Develop Reference

Make sure no duplicate directory paths exist

I'm writing a script wherein the user selects directories, which are then stored in an array property, so that they can be recursively crawled.
{
"archives": [
"C:\\AMD\\Packages",
"C:\\Users",
"C:\\Windows",
"D:\\",
"E:\\Pictures\\Birthday"
]
}
I obviously don't want to be storing duplicate paths or paths that are contained by other paths. For example, if the user were to select a new folder to add to the array, E:\\Pictures, then E:\\Pictures\\Birthday would be discarded and replaced by it since E:\\Pictures contains E:\\Pictures\\Birthday.
{
"archives": [
"C:\\AMD\\Packages",
"C:\\Users",
"C:\\Windows",
"D:\\",
"E:\\Pictures"
]
}
I know this can be done by parsing all of the values being considered (i.e. ['C:', 'AMD', 'Packages'], [...], ... etc) and then comparing them all to one another. However, this seems extremely intensive, especially if the array of paths grows bigger and the directory paths are longer.
You could also do it by comparing the strings with includes. For example, if A includes B or B includes A, split them, and discard the one with a longer length.
for (const dir of dirs){
if (newPath.includes(dir) || dir.includes(newPath)){
if (newPath.split('\\') < dir.split('\\')){
// remove dir from json object and replace it with newPath
}
} else {
pathArray.push(dir)
}
}
After reading one of the answers below, I just realized that the includes method runs into the issue of comparing similar, yet unique paths i.e. C:\Users and C:\User.
Although there's gotta be a better way to do this??

This function will give you your desired results. It first looks to see if the parent of the path exists in the archives, and if so, does nothing. If it doesn't, it then removes any children of the path and then inserts the new path.
Update
I've added a delim input to the function to make it usable for unix/MacOS style filenames as well.
let data = {
"archives": [
"C:\\AMD\\Packages",
"C:\\Users",
"C:\\Windows",
"D:\\",
"E:\\Pictures"
]
};
const add_to_archives = (path, data, delim) => {
// does the parent of this path already exist? if so, nothing to do
if (data.archives.reduce((c, v) =>
c || path.indexOf(v.slice(-1) == delim ? v : (v + delim)) === 0, false)) return data;
// not found. remove any children of this path
data.archives = data.archives.filter(v => v.indexOf(path.slice(-1) == delim ? path : (path + delim)) !== 0);
// and add the new path
data.archives.push(path);
return data;
}
add_to_archives("E:\\Pictures\\Too", data, "\\");
console.log(data);
add_to_archives("E:\\PicturesToo", data, "\\");
console.log(data);
add_to_archives("D:\\Documents", data, "\\");
console.log(data);
add_to_archives("C:\\AMD", data, "\\");
console.log(data);
data = {
"archives": [
"/var/www/html/site",
"/etc",
"/usr/tim",
"/bin"
]
};
add_to_archives("/var/www/html/site2", data, "/");
console.log(data);
add_to_archives("/etc/conf.d", data, "/");
console.log(data);
add_to_archives("/usr", data, "/");
console.log(data);
add_to_archives("/var/www/html", data, "/");
console.log(data);
.as-console-wrapper {
max-height: 100% !important;
}

