Javascript: Turn a nested array into a single array using a nested for loop

This is just a simple javascript exercise that I'm working on.

I'm trying to convert this array...

var array = [

[1,2],

[3,4],

[5,6]

];

into...

array = [1, 2, 3, 4, 5, 6];

by using this nested for loop.

var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

    for (var j = 0; j < array[i].length; j++) {

        series = array[i][j];

        for (var k = 0; k < 6; k++) {

            storage[k] = series;

        };

    };

};



console.log(storage);

With an output of...

//Output: [6, 6, 6, 6, 6, 6]

Why is this the output and how can I fix it?

asked Nov 22 '18 at 16:19

simon

add a comment |

This is just a simple javascript exercise that I'm working on.

I'm trying to convert this array...

var array = [

[1,2],

[3,4],

[5,6]

];

into...

array = [1, 2, 3, 4, 5, 6];

by using this nested for loop.

var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

    for (var j = 0; j < array[i].length; j++) {

        series = array[i][j];

        for (var k = 0; k < 6; k++) {

            storage[k] = series;

        };

    };

};



console.log(storage);

With an output of...

//Output: [6, 6, 6, 6, 6, 6]

Why is this the output and how can I fix it?

asked Nov 22 '18 at 16:19

simon

add a comment |

This is just a simple javascript exercise that I'm working on.

I'm trying to convert this array...

var array = [

[1,2],

[3,4],

[5,6]

];

into...

array = [1, 2, 3, 4, 5, 6];

by using this nested for loop.

var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

    for (var j = 0; j < array[i].length; j++) {

        series = array[i][j];

        for (var k = 0; k < 6; k++) {

            storage[k] = series;

        };

    };

};



console.log(storage);

With an output of...

//Output: [6, 6, 6, 6, 6, 6]

Why is this the output and how can I fix it?

asked Nov 22 '18 at 16:19

simon

This is just a simple javascript exercise that I'm working on.

I'm trying to convert this array...

var array = [

[1,2],

[3,4],

[5,6]

];

into...

array = [1, 2, 3, 4, 5, 6];

by using this nested for loop.

var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

    for (var j = 0; j < array[i].length; j++) {

        series = array[i][j];

        for (var k = 0; k < 6; k++) {

            storage[k] = series;

        };

    };

};



console.log(storage);

With an output of...

//Output: [6, 6, 6, 6, 6, 6]

Why is this the output and how can I fix it?

javascript loops for-loop nested

asked Nov 22 '18 at 16:19

simon

asked Nov 22 '18 at 16:19

simon

asked Nov 22 '18 at 16:19

simon

asked Nov 22 '18 at 16:19

simon

asked Nov 22 '18 at 16:19

simon

add a comment |

8 Answers
8

active

oldest

votes

for (var k = 0; k < 6; k++) { is not required . array[i] will be each of element inside main array , so iterating over array[i] you can access each of the element

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

answered Nov 22 '18 at 16:26

brk

27.9k32142

add a comment |

    series = array[i][j];

    for (var k = 0; k < 6; k++) {

        storage[k] = series;

    };

Seriously, here you set the same value to each element of the resulting array.

You probably need something like

for(let x of array) {

    for(let y of x) {

        storage.push(y)

    }

}

Or, if your JS machine is experimental enough, simply

var storage = array.flat()

answered Nov 22 '18 at 16:23

bipll

8,2361927

add a comment |

You can use a mix of reduce and concat to achieve what you want in one line

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

As for why your code didn't work, it's mainly down to this bit

for (var k = 0; k < 6; k++) {

    storage[k] = series;

};

It would overwrite everything in the array with the last value of series, which in your case would be 6

answered Nov 22 '18 at 16:25

George

4,50811732

add a comment |

Array.concat can do this on its own

var merged = .concat.apply(, array);

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

add a comment |

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

answered Nov 22 '18 at 16:28

Ayon Saha

30427

add a comment |

You can use array.flat()

Reference : flatten

var array = [

  [1,2],

  [3,4],

  [5,6]

];

array.flat();

// 1,2,3,4,5,6....

answered Nov 22 '18 at 16:24

Christheoreo

2269

add a comment |

You could use the ES6 spread syntax like this:

for (let element of array){

    storage.push( ... el )

}

answered Nov 22 '18 at 16:25

weibenfalk

54617

add a comment |

storage = ;

for(var i=0; i<array.length; i++)

    storage = storage.concat(array[i]);

