I have the following code:
function build_all_combinations(input_array){
array = [1,2,3]
limit = array.length - 1
combo = []
for(i = 0; i<= limit; i++){
splice_value = array.splice(0,1)
push_value = splice_value[0]
array.push(push_value)
console.log(array)
combo.push(array)
}
console.log(combo)
}
Which outputs:
[2, 3, 1]
[3, 1, 2]
[1, 2, 3]
[[1, 2, 3], [1, 2, 3], [1, 2, 3]]
The last line should be: [[2, 3, 1],[3, 1, 2],[1, 2, 3]]
I'm obviously not grokking something about the way the array is working. Each individual array is correct, but when I go to push them to the combo array, something is failing along the way. What is it?
Each time you are pushing the same array into the combo array; ie, all the references are pointing to the same instance in memory. So when you update one, you've in reality updated them all.
If you want separate references, you'll need to create separate arrays.
shift and slice are your friends here:
var array = [1,2,3];
var combo = []
for(var i = 0; i<array.length; i++){
array.push(array.shift()); // remove first element (shift) and add it to the end (push)
combo.push(array.slice()); // add a copy of the current array to combo
}
DEMO
jordan002 has given the explanation for this behaviour.
Here's the workaround.
array = [1,2,3]
limit = array.length - 1
combo = []
for(i = 0; i<= limit; i++){
splice_value = array.splice(0,1)
push_value = splice_value[0];
array.push(push_value);
console.log(array);
combo.push(array.concat([]));
}
console.log(combo);
You can store a copy in temp variable.
array = [1,2,3]
limit = array.length - 1
combo = []
for(i = 0; i<= limit; i++){
splice_value = array.splice(0,1)
push_value = splice_value[0];
array.push(push_value);
console.log(array);
var temp = array.slice();
combo.push(temp);
}
console.log(combo)
Reference
Related
i have a problem and i need help for this question.
My reverse function doesn't work the way I want it to.
function reverseArrayInPlace(array){
let old = array;
for (let i = 0; i < old.length; i++){
array[i] = old[old.length - 1 - i];
};
};
let arrayValue = [1, 2, 3, 4, 5];
reverseArrayInPlace(arrayValue);
console.log(arrayValue);
I expect on [5, 4, 3, 2, 1] but i have [5, 4, 3, 4, 5].
Why does it work this way? Please help me understand.
P.S I know about the reverse method.
Variable, with assigned object (array, which is a modified object in fact) - stores just a link to that object, but not the actual object. So, let old = array; here you just created a new link to the same array. Any changes with both variables will cause the change of array.
(demo)
let arr = [0,0,0,0,0];
let bubu = arr;
bubu[0] = 999
console.log( arr );
The simplest way to create an array clone:
function reverseArrayInPlace(array){
let old = array.slice(0); // <<<
for (let i = 0; i < old.length; i++){
array[i] = old[old.length - 1 - i];
};
return array;
};
console.log( reverseArrayInPlace( [1, 2, 3, 4, 5] ) );
P.s. just for fun:
function reverseArrayInPlace(array){
let len = array.length;
let half = (len / 2) ^ 0; // XOR with 0 <==> Math.trunc()
for( let i = 0; i < half; i++ ){
[ array[i], array[len - i-1] ] = [ array[len - i-1], array[i] ]
}
return array;
};
console.log( reverseArrayInPlace( [1, 2, 3, 4, 5] ) );
If you're writing a true in place algorithm, it's wasteful from both a speed and memory standpoint to make an unnecessary copy (as other answers point out--array and old are aliases in the original code).
A better approach is to iterate over half of the array, swapping each element with its length - 1 - i compliment. This has 1/4 of the iterations of the slice approach, is more intuitive and uses constant time memory.
const reverseArrayInPlace = a => {
for (let i = 0; i < a.length / 2; i++) {
[a[i], a[a.length-1-i]] = [a[a.length-1-i], a[i]];
}
};
const a = [1, 2, 3, 4, 5];
reverseArrayInPlace(a);
console.log(a);
Since this is a hot loop, making two array objects on the heap just to toss them out is inefficient, so if you're not transpiling this, you might want to use a traditional swap with a temporary variable.
function reverseArrayInPlace(a) {
for (var i = 0; i < a.length / 2; i++) {
var temp = a[i];
a[i] = a[a.length-i-1];
a[a.length-i-1] = temp;
}
};
var a = [1, 2, 3, 4];
reverseArrayInPlace(a);
console.log(a);
You have in old variable the same array (not by value, but by reference), so if you change any of them you'll have changes in both.
So you should create new array, make whatever you want with them and return it (or just copy values back to your origin array if you don't want to return anything from your function).
It happend like that because:
1) arr[0] = arr[4] (5,2,3,4,5)
2) arr[1] = arr[3] (5,4,3,4,5)
....
n) arr[n] = arr[0] (5,4,3,4,5)
So you can just make a copy (let old = array.slice()) and it'll be woking as you'd expect.
This question already has answers here:
Why does changing an Array in JavaScript affect copies of the array?
(12 answers)
Closed 4 years ago.
This code seems straightforward enough
let arr = [1,2,3,4,5,6]
let store = [];
for(i = 0; i < arr.length; i++){
console.log(arr)
store.push(arr)
arr.push(arr.shift())
}
console.log('store', JSON.stringify(store))
I'm expecting it to return
[[1,2,3,4,5,6],[2,3,4,5,6,1],[3,4,5,6,1,2],[4,5,6,1,2,3],[5,6,1,2,3,4],[6,1,2,3,4,5]]
But when the loop is complete, it shows
[[1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6],[1,2,3,4,5,6]]
When printing the values in the console, they appear correct, but when the store variable is logged, they are all reordered.
[1, 2, 3, 4, 5, 6]
[2, 3, 4, 5, 6, 1]
[3, 4, 5, 6, 1, 2]
[4, 5, 6, 1, 2, 3]
[5, 6, 1, 2, 3, 4]
[6, 1, 2, 3, 4, 5]
Why is this occurring?
There is a similar question here, but it doesn't really provide an answer for my case.
Creating a pair of arrays from an array resulting in circular array
Just make sure you are inserting not the same reference to your array:
let arr = [1,2,3,4,5,6]
let store = [];
for(i = 0; i < arr.length; i++){
store.push(Array.from(arr)) // <-- Make sure not the same ref
arr.push(arr.shift())
}
console.log('store', JSON.stringify(store))
store contains the same instance of arr 6 times. You modify it and print it every time, then add the same object to store, then modify it again.
To get your desired behaviour you need to add a new instance every time e.g. by cloning it when you add it.
You could push a copy which does not keep the same object reference.
let arr = [1, 2, 3, 4, 5, 6]
let store = [];
for (i = 0; i < arr.length; i++) {
store.push(arr.slice()) // just take a copy
arr.push(arr.shift())
}
console.log(store.map(a => a.join(' ')))
Alternate approach leaving original array untouched and using splice() to remove beginning elements and concat() to add to the end
let arr = [1,2,3,4,5,6];
let store = arr.map((_,i)=> {
let a = arr.slice();
return i ? a.concat(a.splice(0,i)) : a;
})
console.log(JSON.stringify(store))
On every loop, you are pushing the reference of the array to store
Example:
Loop - reference of arr - 0x112, then store[ i ] = reference to 0x112
At end:
arr - refer to 0x112 - value = [1, 2, 3, 4, 5, 6]
store[ ] = [ref(0x112), ref(0x112), ref(0x112), ref(0x112), ref(0x112), ref(0x112)]
Use this:
let arr = [1,2,3,4,5,6]
let store = [];
for(i = 0; i < arr.length; i++){
console.log(arr)
let newArr = arr.slice();
store.push(newArr)
arr.push(arr.shift())
}
console.log('store', JSON.stringify(store))
I have 2 array:
var arr1 = [1, 1, 2, 3, 4, 4];
var arr2 = [];
arr1 defines some groups, the group members are defined by the position. So the group members 0 and 1 belong to group 1,
member 2 belong to group 2,
member 3 belong to group 3,
the group member 4 and 5 belong to group 4.
I would like to load the groups with their members into another array, so that I looks like:
arr2 = [[0,1], [2], [3], [4,5]];
So group 1 has the members 0 and 1 and so on...
When doing:
for(i=0; i<arr1.length; i++){
arr2[arr1[i]] = i;
}
I get:
arr2 = [1: 1, 2: 2, 3: 3, 4: 5];
You've almost gotten it right, but you need to create an array for each index in arr2, and then append i to it:
for(i=0; i<arr1.length; i++){
arr2[arr1[i]] = arr2[arr1[i]] || []; // create array here if none exists yet
arr2[arr1[i]].push(i) // add i to array
}
Note that this produces an of-by-one difference from your desired solution (the first group will be at index 1, rather than index 0 in your solution), so you can shift the indicies of arr2 down by one:
for(i=0; i<arr1.length; i++){
arr2[arr1[i]-1] = arr2[arr1[i]-1] || [];
arr2[arr1[i]-1].push(i)
}
Note that you have not specified your desired behavior if there would be a gap in the arr2 result. In this case, the index remains unset in the output array.
You need to create new array for each group and then push values to it. See my comments in code below:
var arr1 = [1, 1, 2, 3, 4, 4];
var arr2 = [];
for(i=0; i<arr1.length; i++) {
// array indexes begins from 0, therefore arr2[arr1[i] - 1]
if (!arr2[arr1[i] - 1]) { // if array for this index does not exist
// then initialize new empty array
arr2[arr1[i] - 1] = [];
}
arr2[arr1[i] - 1].push(i); // push correct value to the array
}
var i;
var arr2 = [];
for(i=0; i<arr1.length; i++){
arr2[arr1[i]-1] = arr2[arr1[i]-1] || [];
arr2[arr1[i]-1].push(i);
}
I have two arrays of complex nexted objects that I'm looking for qualifying values within using loops and if statements as seen below. When I find a qualifying object, I need to filter that object out during the next go around of the loop. I'm trying to do that with an array as you can see here but it isn't working as the array starts over during each iteration of the loop. The following version is a simplified version of my code.
I want to update the values in array2 based on the if statement so that those values are not repeated in the nested loop. Instead my emptyArray remains empty instead of adding values from the array2 as elements of array2 are equal to elements of array.
To be clear, right now emptyArray remains empty and never filters array2. I'd like to see emptyArray collect value 2 at the start of the outer loop's second iteration then I'd like to see emptyArray collect value 4 at the start of the 4th iteration of the outer loop.
I'd want to filter each of these values from array2 as they become part of emptyArray so that they do not set off the if statement during the 6th and 8th iterations of the outer loop. I imagine that emptyArray = [2, 4] and array2 = [6, 8, 10] when the loops are finished.
Bottom line, I need emptyArray to collect the qualifying values and pass them back to var array2 for filtering as the loop processes. Remember this is a simplified version of the arrays, and underscore based solution would be very complicated for me to implement or for you to successfully suggest without much more detail.
My code:
var array = [1, 2, 3, 4, 1, 2, 3, 4];
var array2 = [2, 4, 6, 8, 10];
var emptyArray = [];
for (i = 0; i < array.length; i++){
var something = array[i];
var array2 = _.without(array2, emptyArray);
for (a = 0; a < array2.length; a++){
var value = array2[a];
if(something === value){
emptyArray.push(value);
break;
}
}
}
There are a few things wrong with your code, but the reason why you think that push isn't working is because you are overriding your array2 inside the loop.
The push never gets called because your for loop sees an empty array2 when you are doing var array2 = _.without(array2, emptyArray);
Basically var array2 = _.without(array2 /* this is empty, you just overrode it in this scope */, emptyArray); will always result in an empty array and your for loop will exit because length is array2.length === 0 from the start.
