回答編集履歴
3
コメントを受けてIDをユニークに
answer
CHANGED
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@@ -14,4 +14,27 @@
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)
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)]
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);
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+
```
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+
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+
### コメントを受けてIDをユニークに
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+
```javascript
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(
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( {itemDict, productsIds} ) =>
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[...new Set(
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[].concat(
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...productsIds.map( e => itemDict[e] )
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).map( e => JSON.stringify(e) )
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)].map( e => JSON.parse(e) )
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)({
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itemDict:
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Object.assign( ...itemList.map( ({id,items}) => ({[id]:items}) ) )
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,
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productsIds:
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[...new Set(
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[].concat(
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...orderList.map( ({products}) => products )
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)
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)]
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,
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});
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40
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```
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2
こっちのほうがスマートかな……
answer
CHANGED
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@@ -7,7 +7,7 @@
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7
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...productsIds.map( e => itemDict[e] )
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)
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)(
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-
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Object.assign( ...itemList.map( ({id,items}) => ({[id]:items}) ) ),
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[...new Set(
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[].concat(
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...orderList.map( ({products}) => products )
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1
こっちのほうが読みやすい、かな？
answer
CHANGED
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@@ -2,16 +2,16 @@
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2
2
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あえて工夫するとするなら、とりあえず`itemList`を辞書化するとか？
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3
3
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```javascript
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(
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-
( itemDict,
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+
( itemDict, productsIds ) =>
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[].concat(
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-
...[...new Set(
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[].concat(
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-
...orderList.map( ({products}) => products )
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-
)
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-
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7
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+
...productsIds.map( e => itemDict[e] )
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8
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)
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)(
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itemList.reduce( (d, {id,items}) => ( d[id]=items,d ), {} ),
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[...new Set(
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-
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[].concat(
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...orderList.map( ({products}) => products )
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)
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)]
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);
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```
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