Related
This is a much simplified version of my real problem so ideally I'm looking a simple solution without builtin functions.
I have an array:
let data = [5,5,5,5,5];
I want to start at the second item and compare it to the first. If they are the same I want to increase by one so my array looks like this:
let data = [5,6,5,5,5];
I then want to compare the next array item to all the other ones I've checked and increase by 1 if there are any the same so the array would look like this:
let data = [5,6,6,5,5];
but now there are two values the same in the values I've checked, so I want to check for similar values in the array I've already checked and add one to the item at the current array index if the current value is the same as any that have come before. 6 has been used, so I want to change the array to look like this:
let data = [5,6,7,5,5];
And so on...
I've got this, but it's not producing the result I want:
let data = [5,5,5,5,5]
const looper = () => {
for (let i=-1; i<data.length; i++) {
console.log("i", i)
for (let j=1; j<data.length; j++) {
console.log("j", j)
if (data[i] <= data[i-1] ) {
console.log("DATA", i, data[i])
data[i] += i
console.log("DATA", i, data)
looper()
} else {
continue
}
}
}
}
looper()
You could take an object and take the reference to the largest value.
function fn(array) {
const
getValue = v => {
if (v in values) return getValue(values[v]);
values[v] = v + 1;
return v;
},
values = {};
return array.map(getValue)
}
console.log(...fn([5, 5, 5, 5, 5]));
console.log(...fn([5, 5, 4, 5, 5]));
console.log(...fn([5, 5, 6, 6, 4, 4]));
Funky other data set.
function fn(array) {
const
getValue = (values = {}) => v => {
if (v in values) return getValue(values)(values[v]);
values[v] = v + 1;
return v;
};
return Object
.entries(array.reduce((r, o) => {
Object.entries(o).forEach(([k, v]) => (r[k] ??= []).push(v));
return r;
}, {}))
.map(([k, v]) =>[k, v.map(getValue())])
.reduce((r, [k, a]) => a.map((v, i) => ({ ...r[i], [k]: v })), []);
}
const
data = [{ cx: 2, cy: 3 }, { cx: 2, cy: 3 }, { cx: 2, cy: 3 }, { cx: 2, cy: 3 }, { cx: 2, cy: 3 }],
result = fn(data);
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
A slightly better approach by using only a single loop of the data and keeping the values along with the key.
const
getValue = (values = {}) => {
const
fn = (k, v) => {
const key = [k, v].join('|');
if (key in values) return fn(k, values[key]);
values[key] = v + 1;
return v;
};
return fn;
},
data = [{ cx: 2, cy: 3 }, { cx: 2, cy: 3 }, { cx: 2, cy: 3 }, { cx: 2, cy: 3 }, { cx: 2, cy: 3 }],
getFn = getValue(),
result = data.map(o => Object.fromEntries(Object
.entries(o)
.map(([k, v]) => [k, getFn(k, v)])
));
console.log(result);
.as-console-wrapper { max-height: 100% !important; top: 0; }
Break this down into two parts:
Part 1:
Adjust all elements in array 2 based on array 1
Part 2:
Make numbers in array 2 unique: Here you need to iterate over your array continually, until no changes are found in an iteration.
const fun = ra => {
for (let i = 1; i < ra.length; i++) {
let k = 0;
while (k < i)
ra[i] === ra[k] ? (ra[i]++, k = 0) : k++;
}
return ra;
}
console.log(...fun([3, 3, 3]));
console.log(...fun([4, 3, 3]));
console.log(...fun([4, 3, 3, 5]));
I have an object name config in which i have "from" and "to".
