javascript push() 在 reduce() 中不会按预期工作
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push() won't work as expected in reduce()
提问by Lasse Karagiannis
Why doesn't a.push(b)work in my Array.reduce()? a=a.push(b)where bis a string, turns ato an integer.?!
为什么没有a.push(b)在我的工作，Array.reduce()? a=a.push(b)在那里b是一个字符串，轮流a为整数。？！
getHighestValuesInFrequency: function(frequency) {
//Input:var frequency = {mats: 1,john: 3,johan: 2,jacob: 3};
//Output should become ['John','jacob']
var objKeys = Object.keys(frequency);
var highestVal = objKeys.reduce((a,b)=>
{highestVal = (frequency[b] > a)? frequency[b] : a;
return highestVal;},0);
var winner = objKeys.reduce((a,b)=>
{ a = (frequency[b] === highestVal)? a.push(b) : a;
return a},[]);
return winner;
}
回答by Barmar
Since push()returns the new length of the array, you're assigning the length to a. Instead of a conditional operator, use an ifstatement.
由于push()返回数组的新长度，因此您将长度分配给a. 使用if语句代替条件运算符。
var winner = objKeys.reduce((a, b) => {
if (frequency[b] === highestVal?) {
a.push(b);
}
return a;
}, []);
回答by Alexander Moiseyev
The push() returns the new length. You can use ES2015 spread syntax:
push() 返回新长度。您可以使用 ES2015 扩展语法：
var winner = objKeys.reduce((a, b)=> {
a = (frequency[b] === highestVal)? [...a, b] : a;
return a
}, []);
回答by alexi2
Building on the excellent answer by @Alexander Moiseyev.
基于@Alexander Moiseyev 的出色回答。
You can avoid setting and mutating both variables aand winnerby doing the following:
您可避免设置和不断变化的两个变量a，并winner通过执行以下操作：
return objKeys.reduce((acc, val) =>
frequency[val] === highestVal
? [...acc, val]
: acc
,[])
note: for clarity I have explicitly declared the accumulator and value in this reduce method.
注意：为了清楚起见，我在这个 reduce 方法中明确声明了累加器和值。
回答by Daniel Rosano
Please note that this structure you provide is not clear enough
请注意你提供的这个结构不够清晰
I would use instead an array of objects each having a name and a frecuency
我会使用一个对象数组，每个对象都有一个名称和一个频率
var frequencies = [{name : "mats", frecuency : 1},
{name : "john", frecuency: 3},
{name : "johan", frecuency: 2},
{name : "jacob", frecuency: 3}];
Then you can use a filter operation and map to get what you need
然后你可以使用过滤操作和映射来得到你需要的
var max = Math.max.apply(Math, frequencies.map(function(o){return o.frecuency;}));
var maxElems = frequencies.filter(function(a){return a.frecuency == max}).map(function(a){return a.name;});
maxElemswill give you the names of the people with higher frecuency
maxElems会给你更高的人的名字 frecuency
