How to splice an array out of a nested array

How to splice an array out of a nested array - javascript

I am attempting to splice a nested array out of its parent array. Consider the following array. items.splice(0,1) should give me the first nested array([1,2]), however it seems to give me the first nested array, still nested inside of an array:

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



var item = items.splice(0,1); // should slice first array out of items array



console.log(item); //should log [1,2], instead logs [[1,2]]

However it seems to return the needed array (first item), in another array. And I'm not able to get the full array unless I do item[0]. What in the world am I missing!?

edited Sep 1 '15 at 2:55

user663031

asked Sep 1 '15 at 2:17

Afs35mm

101211

1

No need for splice. Just use items[0] instead.

– PHPglue
Sep 1 '15 at 2:22

Well I want to alter the original array and then chose a value based on an element of the spliced array.

– Afs35mm
Sep 1 '15 at 19:24

add a comment |

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



var item = items.splice(0,1); // should slice first array out of items array



console.log(item); //should log [1,2], instead logs [[1,2]]

However it seems to return the needed array (first item), in another array. And I'm not able to get the full array unless I do item[0]. What in the world am I missing!?

edited Sep 1 '15 at 2:55

user663031

asked Sep 1 '15 at 2:17

Afs35mm

101211

1

No need for splice. Just use items[0] instead.

– PHPglue
Sep 1 '15 at 2:22

Well I want to alter the original array and then chose a value based on an element of the spliced array.

– Afs35mm
Sep 1 '15 at 19:24

add a comment |

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



var item = items.splice(0,1); // should slice first array out of items array



console.log(item); //should log [1,2], instead logs [[1,2]]

However it seems to return the needed array (first item), in another array. And I'm not able to get the full array unless I do item[0]. What in the world am I missing!?

edited Sep 1 '15 at 2:55

user663031

asked Sep 1 '15 at 2:17

Afs35mm

101211

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



var item = items.splice(0,1); // should slice first array out of items array



console.log(item); //should log [1,2], instead logs [[1,2]]

However it seems to return the needed array (first item), in another array. And I'm not able to get the full array unless I do item[0]. What in the world am I missing!?

javascript arrays multidimensional-array splice

edited Sep 1 '15 at 2:55

user663031

asked Sep 1 '15 at 2:17

Afs35mm

101211

edited Sep 1 '15 at 2:55

user663031

asked Sep 1 '15 at 2:17

Afs35mm

101211

edited Sep 1 '15 at 2:55

user663031

edited Sep 1 '15 at 2:55

user663031

edited Sep 1 '15 at 2:55

user663031

asked Sep 1 '15 at 2:17

Afs35mm

101211

asked Sep 1 '15 at 2:17

Afs35mm

101211

asked Sep 1 '15 at 2:17

Afs35mm

101211

1

No need for splice. Just use items[0] instead.

– PHPglue
Sep 1 '15 at 2:22

Well I want to alter the original array and then chose a value based on an element of the spliced array.

– Afs35mm
Sep 1 '15 at 19:24

add a comment |

1

No need for splice. Just use items[0] instead.

– PHPglue
Sep 1 '15 at 2:22

Well I want to alter the original array and then chose a value based on an element of the spliced array.

– Afs35mm
Sep 1 '15 at 19:24

No need for splice. Just use items[0] instead.

– PHPglue
Sep 1 '15 at 2:22

Well I want to alter the original array and then chose a value based on an element of the spliced array.

– Afs35mm
Sep 1 '15 at 19:24

add a comment |

5 Answers
5

active

oldest

votes

The MDN says Array.prototype.splice:

Returns

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

So, it won't return just the deleted element, it will be wrapped in an array.

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

Kind of crazy I never knew that splice automatically places it inside a newly created array, I kind of just always assumed it worked that way because it was being taken from an already existing array. Thinking about it that makes sense because that's how Array.prototype.slice.call(arguments) works considering it's only an array-like object :)

– Afs35mm
Sep 1 '15 at 19:25

add a comment |

splice() changes the contents of an array, and returns an array with the deleted elements.

In your case, your original array has elements that also happen to be arrays. That might be confusing you.

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

add a comment |

.splice() is returning correct array at item ; try selecting index 0 of returned array to return [1,2] ; to "flatten" item , try utilizing Array.prototype.concat() ; see How to flatten array in jQuery?

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

edited May 23 '17 at 12:34

Community♦

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

add a comment |

You should reference the first array in items, and then splice it. Try working snippet below.

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



var item = items[0].splice(0,2);



console.log(item);

edited Jan 20 at 0:14

answered Jan 20 at 0:01

M'Boulas

45212

add a comment |

-1

To replace a two dimensional array you should use array[row][col] for example

for(var row = 0; row < numbers.length; row++) {

    for(var col = 0; col < numbers[row].length; col++) {

        if (numbers[row][col] % 2 === 0) {

            numbers[row][col] = "even";

        } else {

            numbers[row][col] = "odd";

        }  

    }



}

console.log(numbers);

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

add a comment |

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5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

The MDN says Array.prototype.splice:

Returns

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

So, it won't return just the deleted element, it will be wrapped in an array.

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

Kind of crazy I never knew that splice automatically places it inside a newly created array, I kind of just always assumed it worked that way because it was being taken from an already existing array. Thinking about it that makes sense because that's how Array.prototype.slice.call(arguments) works considering it's only an array-like object :)

– Afs35mm
Sep 1 '15 at 19:25

add a comment |

The MDN says Array.prototype.splice:

Returns

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

So, it won't return just the deleted element, it will be wrapped in an array.

