How to create an array using a loop

Question

Cotherstone OP

Created 6d

Replies 2

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Participants 2

Hello,

Please can you tell me how to create an array of dictionaries? This code below should create 4 dictionaries in an array, but I'm getting these errors:

For line "var output = [id: "testID", name: "testName"]":

cannot find 'name' in scope Type '(any AnyObject).Type'

cannot conform to 'Hashable'

For line "return output":

Type '(any AnyObject).Type' cannot conform to 'Hashable'

    var quotes: [(id: String, name: String)] {
      var output = [[(id: String, name: String)]] ()
        for i in 1...4 {
            var output = [id: "testID", name: "testName"]
        }
      return output
    }

Answered by Claude31 in 827336022

There were several mistakes:

[(id: String, name: String)]

This is not a good definition of a dictionary, which is just [key:value], no labels

var output = [id: "testID", name: "testName"]

It does not make sense. If you wrote

var output = ["id": "testID", "name": "testName"]

that would be a dictionary with 2 elements, but that's not what you need here

In your loop, you replace output at each iteration. So at the end you get only the 4th input.

        for i in 1...4 {
            var output = [id: "testID", name: "testName"]
        }

Here is a correct code. I created a type alias for better readability.

typealias MyDict = [String: String] // A dictionary is defined as [key:Value]
var quotes: [MyDict] {
    var allQuotes = [MyDict]()  // Creates an empty array
      for i in 1...4 {
          var newOutput = ["testID" : "testName \(i)", "secondID" : "secondName \(i)"]  // a dict with 2 pairs          allQuotes.append(newOutput)   // We append the new output to the array of Dicts
      }
    return allQuotes
  }
print(quotes)

And you get the expected array:

[["secondID": "secondName 1", "testID": "testName 1"], ["testID": "testName 2", "secondID": "secondName 2"], ["testID": "testName 3", "secondID": "secondName 3"], ["secondID": "secondName 4", "testID": "testName 4"]]

Note that as always in dictionaries, the order of pairs is never guaranteed

If you ask for an item:

print(quotes[0]["testID"])

you get an optional (as the value for the key is nil if there is no value)

Optional("testName 1")

So you should write:

print(quotes[0]["testID"] ?? "not found")

and get:

testName 1

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Answer 1

Claude31 OP

6d

Accepted Answer