const config = {
"0": {
id: 0,
from: 0,
to: 10,
hex: null
},
"1": {
id: 1,
from: 11,
to: 20,
hex: null
},
"2": {
id: 2,
from: 21,
to: 30,
hex: null
},
"3": {
id: 3,
from: 31,
to: 40,
hex: null
},
"4": {
id: 4,
from: 41,
to: 50,
hex: null
}
};
I have to check that now range will contradict with each other eg: form:0 => to:10 and from:5=> to:20
here the from of second is contradict because 5 lies between 0 to 10
i have tried following but doesn't full-fill my requirement
function found(conf) {
let isFound = false;
for (let obj in conf) {
for (let x in conf) {
if (
conf[obj].id !== conf[x].id &&
(conf[x].from >= conf[obj].from && conf[x].to <= conf[obj].from)
) {
console.log(conf[obj], conf[x]);
isFound = true;
break;
}
}
if (isFound) break;
}
return isFound;
}
console.log(found(config));
Create a single array by combining all ranges
const arr = Object.entries(config).map(([a, b]) => b).flatMap(({from, to}) => RANGE(from, to))
where RANGE is method which return array of given ranges:
const RANGE = (a,b) => Array.from((function*(x,y){
while (x <= y) yield x++;
})(a,b));
Then find duplicates in the given arr using the following function:
function findDuplicate(array) {
var object = {};
var result = [];
array.forEach(function(item) {
if (!object[item]) object[item] = 0;
object[item] += 1;
});
for (var prop in object) {
if (object[prop] >= 2) {
result.push(prop);
}
}
return result;
}
const duplicates = findDuplicate(arr)
Then finally check duplicates.length
Try renaming your variables so they make sense.
Your logic is: IDs don't match and inner is after outer, but before outer's from.
There will never be a case where this will return true.
const config = {
"0": { id: 0, from: 0, to: 10, hex: null },
"1": { id: 1, from: 11, to: 20, hex: null },
"2": { id: 2, from: 21, to: 30, hex: null },
"3": { id: 3, from: 31, to: 40, hex: null },
"4": { id: 4, from: 41, to: 50, hex: null }
};
console.log(found(config));
function found(conf) {
for (let outer in conf) {
for (let inner in conf) {
let idsDontMatch = conf[outer].id !== conf[inner].id;
let innerFromGteOuterFrom = conf[inner].from >= conf[outer].from;
let innerToLteOuterFrom = conf[inner].to <= conf[outer].from;
let innerAfterOuterButBeforeOuterFrom = innerFromGteOuterFrom && innerToLteOuterFrom;
if (idsDontMatch && innerAfterOuterButBeforeOuterFrom) {
console.log(conf[outer], conf[inner]);
return true;
}
}
}
return false;
}
.as-console-wrapper { top: 0; max-height: 100% !important; }
I have a JSON file as follows:
[
{
"dog": "lmn",
"tiger": [
{
"bengoltiger": {
"height": {
"x": 4
}
},
"indiantiger": {
"paw": "a",
"foor": "b"
}
},
{
"bengoltiger": {
"width": {
"a": 8
}
},
"indiantiger": {
"b": 3
}
}
]
},
{
"dog": "pqr",
"tiger": [
{
"bengoltiger": {
"width": {
"m": 3
}
},
"indiantiger": {
"paw": "a",
"foor": "b"
}
},
{
"bengoltiger": {
"height": {
"n": 8
}
},
"indiantiger": {
"b": 3
}
}
],
"lion": 90
}
]
I want to transform this to obtain all possible properties of any object at any nesting level. For arrays, the first object should contain all the properties. The values are trivial, but the below solution considers the first encountered value for any property. (For ex. "lmn" is preserved for the "dog" property)
Expected output:
[
{
"dog": "lmn",
"tiger": [
{
"bengoltiger": {
"height": {
"x": 4,
"n": 8
},
"width": {
"a": 8,
"m": 3
}
},
"indiantiger": {
"paw": "a",
"foor": "b",
"b": 3
}
}
],
"lion": 90
}
]
Here's a recursive function I tried before this nesting problem struck me
function consolidateArray(json) {
if (Array.isArray(json)) {
const reference = json[0];
json.forEach(function(element) {
for (var key in element) {
if (!reference.hasOwnProperty(key)) {
reference[key] = element[key];
}
}
});
json.splice(1);
this.consolidateArray(json[0]);
} else if (typeof json === 'object') {
for (var key in json) {
if (json.hasOwnProperty(key)) {
this.consolidateArray(json[key]);
}
}
}
};
var json = [
{
"dog": "lmn",
"tiger": [
{
"bengoltiger": {
"height": {
"x": 4
}
},
"indiantiger": {
"paw": "a",
"foor": "b"
}
},
{
"bengoltiger": {
"width": {
"a": 8
}
},
"indiantiger": {
"b": 3
}
}
]
},
{
"dog": "pqr",
"tiger": [
{
"bengoltiger": {
"width": {
"m": 3
}
},
"indiantiger": {
"paw": "a",
"foor": "b"
}
},
{
"bengoltiger": {
"height": {
"n": 8
}
},
"indiantiger": {
"b": 3
}
}
],
"lion": 90
}
];
consolidateArray(json);
alert(JSON.stringify(json, null, 2));
General logic using this new JNode IIFE with comments - ask someone more clever if you do not understand something as me ;-)
And level starts from 1 as there is no root object #start.
var json;
function DamnDemo() {
json = DemoJSON();
var it = new JNode(json), it2 = it;
var levelKeys = []; /* A bit crazy structure:
[
levelN:{
keyA:[JNode, JNode,...],
keyB:[JNode, JNode,...],
...