We can approach the problem by using a prefix tree
The purpose is to limit the number of paths we check for inclusion or "containment".
That approach may be useful if you have a lot of siblings (tree traversal + lookup as key for each folder).
It is overkill if you often have a root folder specified in archives
algorithm
tree = {}
foreach path
split the path in folders (one may iterate with substring but it is worth it?)
try to match folders of that path while traversing the tree
if you encounter a stop node, skip to next path
if not,
if your path end on an existing node
mark that node as a stop node
drop the children of that node (you can let them be, though)
else
include the remaining folders of the path as node in tree
mark the last node as a stop node
Implem
Note that implem below will fail if path includes a folder named "stop". By subjective order of preference
Use Map and Symbol('stop')
or a real tree (at least do not store folders alongside the boolean stop)
do not suppose any stop node and always drop children if you manage to reach the end of your path
Hope no one tries to outsmart you and rename stop as some obscure -folder will not exist- lolol_xxzz9_stop
function nodupes(archives){
tree = {};
archives.forEach(arch=>{
const folders = arch.split('\\');
folders.splice(1,1);
//case of empty string such as D:\\\
if(folders[folders.length-1].length==0){folders.pop();}
let cur = tree;
let dropped = false;
let lastFolderIndex = 0;
let ok = folders.every((folder,i)=>{
if(cur[folder]){
if(cur[folder].stop){
dropped = true;
return false;
}
cur = cur[folder];
return true;
}
cur[folder] = {}
cur = cur[folder];
lastFolderIndex = i;
return true;
});
if(ok){
cur.stop = true;
//delete (facultatively) the subfolders
if(lastFolderIndex < folders.length-1){
console.log('cleanup', folders, 'on node', cur)
Object.keys(cur).forEach(k=>{
if(k != 'stop'){
delete cur[k];
}
})
}
}
});
//console.log('tree', JSON.stringify(tree,null,1));
//dfs on the tree to get all the paths to... the leaves
let dfs = function(tree,paths,p){
if(tree.stop){
return paths.push(p.join('\\\\'));
}
Object.keys(tree).forEach(k=>{
dfs(tree[k], paths, p.concat(k));
});
}
let paths = [];
dfs(tree, paths,[]);
return paths;
}
let archives = [
'C:\\\\ab',
'D:\\\\', //just some root
'D:\\\\ab',//be dropped
'D:\\\\abc',//dropped as well
'F:\\\\abc\\\\e',//two folder creation
'C:\\\\ab\\c',
'B:\\\\ab\\c',
'B:\\\\ab',//expect B\\\\ab\\c to be dropped
]
console.log(nodupes(archives))

Try this
console.log([
"C:\\AMD\\Packages",
"C:\\Users",
"C:\\User",
"E:\\Pictures",
"E:\\Pictures\\Birthday",
"C:\\Windows",
"D:\\",
"D:\\aabbcc",
"E:\\Pictures\\Birthday"
].sort().reduce(
(acc, cur) =>
acc.length > 0
&& cur.startsWith(acc[acc.length - 1])
&& ( cur.indexOf("\\", acc[acc.length - 1].replace(/\\$/,"").length) !== -1 )
&& acc || acc.concat(cur)
, []
))

Return value for function is not working correctly. How to set up correct return value for function checkIfExist()?

I am writing a script for Premiere pro where I can add markers in the timeline and export a still for each marker in one go. However, when I write a function to check if the still has been previously created, the functions tells me it finds the previously created still, but then still creates a new one.
So basically: Function returns true, but still executes the else{}
//checks if the frame that is about to be exported already exists
if(checkIfExist(app.project.rootItem, outputFile)){
alert("frame already exists");
}else{
//This is where the actual still gets created and imported
activeSequence.exportFramePNG(time, outputFileName);
//here the previously created item gets moved to the appropriate bin (This is working great ATM)
moveToBin(outputFile);
}
}
}
//This function is meant to check if an item exists in the project bin. It does this by looping though all the items in the array from the start.
function checkIfExist(currentItem, name){
for(var i = 0; i<currentItem.children.numItems; i++){
currentChild = currentItem.children[i];
if(currentChild.name.toUpperCase() === name.toUpperCase()){
alert("Found: " + currentChild.name);
return true;
}if(currentChild.type == ProjectItemType.BIN){
checkIfExist(currentChild, name);
}
}
return false;
}

I think it happens because of the recursion you do:
if(currentChild.type == ProjectItemType.BIN){
checkIfExist(currentChild, name);
}
If this one gets kicked off before you can return true, you will start to run the function for a second time.
Now the first run can return a true, while the second (or even 3th, or 4th, etc) can return false and thus creating a new one, while also finding it.
Also if possible try to use arr.find or arr.findIndex and check if the value is -1 (or not found). This will make your code shorter, cleaner and less open for errors :)
But this will not work for nested arrays. Then you need to make an other function to first make a flat copy that includes all nested array before you do the arr.find or arr.findIndex. Still think that is the better solution.
You can use this to make nested array into a flat one:
let arr1 = [1,2,3,[1,2,3,4, [2,3,4]]];
function flattenDeep(arr1) {
return arr1.reduce((acc, val) => Array.isArray(val) ? acc.concat(flattenDeep(val)) : acc.concat(val), []);
}
flattenDeep(arr1);// [1, 2, 3, 1, 2, 3, 4, 2, 3, 4]

JavaScript/React Native array(objects) sort

I'm starting with react-native building an app to track lap times from my RC Cars. I have an arduino with TCP connection (server) and for each lap, this arduino sends the current time/lap for all connected clients like this:
{"tx_id":33,"last_time":123456,"lap":612}
In my program (in react-native), I have one state called dados with this struct:
dados[tx_id] = {
tx_id: <tx_id>,
last_time:,
best_lap:0,
best_time:0,
diff:0,
laps:[]
};
This program connects to arduino and when receive some data, just push to this state. More specific in laps array of each transponder. Finally, I get something like this:
dados[33] = {
tx_id:33,
last_time: 456,
best_lap: 3455,
best_time: 32432,
diff: 32,
laps: [{lap:1,time:1234},{lap:2,time:32323},{lap:3,time:3242332}]
}
dados[34] = {
tx_id:34,
last_time: 123,
best_lap: 32234,
best_time: 335343,
diff: 10,
laps: [{lap:1,time:1234},{lap:2,time:32323},{lap:3,time:3242332}]
}
dados[35] = {
tx_id:35,
last_time: 789,
best_lap: 32234,
best_time: 335343,
diff: 8,
laps: [{lap:1,time:1234},{lap:2,time:32323},{lap:3,time:3242332},{lap:4,time:343232}]
}
This data in rendered to View's using map function (not a FlatList).
My problem now is that I need to order this before printing on screen.
Now, with this code, data are printed using tx_id as order, since it's the key for main array. Is there a way to order this array using number of elements in laps property and the second option to sort, use last_time property of element?
In this case, the last tx of my example (35) would be the first in the list because it has one lap more than other elements. The second item would be 34 (because of last_time). And the third would be tx 33.
Is there any way to to this in JavaScript, or I need to create a custom functions and check every item in recursive way?!

Tks #crackhead420
While waiting for reply to this question, I just found what you said.... :)
This is my final teste/solution that worked:
var t_teste = this.state.teste;
t_teste[33] = {tx_id: 33, last_time:998,best_lap:2,best_time:123,diff:0,laps:[{lap:1,time:123},{lap:2,time:456}]};
t_teste[34] = {tx_id: 34, last_time:123,best_lap:2,best_time:123,diff:0,laps:[{lap:1,time:123},{lap:2,time:456}]};
t_teste[35] = {tx_id: 35, last_time:456,best_lap:2,best_time:123,diff:0,laps:[{lap:1,time:123},{lap:2,time:456},{lap:3,time:423}]};
t_teste[36] = {tx_id: 36, last_time:789,best_lap:2,best_time:123,diff:0,laps:[{lap:1,time:123},{lap:2,time:456}]};
console.log('Teste original: ',JSON.stringify(t_teste));
var saida = t_teste.sort(function(a, b) {
if (a.laps.length > b.laps.length) {
return -1;
}
if (a.laps.length < b.laps.length) {
return 1;
}
// In this case, the laps are equal....so let's check last_time
if (a.last_time < b.last_time) {
return -1; // fastest lap (less time) first!
}
if (a.last_time > b.last_time) {
return 1;
}
// Return the same
return 0;
});
console.log('Teste novo: ',JSON.stringify(saida));

Using some simple helper functions, this is definitely possible:
const data = [{tx_id:33,last_time:456,best_lap:3455,best_time:32432,diff:32,laps:[{lap:1,time:1234},{lap:2,time:32323},{lap:3,time:3242332}]},{tx_id:34,last_time:123,best_lap:32234,best_time:335343,diff:10,laps:[{lap:1,time:1234},{lap:2,time:32323},{lap:3,time:3242332}]},{tx_id:35,last_time:789,best_lap:32234,best_time:335343,diff:8,laps:[{lap:1,time:1234},{lap:2,time:32323},{lap:3,time:3242332},{lap:4,time:343232}]}]
const sortBy = fn => (a, b) => -(fn(a) < fn(b)) || +(fn(a) > fn(b))
const sortByLapsLength = sortBy(o => o.laps.length)
const sortByLastTime = sortBy(o => o.last_time)
const sortFn = (a, b) => -sortByLapsLength(a, b) || sortByLastTime(a, b)
data.sort(sortFn)
// show new order of `tx_id`s
console.log(data.map(o => o.tx_id))
sortBy() (more explanation at the link) accepts a function that selects a value as the sorting criteria of a given object. This value must be a string or a number. sortBy() then returns a function that, given two objects, will sort them in ascending order when passed to Array.prototype.sort(). sortFn() uses two of these functions with a logical OR || operator to employ short-circuiting behavior and sort first by laps.length (in descending order, thus the negation -), and then by last_time if two objects' laps.length are equal.

Its possible to sort an object array by theire values:
dados.sort(function(a, b) {
return a.last_time - b.last_time;
});

rx: unfold array to multiple streams

I have a stream holding an array, each element of which has an id. I need to split this into a stream per id, which will complete when the source stream no longer carries the id.
E.g. input stream sequence with these three values
[{a:1}, {b:1}] [{a:2}, {b:2}, {c:1}] [{b:3}, {c:2}]
should return three streams
a -> 1 2 |
b -> 1 2 3
c -> 1 2
Where a has completed on the 3rd value, since its id is gone, and c has been created on the 2nd value, since its id has appeared.
I'm trying groupByUntil, a bit like
var input = foo.share();
var output = input.selectMany(function (s) {
return rx.Observable.fromArray(s);
}).groupByUntil(
function (s) { return s.keys()[0]; },
null,
function (g) { return input.filter(
function (s) { return !findkey(s, g.key); }
); }
)
So, group by the id, and dispose of the group when the input stream no longer has the id. This seems to work, but the two uses of input look odd to me, like there could a weird order dependency when using a single stream to control the input of the groupByUntil, and the disposal of the groups.
Is there a better way?
update
There is, indeed, a weird timing problem here. fromArray by default uses the currentThread scheduler, which will result in events from that array being interleaved with events from input. The dispose conditions on the group are then evaluated at the wrong time (before the groups from the previous input have been processed).
A possible workaround is to do fromArray(.., rx.Scheduler.immediate), which will keep the grouped events in sync with input.

yeah the only alternative I can think of is to manage the state yourself. I don't know that it is better though.
var d = Object.create(null);
var output = input
.flatMap(function (s) {
// end completed groups
Object
.keys(d)
.filter(function (k) { return !findKey(s, k); })
.forEach(function (k) {
d[k].onNext(1);
d[k].onCompleted();
delete d[k];
});
return Rx.Observable.fromArray(s);
})
.groupByUntil(
function (s) { return s.keys()[0]; },
null,
function (g) { return d[g.key] = new Rx.AsyncSubject(); });

Sort an array by a preferred order

I'd like to come up with a good way to have a "suggested" order for how to sort an array in javascript.
So say my first array looks something like this:
['bob','david','steve','darrel','jim']
Now all I care about, is that the sorted results starts out in this order:
['jim','steve','david']
After that, I Want the remaining values to be presented in their original order.
So I would expect the result to be:
['jim','steve','david','bob','darrel']
I have an API that I am communicating with, and I want to present the results important to me in the list at the top. After that, I'd prefer they are just returned in their original order.
If this can be easily accomplished with a javascript framework like jQuery, I'd like to hear about that too. Thanks!
Edit for clarity:
I'd like to assume that the values provided in the array that I want to sort are not guaranteed.
So in the original example, if the provided was:
['bob','steve','darrel','jim']
And I wanted to sort it by:
['jim','steve','david']
Since 'david' isn't in the provided array, I'd like the result to exclude it.
Edit2 for more clarity:
A practical example of what I'm trying to accomplish:
The API will return something looking like:
['Load Average','Memory Usage','Disk Space']
I'd like to present the user with the most important results first, but each of these fields may not always be returned. So I'd like the most important (as determined by the user in some other code), to be displayed first if they are available.

Something like this should work:
var presetOrder = ['jim','steve','david']; // needn't be hardcoded
function sortSpecial(arr) {
var result = [],
i, j;
for (i = 0; i < presetOrder.length; i++)
while (-1 != (j = $.inArray(presetOrder[i], arr)))
result.push(arr.splice(j, 1)[0]);
return result.concat(arr);
}
var sorted = sortSpecial( ['bob','david','steve','darrel','jim'] );
I've allowed for the "special" values appearing more than once in the array being processed, and assumed that duplicates should be kept as long as they're shuffled up to the front in the order defined in presetOrder.
Note: I've used jQuery's $.inArray() rather than Array.indexOf() only because that latter isn't supported by IE until IE9 and you've tagged your question with "jQuery". You could of course use .indexOf() if you don't care about old IE, or if you use a shim.

var important_results = {
// object keys are the important results, values is their order
jim: 1,
steve: 2,
david: 3
};
// results is the orig array from the api
results.sort(function(a,b) {
// If compareFunction(a, b) is less than 0, sort a to a lower index than b.
// See https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/Array/sort
var important_a = important_results[a],
important_b = important_results[b],
ret;
if (important_a && !important_b) {ret = -1}
else if (important_b && !important_a) {ret = 1}
else if (important_a && important_b) {ret = important_a - important_b}
else {ret = 0}; // keep original order if neither a or b is important
return(ret);
}
)
Use a sorting function that treats the previously known important results specially--sorts them to the head of the results if present in results.
items in important_results don't have to be in the results

Here's a simple test page:
<html>
<head>
<script language="javascript">
function test()
{
var items = ['bob', 'david', 'steve', 'darrel', 'jim'];
items.sort(function(a,b)
{
var map = {'jim':-3,'steve':-2,'david':-1};
return map[a] - map[b];
});
alert(items.join(','));
}
</script>
</head>
<body>
<button onclick="javascript:test()">Click Me</button>
</body>
</html>
It works in most browsers because javascript typically uses what is called a stable sort algorithm, the defining feature of which is that it preserves the original order of equivalent items. However, I know there have been exceptions. You guarantee stability by using the array index of each remaining item as it's a1/b1 value.

http://tinysort.sjeiti.com/
I think this might help. The $('#yrDiv').tsort({place:'start'}); will add your important list in the start.
You can also sort using this function the way you like.

Live demo ( jsfiddle seems to be down)
http://jsbin.com/eteniz/edit#javascript,html
var priorities=['jim','steve','david'];
var liveData=['bob','david','steve','darrel','jim'];
var output=[],temp=[];
for ( i=0; i<liveData.length; i++){
if( $.inArray( liveData[i], priorities) ==-1){
output.push( liveData[i]);
}else{
temp.push( liveData[i]);
}
}
var temp2=$.grep( priorities, function(name,i){
return $.inArray( name, temp) >-1;
});
output=$.merge( temp2, output);

there can be another way of sorting on order base, also values can be objects to work with
const inputs = ["bob", "david", "steve", "darrel", "jim"].map((val) => ({
val,
}));
const order = ["jim", "steve", "david"];
const vMap = new Map(inputs.map((v) => [v.val, v]));
const sorted = [];
order.forEach((o) => {
if (vMap.has(o)) {
sorted.push(vMap.get(o));
vMap.delete(o);
}
});
const result = sorted.concat(Array.from(vMap.values()));
const plainResult = result.map(({ val }) => val);

Have you considered using Underscore.js? It contains several utilities for manipulating lists like this.
In your case, you could:
Filter the results you want using filter() and store them in a collection.
var priorities = _.filter(['bob','david','steve','darrel','jim'],
function(pName){
if (pName == 'jim' || pName == 'steve' || pName == 'david') return true;
});
Get a copy of the other results using without()
var leftovers = _.without(['bob','david','steve','darrel','jim'], 'jim', 'steve', 'david');
Union the arrays from the previous steps using union()
var finalList = _.union(priorities, leftovers);