Using array.reduce to count matches in an array

Question

I am trying to replicate the function below using Array.reduce

private getCount = (str, value) => {
var count = 0;
const everything = ['Key1', 'Key2', 'Key3', 'Key4', 'Key5'];
for (let i = 0; i < everything.length; i++) {
  if (everything === value) {
    count++;
  }
}

return count;}

This is my attempt. However, it gives the correct output only when the value is Key1. Could you please suggest what I could be going wrong?

private getCount = (str, value) => {
  const everything = ['Key1', 'Key2', 'Key3', 'Key4', 'Key5'];
  return everything.reduce((accumulator, currentValue, currentIndex, array) => {
    return (currentValue === value ? accumulator + 1 : accumulator)
  }, 0)
}

If the goal is to check for existence, Array.prototype.includes() is a much better fit. But are you wanting that or are you wanting a count? — Phil
– Phil, Commented Jun 9, 2020 at 3:23
@Phil, for me, it works only searching for the first key. Also noted your response about using .includes. However, I am trying to use the reduce function in this scenario, to understand it. — LearnToImprove
– LearnToImprove, Commented Jun 9, 2020 at 3:33
FYI, your for loop approach should be comparing everything[i] === value — Phil
– Phil, Commented Jun 9, 2020 at 3:38
Cannot reproduce this, it seems to work just fine ~ jsfiddle.net/pkfu09sd — Phil
– Phil, Commented Jun 9, 2020 at 3:42

Devon Page · Accepted Answer · 2020-06-09 03:48:36Z

0

I question why you so badly want to do this using reduce, but the below should get the job done.

everything.reduce((ac, current)=> ac + (current === value ? 1 : 0), 0 );

answered Jun 9, 2020 at 3:48

Devon Page

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