data

    const inventors = [
      { first: 'Albert', last: 'Einstein', year: 1879, passed: 1955 },
      { first: 'Isaac', last: 'Newton', year: 1643, passed: 1727 },
      { first: 'Galileo', last: 'Galilei', year: 1564, passed: 1642 },
      { first: 'Marie', last: 'Curie', year: 1867, passed: 1934 },
      { first: 'Johannes', last: 'Kepler', year: 1571, passed: 1630 },
      { first: 'Nicolaus', last: 'Copernicus', year: 1473, passed: 1543 },
      { first: 'Max', last: 'Planck', year: 1858, passed: 1947 },
      { first: 'Katherine', last: 'Blodgett', year: 1898, passed: 1979 },
      { first: 'Ada', last: 'Lovelace', year: 1815, passed: 1852 },
      { first: 'Sarah E.', last: 'Goode', year: 1855, passed: 1905 },
      { first: 'Lise', last: 'Meitner', year: 1878, passed: 1968 },
      { first: 'Hanna', last: 'Hammarström', year: 1829, passed: 1909 }
    ];

    const people = [
      'Bernhard, Sandra', 'Bethea, Erin', 'Becker, Carl', 'Bentsen, Lloyd', 'Beckett, Samuel', 'Blake, William', 'Berger, Ric', 'Beddoes, Mick', 'Beethoven, Ludwig',
      'Belloc, Hilaire', 'Begin, Menachem', 'Bellow, Saul', 'Benchley, Robert', 'Blair, Robert', 'Benenson, Peter', 'Benjamin, Walter', 'Berlin, Irving',
      'Benn, Tony', 'Benson, Leana', 'Bent, Silas', 'Berle, Milton', 'Berry, Halle', 'Biko, Steve', 'Beck, Glenn', 'Bergman, Ingmar', 'Black, Elk', 'Berio, Luciano',
      'Berne, Eric', 'Berra, Yogi', 'Berry, Wendell', 'Bevan, Aneurin', 'Ben-Gurion, David', 'Bevel, Ken', 'Biden, Joseph', 'Bennington, Chester', 'Bierce, Ambrose',
      'Billings, Josh', 'Birrell, Augustine', 'Blair, Tony', 'Beecher, Henry', 'Biondo, Frank'
    ];

🍺 filter();

1. Filter the list of inventors for those who were born in the 1500's

리턴 조건

    const answer1 = inventors.filter((inventor) => {
      return (1500 <= inventor.year && inventor.year < 1600) ? true : false;
    })
    /*0: {first: 'Galileo', last: 'Galilei', year: 1564, passed: 1642}
1: {first: 'Johannes', last: 'Kepler', year: 1571, passed: 1630}*/

🌴 map()

2. Give us an array of the inventors first and last names

    const answer2 = inventors.map((inventor) => {
      return inventor.first + ' ' + inventor.last
    })
    /*['Albert Einstein', 'Isaac Newton', 'Galileo Galilei', 'Marie Curie', 'Johannes Kepler', 'Nicolaus Copernicus', 'Max Planck', 'Katherine Blodgett', 'Ada Lovelace', 'Sarah E. Goode', 'Lise Meitner', 'Hanna Hammarström']*/

6. Boulevards

😓 문제를 잘못 이해했다...ㅠ

create a list of Boulevards in Paris that contain 'de' anywhere in the name
프랑스 파리의 도로명 처럼 '~의 도로' Boulevards de ~
를 만들어주면 된다고 생각했다.

   const answer6 = inventors.map((inventor) => {
     return `Boulevard de ${inventor.first} ${inventor.last}`
   })
/*
['Boulevard de Ada Lovelace', 'Boulevard de Sarah E. Goode', 'Boulevard de Johannes Kepler', 'Boulevard de Marie Curie', 'Boulevard de Nicolaus Copernicus', 'Boulevard de Albert Einstein', 'Boulevard de Galileo Galilei', 'Boulevard de Hanna Hammarström', 'Boulevard de Katherine Blodgett', 'Boulevard de Isaac Newton', 'Boulevard de Max Planck', 'Boulevard de Lise Meitner']*/

