I have a structure like [[[ ]]]
which I want to convert to []
.
E.g. [ [ [ \"Hi\" ] ] ]
into [ \"Hi\" ]
How
joined() returns (a lazy view of) the elements of an collection, concatenated. This can be applied repeatedly for deeper nested collections:
let arr = [ [ [ "A", "B" ], ["C"] ], [ [ "D", "E" ], ["F"] ] ]
let flattened = Array(arr.joined().joined())
print(flattened) // ["A", "B", "C", "D", "E", "F"]
The outer Array()
constructor builds an array from the sequence.
Apart from that, no intermediate arrays are created.
If you just want to iterate over the nested array then the joined sequence is sufficient:
for elem in arr.joined().joined() {
print(elem)
}
Use reduce(_:_:) with your array this way.
let array = [[["One","Two","Three"],["Four","Five"]],[["Six"]]]
let newArray = array.reduce([]) { $0 + $1.reduce([]){ $0 + $1 } }
print(newArray) //["One", "Two", "Three", "Four", "Five", "Six"]
you can use join as
let numbers = [[1,2,3],[4],[5,6,7,8,9]]
let newArray = Array(numbers.joined().joined())
print(newArray)//[1,2,3,4,5,6,7,8,9]]
This is exactly what flatMap() does:
let arr = [ [ [ "A", "B" ], ["C"] ], [ [ "D", "E" ], ["F"] ] ]
// each call reduces the array by one dimension
let flattened = arr.flatMap{$0}.flatMap{$0}
// returns ["A", "B", "C", "D", "E", "F"]