echarts 桑基图dataLinks的自定义方法

let arr1 = [{ type: "t1", chr: "多躲大定夺" }];
let arr2 = [
  { type: "t2", chr: "多躲大打弟" },
  { type: "d2", chr: "定夺洞" },
];

let arr3 = [
  {source: "t1", target: "t2", value: 3},
  {source: "t1", target: "d2", value: 2}
]

想从arr1和arr2实现arr3。
根据chr属性值:
当arr1的 chr值与arr2[0]中chr相同时，为source赋值't1", target赋值"t2", 并根据相同的数量赋值value;
当arr1的 chr值与arr2[1]中chr相同时，为source赋值't1", target赋值"d2",并根据相同的数量赋值value;

let arr1 = [{type: "t1", chr: "多躲大定夺"}];
let arr2 = [
    {type: "t2", chr: "多躲大打弟"},
    {type: "d2", chr: "定夺洞"},
];

let arr3 = [
    {source: "t1", target: "t2", value: 3},
    {source: "t1", target: "d2", value: 2}
]

/**
 * 计算相似度的函数需要你来实现了，目前的悬赏额度我写不出完美的，所以写死了
 * @param str1
 * @param str2
 * @returns {number}
 */
function test(str1, str2) {
    let char1 = str1.split('');
    let char2 = str2.split('');
    let count = 0;
    for (let i = 0; i < char1.length; i++) {
        for (let j = 0; j < char2.length; j++) {
            if (char1[i] == char2[j]) {
                count++
            }
        }
    }
    return count;
}

let result = arr2.map(item => {
    let filter = arr1.map(subItem => ({
        ...subItem,
        value: test(subItem.chr, item.chr)
    })).filter(subItem => subItem.value > 0);
    filter = filter[0] || {value: 0}
    return ({
        source: filter.type,
        target: item.type,
        value: filter.value
    })
}).filter(item => item.value > 0);

console.log(result)

      let arr1 = [
        { type: "t1", chr: "多躲大定夺" },
        { type: "t1", chr: "定夺洞" }
      ];
      let arr2 = [
        { type: "t2", chr: "多躲大定夺" },
        { type: "t2", chr: "多躲大定夺" },
        { type: "d2", chr: "定夺洞" },
      ];
      let arr3 = [];
      arr1.forEach((i) => {
        arr2.forEach((v) => {
          if (i.chr === v.chr) {
            let existObj= arr3.find((item) => {
              return item.target === v.type;
            });
            if (existObj) {
              existObj.value += 1;
            } else {
              arr3.push({ source: i.type, target: v.type, value: 1 });
            }
          }
        });
      });
      console.log(arr3);

//将数组1和2合并
arr3  = arr1 .concat(arr2 )
//根据chr去重数组3
unique(arr3) {
  const res = new Map()
  return arr3.filter((item) => !res.has(item.chr) && res.set(item.chr, 1))
}

不知道是这个意思么？

let arr1 = [{ type: "t1", chr: "多躲大打弟" },{ type: "t1", chr: "定夺洞" }];
let arr2 = [
  { type: "t2", chr: "多躲大打弟" },
  { type: "t3", chr: "多躲大打弟" },
  { type: "d2", chr: "定夺洞" },
];
      let arr3=[]
     for(var o in arr1){ 
        var array = [];
         for(var i in arr2){            
            if(arr1[o].chr==arr2[i].chr){
                if(array[arr1[o].chr] !== undefined){
                    array[arr1[o].chr].value=array[arr1[o].chr].value+1;
                }
                else{
                array[arr1[o].chr] = {source: arr1[o].type, target: arr2[i].type, value: 1};
                }
                //
            }
        }
        for(key in array){
            arr3.push(array[key]);
        }
        
     }
     console.log(arr3);     

看看有没有帮助
https://b23.tv/mq7xAQp

参考链接