Swift 5 group and split array of objects based on object value

Question

I have a sorted array

let things = [
    Thing(activity: "1", name: "value1"),
    Thing(activity: "1", name: "value2"),
    Thing(activity: "1", name: "value3"),
    Thing(activity: "2", name: "value4"),
    Thing(activity: "2", name: "value5"),
    Thing(activity: "3", name: "value6"),
    Thing(activity: "3", name: "value7"),
    Thing(activity: "1", name: "value8"),
    Thing(activity: "1", name: "value9"),
    Thing(activity: "1", name: "value10")
 ]

I would like to produce array of arrays splitted when the activity value changes similar to the following

[[Thing(activity: "1", name: "value1"),
  Thing(activity: "1", name: "value2"),
  Thing(activity: "1", name: "value3")],
 [Thing(activity: "2", name: "value4"),
  Thing(activity: "2", name: "value5")],
 [Thing(activity: "3", name: "value6"),
  Thing(activity: "3", name: "value7")],
 [Thing(activity: "1", name: "value8"),
  Thing(activity: "1", name: "value9"),
  Thing(activity: "1", name: "value10")]]

Use reduce(into:).

matt
– matt

2019-08-14 20:28:42 +00:00
Commented Aug 14, 2019 at 20:28 — matt
– matt, Commented Aug 14, 2019 at 20:28

Rob · Accepted Answer · 2019-08-15 00:20:50Z

4

A generalized solution would be:

extension Sequence {
    func grouped<T: Equatable>(by block: (Element) throws -> T) rethrows -> [[Element]] {
        return try reduce(into: []) { result, element in
            if let lastElement = result.last?.last, try block(lastElement) == block(element) {
                result[result.index(before: result.endIndex)].append(element)
            } else {
                result.append([element])
            }
        }
    }
}

Then you can do:

let results = things.grouped { $0.activity }

A less elegant (but slightly more efficient) solution would be:

extension Sequence {
    func grouped<T: Equatable>(by block: (Element) throws -> T) rethrows -> [[Element]] {
        var results: [[Element]] = []

        var lastValue: T?
        var index = results.endIndex
        for element in self {
            let value = try block(element)
            if let lastValue = lastValue, lastValue == value {
                results[index].append(element)
            } else {
                results.append([element])
                index = results.index(before: results.endIndex)
                lastValue = value
            }
        }
        return results
    } 
}

edited Aug 15, 2019 at 0:20

answered Aug 14, 2019 at 23:54

Rob

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1 Comment

محمد الاشرف أنور Over a year ago

Good answer . I will use Leo's answer as it solved my problem immediately. Thank you.

Leo Dabus · Accepted Answer · 2019-08-15 11:29:23Z

As already mentioned by @matt in comments you can use collection method reduce(into:) to group your elements by checking if the activity of the last element of the last array is equal to the current element activity, if so just append a new element to the last array, otherwise append a new array with a single element to the outer array:

struct Thing {
    let activity, name: String
}

let things: [Thing] = [
    .init(activity: "1", name: "value1"),
    .init(activity: "1", name: "value2"),
    .init(activity: "1", name: "value3"),
    .init(activity: "2", name: "value4"),
    .init(activity: "2", name: "value5"),
    .init(activity: "3", name: "value6"),
    .init(activity: "3", name: "value7"),
    .init(activity: "1", name: "value8"),
    .init(activity: "1", name: "value9"),
    .init(activity: "1", name: "value10")]

let grouped: [[Thing]] = things.reduce(into: []) {
    $0.last?.last?.activity == $1.activity ?
    $0[$0.index(before: $0.endIndex)].append($1) :
    $0.append([$1])
}

print(grouped)  // "[[__lldb_expr_1.Thing(activity: "1", name: "value1"), __lldb_expr_1.Thing(activity: "1", name: "value2"), __lldb_expr_1.Thing(activity: "1", name: "value3")], [__lldb_expr_1.Thing(activity: "2", name: "value4"), __lldb_expr_1.Thing(activity: "2", name: "value5")], [__lldb_expr_1.Thing(activity: "3", name: "value6"), __lldb_expr_1.Thing(activity: "3", name: "value7")], [__lldb_expr_1.Thing(activity: "1", name: "value8"), __lldb_expr_1.Thing(activity: "1", name: "value9"), __lldb_expr_1.Thing(activity: "1", name: "value10")]]\n"

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