},
levelM:...
]
*/
do {
var el = levelKeys[it.level]; // array of level say LevelN or undefined
el = levelKeys[it.level] = el || {}; // set 2 empty it if does not exist
el = el[it.key] = el[it.key] || []; // key array in say levelN
el.push(it); // save current node indexing by level, key -> array
} while (it = it.DepthFirst()) // traverse all nodes
for(var l1 in levelKeys) { // let start simply by iterating levels
l2(levelKeys[l1]);
}
console.log(JSON.stringify(json, null, 2));
}
function l2(arr) { // fun starts here...
var len = 0, items = []; // size of arr, his items to simple Array
for(var ln in arr) { // It's a kind of magic here ;-) Hate recursion, but who want to rewrite it ;-)
if (arr[ln] instanceof JNode) return 1; // End of chain - our JNode for traverse of length 1
len += l2(arr[ln]);
items.push(arr[ln]);
}
if (len == 2) { // we care only about 2 items to move (getting even 3-5)
//console.log(JSON.stringify(json));
if (!isNaN(items[0][0].key) || (items[0][0].key == items[1][0].key)) { // key is number -> ignore || string -> must be same
console.log("Keys 2B moved:", items[0][0].key, items[1][0].key, "/ level:", items[0][0].level);
var src = items[1][0]; // 2nd similar JNode
moveMissing(items[0][0].obj, src.obj); // move to 1st
//console.log(JSON.stringify(json));
if (src.level == 1) { // top level cleaning
delete src.obj;
delete json[src.key]; // remove array element
if (!json[json.length-1]) json.length--; // fix length - hope it was last one (there are other options, but do not want to overcomplicate logic)
} else {
var parent = src.parent;
var end = 0;
for(var i in parent.obj) {
end++;
if (parent.obj[i] == src.obj) { // we found removed in parent's array
delete src.obj; // delete this empty object
delete parent.obj[i]; // and link on
end = 1; // stupid marker
}
}
if (end == 1 && parent.obj instanceof Array) parent.obj.length--; // fix length - now only when we are on last element
}
} else console.log("Keys left:", items[0][0].key, items[1][0].key, "/ level:", items[0][0].level); // keys did not match - do not screw it up, but report it
}
return len;
}
function moveMissing(dest, src) {
for(var i in src) {
if (src[i] instanceof Object) {
if (!dest[i]) { // uff object, but not in dest
dest[i] = src[i];
} else { // copy object over object - let it bubble down...
moveMissing(dest[i], src[i]);
}
delete src[i];
} else { // we have value here, check if it does not exist, move and delete source
if (!dest[i]) {
dest[i] = src[i];
delete src[i];
}
}
}
}
// JSON_Node_Iterator_IIFE.js
'use strict';
var JNode = (function (jsNode) {
function JNode(json, parent, pred, key, obj, fill) {
var node, pred = null;
if (parent === undefined) {
parent = null;
} else if (fill) {
this.parent = parent;
this.pred = pred;
this.node = null;
this.next = null;
this.key = key;
this.obj = obj;
return this;
}
var current;
var parse = (json instanceof Array);
for (var child in json) {
if (parse) child = parseInt(child);
var sub = json[child];
node = new JNode(null, parent, pred, child, sub, true);
if (pred) {
pred.next = node;
node.pred = pred;
}
if (!current) current = node;
pred = node;
}
return current;
}
JNode.prototype = {
get hasNode() {
if (this.node) return this.node;
return (this.obj instanceof Object);
},
get hasOwnKey() { return this.key && (typeof this.key != "number"); },
get level() {
var level = 1, i = this;
while(i = i.parent) level++;
return level;
},
Down: function() {
if (!this.node && this.obj instanceof Object) {
this.node = new JNode(this.obj, this);
}
return this.node;
},
Stringify: function() { // Raw test stringify - #s are taken same as strings
var res;
if (typeof this.key == "number") {
res = '[';
var i = this;
do {
if (i.node) res += i.node.Stringify();
else res += "undefined";
i = i.next;
if (i) res += ','
} while(i);
res += ']';
