Join array based on another array

Question

I am applying between condition in filtering of array based on another array. The example below works if array length is 1. If I have more than one, let's say var expText = ['1-3', '6-9']. In that case I want (Value >= lowerValue1 && Value < upperValue1) || (Value >= lowerValue2 && Value < upperValue2)

var Annual = [{"Date":1998,"Value":6.5},{"Date":1999,"Value":4},{"Date":2000,"Value":1},{"Date":2001,"Value":3}]
var expText = ['1-3']

expText2 = expText[0].match(/\d+/g);
lowerValue = parseInt(expText2[0]);
upperValue = parseInt(expText2[1]);

result = Annual.filter(function(v) { return (v.Value >= lowerValue && v.Value < upperValue) })

console.log(result);

samanime · Accepted Answer · 2022-03-28 13:52:35Z

2

In your filter(), just loop over your expText with some() and compare it to each one (meaning move your splitting bit into the loop as well). some() will return true if at least one of them match.

Or, to be technical, it will continue until one of them matches, at which point it'll return true it runs out of them to check and returns false.

const Annual = [{"Date":1998,"Value":6.5},{"Date":1999,"Value":4},{"Date":2000,"Value":1},{"Date":2001,"Value":3}];
const expTexts = ['1-3', '6-9'];

const result = Annual.filter(function(v) { 
  return expTexts.some(expText => {
    const expText2 = expText.match(/\d+/g);
    const lowerValue = parseInt(expText2[0]);
    const upperValue = parseInt(expText2[1]);

    return (v.Value >= lowerValue && v.Value < upperValue);
  });
});

console.log(result);

Side note: Obligatory "never, ever, ever use var, always use const or let".

edited Mar 28, 2022 at 13:52

answered Mar 28, 2022 at 13:48

samanime

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5 Comments

pilchard Over a year ago

You can seriously simplify the upper/lower parsing using split() and destructuring const [lowerValue, upperValue] = expText.split('-');

samanime Over a year ago

Completely agree. I just didn't because it wasn't part of the original question, but I'd definitely do the same in my code. Or even just destructing on the return of match() even.

pilchard Over a year ago

That's fair enough

Ujjawal Bhandari Over a year ago

@samanime Thanks a ton. I am new to JS. can you a bit explain "never, ever, ever use var, always use const or let"?

samanime Over a year ago

TL;DR, var gets hoisted which causes it to end up in a weird scope that causes odd and hard-to-debug bugs. let and const act how most people would expect (don't get hoisted) and are less error-prone. Use const for values that don't change, let for values that do. Favor const and only switch to let when needed. This article seems to be a good longer explanation: medium.com/@codingsam/…

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