TypeScript数组批量删除的方法？

hacker

在TypeScript中，我们可以使用多种方法来遍历数组并删除满足条件的元素。需要注意的是，在遍历数组的同时删除元素可能会导致索引错乱，因此需要小心处理。

方法 1：从后向前遍历（推荐）

interface People {
  age: number;
  sex: number;
}

let peoples: People[] = [
  { age: 8, sex: 1 },
  { age: 10, sex: 1 },
  { age: 7, sex: 2 },
  { age: 12, sex: 2 },
  { age: 5, sex: 1 }
];

// 从后向前遍历，避免索引问题
for (let i = peoples.length - 1; i >= 0; i--) {
  if (peoples[i].age > 8) {
    peoples.splice(i, 1);
  }
}

console.log(peoples);
// 输出: [{ age: 8, sex: 1 }, { age: 7, sex: 2 }, { age: 5, sex: 1 }]

方法 2：使用 while 循环和索引控制

let peoples: People[] = [
  { age: 8, sex: 1 },
  { age: 10, sex: 1 },
  { age: 7, sex: 2 },
  { age: 12, sex: 2 },
  { age: 5, sex: 1 }
];

let i = 0;
while (i < peoples.length) {
  if (peoples[i].age > 8) {
    peoples.splice(i, 1);
  } else {
    i++; // 只有不删除时才递增索引
  }
}

console.log(peoples);

方法 3：创建新数组（非破坏性）

let peoples: People[] = [
  { age: 8, sex: 1 },
  { age: 10, sex: 1 },
  { age: 7, sex: 2 },
  { age: 12, sex: 2 },
  { age: 5, sex: 1 }
];

const newPeoples: People[] = [];
for (let i = 0; i < peoples.length; i++) {
  if (peoples[i].age <= 8) {
    newPeoples.push(peoples[i]);
  }
}

peoples = newPeoples;
console.log(peoples);

方法 4：使用 for...of 循环和索引记录

let peoples: People[] = [
  { age: 8, sex: 1 },
  { age: 10, sex: 1 },
  { age: 7, sex: 2 },
  { age: 12, sex: 2 },
  { age: 5, sex: 1 }
];

const indexesToRemove: number[] = [];
let index = 0;

// 先记录要删除的索引
for (const person of peoples) {
  if (person.age > 8) {
    indexesToRemove.push(index);
  }
  index++;
}

// 从后向前删除，避免索引变化
for (let i = indexesToRemove.length - 1; i >= 0; i--) {
  peoples.splice(indexesToRemove[i], 1);
}

console.log(peoples);

方法 5：使用 reduce 构建新数组

let peoples: People[] = [
  { age: 8, sex: 1 },
  { age: 10, sex: 1 },
  { age: 7, sex: 2 },
  { age: 12, sex: 2 },
  { age: 5, sex: 1 }
];

peoples = peoples.reduce((acc: People[], current) => {
  if (current.age <= 8) {
    acc.push(current);
  }
  return acc;
}, []);

console.log(peoples);

方法 6：使用 filter 方法（最简洁）

let peoples: People[] = [
  { age: 8, sex: 1 },
  { age: 10, sex: 1 },
  { age: 7, sex: 2 },
  { age: 12, sex: 2 },
  { age: 5, sex: 1 }
];

peoples = peoples.filter(person => person.age <= 8);
console.log(peoples);