Swift: Removing limited number of Objects from an Array without using index

Question

I have the following array:

let chPizza = ["type": "deep", "Style" : "Chicago", "Size" : 12]
let nyPizza = ["type": "thin", "Style" : "New York", "Size" : 14]
let caPizza = ["type": "thai", "Style" : "California", "Size" : 12]
let gkPizza = ["type": "thick", "Style" : "Greek", "Size" : 16]


var pizzas = [chPizza, chPizza, gkPizza, nyPizza, caPizza, chPizza, chPizza, gkPizza, caPizza, chPizza]

How can I remove the first 3 elements of chPizza? Do I have to use the old for-loop, or is there a high-order function that I can use?

What do you mean by "remove the first 3 elements without knowing the index"? What's your expected result? — Code Different
– Code Different, Commented Mar 4, 2016 at 4:13
do you mean remove the first three instances of "chPizza" in the array "pizzas" — Blake Lockley
– Blake Lockley, Commented Mar 4, 2016 at 4:32

matt · Accepted Answer · 2016-03-04 04:39:19Z

2

Let's create our own higher-order method:

extension Array where Element:Equatable {
    mutating func removeObject(obj:Element) {
        if let ix = self.indexOf(obj) {
            self.removeAtIndex(ix)
        }
    }
}

Okay, here we go; it's now a one-liner (but observe that we must cast to NSDictionary because Swift dictionaries are not Equatable so you can't find one in an array):

let chPizza = ["type": "deep", "Style" : "Chicago", "Size" : 12]
let nyPizza = ["type": "thin", "Style" : "New York", "Size" : 14]
let caPizza = ["type": "thai", "Style" : "California", "Size" : 12]
let gkPizza = ["type": "thick", "Style" : "Greek", "Size" : 16]

var pizzas : [NSDictionary] = [chPizza, chPizza, gkPizza, nyPizza, caPizza, chPizza, chPizza, gkPizza, caPizza, chPizza]

(0..<3).forEach {_ in pizzas.removeObject(chPizza)}

Note that this is inefficient! But we can afford that if the array is small and the number of times we remove is small.

edited Mar 4, 2016 at 4:39

answered Mar 4, 2016 at 4:36

matt

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3 Comments

user1107173 Over a year ago

you're just killing it today!

matt Over a year ago

It's not my fault. You're the one who keeps asking the fun questions!

Bhavuk Jain Over a year ago

This gives me an error: Type ["String":"String"] does not conform to protocol 'Equatable'

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