How can I return a different array?

I need to return an array consisting of the largest number from each provided sub-array. So, I've written the following code and it works but I have:
[[25],[48],[21],[-3]] instead of [25,48,21,-3]. Can I change something to get the right answer?

function largestOfFour(arr) {

    for (let i = 0; i < arr.length; i++) {

        arr[i].sort(function(a,b) {

            return b - a;

        }).map(function() {

            return arr[i].splice(1);

        })

    }

    return arr;

}



var result = largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);



console.log(result);

edited Jan 2 at 14:45

Armel

1,206920

asked Jan 2 at 13:08

Ilya Solodeev

One way is that you can flatten it!

– choz
Jan 2 at 13:10

add a comment |

function largestOfFour(arr) {

    for (let i = 0; i < arr.length; i++) {

        arr[i].sort(function(a,b) {

            return b - a;

        }).map(function() {

            return arr[i].splice(1);

        })

    }

    return arr;

}



var result = largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);



console.log(result);

edited Jan 2 at 14:45

Armel

1,206920

asked Jan 2 at 13:08

Ilya Solodeev

One way is that you can flatten it!

– choz
Jan 2 at 13:10

add a comment |

function largestOfFour(arr) {

    for (let i = 0; i < arr.length; i++) {

        arr[i].sort(function(a,b) {

            return b - a;

        }).map(function() {

            return arr[i].splice(1);

        })

    }

    return arr;

}



var result = largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);



console.log(result);

edited Jan 2 at 14:45

Armel

1,206920

asked Jan 2 at 13:08

Ilya Solodeev

function largestOfFour(arr) {

    for (let i = 0; i < arr.length; i++) {

        arr[i].sort(function(a,b) {

            return b - a;

        }).map(function() {

            return arr[i].splice(1);

        })

    }

    return arr;

}



var result = largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);



console.log(result);

function largestOfFour(arr) {

    for (let i = 0; i < arr.length; i++) {

        arr[i].sort(function(a,b) {

            return b - a;

        }).map(function() {

            return arr[i].splice(1);

        })

    }

    return arr;

}



var result = largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);



console.log(result);

function largestOfFour(arr) {

    for (let i = 0; i < arr.length; i++) {

        arr[i].sort(function(a,b) {

            return b - a;

        }).map(function() {

            return arr[i].splice(1);

        })

    }

    return arr;

}



var result = largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);



console.log(result);

javascript

edited Jan 2 at 14:45

Armel

1,206920

asked Jan 2 at 13:08

Ilya Solodeev

edited Jan 2 at 14:45

Armel

1,206920

asked Jan 2 at 13:08

Ilya Solodeev

edited Jan 2 at 14:45

Armel

1,206920

edited Jan 2 at 14:45

Armel

1,206920

edited Jan 2 at 14:45

Armel

1,206920

asked Jan 2 at 13:08

Ilya Solodeev

asked Jan 2 at 13:08

Ilya Solodeev

asked Jan 2 at 13:08

Ilya Solodeev

One way is that you can flatten it!

– choz
Jan 2 at 13:10

add a comment |

One way is that you can flatten it!

– choz
Jan 2 at 13:10

One way is that you can flatten it!

– choz
Jan 2 at 13:10

add a comment |

6 Answers
6

active

oldest

votes

Loop and sort, then return the first element of each sub-array with map;

function largestOfFour(arr) {

   for (let i = 0; i < arr.length; i++) {

      arr[i].sort(function(a,b) {

         return b - a;

      });

   }

   return arr.map(array => array[0]);

}

Better still, do it in one.

function largestOfFour(arr) {

    return arr.map(array => array.sort((a,b) => b - a)[0]);

}

answered Jan 2 at 13:14

prettyfly

1,23611021

1

Better still, use Math.max instead of sort.

– str
Jan 2 at 13:38

add a comment |

You could use map with spread syntax ... to return 1D array as pointed out by @str .

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

You could also use reduce

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

edited Jan 2 at 13:18

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

3

concat is not even needed, arr.map(a => Math.max(...a)) is enough.

– str
Jan 2 at 13:15

return arr.map(function(a) { return Math.max.apply(Math, a); }); If you need to support IE

– freddyrobinson
Jan 2 at 13:19

add a comment |

get the first character of each array using [0] index like this

function largestOfFour(arr) {

  for (let i = 0; i < arr.length; i++) {



    arr[i].sort(function(a,b) {

      return b - a;



    }).map(function() {

       // HERE

       return arr[i].splice(1)[0];



    })



  }

  return arr;

}



largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);

answered Jan 2 at 13:12

Joey Gough

744419

add a comment |

You could map the first element of the descending sorted elements.

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

answered Jan 2 at 13:14

Nina Scholz

193k15106177

add a comment |

You can use map which will return a new array and then use Math.max to get the maximum value form the array

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

answered Jan 2 at 13:16

brk

29.1k32243

add a comment |

-2

you can use flat function for it

var arr1 = [1, 2, [3, 4]];

    arr1.flat(); 

    // [1, 2, 3, 4]



    var arr2 = [1, 2, [3, 4, [5, 6]]];

    arr2.flat();

    // [1, 2, 3, 4, [5, 6]]



    var arr3 = [1, 2, [3, 4, [5, 6]]];

    arr3.flat(2);

    // [1, 2, 3, 4, 5, 6]

answered Jan 2 at 13:11

stackoverflow

506

1

It's always better to mention the source of our answers: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Armel
Jan 2 at 13:14

1

Note that this is not supported in IE or Edge

– freddyrobinson
Jan 2 at 13:15

add a comment |

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

Loop and sort, then return the first element of each sub-array with map;

function largestOfFour(arr) {

   for (let i = 0; i < arr.length; i++) {

      arr[i].sort(function(a,b) {

         return b - a;

      });

   }

   return arr.map(array => array[0]);

}

Better still, do it in one.

function largestOfFour(arr) {

    return arr.map(array => array.sort((a,b) => b - a)[0]);

}

answered Jan 2 at 13:14

prettyfly

1,23611021

1

Better still, use Math.max instead of sort.

– str
Jan 2 at 13:38

add a comment |

Loop and sort, then return the first element of each sub-array with map;

function largestOfFour(arr) {

   for (let i = 0; i < arr.length; i++) {

      arr[i].sort(function(a,b) {

         return b - a;

      });

   }

   return arr.map(array => array[0]);

}

Better still, do it in one.

function largestOfFour(arr) {

    return arr.map(array => array.sort((a,b) => b - a)[0]);

}

answered Jan 2 at 13:14

prettyfly

1,23611021

1

Better still, use Math.max instead of sort.

– str
Jan 2 at 13:38

add a comment |

Loop and sort, then return the first element of each sub-array with map;

function largestOfFour(arr) {

   for (let i = 0; i < arr.length; i++) {

      arr[i].sort(function(a,b) {

         return b - a;

      });

   }

   return arr.map(array => array[0]);

}

Better still, do it in one.

function largestOfFour(arr) {

    return arr.map(array => array.sort((a,b) => b - a)[0]);

}

answered Jan 2 at 13:14

prettyfly

1,23611021

Loop and sort, then return the first element of each sub-array with map;

function largestOfFour(arr) {

   for (let i = 0; i < arr.length; i++) {

      arr[i].sort(function(a,b) {

         return b - a;

      });

   }

   return arr.map(array => array[0]);

}

Better still, do it in one.

function largestOfFour(arr) {

    return arr.map(array => array.sort((a,b) => b - a)[0]);

}

answered Jan 2 at 13:14

prettyfly

1,23611021

answered Jan 2 at 13:14

prettyfly

1,23611021

answered Jan 2 at 13:14

prettyfly

1,23611021

answered Jan 2 at 13:14

prettyfly

1,23611021

1

Better still, use Math.max instead of sort.

– str
Jan 2 at 13:38

add a comment |

1

Better still, use Math.max instead of sort.

– str
Jan 2 at 13:38

Better still, use Math.max instead of sort.

– str
Jan 2 at 13:38

add a comment |

You could use map with spread syntax ... to return 1D array as pointed out by @str .

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

You could also use reduce

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

edited Jan 2 at 13:18

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

3

concat is not even needed, arr.map(a => Math.max(...a)) is enough.

– str
Jan 2 at 13:15

return arr.map(function(a) { return Math.max.apply(Math, a); }); If you need to support IE

– freddyrobinson
Jan 2 at 13:19

add a comment |

You could use map with spread syntax ... to return 1D array as pointed out by @str .

