昨天，我遇到了这个面试问题。最初，它似乎很容易，至少在逻辑上。 但不知何故，我无法在JavaScript中使用它。
这是学生成绩的二维数组，学生姓名可能会重复多次。如果是这种情况，请将所有分数相加并除以出现次数以找到平均值，如有必要，请运行Math.floor。
var arr = [
["Bobby","87"],
["Charles","100"],
["Eric","65"],
["Charles","22"],
["Charles","37"],
["Eric","49"]]
因此，Charles平均得分为Math.floor((100+22+37)/3) = 53
而且，对于Eric，它将是Math.floor((65+49)/2) = 57。
所以最高平均值为["Bobby","87"]。
到目前为止，我的尝试毫无结果..
var op_arr = [];
arr.each(function(item) {
var sum = 0;
var itemCount = 1;
var checkFlag = isItemInArray(arr,item);
if(checkFlag) {
itemCount++;
sum += item[1];
}
});
function isItemInArray(array,item) {
for(let i = 0;i < array.length; i++) {
if(array[i][0] === item[0]) {
return array[i];
}
}
return false;
}
但这不起作用。 请帮助我并解释逻辑。
答案 0 :(得分：5)
我会使用一些代码将列表转换为哈希映射，因为可能有多个同一个学生。
var arr = [
["Bobby","87"],
["Charles","100"],
["Eric","65"],
["Charles","22"],
["Charles","37"],
["Eric","49"]
];
var scores = {};
for (var i = 0; i < arr.length; i++) {
var student = arr[i];
if (!scores.hasOwnProperty(student[0]))
scores[student[0]] = []
scores[student[0]].push(student[1])
}
结果应为：
{
"Bobby": ["87"],
"Charles": ["100", "22", "37"],
"Eric": ["65", "49"]
}
现在你可以对对象进行第二次传递来计算平均值
for (var key in scores) {
if (!scores.hasOwnProperty(key)) continue;
var total = scores[key].reduce(function(next, cur) {
return next + parseInt(cur);
}, 0);
scores[key] = Math.floor(total / scores[key].length);
}
console.log(scores);
我确信你可以使用ES6功能让它变得更优雅，但这可以让你了解一个解决方案。
答案 1 :(得分：0)
您可以先获取每个人的所有平均值，然后获得临时结果的最高平均值。
如果不止一个人的得分相同，则包括所有人。
var array = [["Bobby", "87"], ["Charles", "100"], ["Eric", "65"], ["Charles", "22"], ["Charles", "37"], ["Eric", "49"]],
highest = array
.reduce(function (r, a) {
var i = r.findIndex(b => a[0] === b[0]);
if (i !== -1) {
r[i][1] = (r[i][1] * r[i][2] + +a[1]) / ++r[i][2];
} else {
r.push(a.concat(1));
}
return r;
}, [])
.reduce(function (r, a, i) {
if (!i || r[0][1] < a[1]) {
return [a.slice(0, 2)];
}
if (r[0][1] === a[1]) {
r.push(a.slice(0, 2));
}
return r;
}, []);
console.log(highest);
答案 2 :(得分：0)
您可以遵循此方法
reduce map使用reduce和再次使用var fnSumAvgArray = (arr) => arr.reduce( ( a, c ) => a + c, 0 )/arr.length;
var arr = [
["Bobby","87"],
["Charles","100"],
["Eric","65"],
["Charles","22"],
["Charles","37"],
["Eric","49"]];
var output = Object.values( arr.reduce( (a,c) => (
a[c[0]] = (a[c[0]] || { name : c[0], scores : [] }), //check if accumulator already has been initialized for this name or else intialize
a[ c[ 0 ] ].scores.push ( +c[1] ), //push the score into the name based score array
a ) , {}) ) //return the accumulator
.map( s => (
s.avg = fnSumAvgArray(s.scores), s
));
console.log( output );计算>得分<强>演示
git clone https://github.com/karelzak/util-linux.git
答案 3 :(得分：0)
var arr = [
["Bobby","87"],
["Charles","100"],
["Eric","65"],
["Charles","22"],
["Charles","37"],
["Eric","49"]];
var x, students=[], counter=[], scoreSums=[], name, i=0;
for(x in arr) {
name = arr[x][0];
j = students.indexOf(name);
if(j < 0) {
students[i] = name;
counter[i] = 1;
scoreSums[i] = parseInt(arr[x][1]);
i++;
} else {
scoreSums[j] += parseInt(arr[x][1]);
counter[j]++;
}
}
var maxMean = 0, mean;
for(i=0; i<students.length; i++) {
mean = scoreSums[i] / counter[i];
if(mean > maxMean) {
name = students[i];
maxMean = mean;
}
}
document.getElementById('maxMean').innerHTML='Highest average is '+ name+' at '+maxMean;
console.log(name, maxMean);<div id="maxMean"></div>
答案 4 :(得分：0)
这是一个围绕房屋的地方＆＃39;做法。这是一个稍微长一点的写出来，但我认为这是人类可读的，应该有助于这个概念。
我评论了下面的代码。
var arr = [
["Bobby", "87"],
["Charles", "100"],
["Eric", "65"],
["Charles", "22"],
["Charles", "37"],
["Eric", "49"]
];
let objects = [];
let names = [];
for (let item of arr) {
// Get unique names
if (names.indexOf(item[0]) <= 0) {
names.push(item[0]);
}
// Create object rather than array
let score = {
name: item[0],
score: parseInt(item[1])
}
objects.push(score);
}
// Work out average based on name
function countScore(name) {
let count = 0;
let total = 0;
for (let object of objects) {
if (object.name === name) {
count += 1;
total += object.score
}
}
let avgScore = total / count;
console.log(name + ': ' + avgScore)
}
// Run function for each unique name
for (let name of names) {
countScore(name);
}
我希望你觉得这很有用。
答案 5 :(得分：0)
let findMaxAvg = (arr) => {
let studmap = new Map();
arr.forEach((item,ind) => {
let [name,score] = [item[0],item[1]];
!studmap.has(name) ? studmap.set(name,[score,1]) : studmap.set(name,
[studmap.get(name)[0]+score,studmap.get(name)[1]+1])
});
return Math.max(...[...studmap.values()].map(sdata =>
Math.round(sdata[0]/sdata[1])));
}
var arr = [
["Bobby","87"],
["Charles","100"],
["Eric","200"],
["Charles","22"],
["Charles","37"],
["Eric","49"]];
console.log(findMaxAvg(arr));
答案 6 :(得分：0)
根据上面提供的解决方案，您可以通过多种方式解决此问题。我之前也遇到过这个问题，我用过
<块引用>收集所有学生和他们的分数并返回带有 [name, average] 的数组。
const scores = [['brian', 80], ['brian', 90], ['brian', 65], ['robhy', 85], ['robhy', 75], ['ryan', 75], ['murphy', 85]];
const calculateHigestAvg = (scores)=>{
let studentObj = new Map();
let score = [];
for (var i = 0; i < scores.length; i++) {
if (studentObj.has(scores[i][0])) {
const studentScoreArr = studentObj.get(scores[i][0])
studentScoreArr.push(scores[i][1]);
studentObj.set(scores[i][0], studentScoreArr);
} else {
studentObj.set(scores[i][0], [scores[i][1]]);
}
}
return Array.from(studentObj.entries())
.map((item, index)=>{
return [item[0], Math.floor(item[1].reduce((a, b) => a + b) / item[1].length)]
}
, 0)
}
calculateHigestAvg(scores);
// results
[["brian", 78],["robhy", 80],["ryan", 75],["murphy", 85]]