Question

我正在尝试将此函数转换为使用reduce。我中途了。当所选值为true时，我想知道该元素的索引。

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];

结果：1

我的尝试是这样：

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

如何更改整个要使用reduce的内容？

Answer 1

仅当option.selected为true时，才将索引添加到结果中，否则不执行任何操作。

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()中使用三进制来设置要插入的值。 Array#concat()返回更新后的数组，因此返回reduce()。

或使用传播返回新数组[...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

如果您希望使用selected = true的索引，则可以使用reduce执行类似的操作。如果上下文中的当前项具有selected === true，则将索引添加到累加器，否则返回累加器。当前，如果条件为-1

，您正在添加false

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)

如果您需要整个对象，则只需使用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class MainActivity : AppCompatActivity(), NavigationView.OnNavigationItemSelectedListener { private lateinit var navController: NavController private lateinit var appBarConfiguration: AppBarConfiguration override fun onCreate(savedInstanceState: Bundle?) { super.onCreate(savedInstanceState) setContentView(R.layout.activity_main) setSupportActionBar(toolbar) navController = findNavController(R.id.nav_controller_fragment) appBarConfiguration = AppBarConfiguration(navController.graph,drawer_layout) setupActionBarWithNavController(navController,appBarConfiguration) drawer() setupNavigationMenu() nav_view.setNavigationItemSelectedListener(this) } private fun drawer() { drawer_layout.addDrawerListener(object :DrawerLayout.DrawerListener{ override fun onDrawerStateChanged(newState: Int) { } override fun onDrawerSlide(drawerView: View, slideOffset: Float) { } override fun onDrawerClosed(drawerView: View) { } override fun onDrawerOpened(drawerView: View) { } }) } private fun setupNavigationMenu() { nav_view.let { NavigationUI.setupWithNavController(it, navController) } } override fun onBackPressed() { if (drawer_layout.isDrawerOpen(GravityCompat.START)) { drawer_layout.closeDrawer(GravityCompat.START) } else { super.onBackPressed() } } override fun onCreateOptionsMenu(menu: Menu): Boolean { menuInflater.inflate(R.menu.main, menu) return true } override fun onOptionsItemSelected(item: MenuItem): Boolean { return when (item.itemId) { R.id.action_settings -> true else -> NavigationUI.onNavDestinationSelected(item,navController) || super.onOptionsItemSelected(item) } } override fun onNavigationItemSelected(item: MenuItem): Boolean { drawer_layout.closeDrawer(GravityCompat.START) return when (item.itemId) { R.id.nav_gallery -> true else -> NavigationUI.onNavDestinationSelected( item,navController) || super.onOptionsItemSelected(item) } } override fun onSupportNavigateUp(): Boolean { return navController.navigateUp(appBarConfiguration) || super.onSupportNavigateUp() }为...，则使用Array#reduce和selected扩展运算符保留现有索引并添加新索引。

true

Answer 5

另一种方法是使用push代替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)

如何转换此对象数组以使用reduce javascript

5 个答案: