如何轉換此對象數組以使用reduce javascript

Question

我正在嘗試將此函數轉換為使用reduce。 我中途了。 當選定的值為true我想知道該元素的索引。

let options = [
    {label: 'foo1', selected: false},
    {label: 'foo2', selected: true},
    {label: 'foo2', selected: false},
    {label: 'foo2', selected: false},
    {label: 'foo2', selected: false},
];
const obj = options
              .map((option, index) => (option.selected ? index : -1))
              .filter((val) => val !== -1)[0];

結果： 1

我的嘗試是這樣的：

const obj = options.reduce((acc, currentValue, index) => {
    const i = currentValue["selected"] ? index : -1;
    return acc.concat(i)
}, []); // [-1,1,-1,-1,-1]

我如何改變整個東西使用減少？

Answer 1

僅當option.selected為true時，才將索引添加到結果中，否則不執行任何操作。

 let options = [ {label: 'foo1', selected: false}, {label: 'foo2', selected: true}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, ]; const selected = options.reduce((arr, option, index) => { return option.selected ? arr.concat(index) : arr; }, []); console.log(selected); 

Answer 2

在concat()使用三元數來設置要插入的值。 Array#concat()返回更新后的數組，因此是reduce()的返回值。

還是使用傳播返回新數組[...accumulator, selsected ? i :-1] [...accumulator, selsected ? i :-1]

 const res = options.reduce((a, {selected:s}, i) => a.concat(s ? i : -1) , []); // OR // const res = options.reduce((a, {selected:s}, i) => [....a,s ? i : -1] , []); console.log(res) 

 <script> let options = [ {label: 'foo1', selected: false}, {label: 'foo2', selected: true}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, ]; </script> 

Answer 3

如果要使用selected = true的索引，則可以使用reduce類似的操作。 如果上下文中的當前項已selected === true，則將索引添加到累加器，否則返回累加器。 當前，如果條件為false ，則您要添加-1

 let options = [ {label: 'foo1', selected: false}, {label: 'foo2', selected: true}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, ]; const selectedIndexes = options.reduce((r, o, i) => o.selected ? r.concat(i) : r, []) console.log(selectedIndexes) 

如果需要整個對象，則可以簡單地使用filter ：

 let options = [{label:'foo1',selected:false},{label:'foo2',selected:true},{label:'foo2',selected:false},{label:'foo2',selected:false},{label:'foo2',selected:false},]; const filtered = options.filter(o => o.selected) console.log(filtered) 

Answer 4

如果selected是true則使用Array#reduce和... spread運算符保留現有索引並添加新索引。

 let options = [ {label: 'foo1', selected: false}, {label: 'foo2', selected: true}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, {label: 'foo2', selected: true}, {label: 'foo2', selected: 1} ]; const obj = options.reduce((acc, currentValue, index) => { // checking only for boolean true not for truthy values return acc = currentValue["selected"] === true ? [...acc, index] : acc; }, []); console.log(obj); 

Answer 5

另一種方法是使用push而不是concat ：

 let options = [ {label: 'foo1', selected: false}, {label: 'foo2', selected: true}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, {label: 'foo2', selected: false}, ]; var obj = options.reduce( (acc, currentValue, index) => { currentValue.selected ? acc.push(index) : ''; return acc} ,[] ) ; console.log(obj)