我有这个:
a1 = [%{id: 1, val: 12}, %{id: 3, val: 7}, %{id: 1, val: 5}, %{id: 2, val: 3}, %{id: 2, val: 5}], %{id: 1, val: 3}]
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我怎么能得到这个?
%{
1 => 20,
2 => 8,
3 => 7
}
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也就是说,按"id"分组的每个项目的"val"之和
我应该先用"id"将它们分组吗?
Enum.group_by a1, &(&1.id)
# =>
%{
1 => [%{id: 1, val: 12}, %{id: 1, val: 3}, %{id: 1, val: 5}],
2 => [%{id: 2, val: 3}, %{id: 2, val: 5}],
3 => [%{id: 3, val: 7}]
}
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然后做map和reduce在每个项目上?或者,还有更好的方法?
我这样做,只需一个Enum.reduce/3电话:
[%{id: 1, val: 12}, %{id: 3, val: 7}, %{id: 1, val: 5}, %{id: 2, val: 3}, %{id: 2, val: 5}, %{id: 1, val: 3}]
|> Enum.reduce(%{}, fn %{id: id, val: val}, map ->
Map.update(map, id, val, &(&1 + val))
end)
|> IO.inspect
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输出:
%{1 => 20, 2 => 8, 3 => 7}
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这应该比group_by+ map+ 更有效reduce.