G Alexander - 1 year ago 80

Swift Question

Is this possible to do a generic conversion of floating points and return arbitrary number of arrays?

Perhaps I don't understand how to do arbitrary recursion with arrays?

example

input vector

`Vector:[CGFloat] = [-0.23,-5.2,0.24,-1.4,4.0,10.2]`

output vector if stride by 2

`VectorOut=[[-0.23,0.24,4.0],[-5.2,-1.4,10.2]]`

output vector if stride by 3

`VectorOut=[[-0.23,-1.4],[-5.2,4.0],[0.24.10.2]]`

func splitArray<T:FloatingPoint>(Vector x:[T], byStride N: Int)->([T],[T],...[byStride])

{

guard (x.count % byStride == 0) else {"BOMB \(#file) \(#line) \(#function)"

return [NaN]

}

var index:[Int] = [Int](repeating: 0.0, count: byStride)

var bigVector:[[T]] = [T](repeating: 0.0, count: byStride)

for idx in index {

for elem in stride(from:idx, to: x.count - 1 - idx, by:byStride){

bigVector[idx]=x[idx]

}

}

return bigVector

}

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Answer Source

You cannot return a variable-sized tuple. What you can do is to return a nested array.

Example (hoping that I interpreted the problem correctly):

```
func splitArray<T>(vector x:[T], byStride N: Int) -> [[T]] {
precondition(x.count % N == 0, "stride must evenly divide the array count")
return (0..<N).map { stride(from: $0, to: x.count, by: N).map { x[$0] } }
}
let a = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0]
print(splitArray(vector: a, byStride: 2))
// [[1.0, 3.0, 5.0], [2.0, 4.0, 6.0]]
print(splitArray(vector: a, byStride: 3))
// [[1.0, 4.0], [2.0, 5.0], [3.0, 6.0]]
```

*Explanation:* `(0..<N).map { ... }`

creates an array of size `N`

.
For each value `$0`

,

```
stride(from: $0, to: x.count, by: N).map { x[$0] }
```

creates the "stride" of `x`

starting at index `$0`

with increment `N`

:

```
[ x[$0], x[$0+N], x[$0+2*N], ..., x[x.count-N+$0] ]
```

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