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1

Problem

I'm trying to create a function that evaluates an array and if every element inside the array is the same, it would return true and otherwise false. I don't want it to return true/false for each individual element, just for the entire array.

Attempt 1

This method works, but it returns true/false for each element in the array:

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 else 
 console.log(true);
 
 

Attempt 2

This method returns true/false, once and then prints true again at the end:

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 
 
 console.log(true);

javascript arrays

share|improve this question

edited Mar 23 at 22:04

Emma

3,94231127

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

add a comment | 

1

Problem

I'm trying to create a function that evaluates an array and if every element inside the array is the same, it would return true and otherwise false. I don't want it to return true/false for each individual element, just for the entire array.

Attempt 1

This method works, but it returns true/false for each element in the array:

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 else 
 console.log(true);
 
 

Attempt 2

This method returns true/false, once and then prints true again at the end:

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 
 
 console.log(true);

javascript arrays

share|improve this question

edited Mar 23 at 22:04

Emma

3,94231127

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

add a comment | 

Problem

I'm trying to create a function that evaluates an array and if every element inside the array is the same, it would return true and otherwise false. I don't want it to return true/false for each individual element, just for the entire array.

Attempt 1

This method works, but it returns true/false for each element in the array:

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 else 
 console.log(true);
 
 

Attempt 2

This method returns true/false, once and then prints true again at the end:

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 
 
 console.log(true);

javascript arrays

share|improve this question

edited Mar 23 at 22:04

Emma

3,94231127

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 else 
 console.log(true);
 
 

function isUniform(arr)
 let first = arr[0];
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 console.log(false);
 
 
 console.log(true);

share|improve this question

edited Mar 23 at 22:04

Emma

3,94231127

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

share|improve this question

edited Mar 23 at 22:04

Emma

3,94231127

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

edited Mar 23 at 22:04

Emma

3,94231127

edited Mar 23 at 22:04

Emma

3,94231127

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

asked Mar 23 at 15:17

Ruchi RoyRuchi Roy

62

5 Answers
5

active

oldest

votes

3

If you want to test if something is true for every element of an array, you don't really need to write much — you can use array.every for this and just compare the first element. every() is nice because it will return early if a false condition is found.

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

This will return true for an empty array, which may or may not be what you want.

share|improve this answer

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is the best answer
  
  – Kunal Mukherjee
  Mar 23 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KunalMukherjee, this is a different approach, but not an answer to the question.
  
  – Nina Scholz
  Mar 23 at 15:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NinaScholz the OP is asking for if all the elements in the array are the same or not
  
  – Kunal Mukherjee
  Mar 23 at 15:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thank you! I'll keep this in mind in the future - right now I'm restricted to just using a for loop or for each, but I'm glad I learned this.
  
  – Ruchi Roy
  Mar 23 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if only the question is to check, then is clearly a duplicate question and a matter of a close request.
  
  – Nina Scholz
  Mar 23 at 19:05
  

  
  
  
  

add a comment | 

2

Add a return statement with false and end the function. The return value could be used later.

function isUniform(arr) 
 let first = arr[0];
 for (let i = 1; i < arr.length; i++) 
 if (arr[0] !== arr[i]) 
 console.log(false);
 return false;
 
 
 console.log(true);
 return true;

For using a return value, you need to return true at the end, too.

share|improve this answer

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  THANK YOU SO MUCH. Ugh, it's always something stupidly obvious and simple like this.
  
  – Ruchi Roy
  Mar 23 at 15:49
  

  
  
  
  

add a comment | 

2

Alternative using the object Set

new Set(arr).size === 1 // This means all the elements are equal.

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

edited Mar 23 at 15:25

answered Mar 23 at 15:20

EleEle

26.2k52354

add a comment | 

1

Try with Array#every .its Checking all other value is same with first index of array

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

answered Mar 23 at 15:28

prasanthprasanth

15k21538

add a comment | 

0

The problem is that you need to stop once you've found the first false element:

function isUniform(arr)
 let first = arr[0]; 
 let uniform = true;
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 uniform = false; 
 break;
 
 
 console.log(uniform);

share|improve this answer

edited Mar 23 at 15:33

answered Mar 23 at 15:21

max carin balicmax carin balic

112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  break does not prevent to show true. it just exits the loop.
  
