回答編集履歴
2
Array.prototype.reduce
test
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@@ -1,3 +1,7 @@
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### new Map で管理する
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同じカテゴリ群を持つデータをグループ化すると、管理性が良くなると思います。
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6
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元のデータ構造を維持する理由がなければ、`new Map` のまま運用します。
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@@ -92,4 +96,74 @@
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### Array.prototype.reduce
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103
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データ構造を変えず、検索の為の中間オブジェクトをその都度生成するならば、`Array.prototype.reduce` が適していると思われます。
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104
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105
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(前節のコードもそうですが、カテゴリ1～カテゴリ3は `String` 型であれば、使用できない文字はありません)
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106
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```JavaScript
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111
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const array = [
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["a","a","a",1,"2019/01/01"],
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["a","b","a",1,"2019/01/01"],
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["b","a","a",1,"2019/01/01"],
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["a","a","c",1,"2019/01/01"],
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["c","a","c",1,"2019/01/01"],
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["b","b","b",1,"2019/01/01"],
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["b","b","a",1,"2019/01/01"],
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["c","a","c",1,"2019/07/01"],
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["b","a","b",1,"2019/01/01"],
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["a","a","a",1,"2019/02/02"]
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];
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const lastUpdate = array.reduce((entries, current) => {
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const keys = entries.keys,
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values = entries.values,
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currentKey = JSON.stringify([current[0],current[1],current[2]]),
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index = keys.indexOf(currentKey);
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if (index === -1) {
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keys.push(currentKey), values.push(current);
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} else if (values[index][4] <= current[4]) {
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keys[index] = currentKey, values[index] = current;
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}
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return entries;
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}, {keys: [], values: []}).values;
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console.log(JSON.stringify(lastUpdate)); // [["a","a","a",1,"2019/02/02"],["a","b","a",1,"2019/01/01"],["b","a","a",1,"2019/01/01"],["a","a","c",1,"2019/01/01"],["c","a","c",1,"2019/07/01"],["b","b","b",1,"2019/01/01"],["b","b","a",1,"2019/01/01"],["b","a","b",1,"2019/01/01"]]
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```
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1
typo修正
test
CHANGED
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@@ -1,4 +1,4 @@
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1
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-
同じカテゴリ群を持つデータをグループ化すると、管理性が良くなると
|
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1
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+
同じカテゴリ群を持つデータをグループ化すると、管理性が良くなると思います。
|
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2
2
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|
3
3
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元のデータ構造を維持する理由がなければ、`new Map` のまま運用します。
|
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4
4
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