} else {
res = '{' + '"' + this.key + '":';
res += (this.node?this.node.Stringify():this.hasNode?"undefined":'"'+this.obj+'"');
var i = this;
while (i = i.next) {
res += ',' + '"' + i.key + '":';
if (i.node) res += i.node.Stringify();
else {
if (i.obj instanceof Object) res += "undefined";
else res += '"' + i.obj + '"';
}
};
res += '}';
}
return res;
},
DepthFirst: function () {
if (this == null) return 0; // exit sign
if (this.node != null || this.obj instanceof Object) {
return this.Down(); // moved down
} else if (this.next != null) {
return this.next;// moved right
} else {
var i = this;
while (i != null) {
if (i.next != null) {
return i.next; // returned up & moved next
}
i = i.parent;
}
}
return 0; // exit sign
}
}
return JNode;
})();
// Fire test
DamnDemo();
function DemoJSON() {
return [
{
"dog": "lmn",
"tiger": [
{
"bengoltiger": {
"height": {
"x": 4
}
},
"indiantiger": {
"paw": "a",
"foor": "b"
}
},
{
"bengoltiger": {
"width": {
"a": 8
}
},
"indiantiger": {
"b": 3
}
}
]
},
{
"dog": "pqr",
"tiger": [
{
"bengoltiger": {
"width": {
"m": 3
}
},
"indiantiger": {
"paw": "a",
"foor": "b"
}
},
{
"bengoltiger": {
"height": {
"n": 8
}
},
"indiantiger": {
"b": 3
}
}
],
"lion": 90
}
]
;}
This was an interesting problem. Here's what I came up with:
// Utility functions
const isInt = Number.isInteger
const path = (ps = [], obj = {}) =>
ps .reduce ((o, p) => (o || {}) [p], obj)
const assoc = (prop, val, obj) =>
isInt (prop) && Array .isArray (obj)
? [... obj .slice (0, prop), val, ...obj .slice (prop + 1)]
: {...obj, [prop]: val}
const assocPath = ([p = undefined, ...ps], val, obj) =>
p == undefined
? obj
: ps.length == 0
? assoc(p, val, obj)
: assoc(p, assocPath(ps, val, obj[p] || (obj[p] = isInt(ps[0]) ? [] : {})), obj)
// Helper functions
function * getPaths(o, p = []) {
if (Object(o) !== o || Object .keys (o) .length == 0) yield p
if (Object(o) === o)
for (let k of Object .keys (o))
yield * getPaths (o[k], [...p, isInt (Number (k)) ? Number (k) : k])
}
const canonicalPath = (path) =>
path.map (n => isInt (Number (n)) ? 0 : n)
const splitPaths = (xs) =>
Object .values ( xs.reduce (
(a, p, _, __, cp = canonicalPath (p), key = cp .join ('\u0000')) =>
({...a, [key]: a [key] || {canonical: cp, path: p} })
, {}
))
// Main function
const canonicalRep = (data) => splitPaths ([...getPaths (data)])
.reduce (
(a, {path:p, canonical}) => assocPath(canonical, path(p, data), a),
Array.isArray(data) ? [] : {}
)
// Test
const data = [{"dog": "lmn", "tiger": [{"bengoltiger": {"height": {"x": 4}}, "indiantiger": {"foor": "b", "paw": "a"}}, {"bengoltiger": {"width": {"a": 8}}, "indiantiger": {"b": 3}}]}, {"dog": "pqr", "lion": 90, "tiger": [{"bengoltiger": {"width": {"m": 3}}, "indiantiger": {"foor": "b", "paw": "a"}}, {"bengoltiger": {"height": {"n": 8}}, "indiantiger": {"b": 3}}]}]
console .log (
canonicalRep (data)
)
The first few functions are plain utility functions that I would keep in a system library. They have plenty of uses outside this code:
isInt is simply a first-class function alias to Number.isInteger
path finds the nested property of an object along a given pathway
path(['b', 1, 'c'], {a: 10, b: [{c: 20, d: 30}, {c: 40}], e: 50}) //=> 40
assoc returns a new object cloning your original, but with the value of a certain property set to or replaced with the supplied one.
assoc('c', 42, {a: 1, b: 2, c: 3, d: 4}) //=> {a: 1, b: 2, c: 42, d: 4}
Note that internal objects are shared by reference where possible.
assocPath does this same thing, but with a deeper path, building nodes as needed.
assocPath(['a', 'b', 1, 'c', 'd'], 42, {a: {b: [{x: 1}, {x: 2}], e: 3})
//=> {a: {b: [{x: 1}, {c: {d: 42}, x: 2}], e: 3}}
Except for isInt, these borrow their APIs from Ramda. (Disclaimer: I'm a Ramda author.) But these are unique implementations.