🤩 다시 풀기

사실은 urlhttps://en.wikipedia.org/wiki/Category:Boulevards_in_Paris에 나와있는 도로들 중 de가 들어있는 것들만 리턴해야했다.

const answer6 = inventors.map((inventor) => {
     return `Boulevard de ${inventor.first} ${inventor.last}`
   })
   console.log(answer6)

   const category = document.querySeletor('.mw-category');
   // const links = category.querySelectorAll('a');

   //const de = links.map(link => link.textContent);
   //이렇게 하면 작동하지 않는다. querySelector로 선택해 NodeList가 반환되므로 배열로 바꾸어줘야 한다.
    
   const links = Array.from(category.querySelectorAll('a'));//전개연산자도 사용가능
   const de = links
     .map(link => link.textContent)
     .filter(streetName => streetName.includes('de'));

🥨 sort()

3. Sort the inventors by birthdate, oldest to youngest

    const answer3 = inventors.sort((a, b) => {
      return a.year - b.year;
      }
    )
    /*0: {first: 'Katherine', last: 'Blodgett', year: 1898, passed: 1979}
1: {first: 'Nicolaus', last: 'Copernicus', year: 1473, passed: 1543}
2: {first: 'Marie', last: 'Curie', year: 1867, passed: 1934}
3: {first: 'Albert', last: 'Einstein', year: 1879, passed: 1955}
4: {first: 'Galileo', last: 'Galilei', year: 1564, passed: 1642}
5: {first: 'Sarah E.', last: 'Goode', year: 1855, passed: 1905}
6: {first: 'Hanna', last: 'Hammarström', year: 1829, passed: 1909}
7: {first: 'Johannes', last: 'Kepler', year: 1571, passed: 1630}
8: {first: 'Ada', last: 'Lovelace', year: 1815, passed: 1852}
9: {first: 'Lise', last: 'Meitner', year: 1878, passed: 1968}
10: {first: 'Isaac', last: 'Newton', year: 1643, passed: 1727}
11: {first: 'Max', last: 'Planck', year: 1858, passed: 1947}*/

5. Sort the inventors by years lived

정렬할 기준 잡기

  const answer5 = inventors.sort((a, b) => {
     return (a.passed - a.year) - (b.passed - b.year);
   })
  
  /*
  0: {first: 'Katherine', last: 'Blodgett', year: 1898, passed: 1979}
1: {first: 'Nicolaus', last: 'Copernicus', year: 1473, passed: 1543}
2: {first: 'Marie', last: 'Curie', year: 1867, passed: 1934}
3: {first: 'Albert', last: 'Einstein', year: 1879, passed: 1955}
4: {first: 'Galileo', last: 'Galilei', year: 1564, passed: 1642}
5: {first: 'Sarah E.', last: 'Goode', year: 1855, passed: 1905}
6: {first: 'Hanna', last: 'Hammarström', year: 1829, passed: 1909}
7: {first: 'Johannes', last: 'Kepler', year: 1571, passed: 1630}
8: {first: 'Ada', last: 'Lovelace', year: 1815, passed: 1852}
9: {first: 'Lise', last: 'Meitner', year: 1878, passed: 1968}
10: {first: 'Isaac', last: 'Newton', year: 1643, passed: 1727}
11: {first: 'Max', last: 'Planck', year: 1858, passed: 1947}*/

7. sort Exercise

String.prototype.localeCompare()

The localeCompare() 메서드는 기준 문자열과 비교했을 때
비교 대상 문자열이 정렬상 전에 오는지, 후에 오는지 혹은 같은 순서에 배치되는지를 알려주는 숫자를 리턴한다.

