Question

我必须为课程分配创建一个Scrabble Scorer。在此程序中，我必须创建一个使用现有对象“ oldScoreKey”的函数，并创建一个newScoreKey，其中26个字母为键，并且为小写字母。每个字母的点就是值。听起来很简单，但是我在完成这部分工作时仍然遇到问题。我希望我可以输入一个新的分数键，但是此分配的要求是要为我做一个函数。我不确定我要执行的功能缺少什么。任何建议或解决方案的解释将不胜感激。我希望我所说的是有道理的，在此先感谢。

在过去的一周中，我尝试了forEach（）和for ... in循环语句。对于小写字母，我使用了.toLowerCase，但是一直出现语法错误，或者它被忽略了。

  1: ["A", "E", "I", "O", "U", "L", "N", "R", "S", "T"],
  2: ["D", "G"],
  3: ["B", "C", "M", "P"],
  4: ["F", "H", "V", "W", "Y"],
  5: ["K"],
  8: ["J", "X"],
  10: ["Q", "Z"]
}

//My most recent attempt
function transform(oldScoreKey){
  for(let key in oldScoreKey){
    for(let i = 0; i < oldScoreKey[key].length; i++){
      for(let items in oldScoreKey[key][i]){
      let newScoreKey =+ oldScoreKey[items][keys];
      newScoreKey.toLowerCase();
      return newScoreKey;  
      }
    }
  }
}

//CLOSE BUT NOT RIGTH: SHows each letter on it's own line but the point value is on it's own line as well right below it.
 function transform(oldScoreKey){
   for(key in oldScoreKey){
     for(let i = 0; i < oldScoreKey[key].length; i++){
       let newScoreKey = {};
       newScoreKey += oldScoreKey[key][i];
      return newScoreKey[key];
    }
  }
 }



My expected results when returning newScoreKey is a new object which shows all 26 letters in lowercase as the keys and their point value as the vale.
Ex) a : 1,
    e : 1,
    and so on

Instead I'm getting undefined or all the letters each with their point values listed under them in the console screen.

Answer 1

在函数内部创建一个新对象，并在每次内部迭代时，将其分配给该对象的属性，然后在函数的最后返回该对象：

const obj = {
  1: ["A", "E", "I", "O", "U", "L", "N", "R", "S", "T"],
  2: ["D", "G"],
  3: ["B", "C", "M", "P"],
  4: ["F", "H", "V", "W", "Y"],
  5: ["K"],
  8: ["J", "X"],
  10: ["Q", "Z"]
}

function transform(oldScoreKey){
  const newObj = {};
  for (const [letterValue, letterArr] of Object.entries(obj)) {
    for (const letter of letterArr) {
      newObj[letter.toLowerCase()] = letterValue;
    }
  }
  return newObj;
}

console.log(transform(obj));

Answer 2

一种方法是获取输入键（使用Object.keys）并在这些键上循环。对于每个键（score），然后遍历每个字母。将该字母分配给具有score值的新对象。返回新创建的对象。

这是这种方法：

const start = {
  1: ["A", "E", "I", "O", "U", "L", "N", "R", "S", "T"],
  2: ["D", "G"],
  3: ["B", "C", "M", "P"],
  4: ["F", "H", "V", "W", "Y"],
  5: ["K"],
  8: ["J", "X"],
  10: ["Q", "Z"]
};

function transform(oldScoreKey) {
  const result = {};

  Object.keys(oldScoreKey).forEach(score => {
    oldScoreKey[score].forEach(letter => {
      result[letter] = score;
    });
  });

  return result;
}

console.log(transform(start));

Answer 3

<com.google.android.material.bottomnavigation.BottomNavigationView

            android:id="@+id/bottomNavigationView"
            android:layout_width="match_parent"
            android:layout_height="match_parent"
            app:labelVisibilityMode="unlabeled"
            app:elevation="0dp"
            app:menu="@menu/bottom_navigation"> 

        </com.google.android.material.bottomnavigation.BottomNavigationView>

Answer 4

有几种方法可以做到这一点；一种方法是通过create or replace function consolidated_ad_traffic() returns table( traffic_source text , ad_month timestamp with time zone , ad_cost numeric(11,2) , ad_sales numeric(11,2) , conversion_cost numeric(11,6) ) language sql AS $$ with ad_sources as ( select 'Google' as traffic_source , "date" as ad_date , round(cast (cost AS numeric ) / 1000000.0,2) as cost , sales , cost_per_conversion from googleads_campaign union all select 'Taboola' , "date" , spent , sales , cost_per_conversion from taboola_campaign union all select 'Microsoft' , "TimePeriod" , "Spend" , sales , cost_per_conversion from microsoft_campaign ) select * from ad_sources; $$;函数，如下所示：

select * from consolidated_ad_traffic();

select distinct on( traffic_source, to_char(ad_month, 'mm'))
       traffic_source
     , to_char(ad_month, 'Mon') "For Month" 
     , to_char(sum(ad_cost) over(partition by traffic_source,  to_char(ad_month, 'Mon')), 'FM99,999,999,990.00') monthly_traffic_cost
     , to_char(sum(ad_cost) over(partition by traffic_source), 'FM99,999,999,990.00')   total_traffic_cost 
  from consolidated_ad_traffic();

select traffic_source, sum(ad_cost) ad_cost
  from consolidated_ad_traffic()
 group by traffic_source
 order by traffic_source;

 select traffic_source
      , to_char(ad_month, 'dd-Mon') "For Month"
      , sum(ad_cost) "Monthly Cost"
   from consolidated_ad_traffic()
  where date_trunc('month',ad_month) = date_trunc('month', date '2019-07-01')
    and traffic_source = 'Google'
  group by traffic_source, to_char(ad_month, 'dd-Mon') ;

Answer 5

您可以简单地做到这一点

function transform(ob){
    var newObject = {};
    Object.keys(ob).forEach((key)=>{
        ob[key].forEach((character)=>{
            newObject[character] = key;
        })
    })
    return newObject;
}

如何创建一个接受对象并创建新对象的函数，其中值是键，键是值

5 个答案: