回答編集履歴
2
Array.prototype.reduce
answer
CHANGED
|
@@ -1,3 +1,5 @@
|
|
|
1
|
+
### new Map で管理する
|
|
2
|
+
|
|
1
3
|
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。
|
|
2
4
|
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。
|
|
3
5
|
|
|
@@ -45,4 +47,39 @@
|
|
|
45
47
|
console.log(JSON.stringify(result)); // [["a","a","a",1,20190202],["a","b","a",1,20190101],["b","a","a",1,20190101],["a","a","c",1,20190101],["c","a","c",1,20190701],["b","b","b",1,20190101],["b","b","a",1,20190101],["b","a","b",1,20190101]]
|
|
46
48
|
```
|
|
47
49
|
|
|
48
|
-
|
|
50
|
+
### Array.prototype.reduce
|
|
51
|
+
|
|
52
|
+
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。
|
|
53
|
+
(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)
|
|
54
|
+
|
|
55
|
+
```JavaScript
|
|
56
|
+
const array = [
|
|
57
|
+
["a","a","a",1,"2019/01/01"],
|
|
58
|
+
["a","b","a",1,"2019/01/01"],
|
|
59
|
+
["b","a","a",1,"2019/01/01"],
|
|
60
|
+
["a","a","c",1,"2019/01/01"],
|
|
61
|
+
["c","a","c",1,"2019/01/01"],
|
|
62
|
+
["b","b","b",1,"2019/01/01"],
|
|
63
|
+
["b","b","a",1,"2019/01/01"],
|
|
64
|
+
["c","a","c",1,"2019/07/01"],
|
|
65
|
+
["b","a","b",1,"2019/01/01"],
|
|
66
|
+
["a","a","a",1,"2019/02/02"]
|
|
67
|
+
];
|
|
68
|
+
|
|
69
|
+
const lastUpdate = array.reduce((entries, current) => {
|
|
70
|
+
const keys = entries.keys,
|
|
71
|
+
values = entries.values,
|
|
72
|
+
currentKey = JSON.stringify([current[0],current[1],current[2]]),
|
|
73
|
+
index = keys.indexOf(currentKey);
|
|
74
|
+
|
|
75
|
+
if (index === -1) {
|
|
76
|
+
keys.push(currentKey), values.push(current);
|
|
77
|
+
} else if (values[index][4] <= current[4]) {
|
|
78
|
+
keys[index] = currentKey, values[index] = current;
|
|
79
|
+
}
|
|
80
|
+
|
|
81
|
+
return entries;
|
|
82
|
+
}, {keys: [], values: []}).values;
|
|
83
|
+
|
|
84
|
+
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"]]
|
|
85
|
+
```
|
1
typo修正
answer
CHANGED
|
@@ -1,4 +1,4 @@
|
|
|
1
|
-
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
|
|
1
|
+
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。
|
|
2
2
|
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。
|
|
3
3
|
|
|
4
4
|
```JavaScript
|