The next function, getPaths, is an adaptation of one from another SO answer. It lists all the paths in your object in the format used by path and assocPath, returning an array of values which are integers if the relevant nested object is an array and strings otherwise. Unlike the function from which is was borrowed, it only returns paths to leaf values.
For your original object, it returns an iterator for this data:
[
[0, "dog"],
[0, "tiger", 0, "bengoltiger", "height", "x"],
[0, "tiger", 0, "indiantiger", "foor"],
[0, "tiger", 0, "indiantiger", "paw"],
[0, "tiger", 1, "bengoltiger", "width", "a"],
[0, "tiger", 1, "indiantiger", "b"],
[1, "dog"],
[1, "lion"],
[1, "tiger", 0, "bengoltiger", "width", "m"],
[1, "tiger", 0, "indiantiger", "foor"],
[1, "tiger", 0, "indiantiger", "paw"],
[1, "tiger", 1, "bengoltiger", "height", "n"],
[1, "tiger", 1, "indiantiger", "b"]
]
If I wanted to spend more time on this, I would replace that version of getPaths with a non-generator version, just to keep this code consistent. It shouldn't be hard, but I'm not interested in spending more time on it.
We can't use those results directly to build your output, since they refer to array elements beyond the first one. That's where splitPaths and its helper canonicalPath come in. We create the canonical paths by replacing all integers with 0, giving us a data structure like this:
[{
canonical: [0, "dog"],
path: [0, "dog"]
}, {
canonical: [0, "tiger", 0, "bengoltiger", "height", "x"],
path: [0, "tiger", 0, "bengoltiger", "height", "x"]
}, {
canonical: [0, "tiger", 0, "indiantiger", "foor"],
path: [0, "tiger", 0, "indiantiger", "foor"]
}, {
canonical: [0, "tiger", 0, "indiantiger", "paw"],
path: [0, "tiger", 0, "indiantiger", "paw"]
}, {
canonical: [0, "tiger", 0, "bengoltiger", "width", "a"],
path: [0, "tiger", 1, "bengoltiger", "width", "a"]
}, {
canonical: [0, "tiger", 0, "indiantiger", "b"],
path: [0, "tiger", 1, "indiantiger", "b"]
}, {
canonical: [0, "lion"],
path: [1, "lion"]
}, {
canonical: [0, "tiger", 0, "bengoltiger", "width", "m"],
path: [1, "tiger", 0, "bengoltiger", "width", "m"]
}, {
canonical: [0, "tiger", 0, "bengoltiger", "height", "n"],
path: [1, "tiger", 1, "bengoltiger", "height", "n"]
}]
Note that this function also removes duplicate canonical paths. We originally had both [0, "tiger", 0, "indiantiger", "foor"] and [1, "tiger", 0, "indiantiger", "foor"], but the output only contains the first one.
It does this by storing them in an object under a key created by joining the path together with the non-printable character \u0000. This was the easiest way to accomplish this task, but there is an extremely unlikely failure mode possible 1 so if we really wanted we could do a more sophisticated duplicate checking. I wouldn't bother.
Finally, the main function, canonicalRep builds a representation out of your object by calling splitPaths and folding over the result, using canonical to say where to put the new data, and applying the path function to your path property and the original object.
Our final output, as requested, looks like this:
[
{
dog: "lmn",
lion: 90,
tiger: [
{
bengoltiger: {
height: {
n: 8,
x: 4
},
width: {
a: 8,
m: 3
}
},
indiantiger: {
b: 3,
foor: "b",
paw: "a"
}
}
]
}
]
What's fascinating for me is that I saw this as an interesting programming challenge, although I couldn't really imagine any practical uses for it. But now that I've coded it, I realize it will solve a problem in my current project that I'd put aside a few weeks ago. I will probably implement this on Monday!
Update
Some comments discuss a problem with a subsequent empty value tries to override a prior filled value, causing a loss in data.
