使用 lodash（或 vanilla javascript）对对象数组进行排序和过滤

Question

I have this object I want to sort and filter by retaining only the 2 highest values by object.

obj={ A :[{
      asset: 9,
      biodiversity: 4,
      infrastructure: 15,
      deflation: 11,
      energy: 9
    }],
  B:[{
      asset: 12,
      biodiversity: 10,
      infrastructure: 9,
      deflation: 7,
      energy: 15
    }],
  C:[{
      asset: 2,
      biodiversity: 12,
      infrastructure: 6,
      deflation: 6,
      energy: 8
    }]}

I would like to sort the objects by their values and filter out the 2 highest eg:

{A :[{
      infrastructure: 15,
      deflation: 11
    }],
  B:[{
      energy: 15,
      asset: 12
    }],
  C:[{
      biodiversity: 12,
      energy: 8
    }]}

I have tried this for sorting:

Object.keys(obj).forEach((a) => _.sortBy(obj[a][0])))

But that is wrong obviously. I am using lodash but will accept vanilla javascript solution as well.

Answer 1

You could get the entries of the inner objects and sort by value descending, get the top two key/value pairs and build a new object from it.

 const data = { A: [{ asset: 9, biodiversity: 4, infrastructure: 15, deflation: 11, energy: 9 }], B: [{ asset: 12, biodiversity: 10, infrastructure: 9, deflation: 7, nergy: 15 }], C: [{ asset: 2, biodiversity: 12, infrastructure: 6, deflation: 6, energy: 8 }]}, result = Object.fromEntries(Object .entries(data) .map(([k, a]) => [k, a.map(o => Object.fromEntries(Object .entries(o) .sort((a, b) => b[1] - a[1]) .slice(0, 2) ))]) ); console.log(result);

 .as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

(Edit/Note: This is based on the code you originally posted. I'm glad to see you updated your question and got rid of the wrapping array.)

Here's a relatively functional approach.

The secondLargestValue function finds the threshold value within each object. The copyWithoutValsBelowThreshold function gives us a modified copy of the object.

We loop though the entries in the top-level object and apply these two functions. See comments in the code for further clarification.

 let json = getArray(); // Identifies the original array // Defines `secondLargestVal` function const secondLargestVal = someObj => // Gets second item of array after sorting numerically in descending order Object.values(someObj).sort((a, b) => b - a)[1]; // Defines `copyWithoutValsBelowThreshold` function const copyWithoutValsBelowThreshold = ( (someObj, threshold) => { // This function doesn't mutate the original value clone = Object.assign({}, someObj); // Copies obj for(let prop in clone){ // Loops through properties, deleting non-qualifying ones if(clone[prop] < threshold){ delete clone[prop]; } } return clone; }); // Defines our main function const mutateJson = () => { let entries = Object.entries(json[0]); entries = entries.map(entry => { // `entry[0]` is the property name (eg: 'A') -- we don't actually use this // `entry[1]` is the value (in this case, an array containing a single object) let obj = entry[1][0]; // Identifies the actual object const threshold = secondLargestVal(obj); // Identifies minimum value // Updates the entry, deleting properties whose values are too low entry[1][0] = copyWithoutValsBelowThreshold(obj, threshold); return entry; }); json[0] = Object.fromEntries(entries); // Replaces the top-level object } // Calls the main function mutateJson(); console.log(json); // Provides the original array function getArray(){ return [{ A :[{ asset: 9, biodiversity: 4, infrastructure: 15, deflation: 11, energy: 9 }], B:[{ asset: 12, biodiversity: 10, infrastructure: 9, deflation: 7, energy: 15 }], C:[{ asset: 2, biodiversity: 12, infrastructure: 6, deflation: 6, energy: 8 }]}] }