CODEWITHSUNDEEP
javascript
Sara Ree
Sara Ree

Reputation: 3543

How to optimize conditional statements in my code

I have a 2D array named extraRank which holds other users arrays named userA, userB, userC and userD whenever I push them into it:

// users arrays:
userA = ["Sara Ree",18, "Doctor", "441", "31"];
userB = ["Marry",18, "Nurse", "41", "1"];
userC = ["John Doe",18, "Woker", "21", "31"];
userD = ["Alex Morph",18, "Musisian", "81", "51"];

// users arrays may be pushed to this array  
let extraRank = [];

what I want is an efficient way to do this:

first, check which users arrays exist in extraRank.

then if for example, userA exist do this:

if extraRank.indexOf(userA) == 0 then assign userA_Index = 10;

if extraRank.indexOf(userA) == 1 then assign userA_Index = 9;

if extraRank.indexOf(userA) == 2 then assign userA_Index = 8;

if extraRank.indexOf(userA) == 3 then assign userA_Index = 7;

I know one way is to this using if statements for each user, but I'm wondering if there is a more efficient way to do it...

Upvotes: 1

Views: 47

Answers (2)

Kevin FONTAINE
Kevin FONTAINE

Reputation: 845

I don't totally understand your requierement, but would a simple substraction do the trick ?

userA_Index = 10 - extraRank.indexOf(userA)

But I'm not sure of what you want to achieve here.

Edit: You say that you want to optimize for execution time. This will be more efficient that a bunch of if. However, I don't think even 10 if statements are going to be noticeable. How slow is your code now ? How many user do you have ? You migth have other place to optimise first, but it is hard to say without a larger view of your app.

Upvotes: 1

gbalduzzi
gbalduzzi

Reputation: 10176

If the assignment is linear, as suggested by your example, you can do a simple subtraction:

userA_index = 10 - extraRank.indexOf(userA)

If it is more complex, you can create a map object that connects the index to the result, such as:

const mapIndexToOutput = {
  0: 10,
  1: 9,
  2: 8,
  3: 7,
  ...
}

userA_index = mapIndexToOutput[extraRank.indexOf(userA)]

Upvotes: 2

Related Questions