Sort the people alphabetically by last name

   const answer7 = inventors.sort((a, b) => {
     return a.last.localeCompare(b.last)
   })
   
   
   /*
   0: {first: 'Katherine', last: 'Blodgett', year: 1898, passed: 1979}
1: {first: 'Nicolaus', last: 'Copernicus', year: 1473, passed: 1543}
2: {first: 'Marie', last: 'Curie', year: 1867, passed: 1934}
3: {first: 'Albert', last: 'Einstein', year: 1879, passed: 1955}
4: {first: 'Galileo', last: 'Galilei', year: 1564, passed: 1642}
5: {first: 'Sarah E.', last: 'Goode', year: 1855, passed: 1905}
6: {first: 'Hanna', last: 'Hammarström', year: 1829, passed: 1909}
7: {first: 'Johannes', last: 'Kepler', year: 1571, passed: 1630}
8: {first: 'Ada', last: 'Lovelace', year: 1815, passed: 1852}
9: {first: 'Lise', last: 'Meitner', year: 1878, passed: 1968}
10: {first: 'Isaac', last: 'Newton', year: 1643, passed: 1727}
11: {first: 'Max', last: 'Planck', year: 1858, passed: 1947}
*/

🍟 reduce()

4. How many years did all the inventors live all together?

풀이1 : 누산기 초기값 명시하기

이렇게 하면 curr의 데이터 타입이 number여도 누산기에는 제대로 합계를 구할 수 없다. 아마도 원본이 객체이기 때문인것 같다.

      const answer4 = inventors.reduce((sum, curr) => {
        const currYear = (curr.passed - curr.year)
        return sum + currYear;
      })

[object Object]78
index-START.html:78 [object Object]7859
index-START.html:78 [object Object]785984
index-START.html:78 [object Object]78598437
index-START.html:78 [object Object]7859843780
index-START.html:78 [object Object]785984378050
index-START.html:78 [object Object]78598437805089
index-START.html:78 [object Object]7859843780508967
index-START.html:78 [object Object]785984378050896790
index-START.html:78 [object Object]78598437805089679076
index-START.html:81 [object Object]7859843780508967907681

초기값을 0 으로 주면 해결된다.
0에서 시작해 '숫자'를 더할 수 있다.

    const answer4 = inventors.reduce((sum, curr) => {
        return sum+ (curr.passed - curr.year)
      },0)
 //861

풀이2 : array 먼저 만들기

다른 풀이로도 풀어봤다.

원본 객체에서 필요한 숫자 - 살아온 기간만 리턴해 배열을 만들어주고, 그 배열에서 reduce로 합계를 구하면 된다.

    const yearArr = inventors.map((inventor) => {
      return inventor.passed - inventor.year
    })
    //[70, 78, 59, 84, 37, 80, 50, 89, 67, 90, 76, 81]

     const answer4 = yearArr.reduce((sum, curr) => {
       return sum + curr;
     })
     //861

8. Reduce Exercise

// Sum up the instances of each of these

const data = ['car', 'car', 'truck', 'truck', 'bike', 'walk', 'car', 'van', 'bike', 'walk', 'car', 'van', 'car', 'truck' ];

😓 문제를 잘못 이해했다...ㅠ

하하.. 이것도 문제를 잘못 이해했다. 중복값을 없애 간결하게 만드는 게 아니었는데, 일단 내 풀이는 아래와 같다.

Reduce로 배열의 중복값 없애기

    const answer8 = data.reduce((sum, curr) => {
      return sum.includes(curr) ? sum : [...sum,curr];
    }, [])
    
    //['car', 'truck', 'bike', 'walk', 'van']

• 누산기 sum에 curr 현재 값이 포함되어 있는지 확인한다.
• 있다면 curr 현재값을 추가하지 않고 누산기 sum값을 그대로 리턴하고,
• 없다면 curr 현재값을 추가해준다. 이 때 sum은 배열이므로 전개연산자를 사용해 펼쳐서 추가했다.
• 배열을 다루기 때문에 누산기 sum의 초기값은 빈 배열로 설정했다.

🤩 다시 풀기

    const transportation = data.reduce(function(obj, item) {
      if(!obj[item]) {
        obj[item] = 0;
      }
      obj[item]++;
      return obj;
    }, {});

/*
bike: 2
car: 5
truck:3
van: 2
walk: 2
*/

Reference

• https://developer.mozilla.org/ko/docs/Web/JavaScript/Reference/Global_Objects/Array/sort
• https://developer.mozilla.org/ko/docs/Web/JavaScript/Reference/Global_Objects/String/localeCompare

  reduce - 배열값 중복 제거
  How to Remove Array Duplicates in ES6
  Dongmin-Jang Array javascript