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Looping and splitting Object

how to loop on object and split it by ":" into separate one
{
labs: [1,2,":",3,4],
level: [1,2,":",3,4]
}
Expected Output:
{
labs: [1,2],
level: [1,2]
}
{
labs: [3,4],
level : [3,4]
}

You can use itertools.groupby from the standard library to group the numbers that are not ":". With that you can use zip to pair off the groups.
from itertools import groupby
d = {
"labs": [1,2,":",3,4],
"level": [10,20,":",30,40]
}
groups = [[(k, list(g)) for b, g in groupby(v, key=lambda n:n != ':') if b]
for k, v in d.items()
]
list(map(dict, zip(*groups)))
# [
# {'labs': [1, 2], 'level': [10, 20]},
# {'labs': [3, 4], 'level': [30, 40]}
# ]
This should work with arbitrary data. For example with input like:
d = {
"labs": [1,2,":",3,4,":", 5, 6],
"level": [10,20,":",30,40,":",50, 60],
"others":[-1,-2,":",-3,-4,":",-5,-6]
}
You will get:
[{'labs': [1, 2], 'level': [10, 20], 'others': [-1, -2]},
{'labs': [3, 4], 'level': [30, 40], 'others': [-3, -4]},
{'labs': [5, 6], 'level': [50, 60], 'others': [-5, -6]}
]
But it does expect the lists to be the same length because the way zip() works. If that's not a good assumption you will need to decide what to do with uneven lists. In that case itertools.zip_longest() will probably be helpful.

Use javaScript to resolve this problem, maybe the code is not the best
const obj = {
labs: [1,2,":",3,4,":",5,6],
level: [1,2,":",3,4,":",7,8],
others: [1,2,":",3,4,":",9,10]
}
const format = (obj = {}, result = []) => {
const keys = Object.keys(obj);
for ( key of keys) {
const itemValues = obj[key].toString().split(':');
let tempRes = {}
itemValues.map((itemValue, index) => {
Object.assign(tempRes, {
[`${keys[index]}`]: itemValue.replace(/^(,)+|(,)+$/g, '').split(',').map(Number)// you can format value here
})
})
result.push(tempRes);
}
return result;
}
console.log(format(obj))
You will get
[
{ labs: [ 1, 2 ], level: [ 3, 4 ], others: [ 5, 6 ] },
{ labs: [ 1, 2 ], level: [ 3, 4 ], others: [ 7, 8 ] },
{ labs: [ 1, 2 ], level: [ 3, 4 ], others: [ 9, 10 ] }
]

Confused about pushing array into array, JavaScript

My code looks like this
var res = [];
var temp = [];
function Permutations(target, size) {
if (size === 0) {
res.push(temp);
console.log(res);
return;
}
for (let i = 0; i < target.length; i++) {
if (target[i] !== null) {
temp.push(target[i]);
target[i] = null;
Permutations(target, size - 1);
target[i] = temp.pop();
}
}
}
Permutations([1, 2, 3], 2);
console.log(res);
When I run my code, I can see my res stores each permutation as it is is being executed. However, when I log it outside the function, all the stored value disappeared.
[ [ 1, 2 ] ]
[ [ 1, 3 ], [ 1, 3 ] ]
[ [ 2, 1 ], [ 2, 1 ], [ 2, 1 ] ]
[ [ 2, 3 ], [ 2, 3 ], [ 2, 3 ], [ 2, 3 ] ]
[ [ 3, 1 ], [ 3, 1 ], [ 3, 1 ], [ 3, 1 ], [ 3, 1 ] ]
[ [ 3, 2 ], [ 3, 2 ], [ 3, 2 ], [ 3, 2 ], [ 3, 2 ], [ 3, 2 ] ]
[ [], [], [], [], [], [] ] // This is my console.log outside the function

The array temp holds is the same array throughout the complete execution of your code. And res.push(temp); adds this same array (not a copy of it) to your res array.
Here a related question about how Objects are handled in JavaScript: Is JavaScript a pass-by-reference or pass-by-value language?
So your code results in res having N times the same array.
You could copy the element stored in temp to a new array using [...temp], and push that to your res array.
var res = [];
var temp = [];
function Permutations(target, size) {
if (size === 0) {
res.push([...temp]);
return;
}
for (let i = 0; i < target.length; i++) {
if (target[i] !== null) {
temp.push(target[i]);
target[i] = null;
Permutations(target, size - 1);
target[i] = temp.pop();
}
}
}
Permutations([1, 2, 3], 2);
console.log(res);

matrix to array with every other row reversed

I have to convert matrix to array. For example
[
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
]
should convert to [1, 2, 3, 6, 5, 4, 7, 8, 9 ].
I write a code
function sortMatrix (matrix) {
let newArr = [];
for(let i = 0; i < matrix.length; i++)
{
newArr = newArr.concat(matrix[i]);
}
return newArr
}
but output is [1, 2, 3, 4, 5, 6, 7, 8, 9 ]. How can I do so that on the second line, it counts from the end