This version attempts to alleviate this with the following main function:
const canonicalRep = (data) => splitPaths ([...getPaths (data)])
.reduce (
(a, {path: p, canonical}, _, __, val = path(p, data)) =>
isEmpty(val) && !isEmpty(path(canonical, a))
? a
: assocPath(canonical, val, a),
Array.isArray(data) ? [] : {}
)
using a simple isEmpty helper function:
const isEmpty = (x) =>
x == null || (typeof x == 'object' && Object.keys(x).length == 0)
You might want to update or expand this helper in various ways.
My first pass worked fine with the alternate data supplied, but not when I switched the two entries in the outer array. I fixed that, and also made sure that an empty value is kept if it's not overridden with actual data (that's the z property in my test object.)
I believe this snippet solves the original problem and the new one:
// Utility functions
const isInt = Number.isInteger
const path = (ps = [], obj = {}) =>
ps .reduce ((o, p) => (o || {}) [p], obj)
const assoc = (prop, val, obj) =>
isInt (prop) && Array .isArray (obj)
? [... obj .slice (0, prop), val, ...obj .slice (prop + 1)]
: {...obj, [prop]: val}
const assocPath = ([p = undefined, ...ps], val, obj) =>
p == undefined
? obj
: ps.length == 0
? assoc(p, val, obj)
: assoc(p, assocPath(ps, val, obj[p] || (obj[p] = isInt(ps[0]) ? [] : {})), obj)
const isEmpty = (x) =>
x == null || (typeof x == 'object' && Object.keys(x).length == 0)
function * getPaths(o, p = []) {
if (Object(o) !== o || Object .keys (o) .length == 0) yield p
if (Object(o) === o)
for (let k of Object .keys (o))
yield * getPaths (o[k], [...p, isInt (Number (k)) ? Number (k) : k])
}
// Helper functions
const canonicalPath = (path) =>
path.map (n => isInt (Number (n)) ? 0 : n)
const splitPaths = (xs) =>
Object .values ( xs.reduce (
(a, p, _, __, cp = canonicalPath (p), key = cp .join ('\u0000')) =>
({...a, [key]: a [key] || {canonical: cp, path: p} })
, {}
))
// Main function
const canonicalRep = (data) => splitPaths ([...getPaths (data)])
.reduce (
(a, {path: p, canonical}, _, __, val = path(p, data)) =>
isEmpty(val) && !isEmpty(path(canonical, a))
? a
: assocPath(canonical, val, a),
Array.isArray(data) ? [] : {}
)
// Test data
const data1 = [{"dog": "lmn", "tiger": [{"bengoltiger": {"height": {"x": 4}}, "indiantiger": {"foor": "b", "paw": "a"}}, {"bengoltiger": {"width": {"a": 8}}, "indiantiger": {"b": 3}}]}, {"dog": "pqr", "lion": 90, "tiger": [{"bengoltiger": {"width": {"m": 3}}, "indiantiger": {"foor": "b", "paw": "a"}}, {"bengoltiger": {"height": {"n": 8}}, "indiantiger": {"b": 3}}]}]
const data2 = [{"d": "Foreign Trade: Export/Import: Header Data", "a": "false", "f": [{"g": "TRANSPORT_MODE", "i": "2"}, {"k": "System.String", "h": "6"}], "l": "true"}, {"a": "false", "f": [], "l": "false", "z": []}]
const data3 = [data2[1], data2[0]]
// Demo
console .log (canonicalRep (data1))
console .log (canonicalRep (data2))
console .log (canonicalRep (data3))
.as-console-wrapper {max-height: 100% !important; top: 0}
Why not change assoc?
This update grew out of discussion after I rejected an edit attempt to do the same sort of empty-checking inside assoc. I rejected that as too far removed from the original attempt. When I learned what it was supposed to do, I knew that what had to be changed was canonicalRep or one of its immediate helper functions.
The rationale is simple. assoc is a general-purpose utility function designed to do a shallow clone of an object, changing the named property to the new value. This should not have complex logic regarding whether the value is empty. It should remain simple.
By introducing the isEmpty helper function, we can do all this with only a minor tweak to canonicalRep.
1That failure mode could happen if you had certain nodes containing that separator, \u0000. For instance, if you had paths [...nodes, "abc\u0000", "def", ...nodes] and [...nodes, "abc", "\u0000def", ...nodes], they would both map to "...abc\u0000\u0000def...". If this is a real concern, we could certainly use other forms of deduplication.