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

Kind of crazy I never knew that splice automatically places it inside a newly created array, I kind of just always assumed it worked that way because it was being taken from an already existing array. Thinking about it that makes sense because that's how Array.prototype.slice.call(arguments) works considering it's only an array-like object :)

– Afs35mm
Sep 1 '15 at 19:25

add a comment |

The MDN says Array.prototype.splice:

Returns

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

So, it won't return just the deleted element, it will be wrapped in an array.

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

The MDN says Array.prototype.splice:

Returns

An array containing the deleted elements. If only one element is removed, an array of one element is returned. If no elements are removed, an empty array is returned.

So, it won't return just the deleted element, it will be wrapped in an array.

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

answered Sep 1 '15 at 2:28

MinusFour

8,74121828

Kind of crazy I never knew that splice automatically places it inside a newly created array, I kind of just always assumed it worked that way because it was being taken from an already existing array. Thinking about it that makes sense because that's how Array.prototype.slice.call(arguments) works considering it's only an array-like object :)

– Afs35mm
Sep 1 '15 at 19:25

add a comment |

Kind of crazy I never knew that splice automatically places it inside a newly created array, I kind of just always assumed it worked that way because it was being taken from an already existing array. Thinking about it that makes sense because that's how Array.prototype.slice.call(arguments) works considering it's only an array-like object :)

– Afs35mm
Sep 1 '15 at 19:25

Kind of crazy I never knew that splice automatically places it inside a newly created array, I kind of just always assumed it worked that way because it was being taken from an already existing array. Thinking about it that makes sense because that's how Array.prototype.slice.call(arguments) works considering it's only an array-like object :)

– Afs35mm
Sep 1 '15 at 19:25

add a comment |

splice() changes the contents of an array, and returns an array with the deleted elements.

In your case, your original array has elements that also happen to be arrays. That might be confusing you.

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

add a comment |

splice() changes the contents of an array, and returns an array with the deleted elements.

In your case, your original array has elements that also happen to be arrays. That might be confusing you.

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

add a comment |

splice() changes the contents of an array, and returns an array with the deleted elements.

In your case, your original array has elements that also happen to be arrays. That might be confusing you.

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

splice() changes the contents of an array, and returns an array with the deleted elements.

In your case, your original array has elements that also happen to be arrays. That might be confusing you.

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

answered Sep 1 '15 at 2:24

cybersam

39.5k43151

add a comment |

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

edited May 23 '17 at 12:34

Community♦

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

add a comment |

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

edited May 23 '17 at 12:34

Community♦

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

add a comment |

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

edited May 23 '17 at 12:34

Community♦

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

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



var item = items.splice(0,1); // should slice first array out of items array

// https://stackoverflow.com/a/7875193/

var arr = .concat.apply(, item);



console.log(item[0], arr);

edited May 23 '17 at 12:34

Community♦

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

edited May 23 '17 at 12:34

Community♦

edited May 23 '17 at 12:34

Community♦

edited May 23 '17 at 12:34

Community♦

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

answered Sep 1 '15 at 2:36

guest271314

79.2k643114

add a comment |

You should reference the first array in items, and then splice it. Try working snippet below.

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



var item = items[0].splice(0,2);



console.log(item);

edited Jan 20 at 0:14

answered Jan 20 at 0:01

M'Boulas

45212

add a comment |

You should reference the first array in items, and then splice it. Try working snippet below.

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



var item = items[0].splice(0,2);



console.log(item);

edited Jan 20 at 0:14

answered Jan 20 at 0:01

M'Boulas

45212

add a comment |

You should reference the first array in items, and then splice it. Try working snippet below.

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



var item = items[0].splice(0,2);



console.log(item);

edited Jan 20 at 0:14

answered Jan 20 at 0:01

M'Boulas

45212

You should reference the first array in items, and then splice it. Try working snippet below.

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



var item = items[0].splice(0,2);



console.log(item);

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



var item = items[0].splice(0,2);



console.log(item);

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



var item = items[0].splice(0,2);



console.log(item);

edited Jan 20 at 0:14

answered Jan 20 at 0:01

M'Boulas

45212

edited Jan 20 at 0:14

answered Jan 20 at 0:01

M'Boulas

45212

answered Jan 20 at 0:01

M'Boulas

45212

answered Jan 20 at 0:01

M'Boulas

45212

add a comment |

-1

To replace a two dimensional array you should use array[row][col] for example

for(var row = 0; row < numbers.length; row++) {

    for(var col = 0; col < numbers[row].length; col++) {

        if (numbers[row][col] % 2 === 0) {

            numbers[row][col] = "even";

        } else {

            numbers[row][col] = "odd";

        }  

    }



}

console.log(numbers);

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

add a comment |

-1

To replace a two dimensional array you should use array[row][col] for example

for(var row = 0; row < numbers.length; row++) {

    for(var col = 0; col < numbers[row].length; col++) {

        if (numbers[row][col] % 2 === 0) {

            numbers[row][col] = "even";

        } else {

            numbers[row][col] = "odd";

        }  

    }



}

console.log(numbers);

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

add a comment |

-1

To replace a two dimensional array you should use array[row][col] for example

for(var row = 0; row < numbers.length; row++) {

    for(var col = 0; col < numbers[row].length; col++) {

        if (numbers[row][col] % 2 === 0) {

            numbers[row][col] = "even";

        } else {

            numbers[row][col] = "odd";

        }  

    }



}

console.log(numbers);

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

To replace a two dimensional array you should use array[row][col] for example

for(var row = 0; row < numbers.length; row++) {

    for(var col = 0; col < numbers[row].length; col++) {

        if (numbers[row][col] % 2 === 0) {

            numbers[row][col] = "even";

        } else {

            numbers[row][col] = "odd";

        }  

    }



}

console.log(numbers);

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

answered Oct 20 '17 at 21:25

Alicia Guzman

2414

add a comment |

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