如何将多个Array.filter合并为单个，或采用更高效的方式优化冗余代码？

问题场景 #

现有重复的数组过滤统计逻辑，存在代码冗余、多次遍历数组带来的不必要性能损耗，需求为将多次filter操作合并为单次遍历，或找到更高效的实现方式，同时确认直接使用for循环实现是否可行。
原有代码如下：

driving += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'driving').length;
breathWork += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'breath work').length;
meditation += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'meditation').length;
cooking += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'cooking').length;
walking += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'walking').length;
other += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'other').length;

优化方案 #

完全可以合并为单次遍历实现，比原有多次filter的性能提升数倍，以下是两种常用实现：

方案1：Array.reduce实现（简洁易读） #

仅遍历数组1次，符合函数式编码风格：

// 初始化统计映射表
const practiceCount = value.reduce((acc, item) => {
  // 跳过非目标类型的项
  if (item.type !== CalendarEventType.MIND) return acc;
  const currentPractice = item.data.practice;
  // 匹配到统计项则计数+1
  if (Object.prototype.hasOwnProperty.call(acc, currentPractice)) {
    acc[currentPractice]++;
  }
  return acc;
}, {
  driving: 0,
  'breath work': 0,
  meditation: 0,
  cooking: 0,
  walking: 0,
  other: 0
});

// 赋值到对应变量
driving += practiceCount.driving;
breathWork += practiceCount['breath work'];
meditation += practiceCount.meditation;
cooking += practiceCount.cooking;
walking += practiceCount.walking;
other += practiceCount.other;

方案2：普通for循环实现（性能最优） #

处理万级以上大数组时性能略高于reduce，无额外函数调用开销：

const practiceCount = {
  driving: 0,
  'breath work': 0,
  meditation: 0,
  cooking: 0,
  walking: 0,
  other: 0
};
const arrLength = value.length;

for (let i = 0; i < arrLength; i++) {
  const item = value[i];
  if (item.type !== CalendarEventType.MIND) continue;
  const currentPractice = item.data.practice;
  if (practiceCount[currentPractice] !== undefined) {
    practiceCount[currentPractice]++;
  }
}

driving += practiceCount.driving;
breathWork += practiceCount['breath work'];
meditation += practiceCount.meditation;
cooking += practiceCount.cooking;
walking += practiceCount.walking;
other += practiceCount.other;

额外优化收益 #

• 可维护性提升：后续新增需要统计的practice类型时，仅需在初始化的统计对象中添加对应key即可，无需新增独立的filter逻辑
• 可复用性提升：可以将该统计逻辑封装为通用工具函数，同类统计场景可直接复用

内容的提问来源于stack exchange，提问作者user2994290