I want to add the numbers together and print every 4 elements, however i cannot wrap my head around using the stride function, if i am using the wrong approach please explain a better method
var numbers = [1,2,3,4,5,6,7,8,9,10,11,12,13]
func addNumbersByStride(){
var output = Stride...
//first output = 1+2+3+4 = 10
//second output = 5+6+7+8 = 26 and so on
print(output)
}
It seems you would like to use stride ...
let arr = [1,2,3,4,5,6,7,8,9,10,11,12,13]
let by = 4
let i = stride(from: arr.startIndex, to: arr.endIndex, by: by)
var j = i.makeIterator()
while let n = j.next() {
let e = min(n.advanced(by: by), arr.endIndex)
let sum = arr[n..<e].reduce(0, +)
print("summ of arr[\(n)..<\(e)]", sum)
}
prints
summ of arr[0..<4] 10
summ of arr[4..<8] 26
summ of arr[8..<12] 42
summ of arr[12..<13] 13
You can first split the array into chunks, and then add the chunks up:
extension Array {
// split array into chunks of n
func chunked(into size: Int) -> [[Element]] {
return stride(from: 0, to: count, by: size).map {
Array(self[$0 ..< Swift.min($0 + size, count)])
}
}
}
// add each chunk up:
let results = numbers.chunked(into: 4).map { $0.reduce(0, +) }
If you would like to discard the last sum if the length of the original array is not divisible by 4, you can add an if statement like this:
let results: [Int]
if numbers.count % 4 != 0 {
results = Array(numbers.chunked(into: 4).map { $0.reduce(0, +) }.dropLast())
} else {
results = numbers.chunked(into: 4).map { $0.reduce(0, +) }
}
This is quite a basic solution and maybe not so elegant. First calculate and print sum of every group of 4 elements
var sum = 0
var count = 0
for n in stride(from: 4, to: numbers.count, by: 4) {
sum = 0
for i in n-4..<n {
sum += numbers[i]
}
count = n
print(sum)
}
Then calculate the sum of the remaining elements
sum = 0
for n in count..<numbers.count {
sum += numbers[n]
}
print(sum)
Related
I hope you guys can check. when I use 5 as x it should be showing me -0.17749282815107623 but it returns -0.2792375. I couldn't where I have been doing the mistake.
var evenNumbers = [Int]()
for i in 2...10 {
if i % 2 == 0 {
evenNumbers.append(i)
}
}
func power(val: Float, power: Int)->Float{
var c:Float = 1
for i in 1...power {
c *= val
}
return c
}
func bessel(x: Float)->Float{
var j0:Float = 0
var counter = 1
var lastDetermVal:Float = 1
for eNumber in evenNumbers {
print(lastDetermVal)
if counter == 1 {
lastDetermVal *= power(val: Float(eNumber), power: 2)
j0 += (power(val: x, power: eNumber))/lastDetermVal
counter = -1
}else if counter == -1{
lastDetermVal *= power(val: Float(eNumber), power: 2)
j0 -= (power(val: x, power: eNumber))/lastDetermVal
counter = 1
}
}
return 1-j0
}
bessel(x: 5)
Function 1:
Your mistake seems to be that you didn't have enough even numbers.
var evenNumbers = [Int]()
for i in 2...10 {
if i % 2 == 0 {
evenNumbers.append(i)
}
}
After the above is run, evenNumbers will be populated with [2,4,6,8,10]. But to evaluate 10 terms, you need even numbers up to 18 or 20, depending on whether you count 1 as a "term". Therefore, you should loop up to 18 or 20:
var evenNumbers = [Int]()
for i in 2...18 { // I think the 1 at the beginning should count as a "term"
if i % 2 == 0 {
evenNumbers.append(i)
}
}
Alternatively, you can create this array like this:
let evenNumbers = (1..<10).map { $0 * 2 }
This means "for each number between 1 (inclusive) and 10 (exclusive), multiply each by 2".
Now your solution will give you an answer of -0.1776034.
