JavaScript objects array filter by minimum value of an attribute

Question

I need to filter this object array by minimum value of 'rest' attribute. This is an one way to do it. Is there any other ways?

'data' variable is a result of chained function. Is there any other way to do this without calling 'data' variable again inside Math.min() function.

 let data = [ { size: 5, qty: 2, rest: 0 }, { size: 2, qty: 5, rest: 0 }, { size: 1, qty: 10, rest: 0 }, { size: 3, qty: 3, rest: 1 }, { size: 4, qty: 2, rest: 2 } ] let result = data.filter(e=> e.rest === Math.min(...data.map(f=>f.rest) ) ); console.log(result); // result is //[ { size: 5, qty: 2, rest: 0 }, // { size: 2, qty: 5, rest: 0 }, // { size: 1, qty: 10, rest: 0 }]

Answer 1

data.map inside of data.filter is O(N^2) ; for an O(N) solution, iterate through data ahead of time to calculate the minimum, then filter by that minimum:

 let data = [ { size: 5, qty: 2, rest: 0 }, { size: 2, qty: 5, rest: 0 }, { size: 1, qty: 10, rest: 0 }, { size: 3, qty: 3, rest: 1 }, { size: 4, qty: 2, rest: 2 } ]; const minRest = Math.min(...data.map(({ rest }) => rest)); let result = data.filter(({ rest }) => rest === minRest); console.log(result); 

Answer 2

The easiest way is to pull the min function out of the filter like this:

let min = Math.min(...data.map(item => item.rest))

This is much more efficient as we are no longer loop over the data to find the min for every iteration of the filter.

We now have n * 2 passes instead of n^2 passes. (n is the size of your data set, 5 in this case)

Full example below:

   let data = [ 
     { size: 5, qty: 2, rest: 0 },
     { size: 2, qty: 5, rest: 0 },
     { size: 1, qty: 10, rest: 0 },
     { size: 3, qty: 3, rest: 1 },
     { size: 4, qty: 2, rest: 2 } 
   ]

  let min = Math.min(...data.map(item => item.rest))
  let result = data.filter(item => item.rest === min)
  console.log(result)

Hope this helps!

Lloyd

Answer 3

It sounds like you want to sort the list. I would do it as following:

const result = data.sort((a, b) => a.rest - b.rest)

Answer 4

imo. the simplest/best solution is the one @CertainPerformance gave you.

Just wanted to add another solution with linear runtime (that truly iterates only once over the Array)

 let data = [ { size: 5, qty: 2, rest: 0 }, { size: 2, qty: 5, rest: 0 }, { size: 1, qty: 10, rest: 0 }, { size: 3, qty: 3, rest: 1 }, { size: 4, qty: 2, rest: 2 } ]; let result = data.reduce((result, item) => { let minRest = result.length? result[0].rest: item.rest; if (item.rest < minRest) { minRest = item.rest; result.length = 0; } if (item.rest === minRest) { result.push(item); } return result; }, []); console.log(result); 

@mathieux51 got me another idea how you can do this inside a method chain, but the readability/clarity/intention is not as good as with the other approaches:

 let data = [ { size: 5, qty: 2, rest: 0 }, { size: 2, qty: 5, rest: 0 }, { size: 1, qty: 10, rest: 0 }, { size: 3, qty: 3, rest: 1 }, { size: 4, qty: 2, rest: 2 } ]; let result = data.sort((a, b) => a.rest - b.rest) .filter((item, index, array) => item.rest === array[0].rest); console.log(result); 

Answer 5

Get Min or Max Since no one mentioned this method I will update it here.

myArray.sort(function (a, b) {
return a.rest - b.rest
})

var min = myArray[0],
max = myArray[myArray.length - 1]

It has good readability/clarity/intentions.