我无法将此 Python 函数转换为 Swift 函数

Question

I'm working on this blackjack app for iOS and I'm having trouble getting the function to work properly. I built it in Python where it works like a charm, but something is wonky when I adapt it. The idea is to get every possible value of the cards. Once I get that working, I'll sum up the numbers. So, a [[1,11],[4],[5]] will output [10,20] .

Python code

init_array = [[1,11],[3],[4],[1,11],[5]]
r = [[]]
for x in init_array:
    t = []
    for y in x:
        for i in r:
            t.append(i + [y])
    r = t
print(r)

Output:

[[1, 3, 4, 1, 5], [11, 3, 4, 1, 5], [1, 3, 4, 11, 5], [11, 3, 4, 11, 5]]

Swift code

var init_array = [[Int]]()
init_array.append([1,11])
init_array.append([3])
init_array.append([4])
init_array.append([1,11])
init_array.append([5])

var r = [[Int]]()
for x in init_array {
    var t = [Int]()
    for y in x{
        for i in r {
            t.append(contentsOf: (i + [y]))
        }
    }
    r = [t]
}
print(r)

Output:

[[3, 4, 1, 3, 4, 11, 5]]

It's part of a larger function, so creating the init_array variables and printing the r variables for both examples is for testing purposes. The init_array has already been generated by the time it gets to my trouble code.

It's probably one simple line of code, but I've been trying to figure this out for days. I'm relatively new to Swift, so this might take someone more advanced two seconds to figure out.

Answer 1

When you initialised r : In python , length of r is 1 where as in swift the length is 0 . This is because, in python r is initialised with an empty list in inside it [[]] . But in swift, it didn't happen [] . This resulted in getting unwanted output.

Finally I found a way to fix it by appending an empty list into r in the beginning.

var r = [[Int]]()
r.append([])

Also t.append(ContentsOf ...) wasn't creating arrays as you wanted.

Here is the final solution

var init_array = [[Int]]()
init_array.append([1,11])
init_array.append([3])
init_array.append([4])
init_array.append([1,11])
init_array.append([5])

var r = [[Int]]()
r.append([])
print(r.count,r) // debuggging part
for x in init_array {
    var t = [[Int]]()
    for y in x{
        for i in r {

            //method 1 
            var dm=[Int]() //for making the kind of arrays u wanted
            dm.append(contentsOf: (i + [y]))  // appending it to array
            t.append(dm)
            // Or
            //method 2
            t+=[(contentsOf: (i + [y]))] as! [[Int]] //HereWeTypeCastToInt
        }
        print(t,y) // debuggging part
    }
    r = t
}
print(r)

Output : [[1, 3, 4, 1, 5], [11, 3, 4, 1, 5], [1, 3, 4, 11, 5], [11, 3, 4, 11, 5]]