如何将含重复元素的字符串数组按流行度排序并生成计数对象数组？

Hey there! Let's walk through how to solve this problem—turning your array of fruits into a sorted list of objects with counts, ordered by popularity (most frequent first). Here's a step-by-step breakdown:

Step 1: Count Occurrences #

First, we need to tally up how many times each fruit appears. The reduce method is ideal for this, as it lets us build a count object as we iterate through the array:

const arr = ['banana', 'orange', 'banana', 'apple', 'apple', 'apple', 'orange', 'apple', 'banana'];

// Build an object where keys are fruits and values are their counts
const countObject = arr.reduce((accumulator, fruit) => {
  // If the fruit is already in the accumulator, increment its count; otherwise set to 1
  accumulator[fruit] = (accumulator[fruit] || 0) + 1;
  return accumulator;
}, {});

This will give us:

{ banana: 3, orange: 2, apple: 4 }

Step 2: Sort by Popularity #

Next, we convert the count object into an array of key-value pairs using Object.entries(), then sort this array in descending order based on the count:

// Convert to [fruit, count] pairs and sort by count (highest first)
const sortedEntries = Object.entries(countObject).sort((a, b) => b[1] - a[1]);

The result here is:

[ ['apple', 4], ['banana', 3], ['orange', 2] ]

Step 3: Convert to Target Object Format #

Finally, we use map() to transform each entry into the object structure you need, with fruit and count properties:

// Map each entry to the desired object shape
const newArr = sortedEntries.map(([fruit, count]) => ({ fruit, count }));

Full Working Code #

You can chain all these steps together for a concise solution:

const arr = ['banana', 'orange', 'banana', 'apple', 'apple', 'apple', 'orange', 'apple', 'banana'];

const newArr = arr
  .reduce((acc, fruit) => {
    acc[fruit] = (acc[fruit] || 0) + 1;
    return acc;
  }, {})
  .entries()
  .sort((a, b) => b[1] - a[1])
  .map(([fruit, count]) => ({ fruit, count }));

console.log(newArr);
// Output: [ { fruit: 'apple', count: 4 }, { fruit: 'banana', count: 3 }, { fruit: 'orange', count: 2 } ]

Bonus: Handling Ties #

If two fruits have the same count, their order will depend on their first occurrence in the original array (modern JavaScript engines use stable sorts). If you want to add a tiebreaker—like sorting alphabetically by fruit name when counts are equal—adjust the comparator function:

.sort((a, b) => b[1] - a[1] || a[0].localeCompare(b[0]))

This will sort fruits with the same count from A-Z.

内容的提问来源于stack exchange，提问作者Kastlej