I have multiple objects that all have the same keys lets say each object has: name and position. The first object will start with position=0. The second object would have position=1. The third object would have position=2 and so on until we get to the 10th object that would have position=9.
I need a way to subtract 1 from every objects position (with only possible values being 0-9 so that 0-1=9)
Looking for a solution that handles all of them mathematically at once, not just re-writing out new values to assign to each key individually.
Suppose you have an array of JavaScript objects, you could use map:
var newObjs = objects.map(function (object) {
object.position = (object.position === 9 ? 0 : object.position--);
return object;
});
A better approach would be:
objects.forEach( function (object) {
object.position--;
object.position = object.position < 0 ? 9 : object.position;
});
I have an array of subscription instances #subscription_valids and an array of subscribed player that look like that :
array_subscribed_players = [{"name0" => "link1"}, {"name1"=>"link2"}, {"name2"=>"link3"}....]
What I need to do is : for each subscription in #subscription_valids :
#subscription_valids.each do |subscription|
I need to check if subscription.user.full_name or if subscription.user.full_name_inversed matches a key in one of the hashes of array_subscribed_player (in the exemple "name0", "name1" or "name2").
If it matches then I should store the relevant subscription as key in a hash in a new array and extract the relevant link as value of this hash. My final outpout should be an array that looks like this :
[{subscription1 => "link1"}, {subscription2 => "link2}, ...]
else if the subscription.user.full_name doesnt match i'll just store the subscription in a failure array.
How can I achieve this result ?
See http://ruby-doc.org/core-2.2.3/Hash.html
A user-defined class may be used as a hash key if the hash and eql?
methods are overridden to provide meaningful behavior. By default,
separate instances refer to separate hash keys.
so I think you should override your .eql? method in #subscription_valids to something meaningful (like a unique string)
I can't think for a Array method so you can go like:
demo
results = []
failures = []
#subscription_valids.each do |subscription|
array_subscribed_players.each do |player|
if player.keys.first == subscription.user.full_name || player.keys.first == subscription.user.full_name_inversed
results << { subscription => player[player.keys.first] }
else
failures << subscription
end
end
end
You can try the following:
valid = array_subscribed_players.select{|x| #subscription_valids.map(&:name).include?(x.keys.first)}
Demo
If you need to store both valid and invalid values somewhere:
valid, invalid = array_subscribed_players.partition{|x| #subscription_valids.map(&:name).include?(x.keys.first)}
In the library of the .fla file I have a square exported as Class "cuad" on frame 1
I want to create an Array with 100 squares so as to move them later
So I do like this:
for (var i:uint = 0; i<100;i++)
{
var cuad_mc = new cuad();
addChild(cuad_mc);
myArray.push("cuad_mc");
trace(myArray[i]);
}
I have a runtime error
The error you experience is
Error #1069: Did not find alpha propiety in the String and there is not any value predetermined
The problem comes from your line
myArray.push("cuad_mc");
What you are doing here is pushing a String Object into your Array, not the cuad Object you want. String Objects don't have Alpha values, or x values.
What you want to do is
myArray.push(cuad_mc);
cuad_mc (without the " quotation marks) is a reference to the object you just created.
This should solve your problem. I also recommend using Vectors instead of Array if you only need to store one type of Object. Like this:
var myArray:Vector<cuad> = new Vector<cuad>();
for(var i:int=0;i<100;i++){
var cuad_mc:cuad = new cuad();
addChild(cuad_mc);
myArray.push(cuad_mc);
trace(myArray[i]);
}
Vectors are just like Arrays, but they only allow one specific type, so that a situation like yours doesn't occur.
I have a Map of type Sobject as a key and Integer as a value. The Sobject has field of type Text, other field of type date and another field of type Number(16,2). When I put data into the map and then debug it, the map returns data in a sorted way. The map sorts it by the Number field present in the Key i.e the number data field of the object. Can I get the map sorted by the date field of the object which is its key? Below is the rough structure of my Map and the key object fields.
Map<Effort_Allocation__c, Double> cellContent;
Effort_Allocation__c.Allocated_Effort_Hours__c; //The Number field by which the map gets sorted
Effort_Allocation__c.Assignment_Date__c; // The date field by which I want the map to get sorted
It's a bad idea to use an object as a key for a map. You should, instead, use the object's ID as the key. The reasons for this are discussed in detail here. Short version - the object's values may change which changes the object's hash value and wrecks your map.
Although maps cannot be sorted directly, lists can be sorted, so they can be used to access map elements in sorted order. You'll need a wrapper class for your object that implements the "Comparable" interface. There's an example of that here. Note that the example sorts by date.
The class is declared with "comparable"
global class AccountHistoryWrapper implements Comparable{
and has the following CompareTo method
global Integer compareTo(Object compareTo) {
// Cast argument to AccountHistoryWrapper
AccountHistoryWrapper aHW = (AccountHistoryWrapper)compareTo;
// The return value of 0 indicates that both elements are equal.