To fix your code - reverse sub-arrays at odd indexes.
Note: since Array.reverse() mutates the original array, we need to shallow clone it using array spread.
function sortMatrix(matrix) {
let newArr = [];
for (let i = 0; i < matrix.length; i++) {
const arr = matrix[i];
newArr = newArr.concat(i % 2 ? [...arr].reverse() : arr);
}
return newArr;
}
const matrix = [
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
];
const result = sortMatrix(matrix);
console.log(result);
Another option is to use Array.flatMap(), and reverse sub-arrays at odd indexes.
function sortMatrix(matrix) {
return matrix.flatMap((arr, i) => i % 2 ? [...arr].reverse() : arr);
}
const matrix = [
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
];
const result = sortMatrix(matrix);
console.log(result);

You can use reduce to create it.
const arr = [
[1,2,3],
[4,5,6],
[7,8,9]
]
const flatArray = (arr) => arr.reverse().reduce((acc, curr)=> {
acc.push(...curr.reverse());
return acc;
},[])
console.log(flatArray(arr))

You can directly use the flat function
array = [
[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ],
]
console.log(array.flat());

How to arrange items in an array in all possible ways?

Items in an array can be arranged in ascending order using sort() method in JavaScript but how to arrange them in all possible ways and show them in our web page.

You describe permutations, one way to implement it:
function permutations(arr, r=[]) {
if (arr.length === 0) {
console.log(r)
} else {
const first = arr[0]
for (let i = 0; i <= r.length; i++) {
permutations(arr.slice(1), r.slice(0, i).concat([first]).concat(r.slice(i)))
}
}
}
permutations([1, 2, 3])
OUTPUT
[ 3, 2, 1 ]
[ 2, 3, 1 ]
[ 2, 1, 3 ]
[ 3, 1, 2 ]
[ 1, 3, 2 ]
[ 1, 2, 3 ]

find and merge duplicate match with array in javascript

Following is an array of product which has been find and process on contact data and creates the possible array of merge occurrences.
In the following arrProduct, I would like to filter duplicate array and merge with existing array and finally create a unique array product called arrFinalProduct.
Array product
arrProduct =
[
[ 0 ],
[ 1, 2 ],
[ 2 ],
[ 3 ],
[ 4, 5, 6, 10 ],
[ 5, 6, 7, 11 ],
[ 6 ],
[ 7, 11 ],
[ 8 ],
[ 9 ],
[ 10 ],
[ 11 ],
[ 12 ],
[ 13, 14 ],
[ 14 ]
]
Final product
arrFinalProduct =
[
[ 0 ],
[ 1, 2 ],
[ 3 ],
[ 4, 5, 6, 7, 10, 11 ],
[ 8 ],
[ 9 ],
[ 12 ],
[ 13, 14 ]
]
arrProduct is the product array and arrFinalProduct is the final product array. The basic logic is that we require merging array, if any occurrences match found from an array.
Let's say value 0 is not found in any index in arrProduct so it does not merge and push into arrFinalProduct,
arrProduct value 2 is found on 1st and 2nd index so it may merge with 1st index and become [1,2] and delete 3rd index of [2].
arrProduct index 5 have two common value 5 and 6 and it also in index 6 so it may merge into 1 and become "4,5,6,7" and so on...
This process is going to process data recursively until I found unique value from an array. So it probably merges array horizontally.
Hope reader may get an enough idea.

Basically you could search for exisitent items in the result set and join the arrays with same items.
The order of items is the same as the appearance.
var array = [[0], [1, 2], [2], [3], [4, 5, 6, 10], [5, 6, 7, 11], [6], [7, 11], [8], [9], [10], [11], [12], [13, 14], [14]],
result = array.reduce(function (r, a) {
r.some(function (b, i, bb) {
if (a.some(c => b.includes(c))) {
bb[i] = [...new Set(b.concat(a))];
return true;
}
}) || r.push(a);
return r;
}, []);
console.log(result);
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