如何避免循环重复？优化集合过滤逻辑的技术问询

原问题场景 #

现有一段集合过滤代码，逻辑是依次执行三次过滤规则，若前一次过滤结果为空则触发下一次，但三次过滤存在大量重复逻辑，且多次遍历集合影响性能，代码如下：

let rows = (collection || []).filter((item: any) => {
  const { country, year, productType, units, productSubtype } = item;
  const cond1 = unitsPlain.length ? unitsPlain.includes(units) : true;
  const cond3 = years.includes(year);
  const cond2 = country === value;
  const cond4 = productType === key && !productSubtype;

  return cond1 && cond2 && cond3 && cond4;
});

if (!rows.length) {
  rows = collection.filter((item: any) => {
    const { country, year, productType, productSubtype } = item;
    const cond3 = years.includes(year);
    const cond2 = country === value;
    const cond4 = productSubtype === key && !productType;

    return cond2 && cond3 && cond4;
  });
}

if (!rows.length) {
  rows = collection.filter((item: any) => {
    const { country, year, productType, productSubtype } = item;
    const cond3 = years.includes(year);
    const cond2 = country === value;
    const cond4 = productSubtype === key || productType === key;

    return cond2 && cond3 && cond4;
  });
}

优化思路 #

核心是尽量减少集合遍历次数，同时保留原逻辑的匹配优先级：先尝试匹配第一级规则，再第二级，若前两级都无匹配项，最后用第三级规则兜底。通过一次遍历完成前两级规则的匹配，仅在前两级无结果时才执行第三级过滤，避免重复遍历。

优化后的代码 #

const targetCollection = collection || [];
const result: any[] = [];
// 标记是否找到前两级规则的匹配项
let hasHighPriorityMatches = false;

// 一次遍历完成前两级规则的匹配
for (const item of targetCollection) {
  const { country, year, productType, units, productSubtype } = item;
  // 提取公共条件，避免重复判断
  const meetsCommonCond = country === value && years.includes(year);
  
  if (!meetsCommonCond) continue;

  // 第一优先级规则：units匹配 + productType为key且无productSubtype
  const isLevel1Match = (unitsPlain.length ? unitsPlain.includes(units) : true) && 
                        productType === key && !productSubtype;
  // 第二优先级规则：productSubtype为key且无productType
  const isLevel2Match = productSubtype === key && !productType;

  if (isLevel1Match || isLevel2Match) {
    result.push(item);
    hasHighPriorityMatches = true;
  }
}

// 前两级无匹配时，用第三级规则兜底
if (!hasHighPriorityMatches) {
  result.push(...targetCollection.filter(item => {
    const { country, year, productType, productSubtype } = item;
    return country === value && years.includes(year) && 
           (productSubtype === key || productType === key);
  }));
}

const rows = result;

优化点说明 #

• 减少遍历次数：原代码最多遍历3次集合，优化后最多2次（前两级一次，第三级一次；若前两级有匹配则仅遍历1次）
• 复用公共逻辑：将country === value和years.includes(year)提取为公共条件，避免重复判断
• 保留优先级逻辑：严格遵循原逻辑的匹配顺序，确保高优先级规则先生效
• 代码更简洁：消除重复的解构和条件代码，逻辑层次更清晰

内容的提问来源于stack exchange，提问作者Dalama