Here's my (rather slow) solution:
func productOfFirstNEvenNumbers(_ n: Int) -> Float {
if n == 0 {
return 1
}
let firstNEvenNumbers = (1...n).map { Float($0) * 2.0 }
// ".reduce(1.0, *)" means "multiply everything"
return firstNEvenNumbers.reduce(1.0, *)
}
func nthTerm(_ n: Int, x: Float) -> Float {
let numerator = pow(x, Float(n) * 2)
// yes, this does recalculate the product of even numbers every time...
let product = productOfFirstNEvenNumbers(n)
let denominator = product * product
return numerator / (denominator) * pow(-1, Float(n))
}
func bessel10Terms(x: Float) -> Float {
// for each number n in the range 0..<10, get the nth term, add them together
(0..<10).map { nthTerm($0, x: x) }.reduce(0, +)
}
print(bessel10Terms(x: 5))
You code is a bit unreadable, however, I have written a simple solution so try to compare your intermediate results:
var terms: [Float] = []
let x: Float = 5
for index in 0 ..< 10 {
guard index > 0 else {
terms.append(1)
continue
}
// calculate only the multiplier for the previous term
// - (minus) to change the sign
// x * x to multiply nominator
// (Float(index * 2) * Float(index * 2) to multiply denominator
let termFactor = -(x * x) / (Float(index * 2) * Float(index * 2))
terms.append(terms[index - 1] * termFactor)
}
print(terms)
// sum the terms
let result = terms.reduce(0, +)
print(result)
One of the errors I see is the fact that you are actually calculating only 5 terms, not 10 (you iterate 1 to 10, but only even numbers).
I'm trying to implement a solution using the backtracking algorithm.
I have some weights [1,2,7,10,20,70,100,200,700...] and I want to return the weights after a given input/
For example input => 12 should return [2,10]
For example input => 8 should return [1,7]
My code seem's not to work well. It works only for some input numbers like 13 or 8
for targetValue in [13] {
var currentValue = 0
var usedWeights: [Int] = []
for weight in weights {
if targetValue > weight {
currentValue += weight
usedWeights.append(weight)
} else if weight > targetValue {
let rememberLast = usedWeights.last ?? 0
usedWeights.remove(at: usedWeights.count-1)
currentValue -= rememberLast
if currentValue > targetValue || currentValue < targetValue {
let last = usedWeights.remove(at: usedWeights.count-1)
currentValue -= last
usedWeights.append(rememberLast)
currentValue -= rememberLast
print(usedWeights) /// [1, 2, 10] Yeah it work's :) but only for some number ..:(
}
}
}
}
The used weights should be unique.
I have some trouble to find the weights.
This is how the algorithm work
Input => 13
1
1+2
1+2+7
1+2+7+10 //currentValue is now 20
1+2+7 // still no solution get the last removed element and remove the current last element
1+2+10 // Correct weights
I hope you can help me and I explain what I'm doing wrong.
Here's one solution. Iterate in reverse through the weights. If the weight is less than or equal to the current total, use the weight.
let weights = [1,2,7,10,20,70,100,200,700] // the weights you have
let needed = 12 // The total weight you want
var total = needed // The current working total
var needs = [Int]() // The resulting weights needed
for weight in weights.reversed() {
if weight <= total {
needs.append(weight)
total -= weight
}
}
if total == 0 {
print("Need \(needs) for \(needed)")
} else {
print("No match for \(needed)")
}
I don't know how you set the weight.
But consider:
[2,3,6,10,20]
and needed = 21
Then the algorithm will not find it (no match), when there is a solution obviously: 2 + 3 + 6 + 10
So, you should call recursively the search algorithm when it fails, after removing from the weights the first you picked.
This is not very clean, but seem to work (in code, some issue in playground)
func searchIt(weightsSearch: [Int], neededSearch: Int) -> (needs: [Int], remainingWeights: [Int], found: Bool) {
var total = neededSearch // The current working total
var needs = [Int]() // The resulting weights needed
var remaining = weightsSearch
var firstNotYetSelected = true
var position = weightsSearch.count - 1
for weight in weightsSearch.reversed() {
if weight <= total {
needs.append(weight)
total -= weight
if firstNotYetSelected {
remaining.remove(at : position)
}
firstNotYetSelected = false
position -= 1
}
}
return (needs, remaining, total == 0)
}
var needs = [Int]() // The resulting weights needed
var remainingWeights = weights
var foundIt: Bool
repeat {
(needs, remainingWeights, foundIt) = searchIt(weightsSearch: remainingWeights, neededSearch: needed)
if foundIt {
print("Need \(needs) for \(needed)")
break
} else {
print("No match yet for \(needed)")
}
} while remainingWeights.count >= 1
with the test case
let weights = [2,3,6,10,20]
let needed = 21
we get
No match yet for 21
Need [10, 6, 3, 2] for 21
If you want ALL the solutions, replace break statement by continue.