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

You could also use reduce

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

edited Jan 2 at 13:18

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

3

concat is not even needed, arr.map(a => Math.max(...a)) is enough.

– str
Jan 2 at 13:15

return arr.map(function(a) { return Math.max.apply(Math, a); }); If you need to support IE

– freddyrobinson
Jan 2 at 13:19

add a comment |

You could use map with spread syntax ... to return 1D array as pointed out by @str .

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

You could also use reduce

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

edited Jan 2 at 13:18

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

You could use map with spread syntax ... to return 1D array as pointed out by @str .

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

You could also use reduce

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

function largestOfFour(arr) {

  return arr.map(a => Math.max(...a))

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

function largestOfFour(arr) {

  return arr.reduce((r, a) => r.concat(Math.max(...a)), )

}



const arr = [[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]

const result = largestOfFour(arr);

console.log(result)

edited Jan 2 at 13:18

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

edited Jan 2 at 13:18

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

answered Jan 2 at 13:13

Nenad Vracar

72.8k126083

3

concat is not even needed, arr.map(a => Math.max(...a)) is enough.

– str
Jan 2 at 13:15

return arr.map(function(a) { return Math.max.apply(Math, a); }); If you need to support IE

– freddyrobinson
Jan 2 at 13:19

add a comment |

3

concat is not even needed, arr.map(a => Math.max(...a)) is enough.

– str
Jan 2 at 13:15

return arr.map(function(a) { return Math.max.apply(Math, a); }); If you need to support IE

– freddyrobinson
Jan 2 at 13:19

concat is not even needed, arr.map(a => Math.max(...a)) is enough.

– str
Jan 2 at 13:15

return arr.map(function(a) { return Math.max.apply(Math, a); }); If you need to support IE

– freddyrobinson
Jan 2 at 13:19

add a comment |

get the first character of each array using [0] index like this

function largestOfFour(arr) {

  for (let i = 0; i < arr.length; i++) {



    arr[i].sort(function(a,b) {

      return b - a;



    }).map(function() {

       // HERE

       return arr[i].splice(1)[0];



    })



  }

  return arr;

}



largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);

answered Jan 2 at 13:12

Joey Gough

744419

add a comment |

get the first character of each array using [0] index like this

function largestOfFour(arr) {

  for (let i = 0; i < arr.length; i++) {



    arr[i].sort(function(a,b) {

      return b - a;



    }).map(function() {

       // HERE

       return arr[i].splice(1)[0];



    })



  }

  return arr;

}



largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);

answered Jan 2 at 13:12

Joey Gough

744419

add a comment |

get the first character of each array using [0] index like this

function largestOfFour(arr) {

  for (let i = 0; i < arr.length; i++) {



    arr[i].sort(function(a,b) {

      return b - a;



    }).map(function() {

       // HERE

       return arr[i].splice(1)[0];



    })



  }

  return arr;

}



largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);

answered Jan 2 at 13:12

Joey Gough

744419

get the first character of each array using [0] index like this

function largestOfFour(arr) {

  for (let i = 0; i < arr.length; i++) {



    arr[i].sort(function(a,b) {

      return b - a;



    }).map(function() {

       // HERE

       return arr[i].splice(1)[0];



    })



  }

  return arr;

}



largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]);

answered Jan 2 at 13:12

Joey Gough

744419

answered Jan 2 at 13:12

Joey Gough

744419

answered Jan 2 at 13:12

Joey Gough

744419

answered Jan 2 at 13:12

Joey Gough

744419

add a comment |

You could map the first element of the descending sorted elements.

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

answered Jan 2 at 13:14

Nina Scholz

193k15106177

add a comment |

You could map the first element of the descending sorted elements.

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

answered Jan 2 at 13:14

Nina Scholz

193k15106177

add a comment |

You could map the first element of the descending sorted elements.

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

answered Jan 2 at 13:14

Nina Scholz

193k15106177

You could map the first element of the descending sorted elements.

