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JavaScript JavaScript Array Iteration Methods Combining Array Methods Combining filter() and reduce()

Ivan Molina
Ivan Molina
2,870 Points
Posted by Ivan Molina
Ivan Molina
2,870 Points

whats the purpose of the Spoiled Alert's answer?

...The instructor's Spoiled Alert code returns $4.49,..shoudn't return $6.99?

// Instructors Code
const products = [
  { name: 'hard drive', price: 59.99 },
  { name: 'lighbulbs', price: 2.59 },
  { name: 'paper towels', price: 6.99 },
  { name: 'flatscreen monitor', price: 159.99 },
  { name: 'cable ties', price: 19.99 },
  { name: 'ballpoint pens', price: 4.49 }
];

const product = products
  .filter(product => product.price < 10)
  .reduce((highest, product) => {
    if (highest.price > product.price) {
      return highest.price; //...---> return highest
    }
    return product.price;  // --->return product
  }, {price: 0});

  console.log(product);

.....my code to return just the Price that is the highest and less than 10:

const product = products
  .filter(product => product.price < 10)
  .reduce((highest, product) => {
    if (highest.price > product.price) {
      return highest.price; 
    }
    return product;  
  }, {price: 0});

1 Answer

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

I believe you've discovered a bug in the code.

I wouldn't expect this to work, since the callback function passed to "reduce" returns a value, but the first parameter (and the dummy argument) are both product objects and not values. But if we make a few changes to account for this, it does work:

const product = products
  .filter(product => product.price < 10)
  .reduce((highestprice, product) => {
    if (highestprice > product.price) {
      return highestprice;
    }
    return product.price;
  }, 0);