我更喜欢下划线框架。它提出了许多有用的对象操作。 你的任务:

var newArray = homes.filter(
    price <= 1000 & 
    sqft >= 500 &
    num_of_beds >=2 & 
    num_of_baths >= 2.5);

可以像这样覆盖:

var newArray = _.filter (homes, function(home) {
    return home.price<=1000 && sqft>=500 && num_of_beds>=2 && num_of_baths>=2.5;
});

希望对大家有用!

你可以使用forEach

const filterOutputs = [];
json.homes.forEach((home) => {
if (
parseInt(home.price) <= 1000 &&
parseInt(home.sqft) >= 500 &&
parseInt(home.num_of_beds) >= 2 &&
parseInt(home.num_of_baths) >= 2
) {
filterOutputs.push(home);
}
});

console.log(filterOutputs);

使用数组过滤数据

const pickupData = [
    {
      id: 2876635,
      pickup_location: "6311cdacf6b493647d86",
      address_type: null,
      address: "999, Jagarati",``
      address_2: "Vihar",
      updated_address: false,
      old_address: "",
      old_address2: "",
      city: "Meerut",
      state: "Uttar Pradesh",
      country: "India",
      pin_code: "250001",
      email: "938@gmail.com",
      is_first_mile_pickup: 0,
      phone: "76898",
      name: "Aa",
      company_id: 2889808,
      gstin: null,
      vendor_name: null,
      status: 2,
      phone_verified: 1,
      lat: null,
      long: null,
      warehouse_code: null,
      alternate_phone: null,
      rto_address_id: 2867270,
      lat_long_status: 0,
      new: 1,
      associated_rto_address: null
    },
    {
      id: 2872407,
      pickup_location: "6311cdad490cf6b493647d82",
      address_type: null,
      address: "Nagar",
      address_2: "Delhi",
      updated_address: false,
      old_address: "",
      old_address2: "",
      city: "Bijnor",
      state: "Uttar Pradesh",
      country: "India",
      pin_code: "246701",
      email: "ima@gmail.com",
      is_first_mile_pickup: 0,
      phone: "75398",
      name: "Amit Sharma",
      company_id: 2889808,
      gstin: null,
      vendor_name: null,
      status: 1,
      phone_verified: 1,
      lat: null,
      long: null,
      warehouse_code: null,
      alternate_phone: null,
      rto_address_id: 2867270,
      lat_long_status: 0,
      new: 1,
      associated_rto_address: null
    }
  ];

const shiprocketData = [ { line1: "999, Jagarati", line2: "Vihar", city: "Meerut", state: "Uttar Pradesh", pincode: 250001, country: "India", isCurrent: true, _id: "6311cdad490cf6b3647d86" }, { line1: "999, Jagarati", line2: "Vihar", city: "Meerut", state: "Uttar Pradesh", pincode: 250001, country: "India", isCurrent: true, _id: "6311cdad490cb493647d82" }, { line1: "999, Jagarati", line2: "Vihar", city: "Meerut", state: "Uttar Pradesh", pincode: 250001, country: "India", isCurrent: true, _id: "6311cdad490cf693647d89" } ];

const updatedData = () => {
    const data = pickupData.filter(
      (item, index) =>
        item.pickup_location === shiprocketData.map((item) => item._id)[index]
    );
    return console.log(data);
  };

我使用ruleOut函数根据特定的不需要的属性值过滤对象。我知道在您的示例中，您希望使用条件而不是值，但我的答案对问题标题有效，因此我想把我的方法留在这里。

function ruleOut(arr, filterObj, applyAllFilters=true) {    
    return arr.filter( row => {            
        for (var field in filterObj) {
            var val = row[field];
            if (val) {                    
                if (applyAllFilters && filterObj[field].indexOf(val) > -1) return false;                
                else if (!applyAllFilters) {                        
                    return filterObj[field].filter(function(filterValue){ 
                        return (val.indexOf(filterValue)>-1);
                    }).length == 0;                 
                }
            }
        }
        return true;
    });
}

假设你有一个演员名单，是这样的:

let actors = [
  {userName:"Mary", job:"star", language:"Turkish"},
  {userName:"John", job:"actor", language:"Turkish"},
  {userName:"Takis", job:"star", language:"Greek"},
  {userName:"Joe", job:"star", language:"Turkish"},
  {userName:"Bill", job:"star", language:"Turkish"}
];

你想找到所有被评为好莱坞明星的演员，他们的国籍不应该是“英国”、“意大利”、“西班牙”、“希腊”之一，加上他们的名字也不应该是“玛丽”、“乔”之一。奇怪的例子，我知道!不管怎样，有了这些条件，你可以创建以下对象:

let unwantedFieldsFilter= { 
  userName: ['Mary', 'Joe'],    
  job: ['actor'],   
  language: ['English', 'Italian', 'Spanish', 'Greek']  
};

现在如果你排除(actors, unwantedFieldsFilter)你只会得到

[{用户名:“比尔”，职位:“明星”，语言:“土耳其语”}]

比尔是你的男人，因为他的名字不是“玛丽”，“乔”之一，他的国籍不包括在['英国'，'意大利'，'西班牙'，'希腊']，加上他是一个明星!

在我的方法中有一个选项，即applyAllFilters，默认为true。 如果你试图将这个参数设置为false，这将作为“或”过滤而不是“与”过滤。 例如:ruleOut(actors， {job:["actor"]， language:["Italian"]}， false)将得到所有不是演员或意大利人的人:

[{用户名:“玛丽”，职位:“明星”，语言:“土耳其语”}， {用户名:“Takis”，职位:“star”，语言:“Greek”}， {用户名:“乔”，工作:“明星”，语言:“土耳其语”}， {用户名:“比尔”，工作:“明星”，语言:“土耳其语”}]

我很惊讶居然没有人发这样的回复:

const filteredHomes = json.homes.filter(x => x.price <= 1000 && x.sqft >= 500 && x.num_of_beds >=2 && x.num_of_baths >= 2.5);

.．.为了便于阅读:

const filteredHomes = json.homes.filter( x => 
  x.price <= 1000 && 
  x.sqft >= 500 && 
  x.num_of_beds >=2 && 
  x.num_of_baths >= 2.5
);

const state.contactList = [{
    name: 'jane',
    email: 'jane@gmail.com'
  },{},{},...]

const fileredArray = state.contactsList.filter((contactItem) => {
  const regex = new RegExp(`${action.payload}`, 'gi');
  return contactItem.nameProperty.match(regex) || 
    contactItem.emailProperty.match(regex);
});


// contactList: all the contacts stored in state
// action.payload: whatever typed in search field