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

function largestOfFour(array) {

    return array.map(a => a.sort((a, b) => b - a)[0]);

}



console.log(largestOfFour([[17, 23, 25, 12], [25, 7, 34, 48], [4, -10, 18, 21], [-72, -3, -17, -10]]));

answered Jan 2 at 13:14

Nina Scholz

193k15106177

answered Jan 2 at 13:14

Nina Scholz

193k15106177

answered Jan 2 at 13:14

Nina Scholz

193k15106177

answered Jan 2 at 13:14

Nina Scholz

193k15106177

add a comment |

You can use map which will return a new array and then use Math.max to get the maximum value form the array

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

answered Jan 2 at 13:16

brk

29.1k32243

add a comment |

You can use map which will return a new array and then use Math.max to get the maximum value form the array

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

answered Jan 2 at 13:16

brk

29.1k32243

add a comment |

You can use map which will return a new array and then use Math.max to get the maximum value form the array

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

answered Jan 2 at 13:16

brk

29.1k32243

You can use map which will return a new array and then use Math.max to get the maximum value form the array

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

function largestOfFour(arr) {

  return arr.map(item => {

    return [Math.max.apply(null, item)]



  })

}



console.log(largestOfFour([

  [17, 23, 25, 12],

  [25, 7, 34, 48],

  [4, -10, 18, 21],

  [-72, -3, -17, -10]

]));

answered Jan 2 at 13:16

brk

29.1k32243

answered Jan 2 at 13:16

brk

29.1k32243

answered Jan 2 at 13:16

brk

29.1k32243

answered Jan 2 at 13:16

brk

29.1k32243

add a comment |

-2

you can use flat function for it

var arr1 = [1, 2, [3, 4]];

    arr1.flat(); 

    // [1, 2, 3, 4]



    var arr2 = [1, 2, [3, 4, [5, 6]]];

    arr2.flat();

    // [1, 2, 3, 4, [5, 6]]



    var arr3 = [1, 2, [3, 4, [5, 6]]];

    arr3.flat(2);

    // [1, 2, 3, 4, 5, 6]

answered Jan 2 at 13:11

stackoverflow

506

1

It's always better to mention the source of our answers: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Armel
Jan 2 at 13:14

1

Note that this is not supported in IE or Edge

– freddyrobinson
Jan 2 at 13:15

add a comment |

-2

you can use flat function for it

var arr1 = [1, 2, [3, 4]];

    arr1.flat(); 

    // [1, 2, 3, 4]



    var arr2 = [1, 2, [3, 4, [5, 6]]];

    arr2.flat();

    // [1, 2, 3, 4, [5, 6]]



    var arr3 = [1, 2, [3, 4, [5, 6]]];

    arr3.flat(2);

    // [1, 2, 3, 4, 5, 6]

answered Jan 2 at 13:11

stackoverflow

506

1

It's always better to mention the source of our answers: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Armel
Jan 2 at 13:14

1

Note that this is not supported in IE or Edge

– freddyrobinson
Jan 2 at 13:15

add a comment |

-2

you can use flat function for it

var arr1 = [1, 2, [3, 4]];

    arr1.flat(); 

    // [1, 2, 3, 4]



    var arr2 = [1, 2, [3, 4, [5, 6]]];

    arr2.flat();

    // [1, 2, 3, 4, [5, 6]]



    var arr3 = [1, 2, [3, 4, [5, 6]]];

    arr3.flat(2);

    // [1, 2, 3, 4, 5, 6]

answered Jan 2 at 13:11

stackoverflow

506

you can use flat function for it

var arr1 = [1, 2, [3, 4]];

    arr1.flat(); 

    // [1, 2, 3, 4]



    var arr2 = [1, 2, [3, 4, [5, 6]]];

    arr2.flat();

    // [1, 2, 3, 4, [5, 6]]



    var arr3 = [1, 2, [3, 4, [5, 6]]];

    arr3.flat(2);

    // [1, 2, 3, 4, 5, 6]

answered Jan 2 at 13:11

stackoverflow

506

answered Jan 2 at 13:11

stackoverflow

506

answered Jan 2 at 13:11

stackoverflow

506

answered Jan 2 at 13:11

stackoverflow

506

1

It's always better to mention the source of our answers: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Armel
Jan 2 at 13:14

1

Note that this is not supported in IE or Edge

– freddyrobinson
Jan 2 at 13:15

add a comment |

1

It's always better to mention the source of our answers: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Armel
Jan 2 at 13:14

1

Note that this is not supported in IE or Edge

– freddyrobinson
Jan 2 at 13:15

It's always better to mention the source of our answers: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…

– Armel
Jan 2 at 13:14

Note that this is not supported in IE or Edge

– freddyrobinson
Jan 2 at 13:15

add a comment |

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