Question

我有一份清单，如此

list = [[title, description, ~N[2018-01-01 23:00:07], comment, user],
       [title, description, ~N[2018-03-02 12:10:18], comment, user]]

现在我需要将每个NaiveDateTime转换为Erlang日期（或Unix时间戳）。

我相信我可以做点像

new_list = Enum.map(list,&modify_date/1)

def modify_date(list) do
  ##
end

但我无法弄清楚如何让modify_date/1只影响第三个元素。有什么想法吗？

Answer 1

 SELECT 
 (SELECT AVG(TicketPrice) FROM ticket 
    WHERE TicketPrice between 552 and 1302
    AND AirlineID=1 
    AND TicketDate between '2016-01-01' and '2016-12-31' 
    GROUP BY TicketDate) as Col1,
(SELECT AVG(TicketPrice) FROM ticket 
    WHERE TicketPrice between 552 and 1302 
    AND AirlineID=1 
    AND TicketDate between '2017-01-01' and '2017-12-31' 
    GROUP BY TicketDate) as Col2

Answer 2

Elixir中解决此问题的最简单，最惯用的方法是直接进行模式匹配。与目前为止提出的其他解决方案相比，这将大大提高可读性。

关于使用modify_date([title, description, date, comment, user])，你是正确的，如果你模仿这样的匹配：

A.run()

您可以轻松选择所需的任何数据并进行相应操作。

下面介绍了一个完整的解决方案，您可以将其作为defmodule A do def run do [["title", "description", ~N[2018-01-01 23:00:07], "comment", "user"], ["title", "description", ~N[2018-03-02 12:10:18], "comment", "user"]] |> process end def process(list) do Enum.map(list, &modify_date/1) end def modify_date([title, description, date, comment, user]) do unix_timestamp = date |> DateTime.from_naive!("Etc/UTC") |> DateTime.to_unix() [title, description, unix_timestamp, comment, user] end end运行：

.*

Answer 3

您也可以将列表转移到元组。

list = [["fds", "fds", ~N[2018-01-01 23:00:07], "sfd", "fdsa"],
       ["fds", "fds", ~N[2018-03-02 12:10:18], "fds", "fdsa"]]
Enum.map(list, fn x ->
  IO.inspect x|> List.to_tuple |> elem(2)
end)

修改列表列表中的列

3 个答案: