我有 3 個不同的測驗用例，每個測驗用例都傳遞給getEarliest下面的函式：
{}
{ '2022-04-29': 1 }
{ '2022-04-29': 3, '2022-04-30': 2, '2022-04-28': 3, '2022-05-02': 2 }
從每個測驗用例中，我想回傳具有最小日期和最大值的鍵：值。我的問題是在第三個測驗用例中，我得到'2022-04-29': 3的是'2022-04-28': 3
getEarliest = dates => {
return Object.keys(dates).reduce((prev, curr) => {
if (dates[curr] > prev.date) {
return {
val: dates[curr],
date: curr
};
} else {
return prev;
}
}, {
val: 0,
date: null
}); }
預期結果測驗用例1： { val: 0, date: null }
預期結果測驗用例2： { val: 1, date: '2022-04-29' }
預期結果測驗用例 3： { val: 3, date: '2022-04-28' }
uj5u.com熱心網友回復：
看起來你只需要sort兩個引數的物件：
const test1 = {};
const test2 = { '2022-04-29': 1 };
const test3 = { '2022-04-29': 3, '2022-04-30': 2, '2022-04-28': 3, '2022-05-02': 2 };
const test4 = { '2022-04-29': 4, '2022-04-28': 4, '2022-04-30': 4, '2022-05-02': 2 };
const getEarliest = (dates) => {
const [date, val] = Object.entries(dates)
.sort(([k1, v1], [k2, v2]) => v2 - v1 || Date.parse(k1) - Date.parse(k2) )
.at(0) ?? [null, 0];
return { val, date };
};
console.log(getEarliest(test1));
console.log(getEarliest(test2));
console.log(getEarliest(test3));
console.log(getEarliest(test4));.as-console-wrapper {max-height: 100% !important; top: 0}和相同的結果reduce：
const test1 = {};
const test2 = { '2022-04-29': 1 };
const test3 = { '2022-04-29': 3, '2022-04-30': 2, '2022-04-28': 3, '2022-05-02': 2 };
const test4 = { '2022-04-29': 4, '2022-04-28': 4, '2022-04-30': 4, '2022-05-02': 2 };
const getEarliest = (dates) => {
const [date, val] = Object.entries(dates)
.reduce((prev, curr) => (curr[1] - prev[1] || Date.parse(prev[0]) - Date.parse(curr[0])) > 0
? curr
: prev
, [null, 0]);
return { val, date };
};
console.log(getEarliest(test1));
console.log(getEarliest(test2));
console.log(getEarliest(test3));
console.log(getEarliest(test4));.as-console-wrapper {max-height: 100%!important;top:0 }uj5u.com熱心網友回復：
在您的代碼prev.date中，值最初為空。所以reduce()總是會回傳初始物件。
試試這個代碼它可以幫助你
getEarliest = (dates) => {
if (Object.keys(dates).length > 0) {
let date = Object.keys(dates).reduce((prev, curr) => {
if (prev > curr) {
return curr;
} else {
return prev;
}
});
return {
val: dates[date],
date: date
}
} else {
return {
val: 0, date: null
}
}
}
uj5u.com熱心網友回復：
因為你有 2 個排序條件，所以它們必須按一定的順序排列。從您的示例中，我假設順序是：“最大值”->“最小日期”。然后你if會變成下面這樣：
getEarliest = (dates) => {
return Object.keys(dates).reduce(
(prev, curr) => {
if (dates[curr] > prev.val || (dates[curr] === prev.val && curr < prev.date)) {
return {
val: dates[curr],
date: curr,
}
} else {
return prev
}
},
{
val: 0,
date: null,
}
)
}
uj5u.com熱心網友回復：
在第一個陳述句中，該
.date屬性從未被定義，所以它應該是undefined。dates是一個陣列而不是一個物件，所以dates[curr]看起來dates['2022-04-28']沒有幫助。以符號形式 ex。array[5]括號中的整數作為索引的第三個引數.reduce()if (dates[curr] > prev.date) {...你得到的原因是因為陣列永遠不會改變，你需要用更簡單的方法
"2022-04-29"對日期進行排序，或者對新手來說太復雜了。.sort().reduce()
細節在下面的例子中注釋
let A = {};
let B = {
'2022-04-29': 1
};
let C = {
'2022-04-29': 3,
'2022-04-30': 2,
'2022-04-28': 3,
'2022-05-02': 2
};
// Optional utility function
const log = data => console.log(JSON.stringify(data));
/*
Can convert from timestamp to YYYY-MM-DD and vice versa
*/
const dX = (number, string) => {
let D;
if (number) {
let date = new Date(number);
D = ` ${date.getFullYear()}-${("0" (date.getMonth() 1)).slice(-2)}-${("0" date.getDate()).slice(-2)}`;
} else if (string) {
D = Date.parse(string);
} else {
return false;
}
return D;
};
// Pass tS = {...}
const getEarliest = tS => {
// if tS is empty...
if (Object.keys(tS).length === 0) { // Return premade object
return {
val: 0,
date: null
};
}
/*
Object.entries(tS) converts {k: v, k: v,...} into [[k,v], [k,v]...]
*/
/*
.map(([k, v]) exposes key/value
pairs by destructuring [k, v]
*/
/*
.sort(a, b) is referencing each
date key and converting it into
a number (ms since epoch)
*/
let table = Object.entries(tS).map(([k, v]) => [k, v]).sort((a, b) => dX(null, a[0]) - dX(null, b[0]));
let obj = {};
obj.val = table[0][1];
obj.date = table[0][0];
return obj;
};
log(getEarliest(A));
log(getEarliest(B));
log(getEarliest(C));轉載請註明出處，本文鏈接：https://www.uj5u.com/caozuo/414922.html
標籤：