How to enumerate a slice using the original indices?

问题

If I want to enumerate an array (say for a map() function where I would need to use the index of an element together with its value), I could use enumerate() function. E.g.:

import Foundation

let array: [Double] = [1, 2, 3, 4]

let powersArray = array.enumerate().map() {
    pow($0.element, Double($0.index))
}

print("array == \(array)")
print("powersArray == \(powersArray)")

// array == [1.0, 2.0, 3.0, 4.0]
// powersArray == [1.0, 2.0, 9.0, 64.0] <- As expected

Now, if I want to use some sub-sequence from the array, I could use a slice, and it would allow me to use just the same indices as I would use in the original array (which is just what I want in case if I would use subscript accessor in a for loop). E.g.:

let range = 1..<(array.count - 1)
let slice = array[range]
var powersSlice = [Double]()

for index in slice.indices {
    powersSlice.append(pow(slice[index], Double(index)))
}

print("powersSlice == \(powersSlice)")

// powersSlice == [2.0, 9.0] <- As expected

However, should I try to use enumerate().map() approach like I did with original array, then I get totally different behaviour. Instead of slice's range of indices I would get a new, 0-based range:

let powersSliceEnumerate = slice.enumerate().map() {
    pow($0.element, Double($0.index))
}

print("powersSliceEnumerate == \(powersSliceEnumerate)")

// powersSliceEnumerate == [1.0, 3.0] <- Not as expected

Question is whether there is a decent way (i.e. without manual adjustments using offsets, or something) to enumerate a slice using its own indices and not the auto-generated 0-based ones?

回答1:

enumerate() returns a sequence of (n, elem) pairs, where the ns are consecutive Ints starting at zero. This makes sense because it is a protocol extension method for SequenceType, and an arbitrary sequence does not necessarily have index associated with the elements.

You would get your expected result with

let powersSlice = slice.indices.map { pow(slice[$0], Double($0)) }

let powersSlice = zip(slice.indices, slice).map { pow($1, Double($0)) }

The latter approach could be generalized to a protocol extension method for arbitrary collections:

extension CollectionType {
    func indexEnumerate() -> AnySequence<(index: Index, element: Generator.Element)> {
        return AnySequence(zip(indices, self))
    }
}

This returns a sequence of (index, elem) pairs where index is an index of the collection and elem the corresponding element. AnySequence is used to "hide" the specific type Zip2Sequence<RangeGenerator<Self.Index>, Self> returned from zip() from the caller.

Example:

let powersSliceEnumerate = slice.indexEnumerate().map() { pow($0.element, Double($0.index)) }
print("powersSliceEnumerate == \(powersSliceEnumerate)")
// powersSliceEnumerate == [2.0, 9.0]

Update for Swift 3:

extension Collection {
    func indexEnumerate() -> AnySequence<(Indices.Iterator.Element, Iterator.Element)> {
        return AnySequence(zip(indices, self))
    }
}

来源：https://stackoverflow.com/questions/32688798/how-to-enumerate-a-slice-using-the-original-indices

标签

swift

slice

enumerate