Also, you want to use _.difference instead of _.without
var array = [1, 2, 3, 4, 1, 2, 3, 4];
var array2 = [2, 4, 6, 8, 10];
var emptyArray = [];
for (var i = 0; i < array.length; i++) {
var something = array[i];
array2 = _.difference(array2, emptyArray);
for (var j = 0; j < array2.length; j++) {
var value = array2[j];
if (something === value) {
emptyArray.push(value);
break;
}
}
}
console.log("array2",array2);
console.log("emptyArray", emptyArray);
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/3.5.0/lodash.js"></script>
array2 [6, 8, 10]
emptyArray [2, 4]
var array = [1, 2, 3, 4, 1, 2, 3, 4];
var array2 = [2, 4, 6, 8, 10];
var emptyArray = [];
for (var i = 0; i < array.length; i++) {
var something = array[i];
for (var j = 0; j < array2.length; j++) {
var value = array2[j];
if (something === value) {
array2 = _.without(array2, value);
break;
}
}
}
I wanted to change the rows into columns of an array.
[
[1],
[1,2],
[1,2,3],
[4,2,3],
[4,5,3],
[4,5,6]
]
to
[
[1,1,1,4,4,4],
[2,2,2,5,5],
[3,3,3,6]
]
I tried
var res = [];
for(i in this.fields) {
for(j in this.fields[i].value) {
if(i === 0) res[j] = [];
res[j][i] = this.fields[i].value[j];
}
}
this gives me empty set.
Create this function:
function transpose(arr) {
return Object.keys(arr[0]).map(function (c) {
return arr.map(function (r) {
return r[c];
});
});
}
and then:
var transposedArray = transpose(originalArray);
What you're asking looks a little weird because you have different lengths and you're ignoring undefined values, but it is still achievable.
Don't use for..in loops for Array, use a normal for. Also, you'll need to know how many items you'll have in your new parent Array, which is the max of the lengths of the original child Arrays.
var arrR = [ // will refer to "down" and "across" as in this literal
[1],
[1, 2],
[1, 2, 3],
[4, 2, 3],
[4, 5, 3],
[4, 5, 6]
];
function r2c(arr) {
var arrC = [], // next get the longest sub-array length
x = Math.max.apply(Math, arr.map(function (e) {return e.length;})),
y = arr.length,
i, j;
for (i = 0; i < x; ++i) { // this is the loop "down"
arrC[i] = [];
for (j = 0; j < y; ++j) // and this is the loop "across"
if (i in arr[j])
arrC[i].push(arr[j][i]);
}
return arrC;
}
var arrC = r2c(arrR);
/* [
[1, 1, 1, 4, 4, 4],
[2, 2, 2, 5, 5],
[3, 3, 3, 6]
] */
You should still consider if you're happy with [[1], [1, 2], [1]] becoming [[1, 1, 1], [2]], which I would consider unexpected (the position of 2 is completely lost), but seems to be what you intend.
Similar to Pauls but doesn't need to get the max length first:
function transpose(arr) {
// Loop over arrays as long as one has values
// Arrays should be contiguous, may fail if sparse
for (var result = [], i=0, more; more; i++) {
more = false;
// Get the ith element of each array (if there is one)
for (var j=0, jLen=arr.length; j<jLen; j++) {
// Don't add missing members
if (arr[j].hasOwnProperty(i)) {
// Add array for result if not already there
result[i] = result[i] || [];
// Do transpose
result[i][j] = arr[j][i];
// Only keep going while there is data
more = true;
}
}
}
return result;
}
BTW, a fixed version of your original function is:
function transpose2(fields) {
// Make sure the result array is initialised
var res = [];
// Don't forget to keep counters local - declare them
// I've removed *this* as it's a plain function, use it if
// it's an instance method
for(var i in fields) {
// Values are read directly, there is no "value" accessor
for(var j in fields[i]) {
// Don't rely on order of enumeration - may not start at 0
if(!res[j]) res[j] = [];
// Do the transpose
res[j][i] = fields[i][j];
}
}
return res;
}
But as noted above, for..in is not liked for arrays, particularly as there are many libraries that extend built-ins like Array.prototype so you will traverse those properties too. But if you're cool with that, this is a good way to deal with sparse arrays. You can add a hasOwnProperty test to avoid inherited enumerables.
Note also that the order of enumeration isn't necessarily from '0' or in any particular order, hence changed way of initialising res[j].