  – Nina Scholz
  Mar 23 at 15:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes, my bad, fixed it
  
  – max carin balic
  Mar 23 at 15:30
  

  
  
  
  

add a comment | 

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3

If you want to test if something is true for every element of an array, you don't really need to write much — you can use array.every for this and just compare the first element. every() is nice because it will return early if a false condition is found.

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

This will return true for an empty array, which may or may not be what you want.

share|improve this answer

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is the best answer
  
  – Kunal Mukherjee
  Mar 23 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KunalMukherjee, this is a different approach, but not an answer to the question.
  
  – Nina Scholz
  Mar 23 at 15:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NinaScholz the OP is asking for if all the elements in the array are the same or not
  
  – Kunal Mukherjee
  Mar 23 at 15:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thank you! I'll keep this in mind in the future - right now I'm restricted to just using a for loop or for each, but I'm glad I learned this.
  
  – Ruchi Roy
  Mar 23 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if only the question is to check, then is clearly a duplicate question and a matter of a close request.
  
  – Nina Scholz
  Mar 23 at 19:05
  

  
  
  
  

add a comment | 

3

If you want to test if something is true for every element of an array, you don't really need to write much — you can use array.every for this and just compare the first element. every() is nice because it will return early if a false condition is found.

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

This will return true for an empty array, which may or may not be what you want.

share|improve this answer

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This is the best answer
  
  – Kunal Mukherjee
  Mar 23 at 15:25
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @KunalMukherjee, this is a different approach, but not an answer to the question.
  
  – Nina Scholz
  Mar 23 at 15:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @NinaScholz the OP is asking for if all the elements in the array are the same or not
  
  – Kunal Mukherjee
  Mar 23 at 15:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thank you! I'll keep this in mind in the future - right now I'm restricted to just using a for loop or for each, but I'm glad I learned this.
  
  – Ruchi Roy
  Mar 23 at 15:47
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  if only the question is to check, then is clearly a duplicate question and a matter of a close request.
  
  – Nina Scholz
  Mar 23 at 19:05
  

  
  
  
  

add a comment | 

If you want to test if something is true for every element of an array, you don't really need to write much — you can use array.every for this and just compare the first element. every() is nice because it will return early if a false condition is found.

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

This will return true for an empty array, which may or may not be what you want.

share|improve this answer

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

var arr1 = [1, 1, 1, 1, 1, 1, 1]
var arr2 = [1, 1, 1, 2, 1, 1, 1]

console.log(arr1.every((n, _, self) => n === self[0]))
console.log(arr2.every((n, _, self) => n === self[0]))

share|improve this answer

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

answered Mar 23 at 15:20

Mark MeyerMark Meyer

43.5k33767

2

Add a return statement with false and end the function. The return value could be used later.

function isUniform(arr) 
 let first = arr[0];
 for (let i = 1; i < arr.length; i++) 
 if (arr[0] !== arr[i]) 
 console.log(false);
 return false;
 
 
 console.log(true);
 return true;

For using a return value, you need to return true at the end, too.

share|improve this answer

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  THANK YOU SO MUCH. Ugh, it's always something stupidly obvious and simple like this.
  
  – Ruchi Roy
  Mar 23 at 15:49
  

  
  
  
  

add a comment | 

2

Add a return statement with false and end the function. The return value could be used later.

function isUniform(arr) 
 let first = arr[0];
 for (let i = 1; i < arr.length; i++) 
 if (arr[0] !== arr[i]) 
 console.log(false);
 return false;
 
 
 console.log(true);
 return true;

For using a return value, you need to return true at the end, too.

share|improve this answer

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  THANK YOU SO MUCH. Ugh, it's always something stupidly obvious and simple like this.
  
  – Ruchi Roy
  Mar 23 at 15:49
  

  
  
  
  

add a comment | 

Add a return statement with false and end the function. The return value could be used later.

function isUniform(arr) 
 let first = arr[0];
 for (let i = 1; i < arr.length; i++) 
 if (arr[0] !== arr[i]) 
 console.log(false);
 return false;
 
 
 console.log(true);
 return true;

For using a return value, you need to return true at the end, too.

share|improve this answer

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

function isUniform(arr) 
 let first = arr[0];
 for (let i = 1; i < arr.length; i++) 
 if (arr[0] !== arr[i]) 
 console.log(false);
 return false;
 
 
 console.log(true);
 return true;

share|improve this answer

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

answered Mar 23 at 15:18

Nina ScholzNina Scholz

204k16117188

2

Alternative using the object Set

new Set(arr).size === 1 // This means all the elements are equal.

