using a second array count unique values in array

Question

There are many posts on counting unique values in arrays, but I want to count the unique values in an array based on a second array. Specifically: I want to count the distinct occurrences within arr, but of all the values within grade

arr = [
  "12D","12D","12D",
  "12C","12C","12C","12C","12C",
  "12B","12B","12B","12B","12B",
  "12B","12B",
  "12A","12A","12A","12A","12A",
  "12A","12A","12A","12A","12A",
  "12A","12A"
]

grade = ["13B", "13A", "12D", "12C", "12B", "12A"]

result = arr.filter(item => grade.includes(item))
  .reduce((acc, val) => {
    acc[val] = acc[val] === undefined ? 1 : acc[val] += 1;
    return acc;
  }, {});

Current Result

My current result does count up the unique occurances

{ '12D': 3, '12C': 5, '12B': 7, '12A': 12 }

Desired Output

but I want to include ALL the values within grade inside my final object (and in the order of grade]

desired_result = {
    '13B': 0,
    '13A': 0,
    '12D': 3,
    '12C': 5,
    '12B': 7,
    '12A': 12
}

Is it possible to perform this reduce using a second vector?

I think some context is missing but if thats the whole thing, I would go at it from another direction, I would first convert the "grade" array to an object with all 0 as a value and key would be the grade from array. Then run on "arr" and just add the counter where it fits. — Shai
– Shai, Commented Nov 21, 2021 at 16:20

PeterKA · Accepted Answer · 2021-11-21 16:30:35Z

1

Another approach would be to start off with the grades and build a frequency table as follows. Then you can invoke Object.fromEntries() on the resulting array of array pairs:

const result = Object.fromEntries(
    grade.map(grade => [grade, arr.filter(g => g === grade).length])
);

DEMO

const arr = ["12D","12D","12D","12C","12C","12C","12C","12C","12B","12B","12B","12B","12B","12B","12B","12A","12A","12A","12A","12A","12A","12A","12A","12A","12A","12A","12A"];

const grade = ["13B", "13A", "12D", "12C", "12B", "12A"];

const result = Object.fromEntries(
    grade.map(grade => [grade, arr.filter(g => g === grade).length])
);

console.log( result );

answered Nov 21, 2021 at 16:30

PeterKA

24.7k5 gold badges28 silver badges52 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Nina Scholz · Accepted Answer · 2021-11-21 16:34:13Z

1

You could map grades as entries and take the object as start value of reduce.

const
    array = ["12D", "12D", "12D", "12C", "12C", "12C", "12C", "12C", "12B", "12B", "12B", "12B", "12B", "12B", "12B", "12A", "12A", "12A", "12A", "12A", "12A", "12A", "12A", "12A", "12A", "12A", "12A"],
    grade = ["13B", "13A", "12D", "12C", "12B", "12A"],
    result = array.reduce(
        (acc, val) => {
            if (val in acc) acc[val]++;
            return acc;
        },
        Object.fromEntries(grade.map(g => [g, 0]))
    );

console.log(result);

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

edited Nov 21, 2021 at 16:34

answered Nov 21, 2021 at 16:17

Nina Scholz

388k26 gold badges367 silver badges417 bronze badges

2 Comments

charlietfl Over a year ago

Not sure if it's critical to OP but doesn't filter out values that aren't in grade. Example doesn't have any but real cases may do

Nina Scholz Over a year ago

@charlietfl, now it counts only known props.

Collectives™ on Stack Overflow

using a second array count unique values in array

Current Result

Desired Output

2 Answers 2

Comments

2 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

Current Result

Desired Output

2 Answers 2

Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Related