With the test case
let weights = [2,3,6,10,15,20]
let needed = 21
we get the 2 solutions
No match for 21
No match for 21
Need [15, 6] for 21
Need [10, 6, 3, 2] for 21
No match for 21
No match for 21
No match for 21
I am trying to solve the task
Using a standard for-in loop add all odd numbers less than or equal to 100 to the oddNumbers array
I tried the following:
var oddNumbers = [Int]()
var numbt = 0
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
print(oddNumbers)
This results in:
1,3,5,7,9,...199
My question is: Why does it print numbers above 100 although I specify the range between 0 and <100?
You're doing a mistake:
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
The variable newNumt defined inside the loop does not affect the variable newNumt declared in the for statement. So the for loop prints out the first 100 odd numbers, not the odd numbers between 0 and 100.
If you need to use a for loop:
var odds = [Int]()
for number in 0...100 where number % 2 == 1 {
odds.append(number)
}
Alternatively:
let odds = (0...100).filter { $0 % 2 == 1 }
will filter the odd numbers from an array with items from 0 to 100. For an even better implementation use the stride operator:
let odds = Array(stride(from: 1, to: 100, by: 2))
If you want all the odd numbers between 0 and 100 you can write
let oddNums = (0...100).filter { $0 % 2 == 1 }
or
let oddNums = Array(stride(from: 1, to: 100, by: 2))
Why does it print numbers above 100 although I specify the range between 0 and <100?
Look again at your code:
for newNumt in 0..<100 {
var newNumt = numbt + 1; numbt += 2; oddNumbers.append(newNumt)
}
The newNumt used inside the loop is different from the loop variable; the var newNumt declares a new variable whose scope is the body of the loop, so it gets created and destroyed each time through the loop. Meanwhile, numbt is declared outside the loop, so it keeps being incremented by 2 each time through the loop.
I see that this is an old question, but none of the answers specifically address looping over odd numbers, so I'll add another. The stride() function that Luca Angeletti pointed to is the right way to go, but you can use it directly in a for loop like this:
for oddNumber in stride(from:1, to:100, by:2) {
// your code here
}
stride(from:,to:,by:) creates a list of any strideable type up to but not including the from: parameter, in increments of the by: parameter, so in this case oddNumber starts at 1 and includes 3, 5, 7, 9...99. If you want to include the upper limit, there's a stride(from:,through:,by:) form where the through: parameter is included.
If you want all the odd numbers between 0 and 100 you can write
for i in 1...100 {
if i % 2 == 1 {
continue
}
print(i - 1)
}
For Swift 4.2
extension Collection {
func everyOther(_ body: (Element) -> Void) {
let start = self.startIndex
let end = self.endIndex
var iter = start
while iter != end {
body(self[iter])
let next = index(after: iter)
if next == end { break }
iter = index(after: next)
}
}
}
And then you can use it like this:
class OddsEvent: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
(1...900000).everyOther{ print($0) } //Even
(0...100000).everyOther{ print($0) } //Odds
}
}
This is more efficient than:
let oddNums = (0...100).filter { $0 % 2 == 1 } or
let oddNums = Array(stride(from: 1, to: 100, by: 2))
because supports larger Collections
Source: https://developer.apple.com/videos/play/wwdc2018/229/
I got a simple code which works and which I am programming in and old fashioned way and I am sure there is a more elegant way of doing this in swift. Here is the code:
var cardsInCompartment1:Int = 0
var cardsInCompartment2:Int = 0
for card in cards{
if card.compartment == 1{
cardsInCompartment1 += 1
print(cardsInCompartment1)
}
if card.compartment == 2{
cardsInCompartment2 += 1
print(cardsInCompartment2)
}
}
I basically got cards in different compartments and now I want to count how many cards are in each compartment.
