Use contains instead:

let arr = ["Hello","Bye","Halo"]
let filtered = arr.filter { $0.contains("lo") }
print(filtered)

Output

["Hello", "Halo"]

Thanks to @user3441734 for pointing out that functionality is of course only available when you import Foundation

[Updated for Swift 2.0]

As NSString is toll-free bridged to Swift String, just avoid the coercions with:

  3> ["abc", "bcd", "xyz"].filter() { nil != $0.rangeOfString("bc") }
$R1: [String] = 2 values {
  [0] = "abc"
  [1] = "bcd"
}

But, if you think allValues aren't strings:

(keywords.allValues as? [String]).filter() { nil != $0.rangeOfString("bc") }

which returns an optional array.

Use this code:

let yearArray: [String] = ... // your array of type [String]

let filteredArray = yearArray.filter {
    !$0.isEmpty
}

Look at the picture for output:

You are getting this error because unfortunately range is not available in Linux. If you had access to a machine with macOS this would work.

As @Hamish said, I was completely wrong, you have to import Foundation in order to use range or contains on String (even though you can call contains on string.characters without importing Foundation).

Add this line to the beginning of your code:

import Foundation

I would write this something like this I think...

// make it a function that takes everything it needs and returns a result array
func filteredCountryArray(countryArray: [[String]], searchString: String) -> [[String]] {
    guard !searchString.isEmpty else {
        // search string is blank so return entire array...
        return countryArray
    }

    // Filter the data array and get only those countries that match the search text.
    return countryArray.filter { match in
        // check array is long enough to get [1] out of it
        guard match.count >= 2 else {
            return false
        }

        let matchText = match[1]
        return matchText.range(of: searchString, options: .caseInsensitive) != nil
    }
}

plazaDictionary is a very odd name to give to an array. It is an array, not a dictionary.

What you should probably do here is create some data objects (structs, classes, enums, etc...) that hold the data in a better format than a nested array.

This should do what you want for now though.

EDIT

Having another think about it, I would also change the serachString and the array into input parameters...

One approach is to update your filter to see if any value in pets is in the petArr array:

users = users.filter { $0.pets.contains(where: { petArr.contains($0) }) }

The first $0 is from the filter and it represents each User.

The second $0 is from the first contains and it represents each pet within the pets array of the current User.

How about

let filteredMenuItems = allMenuItems.filter { anArrayOfIds.contains($0.id)}

That gives me the items with 1 and 3.

I've taken the liberty of updating your code a little:

struct MenuItems {
    let id: String
    let name: String
}

let m1 = MenuItems(id: "1", name: "Name 1")
let m2 = MenuItems(id: "2", name: "Name 2")
let m3 = MenuItems(id: "3", name: "Name 3")

let allMenuItems = [m1, m2, m3]
let anArrayOfIds: [String] = ["1", "3"]

let filteredMenuItems = allMenuItems.filter { anArrayOfIds.contains($0.id)}

Hope that helps