根据时间戳条件过滤数组中的唯一对象

Question

I have the following array:

let arr = [
    {"id": 123, "lastUpdate": 1543229793},
    {"id": 456, "lastUpdate": 1545269320},
    {"id": 123, "lastUpdate": 1552184795}
]

I need to filter the array based on the same ID, but also check the "lastUpdate" timestamp and keep only the newer entries. The result should be:

[
    {"id": 456, "lastUpdate": 1545269320},
    {"id": 123, "lastUpdate": 1552184795}
]

I have tried the following:

arr = arr.filter((e, index, self) =>
    index === self.findIndex((t) => (
        t.id === intent.id && t.lastUpdate > e.lastUpdate
    ))
)

However, this filters everything for me and the resulting array is empty. I think something is wrong with the last part of above && t.lastUpdate > e.lastUpdate .

Many thanks for any tips!

Answer 1

Hi there if you are looking for a performant solution you can use an object :)

let arr = [{"id": 123,"lastUpdate": 1543229793},
{"id": 456,"lastUpdate": 1545269320},
{"id": 123, "lastUpdate": 1552184795}];

let newArr = {}
arr.forEach(el => {
  if(!newArr[el.id] || newArr[el.id].lastUpdate < el.lastUpdate){
      newArr[el.id] = el
  }
})

console.log(Object.values(newArr));

Answer 2

You can achieve it by looking for items that don't have an item2 where the update was later

   arr.filter(item => 
                 { return !arr.some(item2 => 
                  item.id === item2.id && item.lastUpdate < item2.lastUpdate)
            });

What that code does is :

For each item in the array it look if in the array there is an item with the same id where the lastUpdate is superior to its own. If there is one it return true (Array.some returns a boolean). We negate that value and use it to filter.

Answer 3

You could do it step by step by converting to a set, sorting then getting the first item for each id:

 let arr = [ {"id": 123, "lastUpdate": 1543229793}, {"id": 456, "lastUpdate": 1545269320}, {"id": 123, "lastUpdate": 1552184795} ] // Get the ids by making a set of ids and then converting to array let ids = [ ...new Set(arr.map((e) => e.id)) ]; // Sort the original by lastUpdate descending arr.sort((a, b) => b.lastUpdate - a.lastUpdate); // Get array of first item from arr by id let res = ids.map(id => arr.find((e) => e.id == id)); console.log(res); 

Answer 4

Silvio's approach packed into reusable code, reduces the OP's problem to eg something like that ...

 function collectUniqueItemByIdWithMostRecentUpdate (collector, item, idx, arr) { const store = collector.store; const storedItem = store[item.id]; if (!storedItem || (storedItem.lastUpdate < item.lastUpdate)) { store[item.id] = item; } if (idx >= (arr.length - 1)) { collector.list = Object.values(store); } return collector; } let arr = [ {"id": 123, "lastUpdate": 1543229793}, {"id": 456, "lastUpdate": 1555269320}, {"id": 123, "lastUpdate": 1552184795}, {"id": 456, "lastUpdate": 1545269320}, {"id": 123, "lastUpdate": 1553229793} ]; console.log(arr.reduce(collectUniqueItemByIdWithMostRecentUpdate, { store: {}, list: [] }).list); 

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