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I've got an array that looks like:
Is there a way that when another object "AAPL" gets added to the array, the fees and amount gets added and AAPL stays one object in the array?
Thank you in advance!
I wouldn't want to assume that "AAPL" is always within the array.
Check first if the new item is within the array. If not, I will return and end the call (but you may wish to append this to the array instead as it is a new item).
If it is, replace the current object (or parts of the object, in this case I have done just fees & amount)
function addPortfolioItem(item) {
const index = returnPortfolioHoldingIndex(item);
if (index = -1) {
// Returning if not found - You may wish to append to the array as mentioned above
return;
} else {
// Overwrite the object within the array with the new object
portfolioHoldings[index].fees = item.fees;
portfolioHoldings[index].amount = item.amount;
}
}
function returnPortfolioHoldingIndex() {
const tickerSymbolToSearch = "AAPL";
return portfolioHoldings.findIndex(item => item.ticker_symbol === tickerSymbolToSearch);
}
First identify the entry then just add another integer to the value.
It depends on how you are updating that object now - if it's always in 0 position then just portfolioHoldings[0][amount] += number;
if not, then loop through all and check if ticker_symbol == AAPL, then update that array value, by adding needed amount.
I'm having a tough time figuring out how to loop through an array and if certain items do exist within the array, i'd like to perform a .slice(0, 16) to kind of filter an already existing array (lets call that existing array "routes").
For example, a previous process will yield the following array:
points = ['=00ECY20WA200_RECV_P1SEL',
'=00ECY20WA200_RECV_P2SEL',
'=00RECV_C1A_EINCSCMPP1',
'=00RECV_C1A_EINCSCMPP2',
'=00BYPS_C1A_EINCSCMP',
'=00ECY20WA200_BYPS_SPSL1',
'=00ECC92AG184YB01',
'=00ECC92AG185YB01',
'=00ECC92AG186YB01',
'=00ECC92AG187YB01',
]
So if any of the above items exist in the "points" Array, which in this case they all do (but in some cases it could just be 1 of the 10 items existing there), I'm trying to perform routes.slice(0, 16) to the other already existing array.
I've tried lots of different ways (for loops with if statements) and at this point I'm not sure if its my syntax or what, but I'm back at square 0 and I don't even have a competent piece of code to show for. Any direction would be greatly appreciated.
You could use a hash table for checking and filtering.
var points = ['=00ECY20WA200_RECV_P1SEL', '=00ECY20WA200_RECV_P2SEL', '=00RECV_C1A_EINCSCMPP1', '=00RECV_C1A_EINCSCMPP2', '=00BYPS_C1A_EINCSCMP', '=00ECY20WA200_BYPS_SPSL1', '=00ECC92AG184YB01', '=00ECC92AG185YB01', '=00ECC92AG186YB01', '=00ECC92AG187YB01'],
hash = Object.create(null),
filtered = points.filter(function (a) {
if (!hash[a.slice(0, 16)]) {
hash[a.slice(0, 16)] = true;
return true;
}
});
console.log(filtered);
ES6 with Set
var points = ['=00ECY20WA200_RECV_P1SEL', '=00ECY20WA200_RECV_P2SEL', '=00RECV_C1A_EINCSCMPP1', '=00RECV_C1A_EINCSCMPP2', '=00BYPS_C1A_EINCSCMP', '=00ECY20WA200_BYPS_SPSL1', '=00ECC92AG184YB01', '=00ECC92AG185YB01', '=00ECC92AG186YB01', '=00ECC92AG187YB01'],
pSet = new Set,
filtered = points.filter(a => !pSet.has(a.slice(0, 16)) && pSet.add(a.slice(0, 16)));
console.log(filtered);
EDIT: So it seems like you want to remove an element from an array called routes for each element in the points array. This is how you could do this:
function removeBrokenRoutes(brokenPoints, routes){
for(let pt of brokenPoints){
let index = routes.indexOf(pt);
if(index !== -1) routes.splice(index,1);
}
return routes;
}
Keep in mind that the larger the arrays, the more time this is going to take to complete.
You could use the filter and indexOf methods in combination:
var arr = [/* all the data you're checking against */];
var points = [/* the data you're checking for */];
var filteredArr = arr.filter(function(x) {
// will return -1 if the point is not found
return points.indexOf(x) !== -1;
});
filteredArr will contain all the points that appear in both arrays. The filter function works by taking a function with one argument x, which represents each item in the array. if the function returns true, the item will be added to the new array (filteredArr), and if false the function will move on to the next item. indexOf will check if the item is found in the other array. Also it is important to note that you will need a more complex solution (such as a hashtable) if the data set is very, very large as this is not necessarily the most performant method. But it's a good place to start as it is easy to understand.
Title is pretty much self explanatory...
I want to be able to find duplicated values from JavaScript array.
The array keys can be duplicated so I need to validate only the array values.
Here is an example :
var arr=[
Ibanez: 'JoeSatriani',
Ibanez: 'SteveVai',
Fender: 'YngwieMalmsteen',
Fender: 'EricJohnson',
Gibson: 'EricJohnson',
Takamine: 'SteveVai'
];
In that example:
the key is the guitar brand
the value is the guitar player name.
So:
If there is duplicated keys (like: Ibanez or Fender) as on that current example that is OK :-)
But
If there is duplicated values (like: EricJohnson or SteveVai) I'm expecting to get (return) that error:
EricJohnson,SteveVai
You can't have associative arrays in Javascript. You can create an array of objects, like:
var arr=[
{Ibanez: 'JoeSatriani'},
{Ibanez: 'SteveVai'},
{Fender: 'YngwieMalmsteen'},
{Fender: 'EricJohnson'},
{Gibson: 'EricJohnson'},
{Takamine: 'SteveVai'}
];
Then you'll need a for...in loop to go over every object in the array, create a new array of values and check that for duplicates, which is also not very straightforward - basically you'll want to sort the array and make sure no value is the same as the one after it.
var arrayOfValues = [];
arr.forEach(function(obj){
for(var prop in obj)
arrayOfValues.push(obj[prop]);
});
arrayOfValues.sort(); // by default it will sort them alphabetically
arrayOfValues.forEach(function(element,index,array){
if(array[index+1] && element==array[index+1])
alert("Duplicate value found!");
});
First of all, object keys can not be repeated.
This means that:
({
"Fender": "Jim",
"Fender": "Bob"
})["Fender"]
Would simply return: "Bob".
However, I did make a code that could allow you to find duplicates in values, but as I said, the key will have to be unique:
var arr = {
Ibanez: 'EricJohnson',
Fender: 'YngwieMalmsteen',
Gibson: 'EricJohnson',
Takamine: 'SteveVai',
"Takamine2": 'SteveVai'
};
function contains(a, obj) {
for (var i = 0; i < a.length; i++) {
if (a[i] === obj) {
return true;
}
}
return false;
}
var track = [];
var exists = [];
for (var val in arr) {
if (contains(track, arr[val])) {
exists.push(arr[val]);
} else {
track.push(arr[val])
}
}
alert(exists)
You can see it working here: http://jsfiddle.net/dr09sga6/2/
As others have commented, the example array you provided isn't a valid JavaScript array. You could, however, keep a list for each guitar type:
var mapping = {
Ibanez: ['JoeSatriani','SteveVai'],
Fender: ['YngwieMalmsteen','EricJohnson']
Gibson: ['EricJohnson'],
Takamine: ['SteveVai']
];
Or a list of each guitar/musician pair:
var pairs = [
['Ibanez','JoeSatriani'],
['Ibanez','SteveVai'],
['Fender','YngwieMalmsteen'],
['Fender','EricJohnson'],
['Gibson','EricJohnson'],
['Takamine','SteveVai']
];
Your solution is going to depend on which pattern you go with. However, in the second case it can be done in one chained functional call:
pairs.map(function(e) {return e[1]}) // Discard the brand names
.sort() // Sort by artist
.reduce(function(p,c,i,a){
if (i>0 && a[i]==a[i-1] && !p.some(function(v) {return v == c;})) p.push(c);
return p;
},[]); //Return the artist names that are duplicated
http://jsfiddle.net/mkurqmqd/1/
To break that reduce call down a bit, here's the callback again:
function(p,c,i,a){
if (i>0
&& a[i]==a[i-1]
&& !p.some(function(v) {
return v == c;
}))
p.push(c);
return p;
}
reduce is going to call our callback for each element in the array, and it's going to pass the returned value for each call into the next call as the first parameter (p). It's useful for accumulating a list as you move across an array.
Because we're looking back at the previous item, we need to make sure we don't go out of bounds on item 0.
Then we're checking to see if this item matches the previous one in the (sorted) list.
Then we're checking (with Array.prototype.some()) whether the value we've found is ALREADY in our list of duplicates...to avoid having duplicate duplicates!
If all of those checks pass, we add the name to our list of duplicate values.
I currently have a list of objects in javascript indexed by a key:
var list = [];
list['a'] = [];
list['a'].push({obj: 'test'});
list['a'].push({obj: 'test2'});
list['b'] = [];
list['b'].push({obj: 'test'});
list['b'].push({obj: 'test2'});
I would list to remove the entry based on the key (a/b)
I have tried the following:
for(var x in list) { delete list[x]; }
that works but it actually leaves an undefined entry in the list.
