Question

我有以下小提琴使用测试数据，我正在使用$ .each

查看数据

我可以循环没有问题，但我想循环遍历数据，然后随机从中获取3个对象。

任何提示或技巧都会有所帮助：

https://jsfiddle.net/inkedraskal/pah44qv6/

$.each(testData,function(x, blah){
    //console.log(blah._id);

  //this gets each objects id & picture, but I want to get 3 random objects, and their corresponding data
  var activeValue = blah._id,
        pictureValue = blah.picture;


  var markUp = '';

  markUp += activeValue + pictureValue;

  console.log(markUp);
});

如下问题，它们必须是唯一的**

Answer 1

我们可以创建Math.random()函数，它将使用function getRandomEntry() { return testData[Math.round(Math.random() * (testData.length - 1))]; } for (var i=0; i<3; i++) { var entry = getRandomEntry(); console.log(entry._id, entry.picture); }函数返回一个数组元素。我们将创建一个循环，每次迭代时都会随机输入。

var randomEntries = [];

function getRandomEntry() {
  return testData[Math.round(Math.random() * (testData.length - 1))];
}

function entryExists(entry) {
  return randomEntries.indexOf(entry) > -1;
}

for (var i=0; i<3; i++) {
  var entry;

  do {

    entry = getRandomEntry();

  } while(entryExists(entry))

  randomEntries.push(entry);
  console.log(entry._id, entry.picture);
}

如果您每次都需要一个独特的条目。您可以将随机条目保存在单独的数组中，并检查新的是否为unqiue。

question = raw_input
y = "Blah"
c = "Blahblahb"
y_counter = 0
c_counter = 0


print "Is bacon awesome"
if question() = "Yes":
    print y
    y_counter = y_counter + 1
else:
    print c
    c_counter = c_counter + 1

print "Blah"
if question() = "Yes":
    print y
    y_counter = y_counter + 1
else:
    print c
    c_counter = c_counter + 1

print "y was used " + str(y_counter) + " times!"
print "c was used " + str(c_counter) + " times!"

Answer 2

下面的代码段中的函数获取一个数组和一个数字（X）的项目，并返回一个新数组，其中包含原始数组中的X个唯一随机项：

function getRandomItems(arr, items) {
  var ret = [];
  var indexes = [];
  var arr_length = arr.length;
  
  // If we don't have enough items to return - return the original array
  if (arr_length < items) {
    return arr;
  }
  
  while (ret.length < items) {
    i = Math.floor(Math.random() * arr_length);
    if (indexes.indexOf(i) == -1) {
      indexes[indexes.length] = i;
      ret[ret.length] = arr[i];
    }
  }
  return ret;
}

arr = ['a', 'b', 'c', 'd', 'e']
console.log(getRandomItems(arr, 2))

如果您希望在代码中的每个Array.prototype上将该函数用作“本机”函数，也可以将该函数添加到Array：

Array.prototype.getRandomItems = function(items) {
  var ret = [];
  var indexes = [];
  var arr_length = this.length;
  
  // If we don't have enough items to return - return the original array
  if (arr_length < items) {
    return this;
  }
  
  while (ret.length < items) {
    i = Math.floor(Math.random() * arr_length);
    if (indexes.indexOf(i) == -1) {
      indexes[indexes.length] = i;
      ret[ret.length] = this[i];
    }
  }
  return ret;
}

arr1 = ['a', 'b', 'c', 'd', 'e']
arr2 = ['aaa', 'bbb', 'ccc', 'ddd', 'eee']

console.log(arr1.getRandomItems(1))
console.log(arr1.getRandomItems(2))
console.log(arr2.getRandomItems(3))
console.log(arr2.getRandomItems(4))

Answer 3

您可以使用如下的递归函数。

请注意，此实施可防止结果中的重复

function getRandomObjects(array,selected,needed){
    /*
     *@param array array The array to pull from
     *@param selected array The array of results pulled so far
     *@param needed int The number of results we want
     */
    var length = array.length;
    var num = Math.floor(Math.random() * length) + 1; // get random number in bounds of array
    var exists=false; // make sure we didnt already pick this object
    $.each(selected,function(i,obj){
        if(obj.index==num)exists=true;
    })
    if(exists) getRandomObjects(array,selected,needed); // get a new one if this was a duplicate
    else selected.push(array[num]);
    if(selected.length!=needed) return getRandomObjects(array,selected,needed); // get another object if we need more
    else return selected; // return the final result set
}




var testData = [
  {
    "_id": "57e5d1a90c4206b128cd8654",
    "index": 0,
    "guid": "1f3269fc-0822-4c5a-9c52-8055155b407e",
    "isActive": true,
    "balance": "$3,026.95",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a9a986ccb2f41cf7b9",
    "index": 1,
    "guid": "a6b726b6-6466-4e48-8697-1c6bd7b1c79e",
    "isActive": true,
    "balance": "$2,642.74",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a9f98f8b2f6880de32",
    "index": 2,
    "guid": "e7d736cc-19e0-4bcb-8d0a-4d17442d8cee",
    "isActive": true,
    "balance": "$3,341.64",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a9e40ded5b017e45cd",
    "index": 3,
    "guid": "64230ca8-05c0-4c39-a931-794172475a32",
    "isActive": true,
    "balance": "$2,196.13",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a90cc30be769a06d7c",
    "index": 4,
    "guid": "d6618b78-753a-4ad0-bc14-3687d0b99196",
    "isActive": true,
    "balance": "$1,611.62",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a92481a43f50607415",
    "index": 5,
    "guid": "35ec8186-9494-4f89-ab89-bed7f39872c3",
    "isActive": true,
    "balance": "$3,148.87",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a9164f17c558ba7ce1",
    "index": 6,
    "guid": "244970a0-1ce2-405a-8d69-c7903f9bf5eb",
    "isActive": false,
    "balance": "$3,758.13",
    "picture": "http://placehold.it/32x32"
  },
  {
    "_id": "57e5d1a95afde31c5cf592a8",
    "index": 7,
    "guid": "aa30c82d-dd2b-420c-8b30-7d66cec8d10b",
    "isActive": true,
    "balance": "$1,311.40",
    "picture": "http://placehold.it/32x32"
  }
]


var randomObjects=getRandomObjects(testData,[],3);

console.log(randomObjects);

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer 4

您可以做到

var items = [1, 2, 3, 4, 5];
var newItems = [];

for(var i = 0; i < 3; i++) {
    var idx = Math.floor(Math.random() * items.length);
    newItems.push(items[idx]);
    items.splice(idx, 1);
}

console.log(newItems);

在javascript中从数组中获取随机X项

4 个答案: