Swift: Filter a Dictionary with Array as Value

Question

I'm new to Swift programming. For my particular project, I'm trying to filter a dictionary with some user input, and the dictionary's value consists of an array.

Here is some sample code, and what I'm trying to accomplish:

var dictionary = ["a": ["aberration", "abc"], "b" : ["babel", "bereft"]]

var filteredDictionary = [String: [String]]()

var searchText = "aberration"

//getting the first letter of string
var firstLetter = searchText[searchText.startIndex]

With this particular searchText, I'm trying to get:

filteredDictionary = ["a": ["aberration"]]

Edit: I want the dictionary to return with the first letter as its key, and the values with what searchText matches up with. Sorry if it I wasn't clear.

Here is some code I have tried, but obviously, I can't get it to work:

filteredDictionary = dictionary.filter{$0.key == firstLetter && for element in $0.value { element.hasPrefix(searchText) }}

Any help would be appreciated. Thanks.

I updated the question. I want to return a dictionary with the letter as its key, and the matching searchText as an array as its value. — Alan Dolan
– Alan Dolan, Commented Jan 4, 2018 at 17:30
The part you really need to clarify is what values in the array should be returned. It seems you want any word that has a prefix matching searchText but you don't make that clear (it's implied based on your code though). What if searchText is abe? Do you still want just aberration? What if searchText is ab? Do you want both aberration` and abc since both start with ab? — rmaddy
– rmaddy, Commented Jan 4, 2018 at 18:48

rmaddy · Accepted Answer · 2018-01-04 05:56:38Z

9

Here's a solution that maps the values based on the search and then filters out the empty results.

var dictionary = ["a": ["aberration", "abc"], "b" : ["babel", "bereft"]]
var searchText = "aberration"
let filteredDictionary = dictionary.mapValues { $0.filter { $0.hasPrefix(searchText) } }.filter { !$0.value.isEmpty }
print(filteredDictionary)

Output:

["a": ["aberration"]]

answered Jan 4, 2018 at 5:56

rmaddy

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3 Comments

Alan Dolan Over a year ago

As a followup question, how can you filter it so that it does not filter out the empty results, such that filteredDictionary keeps its keys? ["a": ["aberration"], "b" : [], "c" : [], "d" : [] ] etc.

Alan Dolan Over a year ago

Nevermind, I think I got it by getting rid of the last segment like so: filteredDictionary = dictionary.mapValues { $0.filter { $0.hasPrefix(searchText) } }

rmaddy Over a year ago

Correct, just drop the .filter { !$0.value.isEmpty } if you want to keep the empty keys.

iWheelBuy · Accepted Answer · 2018-01-04 06:44:43Z

2

let firstLetter = String(searchText[searchText.startIndex])
let filteredDictionary = dictionary
    .reduce(into: [String: [String]]()) { (result, object) in
        if object.key == firstLetter {
            let array = object.value.filter({ $0.hasPrefix(searchText) })
            if array.count > 0 {
                result[object.key] = array
            }
        }
    }

Output:

["a": ["aberration"]]

edited Jan 4, 2018 at 6:44

answered Jan 4, 2018 at 5:51

iWheelBuy

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Comments

Alexander · Accepted Answer · 2018-01-04 06:01:19Z

1

Try this:

    var dictionary = ["a": ["aberration", "abc"], "b" : ["babel", "bereft"]]
    var searchText = "aberration"
    var filteredDictionary = dictionary.filter { (key, value) -> Bool in
            return (value as! [String]).contains(searchText)
        }.mapValues { (values) -> [String] in
            return [searchText]
        }
    print(filteredDictionary)

You can use a combination of filter and map to achieve the desired result.

Output:

["a": ["aberration"]]

edited Jan 4, 2018 at 6:01

Alexander

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answered Jan 4, 2018 at 5:52

PGDev

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5 Comments

rmaddy Over a year ago

The values returned by this are not filtered as indicated in the question.

rmaddy Over a year ago

This still isn't correct since the search text needs to be a prefix and can match more than one value in each array.

PGDev Over a year ago

It isn't mentioned anywhere in the question.

rmaddy Over a year ago

It's implied based on the use of hasPrefix in the question's code.

PGDev Over a year ago

Requirements must be specified clearly in the question. Just adding the code isn't enough.

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