I have also tried splicing the array, but that does not seems to work in this case.
Any thoughts on how to remove the entry in javascript or jQuery?
Thanks.
The Fix:
After reading some of the comments, i was able to better understand what my list is consistent of. Therefor, i was able to do the removal by doing the following:
delete list.b;
I'm not sure if my list is best way to organize my structure, but doing a delete on the list and treating it like an object property did the trick.
Thanks for all the feedback.
I'll assume list is an object, not an array.
If you want to reset a or (or b it's done the same way)
list.a.length = 0;
If you want to delete an element from a at a known index (let index)
list.a.splice(index, 1);
You're attempting to add the elements to the array object as object properties and not as array elements. You can verify this by inspecting the value of list.length (will be 0).
So when doing something such as the following:
function removeProperty(id) {
if (list.hasOwnProperty(id)) {
delete list[id];
}
}
removeProperty('a');
it's really the same as:
delete list.a;
which is why you think it leaves an undefined 'entry' in the 'list'.
You'll need to use a literal object instead:
var list = {};
list['a'] = [];
...
list['b' = [];
...
which would allow you to use delete and have it behave as you expect. Of course you'll lose the .length property on the array but you never had that anyway.
Create a simple prototype for the Array class
Array.prototype.remove = function() {
// Helper function to remove a single element from a list if exists
item = arguments[0]
if (this.includes(item)) {
index = this.indexOf(item)
this.splice(index, 1)
}
}
// Now we can call
myList.remove(YourObject)
The above code will add the remove method to all your lists, so this will help you not just for objects but also strings, integers, or any data type
var list = {1: [{},{}], 2: [{},{}]};
function removeProperty(obj, prop){
if(obj[prop]){
delete obj[prop];
}
}
removeProperty(list,"1");
console.log(list);
If this quote:
I would list to remove the entry based on the key (a/b)
means you would like to select the list to consider based off the key (a/b), then remove elements in the list (or all of them), you can try this:
var list = [];
list['a'] = [];
list['a'].push({obj: 'test4'});
list['a'].push({obj: 'test5'});
list['b'] = [];
list['b'].push({obj: 'test'});
list['b'].push({obj: 'test2'});
var toRemove = 'test4';
var removeFrom = "a";
var consideredList;
for (var prop in list) {
if (prop == removeFrom) {
consideredList = list[prop];
}
}
//Remove everything from the considered list
consideredList.splice(0, consideredList.length);
//Remove based off value, if you know the property name
// for(var pos in consideredList) {
// if(consideredList[pos].obj == toRemove) {
// consideredList.splice(pos, 1);
// }
// }
I made a Plunker of a few different cases (check the script.js file). There seems to be a bit of confusion on what you are after and hopefully this is helpful to you somehow. Good luck.
I want to remove an element in an array with multiple occurrences with a function.
var array=["hello","hello","world",1,"world"];
function removeItem(item){
for(i in array){
if(array[i]==item) array.splice(i,1);
}
}
removeItem("world");
//Return hello,hello,1
removeItem("hello");
//Return hello,world,1,world
This loop doesn't remove the element when it repeats twice in sequence, only removes one of them.
Why?
You have a built in function called filter that filters an array based on a predicate (a condition).
It doesn't alter the original array but returns a new filtered one.
var array=["hello","hello","world",1,"world"];
var filtered = array.filter(function(element) {
return element !== "hello";
}); // filtered contains no occurrences of hello
You can extract it to a function:
function without(array, what){
return array.filter(function(element){
return element !== what;
});
}
However, the original filter seems expressive enough.
Here is a link to its documentation
Your original function has a few issues:
It iterates the array using a for... in loop which has no guarantee on the iteration order. Also, don't use it to iterate through arrays - prefer a normal for... loop or a .forEach
You're iterating an array with an off-by-one error so you're skipping on the next item since you're both removing the element and progressing the array.
That is because the for-loop goes to the next item after the occurrence is deleted, thereby skipping the item directly after that one.
For example, lets assume item1 needs to be deleted in this array (note that <- is the index of the loop):
item1 (<-), item2, item3
after deleting:
item2 (<-), item3
and after index is updated (as the loop was finished)
item2, item3 (<-)
So you can see item2 is skipped and thus not checked!
Therefore you'd need to compensate for this by manually reducing the index by 1, as shown here:
function removeItem(item){
for(var i = 0; i < array.length; i++){
if(array[i]==item) {
array.splice(i,1);
i--; // Prevent skipping an item
}
}
}
Instead of using this for-loop, you can use more 'modern' methods to filter out unwanted items as shown in the other answer by Benjamin.