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

answered Nov 22 '18 at 16:32

bugpulver

add a comment |

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8 Answers
8

active

oldest

votes

8 Answers
8

active

oldest

votes

for (var k = 0; k < 6; k++) { is not required . array[i] will be each of element inside main array , so iterating over array[i] you can access each of the element

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

answered Nov 22 '18 at 16:26

brk

27.9k32142

add a comment |

for (var k = 0; k < 6; k++) { is not required . array[i] will be each of element inside main array , so iterating over array[i] you can access each of the element

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

answered Nov 22 '18 at 16:26

brk

27.9k32142

add a comment |

for (var k = 0; k < 6; k++) { is not required . array[i] will be each of element inside main array , so iterating over array[i] you can access each of the element

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

answered Nov 22 '18 at 16:26

brk

27.9k32142

for (var k = 0; k < 6; k++) { is not required . array[i] will be each of element inside main array , so iterating over array[i] you can access each of the element

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];





var series;

var storage = ;

for (var i = 0; i < array.length; i++) {

  for (var j = 0; j < array[i].length; j++) {

    storage.push(array[i][j])

  };

};



console.log(storage);

answered Nov 22 '18 at 16:26

brk

27.9k32142

answered Nov 22 '18 at 16:26

brk

27.9k32142

answered Nov 22 '18 at 16:26

brk

27.9k32142

answered Nov 22 '18 at 16:26

brk

27.9k32142

add a comment |

    series = array[i][j];

    for (var k = 0; k < 6; k++) {

        storage[k] = series;

    };

Seriously, here you set the same value to each element of the resulting array.

You probably need something like

for(let x of array) {

    for(let y of x) {

        storage.push(y)

    }

}

Or, if your JS machine is experimental enough, simply

var storage = array.flat()

answered Nov 22 '18 at 16:23

bipll

8,2361927

add a comment |

    series = array[i][j];

    for (var k = 0; k < 6; k++) {

        storage[k] = series;

    };

Seriously, here you set the same value to each element of the resulting array.

You probably need something like

for(let x of array) {

    for(let y of x) {

        storage.push(y)

    }

}

Or, if your JS machine is experimental enough, simply

var storage = array.flat()

answered Nov 22 '18 at 16:23

bipll

8,2361927

add a comment |

    series = array[i][j];

    for (var k = 0; k < 6; k++) {

        storage[k] = series;

    };

Seriously, here you set the same value to each element of the resulting array.

You probably need something like

for(let x of array) {

    for(let y of x) {

        storage.push(y)

    }

}

Or, if your JS machine is experimental enough, simply

var storage = array.flat()

answered Nov 22 '18 at 16:23

bipll

8,2361927

    series = array[i][j];

    for (var k = 0; k < 6; k++) {

        storage[k] = series;

    };

Seriously, here you set the same value to each element of the resulting array.

You probably need something like

for(let x of array) {

    for(let y of x) {

        storage.push(y)

    }

}

Or, if your JS machine is experimental enough, simply

var storage = array.flat()

answered Nov 22 '18 at 16:23

bipll

8,2361927

answered Nov 22 '18 at 16:23

bipll

8,2361927

answered Nov 22 '18 at 16:23

bipll

8,2361927

answered Nov 22 '18 at 16:23

bipll

8,2361927

add a comment |

You can use a mix of reduce and concat to achieve what you want in one line

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

As for why your code didn't work, it's mainly down to this bit

for (var k = 0; k < 6; k++) {

    storage[k] = series;

};

It would overwrite everything in the array with the last value of series, which in your case would be 6

answered Nov 22 '18 at 16:25

George

4,50811732

add a comment |

You can use a mix of reduce and concat to achieve what you want in one line

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

As for why your code didn't work, it's mainly down to this bit

for (var k = 0; k < 6; k++) {

    storage[k] = series;

};

It would overwrite everything in the array with the last value of series, which in your case would be 6

answered Nov 22 '18 at 16:25

George

4,50811732

add a comment |

You can use a mix of reduce and concat to achieve what you want in one line

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

As for why your code didn't work, it's mainly down to this bit

for (var k = 0; k < 6; k++) {

    storage[k] = series;

};

It would overwrite everything in the array with the last value of series, which in your case would be 6

answered Nov 22 '18 at 16:25

George

4,50811732

You can use a mix of reduce and concat to achieve what you want in one line

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

As for why your code didn't work, it's mainly down to this bit

for (var k = 0; k < 6; k++) {

    storage[k] = series;

};

It would overwrite everything in the array with the last value of series, which in your case would be 6

