キー名は異なるが対応する値が同じであるオブジェクトの2つの配列を組み合わせる方法は？

const cafes = [
  {
    name: "Bazaar Cafe",
    place_id: "kjk234g4gcvfx8usg1l33pi",
  },
  {
    name: "Ashley's Cafe",
    place_id: "12hydbdf76sljfts87sbfis",
  },
  {
    name: "Avenue Cafe",
    place_id: "skjd86svvfdrsv55svbvf3f",
  },
  {
    name: "Hi-Lo Cafe",
    place_id: "mjdhgetr4pojfyts22fzfsh",
  },
  {
    name: "California Chicken Cafe",
    place_id: "12hydbdf76sljfts87sbfis",
  },
  {
    name: "Avenue Bakery Cafe",
    place_id: "jahgde7wgdiau8ewsahgosd",
  },
  {
    name: "Philz Coffee",
    place_id: "urhw3837ehalod7w02b7835",
  },
];

const places = [
  {
    id: "jahgde7wgdiau8ewsahgosd",
    street_no: "60H",
    locality: "Solomos Island Road",
    postal_code: "20688",
    lat: "36.7783 N",
    long: "119.4179 W",
  },
  {
    id: "12hydbdf76sljfts87sbfis",
    street_no: "1B",
    locality: "Macarthur Blvd",
    postal_code: "20619",
    lat: "38.1781 N",
    long: "118.4171 W",
  },
  {
    id: "kjk234g4gcvfx8usg1l33pi",
    street_no: "45250",
    locality: "Worth Avenue, Unit A",
    postal_code: "20619",
    lat: "36.1152",
    long: "117.521",
  },
  {
    id: "saswe3s6yydtdr52hsd72yst",
    street_no: "1X",
    locality: "Macarthur Blvd",
    postal_code: "20687",
    lat: "36.7783",
    long: "119.4179",
  },
  {
    id: "skjd86svvfdrsv55svbvf3f",
    street_no: "7S",
    locality: "Three Notch Road",
    postal_code: "20619",
    lat: "36.83",
    long: "119.6",
  },
  {
    id: "mjdhgetr4pojfyts22fzfsh",
    street_no: "22803",
    locality: "Gunston Dr Lexington Park",
    postal_code: "20688",
    lat: "35.7788",
    long: "119.979",
  },
  {
    id: "urhw3837ehalod7w02b7835",
    street_no: "225",
    locality: "Macarthur Blvd",
    postal_code: "20687",
    lat: "35.77813",
    long: "119.41791",
  },
];

最終的な結合配列を次のようにしたい-

finalArr = [
  {
    id: "kjk234g4gcvfx8usg1l33pi",
    street_no: "45250",
    locality: "Worth Avenue, Unit A",
    postal_code: "20619",
    lat: "36.1152",
    long: "117.521",
    name: "Bazaar Cafe"
  },
  {
    id: "12hydbdf76sljfts87sbfis",
    street_no: "1B",
    locality: "Macarthur Blvd",
    postal_code: "20619",
    lat: "38.1781 N",
    long: "118.4171 W",
    name: "Ashley's Cafe"
  },
  {
    id: "12hydbdf76sljfts87sbfis",
    street_no: "1B",
    locality: "Macarthur Blvd",
    postal_code: "20619",
    lat: "38.1781 N",
    long: "118.4171 W",
    name: "California Chicken Cafe"
  },
  {
    id: "skjd86svvfdrsv55svbvf3f",
    street_no: "7S",
    locality: "Three Notch Road",
    postal_code: "20619",
    lat: "36.83",
    long: "119.6",
    name: "Avenue Cafe"
  },
  {
    id: "mjdhgetr4pojfyts22fzfsh",
    street_no: "22803",
    locality: "Gunston Dr Lexington Park",
    postal_code: "20688",
    lat: "35.7788",
    long: "119.979",
    name: "Hi-Lo Cafe"
  },
  {
    id: "jahgde7wgdiau8ewsahgosd",
    street_no: "60H",
    locality: "Solomos Island Road",
    postal_code: "20688",
    lat: "36.7783 N",
    long: "119.4179 W",
    name: "Avenue Bakery Cafe"
  }, 
  {
    id: "urhw3837ehalod7w02b7835",
    street_no: "225",
    locality: "Macarthur Blvd",
    postal_code: "20687",
    lat: "35.77813",
    long: "119.41791",
    name: "Philz Coffee"
  }
];

私はそれをこのように実装しようとしましたが、最終的な配列の最後のオブジェクトのみを取得します-

function combineById (cafes, places) {
  const finalArr = [];
  const wholeObj = {};
  const set = new Set();

  for (const cafe of cafes) { 
    let cafeId = cafe.place_id;
    let cafeName = cafe.name;
    for (const place of places) {
      if (cafeId === place.id && !set.has(cafeId)) {
        set.add(cafeId);
        wholeObj.id = place.id;
        wholeObj.streetNo = place.street_no;
        wholeObj.locality = place.locality;
        wholeObj.postalCode = place.postal_code;
        wholeObj.lat = place.lat;
        wholeObj.long = place.long;
        wholeObj.name = cafeName;
      }
    }
    finalArr.push(wholeObj);
  }
  return finalArr;
}

const resultArr = combineById(cafes, places);
console.log(resultArr);

この最後の配列と検索語という2つの引数を受け入れる検索関数を実装しようとしています。これまでのところ、検索関数は配列で正常に機能していますcafesが、placesとcafes配列を組み合わせるのに問題がありcombineByIdます。この関数は、最終的な配列の最後に結合されたオブジェクトの結果のみを返します。後でさらに操作するためにfinal配列を直接呼び出すことができるように、final配列内のオブジェクトのすべての組み合わせが必要です。解決策を見つけるのを手伝ってください。ありがとうございました。