None of these answers are very optimal. The accepted answer with the filter will result in a new instance of an array. The answer with the second most votes, the for loop that takes a step back on every splice, is unnecessarily complex.
If you want to do the for loop loop approach, just count backward down to 0.
for (var i = array.length - 0; i >= 0; i--) {
if (array[i] === item) {
array.splice(i, 1);
}
}
However, I've used a surprisingly fast method with a while loop and indexOf:
var itemIndex = 0;
while ((itemIndex = valuesArray.indexOf(findItem, itemIndex)) > -1) {
valuesArray.splice(itemIndex, 1);
}
What makes this method not repetitive is that after the any removal, the next search will start at the index of the next element after the removed item. That's because you can pass a starting index into indexOf as the second parameter.
In a jsPerf test case comparing the two above methods and the accepted filter method, the indexOf routinely finished first on Firefox and Chrome, and was second on IE. The filter method was always slower by a wide margin.
Conclusion: Either reverse for loop are a while with indexOf are currently the best methods I can find to remove multiple instances of the same element from an array. Using filter creates a new array and is slower so I would avoid that.
You can use loadash or underscore js in this case
if arr is an array you can remove duplicates by:
var arr = [2,3,4,4,5,5];
arr = _.uniq(arr);
Try to run your code "manually" -
The "hello" are following each other. you remove the first, your array shrinks in one item, and now the index you have follow the next item.
removing "hello""
Start Loop. i=0, array=["hello","hello","world",1,"world"] i is pointing to "hello"
remove first item, i=0 array=["hello","world",1,"world"]
next loop, i=1, array=["hello","world",1,"world"]. second "hello" will not be removed.
Lets look at "world" =
i=2, is pointing to "world" (remove). on next loop the array is:
["hello","hello",1,"world"] and i=3. here went the second "world".
what do you wish to happen? do you want to remove all instances of the item? or only the first one? for first case, the remove should be in
while (array[i] == item) array.splice(i,1);
for second case - return as soon as you had removed item.
Create a set given an array, the original array is unmodified
Demo on Fiddle
var array=["hello","hello","world",1,"world"];
function removeDups(items) {
var i,
setObj = {},
setArray = [];
for (i = 0; i < items.length; i += 1) {
if (!setObj.hasOwnProperty(items[i])) {
setArray.push(items[i]);
setObj[items[i]] = true;
}
}
return setArray;
}
console.log(removeDups(array)); // ["hello", "world", 1]
I must say that my approach does not make use of splice feature and you need another array for this solution as well.
First of all, I guess your way of looping an array is not the right. You are using for in loops which are for objects, not arrays. You'd better use $.each in case you are using jQuery or Array.prototype.forEach if you are using vanila Javascript.
Second, why not creating a new empty array, looping through it and adding only the unique elements to the new array, like this:
FIRST APPROACH (jQuery):
var newArray = [];
$.each(array, function(i, element) {
if ($.inArray(element, newArray) === -1) {
newArray.push(region);
}
});
SECOND APPROACH (Vanila Javascript):
var newArray = [];
array.forEach(function(i, element) {
if (newArray.indexOf(element) === -1) {
newArray.push(region);
}
});
I needed a slight variation of this, the ability to remove 'n' occurrences of an item from an array, so I modified #Veger's answer as:
function removeArrayItemNTimes(arr,toRemove,times){
times = times || 10;
for(var i = 0; i < arr.length; i++){
if(arr[i]==toRemove) {
arr.splice(i,1);
i--; // Prevent skipping an item
times--;
if (times<=0) break;
}
}
return arr;
}
An alternate approach would be to sort the array and then playing around with the indexes of the values.
function(arr) {
var sortedArray = arr.sort();
//In case of numbers, you can use arr.sort(function(a,b) {return a - b;})
for (var i = 0; sortedArray.length; i++) {
if (sortedArray.indexOf(sortedArray[i]) === sortedArray.lastIndexOf(sortedArray[i]))
continue;
else
sortedArray.splice(sortedArray.indexOf(sortedArray[i]), (sortedArray.lastIndexOf(sortedArray[i]) - sortedArray.indexOf(sortedArray[i])));
}
}
You can use the following piece of code to remove multiple occurrences of value val in array arr.
while(arr.indexOf(val)!=-1){
arr.splice(arr.indexOf(val), 1);
}
I thinks this code much simpler to understand and no need to pass manually each element that what we want to remove
ES6 syntax makes our life so simpler, try it out
const removeOccurences = (array)=>{
const newArray= array.filter((e, i ,ar) => !(array.filter((e, i ,ar)=> i !== ar.indexOf(e)).includes(e)))
console.log(newArray) // output [1]
}
removeOccurences(["hello","hello","world",1,"world"])