Swift 1D Dictionary to 2D - ios

Question

I've got a little problem in understanding two-dimensional dictionaries. My function has to return dictionary for UITableView with sections. 1 template type can have multiple template strings. So when in fetchedData there are 2 or more texts with similar types, they have to be in array [String] with 1 key - String.
The code below is absolutely correct from the complier point of view. As for me smth is wrong, but nice auto-completions make me think that everything is OK.
Obviously it returns an empty dictionary [:]
func fetchTemplates() -> Dictionary<String, [String]> {
var templates: Dictionary<String, [String]> = [:]
let fetchRequest: NSFetchRequest<Template> = Template.fetchRequest()
fetchRequest.sortDescriptors = [SortDescriptor.init(key: "templateType", ascending: true)]
let fetchedData = try! context.fetch(fetchRequest)
if (!fetchedData.isEmpty) {
for templateItem in fetchedData {
templates[templateItem.templateType!]?.append(templateItem.templateText!)
}
return templates
}
else {
return templates
}
}
P.S. fetchedData returns:
<Template: 0x003281h4> (entity: Template; id: 0x003281h4 <x-coredata:///Template/> ; data: {
templateText = "Example";
templateType = "First";
})

Answer 1

The issue lies on this line:
templates[templateItem.templateType!]?.append(templateItem.templateText!)
templates was initialized with this line: var templates: Dictionary<String, [String]> = [:]. At this point, templates is an empty dictionary.
Let's break that line down into the steps that happen, in chronological order:
templateItem.templateType is accessed, and force unwrapped. A crash will occur if it's nil.
templateItem.templateType! is used as a key into the templates dictionary. This will always return nil. The dictionary is empty, thus is has no values for any keys, including this one.
?.append() is called, on the condition that it's not being called on nil. If it's called on nil, nothing will happen.
3 is the cause of your issue. You need to initialize a new array if one doesn't exist for the key yet:
func fetchTemplates() -> Dictionary<String, [String]> {
var templates: Dictionary<String, [String]> = [:]
let fetchRequest: NSFetchRequest<Template> = Template.fetchRequest()
fetchRequest.sortDescriptors = [SortDescriptor.init(key: "templateType", ascending: true)]
let fetchedData = try! context.fetch(fetchRequest)
if (!fetchedData.isEmpty) { //see note 2
for templateItem in fetchedData {
let type = templateItem.templateType!
var array = templates[type] ?? [] //see note 1
array!.append(templateItem.templateText!)
templates[type] = array
}
return templates
}
else {
return templates
}
}
This function can be simplified:
func fetchTemplates() -> [String : [String]] {
let fetchRequest = Template.fetchRequest()
fetchRequest.sortDescriptors = [SortDescriptor(key: "templateType", ascending: true)]
let fetchedData = try! context.fetch(fetchRequest)
var templates = [String, [String]]()
for templateItem in fetchedData {
let type = templateItem.templateType!
templates[type] = (templates[text] ?? []) + [templateItem.templateText!]
}
return templates
}
and reduce can be used instead:
func fetchTemplates() -> [String : [String]] { //see note 3
let fetchRequest = Template.fetchRequest() //see note 4
fetchRequest.sortDescriptors = [SortDescriptor(key: "templateType", ascending: true)] //see note 5
let fetchedData = try! context.fetch(fetchRequest)
return fetchedData.reduce([String, [String]]()){templates, templateItem in
(templates[templateItem.tempalteText!] ?? []) + [templateItem.templateText!]
} //see note 6
}
Notes
if template[text] is not nil, it's assigned to array. Otherwise, a new array ([]) is assigned to `array.
this check is unnecessary
Dictionary<String, [String]> can be written as just [String : [String]]
No need for an explicit type signiture
X.init() can be written as just X()
The emptiness check is unnecessary, and the whole for loop can be changed into a reduce call.

Answer 2

The issue is:
templates[templateItem.templateType!] is always nil because the dictionary is empty.
Therefore nothing can be appended.