回答編集履歴
2
テキスト修正
test
CHANGED
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@@ -54,9 +54,7 @@
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```javascript
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const zip = (ary1, ary2) =>
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ary1.map((e, i) =>
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const zip = (ary1, ary2) => ary1.map((e, i) => [e, ary2[i]]);
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```
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テキスト修正
test
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@@ -20,7 +20,81 @@
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-
- 動作確認用 CodePen: [https://codepen.io/jun68ykt/pen/VwwVBPx?editors=0012](https://codepen.io/jun68ykt/pen/VwwVBPx?editors=0012)
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- **動作確認用 CodePen:** [https://codepen.io/jun68ykt/pen/VwwVBPx?editors=0012](https://codepen.io/jun68ykt/pen/VwwVBPx?editors=0012)
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### 別解
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lodash を使わない案も挙げておきます。
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まず、以下の２つの配列
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- `[ "name", "address", "tel" ]`
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- `[ "田中", "東京", "000000000" ]`
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から、下記の配列
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- `[ [ "name", "田中" ], [ "address", "東京" ], [ "tel", "000000000" ] ]`
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を得る関数を作ります。このような処理をする関数は、慣例的に zip と呼ばれるので、それに倣います。
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```javascript
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const zip = (ary1, ary2) =>
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ary1.map((e, i) => ([e, ary2[i]]));
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```
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上記の zip によって得られる
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`[ [ "name", "田中" ], [ "address", "東京" ], [ "tel", "000000000" ] ]`
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という形式の配列を
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- [Object.fromEntries()](https://developer.mozilla.org/ja/docs/Web/JavaScript/Reference/Global_Objects/Object/fromEntries)
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に渡せば、オブジェクトが作られます。以下はこれを使ったコードです。
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```javascript
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const objlist = valuelist.map(v =>
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Object.fromEntries(zip(key, v))
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);
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```
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- **動作確認用 CodePen:** [https://codepen.io/jun68ykt/pen/MWWzBVb?editors=0012](https://codepen.io/jun68ykt/pen/MWWzBVb?editors=0012)
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