Integer returnValue = 0;
if ( aHW.account.CreatedDate > aHW.account.CreatedDate) {
// Set return value to a positive value.
returnValue = 1;
} else if ( aHW.account.CreatedDate < aHW.account.CreatedDate) {
// Set return value to a negative value.
returnValue = -1;
}
return returnValue;
}
I have been working on a project and Stack Overflow has helped me with a few problems so far, so I am very thankful!
My question is this:
I have an array like this:
var records:Object = {};
var arr:Array = [
records["nh"] = { medinc:66303, statename:"New Hampshire"},
records["ct"] = { medinc:65958, statename:"Connecticut"},
records["nj"] = { medinc:65173, statename:"New Jersey"},
records["md"] = { medinc:64596, statename:"Maryland"},
etc... for all 50 states. And then I have the array sorted reverse numerically (descending) like this:
arr.sortOn("medinc", Array.NUMERIC);
arr.reverse();
Can I call the name of the record (i.e. "nj" for new jersey) and then get the value from the numeric position above and below the record in the array?
Basically, medinc is medium income of US states, and I am trying to show a ranking system... a user would click Texas for example, and it would show the medinc value for Texas, along with the state the ranks one position below and the state that ranks one position above in the array.
Thanks for your help!
If you know the object, you can use the array.indexOf().
var index:int = records.indexOf(records["nj"]);
var above:Object;
var below:Object;
if(index + 1 < records.length){ //make sure your not already at the top
above = records[index+1];
}
if(index > 0){ //make sure your not already at the bottom
below = records[index-1];
}
I think this is the answer based on my understanding of your data.
var index:int = arr.indexOf(records["nh"]);
That will get you the index of the record that was clicked on and then for find the ones below and above just:
var clickedRecord:Object = arr[index]
var higherRecord:Object = arr[index++]
var lowerRecord:Object = arr[index--]
Hope that answers your question
Do you really need records to be hash?
If no, you can simply move key to record field and change records to simple array:
var records: Array = new Array();
records.push({ short: "nh", medinc:66303, statename:"New Hampshire"}),
records.push({ short: "ct", medinc:65958, statename:"Connecticut"}),
....
This gives you opportunity to create class for State, change Array to Vector and make all of this type-safe, what is always good.
If you really need those keys, you can add objects like above (with "short" field) in the same way you are doing it now (maybe using some helper function which will help to avoid typing shortname twice, like addState(records, data) { records[data.short] = data }).
Finally, you can also keep those records in two objects (or an object and an array or whatever you need). This will not be expensive, if you will create state object once and keep references in array/object/vector. It would be nice idea if you need states sorted on different keys often.
This is not really a good way to have your data set up - too much typing (you are repeating "records", "medinc", "statename" over and over again, while you definitely could've avoided it, for example:
var records:Array = [];
var states:Array = ["nh", "ct", "nj" ... ];
var statenames:Array = ["New Hampshire", "Connecticut", "New Jersey" ... ];
var medincs:Array = [66303, 65958, 65173 ... ];
var hash:Object = { };
function addState(state:String, medinc:int, statename:String, hash:Object):Object
{
return hash[state] = { medinc: medinc, statename: statename };
}
for (var i:int; i < 50; i++)
{
records[i] = addState(states[i], medincs[i], statenames[i], hash);
}
While you have done it already the way you did, that's not essential, but this could've saved you some keystrokes, if you haven't...
Now, onto your search problem - first of all, true, it would be worth to sort the array before you search, but if you need to search an array by the value of the parameter it was sorted on, there is a better algorithm for that. That is, if given the data in your example, your specific task was to find out in what state the income is 65958, then, knowing that array is sorted on income you could employ binary search.
Now, for the example with 50 states the difference will not be noticeable, unless you do it some hundreds of thousands times per second, but in general, the binary search would be the way to go.
If the article in Wiki looks too long to read ;) the idea behind the binary search is that at first you guess that the searched value is exactly in the middle of the array - you try that assumption and if you guessed correct, return the index you just found, else - you select the interval containing the searched value (either one half of the array remaining) and do so until you either find the value, or check the same index - which would mean that the value is not found). This reduces asymptotic complexity of the algorithm from O(n) to O(log n).
Now, if your goal was to find the correspondence between the income and the state, but it wasn't important how that scales with other states (i.e. the index in the array is not important), you could have another hash table, where the income would be the key, and the state information object would be the value, using my example above:
function addState(state:String, medinc:int, statename:String,
hash:Object, incomeHash:Object):Object
{
return incomeHash[medinc] =
hash[state] = { medinc: medinc, statename: statename };
}
Then incomeHash[medinc] would give you the state by income in O(1) time.