Pad an array with 0's swift [closed] - arrays

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I am attempting to pad an an array with zeros but cannot find a succinct way to do this in swift.
If I have array x = [1,2,3 ; 4,5,6 ; 7,8,9]
in matlab I could use the command
y = [zeros(1,3+2);zeros(3,1),x,zeros(3,1);zeros(1,3+2)]
giving the desired output array of [0,0,0,0,0; 0,1,2,3,0 ; 0,4,5,6,0 ; 0,7,8,9,0 ; 0,0,0,0,0]
However so far in a swift playground I have only been able to refer to each element individually to correctly form the new array.
The methods I have tried so far using x as the input and y as the output, the first similar to matlab,
var x = [[1,2,3,],[4,5,6],[7,8,9]]
var y = [[0,0,0,0,0],[0,x[0],0],[0,x[1],0],[0,x[2],0],[0,0,0,0,0]]
The second being a loop
for i in 0 ..< x.count + 1 {
if i == 0 || i == x.count - 1 {
y[i] = [0,0,0,0,0]
}
else{
y[i] = [0, x[i-1] ,0]
}
}
Rather than looking like a standard array in the Xcode playground preview this is the output given.
[[0, 0, 0, 0, 0], [0, […], 0], [0, 0, 0, 0, 0], [0, […], 0], [0, 0, 0, 0, 0]]
which also prints to the console very strangely, using the code
for i in 0 ..< y.count {
print("\(y[i])")
}
the output is
(
0,
0,
0,
0,
0
)
(
0,
(
1,
2,
3
),
0
)
(
0,
0,
0,
0,
0
)
(
0,
(
7,
8,
9
),
0
)
(
0,
0,
0,
0,
0
)
as opposed to the expected
[0, 0, 0, 0, 0]
[0, 1, 2, 3, 0]
[0, 4, 5, 6, 0]
[0, 7, 8, 9, 0]
[0, 0, 0, 0, 0]
What is the best way to do this?

Answer 1

I made a generic version of appzYourLife's answer that can take any arbitrary nested array type. It also adds support for top and bottom padding
extension Array where Element: _ArrayType {
typealias InnerElement = Element.Generator.Element
func pad2DArray(with padding: InnerElement,
top: Int = 0, left: Int = 0,
right: Int = 0, bottom: Int = 0) -> [[InnerElement]] {
let newHeight = self.count + top + bottom
let newWidth = (self.first?.count ?? 0) + left + right
var paddedArray = [[InnerElement]](count: newHeight, repeatedValue:
[InnerElement](count: newWidth, repeatedValue: padding))
for (rowIndex, row) in self.enumerate() {
for (columnIndex, element) in row.enumerate() {
paddedArray[rowIndex + top][columnIndex + left] = element
}
}
return paddedArray
}
}
var input = [
[1, 2, 3],
[4, 5, 6],
[7, 8, 9]
]
let result = input.pad2DArray(with: 0, top: 1, left: 1, right: 1, bottom: 1)
/*
result:
[
[0, 0, 0, 0, 0],
[0, 1, 2, 3, 0],
[0, 4, 5, 6, 0],
[0, 7, 8, 9, 0],
[0, 0, 0, 0, 0],
]
*/

Answer 2

Extension
If you define this extension
extension _ArrayType where Element == Int {
func pad(left left: Int, right: Int) -> [Int] {
let leftSide = [Int](count: left, repeatedValue: 0)
let rightSide = [Int](count: right, repeatedValue: 0)
return leftSide + (self as! [Int]) + rightSide
}
}
you can then write
[1,2,3].pad(left: 1, right: 1) // [0, 1, 2, 3, 0]

Answer 3

Definitely not as elegant as the accepted answer, but it works nonetheless:
var x = [[1,2,3,],[4,5,6],[7,8,9]]
var y = [[Int]]()
y.insert([0,0,0,0,0], atIndex: 0)
for i in 0 ..< x.count {
var intArray: [Int] = []
for number in x[i] {
intArray.append(number)
}
intArray.insert(0, atIndex: 0)
intArray.insert(0, atIndex: intArray.count)
y.append(intArray)
}
y.insert([0,0,0,0,0], atIndex: x.count + 1)
The output from
for i in 0 ..< y.count {
print("\(y[i])")
}
is
[0, 0, 0, 0, 0]
[0, 1, 2, 3, 0]
[0, 4, 5, 6, 0]
[0, 7, 8, 9, 0]
[0, 0, 0, 0, 0]