Sorting a multidimensional array in javascript?
1
I am new to programming and I have just worked out a multidimensional array problem by using pure javascript but I think I made it too complicated and just wondering can anyone tell me how do you think about my method and do you have a better way to do it.
the problem is to console.log all the numbers in the arrays and the array is [[[1,2],[3,4]],[[5,6]]]. My code is as below, please post your suggestions,thanks
var nestedArr = [[[1,2],[3,4]],[[5,6]]];
var nestedArrOne = nestedArr[0];
var nestedArrTwo = nestedArr[1];
var newArr = nestedArrOne.concat(nestedArrTwo);
function showAll(){
for(var i=0; i<newArr.length; i++){
for(var j=0; j<newArr[i].length; j++){
console.log(newArr[i][j]);
}
}
}
showAll();
javascript arrays sorting for-loop multidimensional-array
edited Nov 20 at 2:27
Cœur
17.4k9102143
asked Jun 8 '17 at 8:53
Sen123
185
add a comment |
1
I am new to programming and I have just worked out a multidimensional array problem by using pure javascript but I think I made it too complicated and just wondering can anyone tell me how do you think about my method and do you have a better way to do it.
the problem is to console.log all the numbers in the arrays and the array is [[[1,2],[3,4]],[[5,6]]]. My code is as below, please post your suggestions,thanks
var nestedArr = [[[1,2],[3,4]],[[5,6]]];
var nestedArrOne = nestedArr[0];
var nestedArrTwo = nestedArr[1];
var newArr = nestedArrOne.concat(nestedArrTwo);
function showAll(){
for(var i=0; i<newArr.length; i++){
for(var j=0; j<newArr[i].length; j++){
console.log(newArr[i][j]);
}
}
}
showAll();
javascript arrays sorting for-loop multidimensional-array
edited Nov 20 at 2:27
Cœur
17.4k9102143
asked Jun 8 '17 at 8:53
Sen123
185
This is called "flatten", not "sort" around these parts ;) Look it up.
– georg
Jun 8 '17 at 8:57
Hi @Sen123 I believe, you should post the question in LINK
– Sudipta Mondal
Jun 8 '17 at 8:57
You have a couple of issues with your code, but here's one to get started with: When you pass an index to an array, you get the next item in the array, not the next nested level. SonestedArronly has one item in it[[1,2],[3,4],[5,6]]which isnestedArr[0]-nestedArr[1]isn't anything.
– Brett East
Jun 8 '17 at 8:59
btw, you don't sort something.
– Nina Scholz
Jun 8 '17 at 9:01
See also stackoverflow.com/a/44103808/1647737
– le_m
Jun 8 '17 at 10:05
add a comment |
1
1
1
I am new to programming and I have just worked out a multidimensional array problem by using pure javascript but I think I made it too complicated and just wondering can anyone tell me how do you think about my method and do you have a better way to do it.
the problem is to console.log all the numbers in the arrays and the array is [[[1,2],[3,4]],[[5,6]]]. My code is as below, please post your suggestions,thanks
var nestedArr = [[[1,2],[3,4]],[[5,6]]];
var nestedArrOne = nestedArr[0];
var nestedArrTwo = nestedArr[1];
var newArr = nestedArrOne.concat(nestedArrTwo);
function showAll(){
for(var i=0; i<newArr.length; i++){
for(var j=0; j<newArr[i].length; j++){
console.log(newArr[i][j]);
}
}
}
showAll();
javascript arrays sorting for-loop multidimensional-array
edited Nov 20 at 2:27
Cœur
17.4k9102143
asked Jun 8 '17 at 8:53
Sen123
185
I am new to programming and I have just worked out a multidimensional array problem by using pure javascript but I think I made it too complicated and just wondering can anyone tell me how do you think about my method and do you have a better way to do it.
