Using one array filter instead of multiple ones

Question

Is there a way to merge all these filters into one? Or another way to make it more effective? Or just use for loop?

 driving += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'driving').length;
    breathWork += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'breath work').length;
    meditation += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'meditation').length;
    cooking += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'cooking').length;
    walking += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'walking').length;
    other += value.filter((obj) => obj.type === CalendarEventType.MIND && obj.data.practice === 'other').length;

They look to me very redudant.

Why do you have 6 different variables? Isn't that redundant in the first place? — Salman Arshad
– Salman Arshad, Commented Nov 10, 2021 at 18:09
I need to know the count of each "obj.data.practice" separatly. — user2994290
– user2994290, Commented Nov 10, 2021 at 18:12
Okay. It would have been better if you posted example of obj itself. The variables wouldn't be required at all — Salman Arshad
– Salman Arshad, Commented Nov 10, 2021 at 18:15

score 3 · Accepted Answer · 2021-11-10 18:18:45Z

3

Currently you're looping 6 times over the same array. You can reduce the runtime with a simple for-loop or reduce:

const counts = value.reduce((acc, el) => {
  if (el.type !== CalendarEventType.MIND) return acc;
  acc[el.data.practice] = (acc[el.data.practice] ?? 0) + 1
  return acc;
}, {});

Example:

const CalendarEventType = {
  MIND: 1,
  OTHER: 2
};

const value = [
  {type: CalendarEventType.MIND, data: { practice: 'driving' } },
  {type: CalendarEventType.MIND, data: { practice: 'breath work' } },
  {type: CalendarEventType.OTHER, data: { practice: 'breath work' } },
  {type: CalendarEventType.MIND, data: { practice: 'other' } },
  {type: CalendarEventType.MIND, data: { practice: 'other' } },
  {type: CalendarEventType.OTHER, data: { practice: 'other' } },
];

const counts = value.reduce((acc, el) => {
  if (el.type !== CalendarEventType.MIND) return acc;
  acc[el.data.practice] = (acc[el.data.practice] ?? 0) + 1
  return acc;
}, {});

console.log(counts);

edited Nov 10, 2021 at 18:18

answered Nov 10, 2021 at 18:13

user16540390

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3 Comments

Donut Over a year ago

This is really good.

user16540390 Over a year ago

@Donut Thank you.

user2994290 Over a year ago

This is perfect! Thanks you

Donut · Accepted Answer · 2021-11-10 18:13:40Z

1

One possible way would be to group by data.practice. Since there's no native "group by" function, you could implement it yourself (as described here), or you could use a library like Lodash or Ramda.

Example, using Lodash's groupBy:

const filteredItems = value.filter((obj) => obj.type === CalendarEventType.MIND);
const groupedItems = groupBy(filteredItems, obj => obj.data.practice);

Then you can do whatever you want with groupedItems, e.g.:

driving += groupedItems['driving'] ? groupedItems['driving'].length : 0;

Note that if a given value for data.practice is not represented in filteredItems, then that key won't be present in groupedItems - hence the need to check for that key.

answered Nov 10, 2021 at 18:13

Donut

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Comments

Nina Scholz · Accepted Answer · 2021-11-10 18:15:55Z

1

You could take an array of values and use a single filter.

const
    practices = ['driving', 'breath work', 'meditation', 'cooking', 'walking', 'other'];
    filter = ({ type, data: { practice } }) => type === CalendarEventType.MIND && practices.includes(practice),
    result = value.filter(filter),
    count = values.reduce((sum, o) => sum + filter(o), 0);

edited Nov 10, 2021 at 18:15

answered Nov 10, 2021 at 18:09

Nina Scholz

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