In the below function I am attempting to get an output which resembles this:
[[1,1,1,1],[2,2,2], 4,5,10,[20,20], 391, 392,591].
I can see that the problem I have embedded is that I am always adding the temp array with a push to the functions return, as a result, all of the individual numbers apart from the last number in the for each function are being pushed into the target array with the array object also.
I feel as though I need a further conditonal check but for the life of me I am unable to come up with solution which works.
Any suggestions would be much appreciated.
const sortme = (unsortedArr)=> {
let tempArr = [];
let outputArr = [];
const reorderedArr = unsortedArr.sort((a,b) => a-b);
reorderedArr.forEach((number, i) => {
if ((i === 0) || (reorderedArr[i] === reorderedArr[i-1])) {
tempArr.push(number);
}
else {
outputArr.push(tempArr);
tempArr = [];
tempArr.push(number);
}
})
outputArr.push(tempArr[0]);
return outputArr;
}
const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
sortme(unsortedArr);
i would make a deduped copy and .map() it to transform the values into arrays containing values from the original ( sorted ) array that you get using a .forEach :
const unsortedArr = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20];
const sortMe = (arr) => {
arr = arr.sort((a, b) => a - b);
// a short way to dedupe an array
// results in : 1, 2, 4, 5, 10, 20, 391, 392, 591
let dedupe = [...new Set(arr)];
let tmpArr;
return dedupe.map(e => {
tmpArr = []; // empty tmpArr on each iteration
// for each element of the deduped array, look for matching elements in the original one and push them in the tmpArr
arr.forEach(a => {
if (a === e)
tmpArr.push(e);
})
if(tmpArr.length === 1)
return tmpArr[0]; // in case you have [4] , just return the 4
else
return tmpArr; // in case you have [1,1,1,1]
// shorthand for the if/else above
// return tmpArr.length === 1 ? tmpArr[0] : tmpArr;
});
}
const result = sortMe(unsortedArr);
console.log(result);
This should work (using reduce):
const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
let lastValue = null;
var newArr = unsortedArr.sort((a,b) => a-b).reduce((acc, value) => {
if (acc.length == 0 || ((acc.length > 0 || !acc[acc.length-1].length) && lastValue !== value)) {
acc.push(value);
} else if (acc.length > 0 && lastValue === value) {
acc[acc.length-1] = (acc[acc.length-1].length ? acc[acc.length-1].concat([value]): [value, value]);
}
lastValue = value;
return acc;
}, []);
console.log(newArr);
And another approach, just for fun:
const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
var arr = unsortedArr.sort((a,b) => a-b).reduce((acc, value) => {
if (acc.length > 0 && acc[acc.length-1].includes(value)) {
acc[acc.length-1].push(value);
} else {
acc.push([value])
}
return acc;
}, []).map((v) => v.length > 1 ? v: v[0]);
console.log(arr);
I hope the below one is quite simple;
function findSame(pos, sortedArr){
for(let i =pos; i<sortedArr.length; i++){
if(sortedArr[i] !== sortedArr[pos]){
return i
}
}
}
function clubSameNumbers(unsortedArr){
let sortedArr = unsortedArr.sort((a,b)=>a-b)
//[ 1, 1, 1, 1, 2, 2, 2, 4, 5, 10, 20, 20, 391, 392, 591 ]
let result = []
for(let i = 0; i < sortedArr.length; i = end){
let start = i
var end = findSame(i, sortedArr)
let arr = sortedArr.slice(i, end)
arr.length > 1 ? result.push(arr) : result.push(...arr)
}
return result
}
console.log(clubSameNumbers([1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]))
//[ [ 1, 1, 1, 1 ], [ 2, 2, 2 ], 4, 5, 10, [ 20, 20 ], 391, 392, 591 ]
I have the following javascript code which produces the desired results, i.e. returns both the 3rd and 4th objects in objectsArray since they both contain the max distance. However, I'm wondering if there is a way to not have to retype the name of the array when calling objectsArray.filter? I'm not trying to be lazy, just avoiding redundancy and the possibility of introducing a typo.
function meetsMax(obj) {
return obj.distance === Math.max.apply(Math, this.map(function(o) { return o.distance; }));
}
const objectsArray = [{ "distance": 1, "name": "first" }, { "distance": 2, "name": "second" }, { "distance": 3, "name": "third" }, { "distance": 3, "name": "fourth" }];
const objMax = objectsArray.filter(meetsMax, objectsArray);
console.log("objMax:", objMax);
I certainly wouldn't mind any other pointers on making the code more efficient and performant.
Function calls in JavaScript have some overhead, so native code is more efficient and performant:
var a = [ { "distance": 1, "name": "first" }, { "distance": 2, "name": "second" },
{ "distance": 3, "name": "third" }, { "distance": 3, "name": "fourth" } ]
for (var o = a[0], objMax = [o], m = o.distance, d, i = 1; i < a.length; i++)
if ((d = (o = a[i]).distance) > m) { objMax = [o]; m = d }
else if (d === m) objMax[objMax.length] = o
console.log(JSON.stringify(objMax))
There are also shorter and less efficient alternatives:
var a = [ { "distance": 1, "name": "first" }, { "distance": 2, "name": "second" },
{ "distance": 3, "name": "third" }, { "distance": 3, "name": "fourth" } ]
var d, b = []; a.forEach(o => (b[d = o.distance] = b[d] || []).push(o))
console.log(JSON.stringify(b[b.length - 1]))
Why don't you use for loop? It will be faster than your code.
"use strict";
let start = performance.now();
for (let z = 0; z < 1000; z++) {
function meetsMax(obj) {
return obj.distance === Math.max.apply(Math, this.map(function(o) { return o.distance; }));
}
const objectsArray = [{ "distance": 1, "name": "first" }, { "distance": 2, "name": "second" }, { "distance": 3, "name": "third" }, { "distance": 3, "name": "fourth" }];
const objMax = objectsArray.filter(meetsMax, objectsArray);
}
let fin = performance.now() - start;
console.log(fin); // 3.25ms
"use strict";
let start = performance.now();
for (let z = 0; z < 1000; z++) {
let a = [{ "distance": 1, "name": "first" }, { "distance": 2, "name": "second" }, { "distance": 3, "name": "third" }, { "distance": 3, "name": "fourth" }];
let maxDistance = 0;
let result = [];
for (let i = 0, max = a.length; i < max; i++) {
if (a[i].distance > maxDistance) {
maxDistance = a[i].distance;
}
}
for (let i = 0, max = a.length; i < max; i++) {
if (a[i].distance === maxDistance) {
result.push(a[i]);
}
}
}
let fin = performance.now() - start;
console.log(fin); // 1.28ms
.filter passes three arguments to the array: the current value, the index of the current value and the array itself. So you can change your filter function to:
function meetsMax(obj, index, objectsArray) {
return obj.distance === Math.max.apply(Math, objectsArray.map(function(o) { return o.distance; }));
}
and call .filter with
objectsArray.filter(meetsMax);
Always read the documentation of the functions you are using.
I certainly wouldn't mind any other pointers on making the code more efficient and performant.
If you, compute the maximum distance only once instead of in every iteration of the array. E.g. you could do:
function filterMax(arr, extractor) {
const max = arr.reduce(function(max, item) {
return max < extractor(item) ? extractor(item) : max;
}, extractor(arr[0]));
return arr.filter(function(item) {
return extractor(item) === max;
});
}
and call it as
filterMax(objectsArray, function(obj) { return obj.distance; });
function filterMax(arr, extractor) {
const max = arr.reduce(function(max, item) {
return max < extractor(item) ? extractor(item) : max;
}, extractor(arr[0]));
return arr.filter(function(item) {
return extractor(item) === max;
});
}
const objectsArray = [{ "distance": 1, "name": "first" }, { "distance": 2, "name": "second" }, { "distance": 3, "name": "third" }, { "distance": 3, "name": "fourth" }];
console.log(filterMax(objectsArray, function(obj) {
return obj.distance;
}));
According to MDN's Array.prototype.filter(), the array name is an optional override to the internal value of this.
So to answer the original question:
I'm wondering if there is a way to not have to retype the name of the array when calling objectsArray.filter?
Yes, you can safely leave it out.
var filter = function(x) { if (x > 5) return true; };
var arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10];
alert(arr.filter(filter).join(","));
or even simpler (albeit harder to read):
alert([ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10].filter(function(x) { if (x > 5) return true; }));
You specifically asked about an object, but you're not filtering objects, you're filtering an array of objects, so same applies.
console.log([ {foo: 1}, {foo: 2}, {foo: 3}, {foo: 4}, {foo: 5}, {foo: 6}, {foo: 7}, {foo: 8}, {foo: 9}, {foo: 10}].filter(function(x) { if (x.foo > 5) return true; }));