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

var array = [

  [1, 2],

  [3, 4],

  [5, 6]

];



console.log(array.reduce((a, v) => a.concat(v), ));

answered Nov 22 '18 at 16:25

George

4,50811732

answered Nov 22 '18 at 16:25

George

4,50811732

answered Nov 22 '18 at 16:25

George

4,50811732

answered Nov 22 '18 at 16:25

George

4,50811732

add a comment |

Array.concat can do this on its own

var merged = .concat.apply(, array);

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

add a comment |

Array.concat can do this on its own

var merged = .concat.apply(, array);

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

add a comment |

Array.concat can do this on its own

var merged = .concat.apply(, array);

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

Array.concat can do this on its own

var merged = .concat.apply(, array);

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

answered Nov 22 '18 at 16:26

SpeedOfRound

629314

add a comment |

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

answered Nov 22 '18 at 16:28

Ayon Saha

30427

add a comment |

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

answered Nov 22 '18 at 16:28

Ayon Saha

30427

add a comment |

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

answered Nov 22 '18 at 16:28

Ayon Saha

30427

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

var array = [

[1,2],

[3,4],

[5,6]

];

var newArray = ;

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

   newArray = newArray.concat(array[i]); 



}

console.log(newArray)

answered Nov 22 '18 at 16:28

Ayon Saha

30427

answered Nov 22 '18 at 16:28

Ayon Saha

30427

answered Nov 22 '18 at 16:28

Ayon Saha

30427

answered Nov 22 '18 at 16:28

Ayon Saha

30427

add a comment |

You can use array.flat()

Reference : flatten

var array = [

  [1,2],

  [3,4],

  [5,6]

];

array.flat();

// 1,2,3,4,5,6....

answered Nov 22 '18 at 16:24

Christheoreo

2269

add a comment |

You can use array.flat()

Reference : flatten

var array = [

  [1,2],

  [3,4],

  [5,6]

];

array.flat();

// 1,2,3,4,5,6....

answered Nov 22 '18 at 16:24

Christheoreo

2269

add a comment |

You can use array.flat()

Reference : flatten

var array = [

  [1,2],

  [3,4],

  [5,6]

];

array.flat();

// 1,2,3,4,5,6....

answered Nov 22 '18 at 16:24

Christheoreo

2269

You can use array.flat()

Reference : flatten

var array = [

  [1,2],

  [3,4],

  [5,6]

];

array.flat();

// 1,2,3,4,5,6....

answered Nov 22 '18 at 16:24

Christheoreo

2269

answered Nov 22 '18 at 16:24

Christheoreo

2269

answered Nov 22 '18 at 16:24

Christheoreo

2269

answered Nov 22 '18 at 16:24

Christheoreo

2269

add a comment |

You could use the ES6 spread syntax like this:

for (let element of array){

    storage.push( ... el )

}

answered Nov 22 '18 at 16:25

weibenfalk

54617

add a comment |

You could use the ES6 spread syntax like this:

for (let element of array){

    storage.push( ... el )

}

answered Nov 22 '18 at 16:25

weibenfalk

54617

add a comment |

You could use the ES6 spread syntax like this:

for (let element of array){

    storage.push( ... el )

}

answered Nov 22 '18 at 16:25

weibenfalk

54617

You could use the ES6 spread syntax like this:

for (let element of array){

    storage.push( ... el )

}

answered Nov 22 '18 at 16:25

weibenfalk

54617

answered Nov 22 '18 at 16:25

weibenfalk

54617

answered Nov 22 '18 at 16:25

weibenfalk

54617

answered Nov 22 '18 at 16:25

weibenfalk

54617

add a comment |

storage = ;

for(var i=0; i<array.length; i++)

    storage = storage.concat(array[i]);

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

answered Nov 22 '18 at 16:32

bugpulver

add a comment |

storage = ;

for(var i=0; i<array.length; i++)

    storage = storage.concat(array[i]);

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

answered Nov 22 '18 at 16:32

bugpulver

add a comment |

storage = ;

for(var i=0; i<array.length; i++)

    storage = storage.concat(array[i]);

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

answered Nov 22 '18 at 16:32

bugpulver

storage = ;

for(var i=0; i<array.length; i++)

    storage = storage.concat(array[i]);

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

answered Nov 22 '18 at 16:32

bugpulver

answered Nov 22 '18 at 16:32

bugpulver

answered Nov 22 '18 at 16:32

bugpulver

answered Nov 22 '18 at 16:32

bugpulver

add a comment |

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