the problem is to console.log all the numbers in the arrays and the array is [[[1,2],[3,4]],[[5,6]]]. My code is as below, please post your suggestions,thanks
var nestedArr = [[[1,2],[3,4]],[[5,6]]];
var nestedArrOne = nestedArr[0];
var nestedArrTwo = nestedArr[1];
var newArr = nestedArrOne.concat(nestedArrTwo);
function showAll(){
for(var i=0; i<newArr.length; i++){
for(var j=0; j<newArr[i].length; j++){
console.log(newArr[i][j]);
}
}
}
showAll();
javascript arrays sorting for-loop multidimensional-array
javascript arrays sorting for-loop multidimensional-array
edited Nov 20 at 2:27
Cœur
17.4k9102143
asked Jun 8 '17 at 8:53
Sen123
185
edited Nov 20 at 2:27
Cœur
17.4k9102143
asked Jun 8 '17 at 8:53
Sen123
185
edited Nov 20 at 2:27
Cœur
17.4k9102143
edited Nov 20 at 2:27
Cœur
17.4k9102143
edited Nov 20 at 2:27
Cœur
17.4k9102143
17.4k9102143
asked Jun 8 '17 at 8:53
Sen123
185
asked Jun 8 '17 at 8:53
Sen123
185
asked Jun 8 '17 at 8:53
Sen123
185
185
This is called "flatten", not "sort" around these parts ;) Look it up.
– georg
Jun 8 '17 at 8:57
Hi @Sen123 I believe, you should post the question in LINK
– Sudipta Mondal
Jun 8 '17 at 8:57
You have a couple of issues with your code, but here's one to get started with: When you pass an index to an array, you get the next item in the array, not the next nested level. SonestedArronly has one item in it[[1,2],[3,4],[5,6]]which isnestedArr[0]-nestedArr[1]isn't anything.
– Brett East
Jun 8 '17 at 8:59
btw, you don't sort something.
– Nina Scholz
Jun 8 '17 at 9:01
See also stackoverflow.com/a/44103808/1647737
– le_m
Jun 8 '17 at 10:05
add a comment |
This is called "flatten", not "sort" around these parts ;) Look it up.
– georg
Jun 8 '17 at 8:57
Hi @Sen123 I believe, you should post the question in LINK
– Sudipta Mondal
Jun 8 '17 at 8:57
You have a couple of issues with your code, but here's one to get started with: When you pass an index to an array, you get the next item in the array, not the next nested level. SonestedArronly has one item in it[[1,2],[3,4],[5,6]]which isnestedArr[0]-nestedArr[1]isn't anything.
– Brett East
Jun 8 '17 at 8:59
btw, you don't sort something.
– Nina Scholz
Jun 8 '17 at 9:01
See also stackoverflow.com/a/44103808/1647737
– le_m
Jun 8 '17 at 10:05
This is called "flatten", not "sort" around these parts ;) Look it up.
– georg
Jun 8 '17 at 8:57
This is called "flatten", not "sort" around these parts ;) Look it up.
– georg
Jun 8 '17 at 8:57
Hi @Sen123 I believe, you should post the question in LINK
– Sudipta Mondal
Jun 8 '17 at 8:57
Hi @Sen123 I believe, you should post the question in LINK
– Sudipta Mondal
Jun 8 '17 at 8:57
You have a couple of issues with your code, but here's one to get started with: When you pass an index to an array, you get the next item in the array, not the next nested level. So
nestedArr only has one item in it [[1,2],[3,4],[5,6]] which is nestedArr[0] - nestedArr[1] isn't anything.– Brett East
Jun 8 '17 at 8:59
You have a couple of issues with your code, but here's one to get started with: When you pass an index to an array, you get the next item in the array, not the next nested level. So
nestedArr only has one item in it [[1,2],[3,4],[5,6]] which is nestedArr[0] - nestedArr[1] isn't anything.– Brett East
Jun 8 '17 at 8:59
btw, you don't sort something.
– Nina Scholz
Jun 8 '17 at 9:01
btw, you don't sort something.
