要为实际更改的数字提供映射，因为如果找不到密钥，dict.get(key, default)将返回给定的默认值。

Answer 2

对于您的确切问题（如果您拒绝使用地图，列表推导，词典或Numpy之外的任何其他简单和Pythonic方法）为什么不继续使用两个额外的行？

given[given==4]=3
given[given==5]=4
given[given==9]=5

Answer 3

使用从l1和l2构建的替换函数：

>>> given = [[0,5,4,9,0], [2,1,4,9,9]]
>>> l1 = [0,1,2,3,4,5,9]
>>> l2 = [0,1,2,3,3,4,5]

>>> change = dict(zip(l1, l2)).get
>>> [map(change, g) for g in given]
[[0, 4, 3, 5, 0], [2, 1, 3, 5, 5]]

如何更改列表编号

3 个答案: