问题
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 }]
回答1:
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);
回答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
回答3:
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);
回答4:
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)
来源:https://stackoverflow.com/questions/53097817/javascript-objects-array-filter-by-minimum-value-of-an-attribute