How about using filter to select the cards you want? Then you can just count them:
let cardsInCompartment1 = cards.filter { $0.compartment == 1 }.count
let cardsInCompartment2 = cards.filter { $0.compartment == 2 }.count
If you have a bunch of compartments, you could store the counts in a dictionary:
var compartmentCounts = [Int:Int]()
cards.forEach {
compartmentCounts[$0.compartment] = (compartmentCounts[$0.compartment] ?? 0) + 1
}
In this case, the key would be the compartment#, and the value would be the card count. Something like [1: 32, 2: 42] if there are 32 and 42 cards in each respective compartment.
Try this:
var cardsInCompartment1:Int = 0
var cardsInCompartment2:Int = 0
for card in cards {
(card.compartment == 1) ? (cardsInCompartment1 += 1) : (cardsInCompartment2 += 1)
}
I think you should store the cardsInCompartment as arrays:
var cardsInCompartment = [0, 0] // you can add more to this array
Then you can just loop through the cards and add the values to the array elements:
for card in cards {
cardsInCompartment[card.compartment - 1] += 1
print(cardsInCompartment[card.compartment - 1])
}
What about a switch statement? Something like this?
var card:Int = 1
var CardsInCompartment:Int = 0
switch (card) {
case 1:
CardsInCompartment += 1
print("CardsInCompartment \(CardsInCompartment)")
case 2:
CardsInCompartment += 2
print("CardsInCompartment \(CardsInCompartment)")
default:
}
Or, use an Array to keep your counts:
var counts = [ 0, 0, 0 ] // create an array of integers, where the Ints in the array represent the count of cards in each compartment
cards.forEach { counts[ $0.compartment ] += 1 } // for each card, increment the count in array corresponding to the compartment of the card. (if card.compartment == 1, increment counts[1], and so on
print("cards in compartment 1 \(counts[1])")
print("cards in compartment 2 \(counts[2])")
(This assumes your only compartments are integers 1 and 2)
I like Aaron Brager's idea which counts values into dictionary. I am using reduce to eliminate mutable dictionary outside the 'loop' (more functional)
let d = cards.reduce([:]) { (d, card) -> [Int:Int] in
var d = d
let s = d[card.compartment] ?? 0
d[card.compartment] = s + 1
return d
}
I receive an Int from my server which I’d like to explode in to an array of bit masks. So for example, if my server gives me the number 3, we get two values, a binary 1 and a binary 2.
How do I do this in Swift?
You could use:
let number = 3
//radix: 2 is binary, if you wanted hex you could do radix: 16
let str = String(number, radix: 2)
println(str)
prints "11"
let number = 79
//radix: 2 is binary, if you wanted hex you could do radix: 16
let str = String(number, radix: 16)
println(str)
prints "4f"
I am not aware of any nice built-in way, but you could use this:
var i = 3
let a = 0..<8
var b = a.map { Int(i & (1 << $0)) }
// b = [1, 2, 0, 0, 0, 0, 0, 0]
Here is a straightforward implementation:
func intToMasks(var n: Int) -> [Int] {
var masks = [Int]()
var mask = 1
while n > 0 {
if n & mask > 0 {
masks.append(mask)
n -= mask
}
mask <<= 1
}
return masks
}
println(intToMasks(3)) // prints "[1,2]"
println(intToMasks(1000)) // prints "[8,32,64,128,256,512]"
public extension UnsignedInteger {
/// The digits that make up this number.
/// - Parameter radix: The base the result will use.
func digits(radix: Self = 10) -> [Self] {
sequence(state: self) { quotient in
guard quotient > 0
else { return nil }
let division = quotient.quotientAndRemainder(dividingBy: radix)
quotient = division.quotient
return division.remainder
}
.reversed()
}
}
let digits = (6 as UInt).digits(radix: 0b10) // [1, 1, 0]
digits.reversed().enumerated().map { $1 << $0 } // [0, 2, 4]
Reverse the result too, if you need it.