update a property in object array

Question

Is there a way to update a property in object array based on the number of times some other property is present as element in some other array

I have 2 arrays, array1 and array2:

var array1 = ["JOHN", "JACK", "JACK"]; var array2 = [ {count: 9, value: "JACK"}, {count: 9, value: "JOHN"}, {count: 2, value: "TEST"} ];

Expected output :

[ {count: 7, value: "JACK"}, // count = 9 - 2 {count: 8, value: "JOHN"}, // count = 9 - 1 {count: 2, value: "TEST"} ]

In array1, "JACK" is present twice, so I need to reduce count by 2, similarly "JOHN" is present once and hence its reduced by 1, "TEST" is not present so unchanged.

I tried the following

array1.map(item => { return array2.find( p => p["value"] === item); });

With this, I am getting the below output,

[ {count: 9, value: "JOHN"}, {count: 9, value: "JACK"}, {count: 9, value: "JACK"} ]

I am not sure whether it can be achieved using single lambda expression.

Thanks in advance!

palaѕн · Accepted Answer · 2020-04-07 11:14:33Z

You can get the result using array .map() and .filter() methods, assuming you are not allowed to change the original array:

var array1 = ["JOHN", "JACK", "JACK"]; var array2 = [{count: 9, value: "JACK"}, {count: 9, value: "JOHN"}, {count: 2, value: "TEST"}] var result = array2.map(({value, count}) => ({value, count: count - array1.filter(a=>a===value).length})) console.log(result)

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

Using .filter() check how many time a value is present in array1
Then subtract that result from current array2 count.
Then just return a new array of objects with the updated count.

Hi Palash, I am a newbie here. Just wanted to confirm whether I have sucessfully marked this as answer ?
No problem, but no it is still not showing as the accepted answer. Maybe this link will help.

Code Maniac · Accepted Answer · 2020-04-07 11:05:59Z

You can use reduce and map

Loop over array1 and create a mapper object with names as key and repetition of names as value
Loop over array2 and see if the name is present in mapper subtract respective value from

const array1 = ["JOHN", "JACK", "JACK"]; const array2 = [{count: 9, value: "JACK"},{count: 9, value: "JOHN"},{count: 2, value: "TEST"}] const mapper = array1.reduce((op, inp) => { op[inp] = op[inp] || 0; op[inp]++; return op; }, Object.create(null)) let final = array2.map(({count,value}) =>({ value, count: count - (mapper[value] || 0) })) console.log(final)

mplungjan · Accepted Answer · 2020-04-07 11:26:43Z

Assuming you are allowed to change the original array. ALSO assuming we are not talking 10s of thousand entries since I look up the value in the name array each time:

var array1 = ["JOHN", "JACK", "JACK"]; var array2 = [{count: 9, value: "JACK"}, {count: 9, value: "JOHN"}, {count: 2, value: "TEST"}] array2.forEach(item => item.count -= array1.filter(val => val === item.value).length); console.log(array2);

Less resources:

var array1 = ["JOHN", "JACK", "JACK"]; var array2 = [{count: 9, value: "JACK"}, {count: 9, value: "JOHN"}, {count: 2, value: "TEST"}] // create lookup table const names = array1.reduce((arr,cur) => { arr[cur] = (arr[cur]||0) + 1; return arr;},{}) // subtract if present array2.forEach(item => item.count -= (names[item.value] || 0)); console.log(array2);

Yeah, I will not have many entries, at the max 100 in each array

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