– Nina Scholz
Jun 8 '17 at 9:01
See also stackoverflow.com/a/44103808/1647737
– le_m
Jun 8 '17 at 10:05
See also stackoverflow.com/a/44103808/1647737
– le_m
Jun 8 '17 at 10:05
add a comment |
3 Answers
3
active
oldest
votes
0
You could iterate the nested array and check if the item is an array, then call the function again for the inner array.
function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
add a comment |
0
The recursive function could be used to flatten the array.
let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)answered Jun 8 '17 at 9:35
shuaibird
36133
add a comment |
0
A fancy Haskellesque approach with "pattern matching by rest operator" could be.
var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);answered Jun 8 '17 at 11:24
Redu
12.6k22435
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
You could iterate the nested array and check if the item is an array, then call the function again for the inner array.
function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
add a comment |
0
You could iterate the nested array and check if the item is an array, then call the function again for the inner array.
function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
add a comment |
0
0
0
You could iterate the nested array and check if the item is an array, then call the function again for the inner array.
function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
You could iterate the nested array and check if the item is an array, then call the function again for the inner array.
function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);function iter(array) {
var i; // declare index
for (i = 0; i < array.length; i++) { // iterate array
if (Array.isArray(array[i])) { // check if item is an array
iter(array[i]); // if so, call iter with item
continue; // and continue the loop
}
console.log(array[i]); // the wanted output
}
}
var nestedArr = [[[1, 2], [3, 4]], [[5, 6]]];
iter(nestedArr);answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
answered Jun 8 '17 at 8:59
Nina Scholz
175k1388152
175k1388152
add a comment |
add a comment |
0
The recursive function could be used to flatten the array.
let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)answered Jun 8 '17 at 9:35
shuaibird
36133
add a comment |
0
The recursive function could be used to flatten the array.
let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)answered Jun 8 '17 at 9:35
shuaibird
36133
add a comment |
0
0
0
The recursive function could be used to flatten the array.
let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)answered Jun 8 '17 at 9:35
shuaibird
36133
The recursive function could be used to flatten the array.
let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)let isArray = val => val.constructor === Array
let log = val => console.log(val)
let flatten = val =>
val
.reduce((acc, cur) =>
isArray(cur) ? [...acc, ...flatten(cur)] : [...acc, cur],
)
// Running examples
let arrA = [[[1,2],[3,4]],[[5,6]]]
flatten(arrA).forEach(log)
log('-------------------')
let arrB = [[[1,2],[3,4]],[[5,6]],[7],[[[[9]]]]]
flatten(arrB).forEach(log)answered Jun 8 '17 at 9:35
shuaibird
36133
edited Jun 8 '17 at 9:48
answered Jun 8 '17 at 9:35
shuaibird
36133
answered Jun 8 '17 at 9:35
shuaibird
36133
answered Jun 8 '17 at 9:35
shuaibird
36133
36133
add a comment |
add a comment |
0
A fancy Haskellesque approach with "pattern matching by rest operator" could be.
var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);answered Jun 8 '17 at 11:24
Redu
12.6k22435
add a comment |
0
A fancy Haskellesque approach with "pattern matching by rest operator" could be.
var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);answered Jun 8 '17 at 11:24
Redu
12.6k22435
add a comment |
0
0
0
A fancy Haskellesque approach with "pattern matching by rest operator" could be.
var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);answered Jun 8 '17 at 11:24
Redu
12.6k22435
A fancy Haskellesque approach with "pattern matching by rest operator" could be.
var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);var flat = ([x,...xs]) => x ? [...Array.isArray(x) ? flat(x) : [x], ...flat(xs)] : ;
na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
fa = flat(na);
console.log(...fa);answered Jun 8 '17 at 11:24
Redu
12.6k22435
answered Jun 8 '17 at 11:24
Redu
12.6k22435
answered Jun 8 '17 at 11:24
Redu
12.6k22435
answered Jun 8 '17 at 11:24
Redu
12.6k22435
12.6k22435
add a comment |
add a comment |
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This is called "flatten", not "sort" around these parts ;) Look it up.
– georg
Jun 8 '17 at 8:57
Hi @Sen123 I believe, you should post the question in LINK
– Sudipta Mondal
Jun 8 '17 at 8:57
You have a couple of issues with your code, but here's one to get started with: When you pass an index to an array, you get the next item in the array, not the next nested level. So
nestedArronly has one item in it[[1,2],[3,4],[5,6]]which isnestedArr[0]-nestedArr[1]isn't anything.– Brett East
Jun 8 '17 at 8:59
btw, you don't sort something.
– Nina Scholz
Jun 8 '17 at 9:01
See also stackoverflow.com/a/44103808/1647737
– le_m
Jun 8 '17 at 10:05