CODEWITHSUNDEEP
javascriptarraysmap-function
webdesignnoob
webdesignnoob

Reputation: 391

map function to return 1 boolean value, instead of an array of boolean values

Say you have an array like this:

arrayExample = [1, 2, 3, 4, 5, 6, 7]

I want to use the map function to iterate through arrayExample and return true if all the numbers are less than 8, false if they are not. However, when I do this I get an array of trues (like: [true, true, true... etc])

Would I be able to return just 1 value?

Here is my code so far:

var testBoolean = true;
var array = [1,2,3,4,5,6,7];

testBoolean = array.map(m => { 
    //if a number in the array is >=8, changes the testBoolean to false and return false only
    if(m >= 8) 
    { 
        return false;
    }
    //VS code said I had to have a return here, I believe its because I need to have in case m < 8 (edited after reading a comment)
    return true;
 })
 
 //prints an array [true,true,true,true,true,true,true]
document.write(testBoolean);

I'm a bit new to "map" but I believe it does this since it returns a value for every element, just confused on how to make it so it returns 1 true or false.

Upvotes: 3

Views: 27884

Answers (3)

Nick Parsons
Nick Parsons

Reputation: 50684

For something like this .map() isn't the right tool. .map() is for converting all the elements in your array to new elements (ie: mapping each element to a new transformed value). As a result, .map() will always return an array (unless this behaviour is intentionally modified). Instead, you can use .every() for this, which will tell you if all elements in your array match your condition (ie: if your function returns true for every element, then you'll get true as your result, otherwise you'll get false). The .every() method will terminate early as soon as it finds one element which your callback function returns false for, which can help with efficiency:

const array = [1, 2, 3, 4, 5, 6, 7];
const testBoolean = array.every(m => {
  if (m >= 8) {
    return false;
  }
  return true;
});
console.log(testBoolean);

This can be written more concisely, by just returning the result of m < 8 (this will either evaluate to true or false)

const array = [1,2,3,4,5,6,7];
const testBoolean = array.every(m => m < 8);
console.log(testBoolean);

Upvotes: 6

Screensavers
Screensavers

Reputation: 36

From a more general perspective, if you want to return a single value from an Array, .reduce() is your friend.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce

Specifically in your case, you could do

array.reduce(function(accumulator, currentVal, i, arr) {
  return accumulator && currentVal < 8;
}, true));

So .reduce() iterates through your array with an initial value true and returns the "previous" value (accumulator) AND whether the current value is less than 8.

You could think of it as

(true && (a[0] < 8 && (a[1] < 8 ... )))

The returned value at each iteration becomes the accumulator of the next. In this way you could not only do math operations but also change an array of a certain shape (e.g. array of {w:0, h:0} objects) to an output of another (e.g. a single number being the sum of all the hypotenuses calculated from each w and h).

Upvotes: 1

ecoplaneteer
ecoplaneteer

Reputation: 1984

The simplest solution is to use .some().

We don't need to check every value. We need to find the first value not less than 8.

const array = [1, 2, 3, 4, 5, 6, 7]
const testBoolean = !array.some(m => m >= 8)
console.log(testBoolean)

Upvotes: 8

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