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

edited Mar 23 at 15:25

answered Mar 23 at 15:20

EleEle

26.2k52354

add a comment | 

2

Alternative using the object Set

new Set(arr).size === 1 // This means all the elements are equal.

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

edited Mar 23 at 15:25

answered Mar 23 at 15:20

EleEle

26.2k52354

add a comment | 

Alternative using the object Set

new Set(arr).size === 1 // This means all the elements are equal.

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

edited Mar 23 at 15:25

answered Mar 23 at 15:20

EleEle

26.2k52354

new Set(arr).size === 1 // This means all the elements are equal.

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

let isUniform = (arr) => new Set(arr).size === 1;

console.log(isUniform([4,4,4,4,4]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

edited Mar 23 at 15:25

answered Mar 23 at 15:20

EleEle

26.2k52354

answered Mar 23 at 15:20

EleEle

26.2k52354

answered Mar 23 at 15:20

EleEle

26.2k52354

1

Try with Array#every .its Checking all other value is same with first index of array

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

answered Mar 23 at 15:28

prasanthprasanth

15k21538

add a comment | 

1

Try with Array#every .its Checking all other value is same with first index of array

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

answered Mar 23 at 15:28

prasanthprasanth

15k21538

add a comment | 

Try with Array#every .its Checking all other value is same with first index of array

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

answered Mar 23 at 15:28

prasanthprasanth

15k21538

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

function isUniform(arr) 
 return arr.every(a=> a === arr[0])


console.log(isUniform([2,2,2,2]));
console.log(isUniform([4,4,4,4,4,5]));

share|improve this answer

answered Mar 23 at 15:28

prasanthprasanth

15k21538

answered Mar 23 at 15:28

prasanthprasanth

15k21538

answered Mar 23 at 15:28

prasanthprasanth

15k21538

0

The problem is that you need to stop once you've found the first false element:

function isUniform(arr)
 let first = arr[0]; 
 let uniform = true;
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 uniform = false; 
 break;
 
 
 console.log(uniform);

share|improve this answer

edited Mar 23 at 15:33

answered Mar 23 at 15:21

max carin balicmax carin balic

112

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  break does not prevent to show true. it just exits the loop.
  
  – Nina Scholz
  Mar 23 at 15:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes, my bad, fixed it
  
  – max carin balic
  Mar 23 at 15:30
  

  
  
  
  

add a comment | 

0

The problem is that you need to stop once you've found the first false element:

function isUniform(arr)
 let first = arr[0]; 
 let uniform = true;
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 uniform = false; 
 break;
 
 
 console.log(uniform);

share|improve this answer

edited Mar 23 at 15:33

answered Mar 23 at 15:21

max carin balicmax carin balic

112

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  break does not prevent to show true. it just exits the loop.
  
  – Nina Scholz
  Mar 23 at 15:22
  
  

  
  
  
  


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  yes, my bad, fixed it
  
  – max carin balic
  Mar 23 at 15:30
  

  
  
  
  

add a comment | 

The problem is that you need to stop once you've found the first false element:

function isUniform(arr)
 let first = arr[0]; 
 let uniform = true;
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 uniform = false; 
 break;
 
 
 console.log(uniform);

share|improve this answer

edited Mar 23 at 15:33

answered Mar 23 at 15:21

max carin balicmax carin balic

112

function isUniform(arr)
 let first = arr[0]; 
 let uniform = true;
 for (let i = 1; i <arr.length; i++)
 if (arr[0] !== arr[i])
 uniform = false; 
 break;
 
 
 console.log(uniform);

share|improve this answer

edited Mar 23 at 15:33

answered Mar 23 at 15:21

max carin balicmax carin balic

112

answered Mar 23 at 15:21

max carin balicmax carin balic

112

answered Mar 23 at 15:21

max carin balicmax carin balic

112