how to filter two fields from nested array in JavaScript

Question

Given edges, I would like exclude any edge which have both nodes points.

nodes = ['1', '2', '3', '4']

edges = [
    {id: 'a', src: '1', target: '2'}, <<exclude cuz '1', '2' are in nodes
    {id: 'b', src: '4', target: '3'}, <<exclude
    {id: 'c', src: '1', target: '5'}, <<DO NOT exclude cuz only '1' is in nodes
    {id: 'd', src: '6', target: '2'}, <<DO NOT
    {id: 'e', src: '6', target: '8'}, <<DO NOT
]

output = ['c', 'd', 'e']

Can you complete the below statement..?? Or do you have any efficient solution?

nodes = ['1', '2', '3', '4']

edges = [
    {id: 'a', src: '1', target: '2'}, //exclude cuz '1', '2' are in nodes
    {id: 'b', src: '4', target: '3'}, //exclude
    {id: 'c', src: '1', target: '5'}, //DO NOT exclude cuz only '1' is in nodes
    {id: 'd', src: '6', target: '2'}, //DO NOT
    {id: 'e', src: '6', target: '8'}, //DO NOT
]

console.log(edges.filter({src, target} => nodes.includes(src) && nodes.includes(target)).map({id}=>id));

If I've understood your intentions correctly, all you need to do is wrap the nodes.includes(src) && nodes.includes(target) in parentheses and use the ! operator, since this describes exactly the condition you want to exclude. — Robin Zigmond
– Robin Zigmond, Commented Apr 22, 2020 at 20:13
Fix the syntax errors in your arrow functions and after then do the opposite in your filter since you are picking out the elements that match. — epascarello
– epascarello, Commented Apr 22, 2020 at 20:14
When you use a destructuring argument list in an arrow function, you have to put () around it. — Barmar
– Barmar, Commented Apr 22, 2020 at 20:14
Sort the nodes array and binary search instead of using [].includes to achieve O(N log N) — Washington Guedes
– Washington Guedes, Commented Apr 22, 2020 at 20:18

Barmar · Accepted Answer · 2020-04-22 20:17:36Z

3

Just invert the test with ! to remove elements instead of keeping them when they match the criteria.

And I've fixed the syntax of your destructuring argument lists.

nodes = ['1', '2', '3', '4']

edges = [
    {id: 'a', src: '1', target: '2'}, //exclude cuz '1', '2' are in nodes
    {id: 'b', src: '4', target: '3'}, //exclude
    {id: 'c', src: '1', target: '5'}, //DO NOT exclude cuz only '1' is in nodes
    {id: 'd', src: '6', target: '2'}, //DO NOT
    {id: 'e', src: '6', target: '8'}, //DO NOT
]

console.log(edges.filter(({src, target}) => !(nodes.includes(src) && nodes.includes(target))).map(({id})=>id));

If nodes is large, you can improve efficiency significantly by changing it to a Set instead of an array.

answered Apr 22, 2020 at 20:17

Barmar

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1 Comment

merry-go-round Over a year ago

Thanks Barmar! Follow up problem: stackoverflow.com/questions/61400327/…

epascarello · Accepted Answer · 2020-04-22 20:17:09Z

1

When you destructure you need to have parenthesis around it. And your check is doing the opposite of you want in the filter. So invert the boolean.

nodes = ['1', '2', '3', '4']

edges = [
    {id: 'a', src: '1', target: '2'}, // <<exclude cuz '1', '2' are in nodes
    {id: 'b', src: '4', target: '3'}, // <<exclude
    {id: 'c', src: '1', target: '5'}, // <<DO NOT exclude cuz only '1' is in nodes
    {id: 'd', src: '6', target: '2'}, // <<DO NOT
    {id: 'e', src: '6', target: '8'}, // <<DO NOT
]


var result = edges.filter(({
  src,
  target
}) => !(nodes.includes(src) && nodes.includes(target))).map(({
  id
}) => id);

console.log(result)

answered Apr 22, 2020 at 20:17

epascarello

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Comments

Nina Scholz · Accepted Answer · 2020-04-22 20:26:57Z

1

This is a more efficient approach with a short circuit if a node is not found.

var nodes = ['1', '2', '3', '4'],
    nodesSet = new Set(nodes),
    edges = [{ id: 'a', src: '1', target: '2' }, { id: 'b', src: '4', target: '3' }, { id: 'c', src: '1', target: '5' }, { id: 'd', src: '6', target: '2' }, { id: 'e', src: '6', target: '8' }],
    result = edges
        .filter(({ src, target }) => !nodesSet.has(src) || !nodesSet.has(target))
        .map(({ id }) => id)

console.log(result);

answered Apr 22, 2020 at 20:26

Nina Scholz

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Comments

Pedro Lima · Accepted Answer · 2020-04-22 20:19:26Z

0

Your code is quite efficient and is nearly correct. Here is how it should be correct:

When making arrow functions, parenthesis are always required around the argument list except when your function has 1 simple argument. You are using destructuring arguments and they always require parenthesis.
Your function for the filter should return true when you want to maintain the element in the resulting array. As it is, you have the opposite.

So the following should work for you:

edges.filter(({src, target}) => !nodes.includes(src) || !nodes.includes(target)).map(({id})=>id)

answered Apr 22, 2020 at 20:19

Pedro Lima

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