I'm trying to get the average of an array of Ints using the following code:
let numbers = [1,2,3,4,5]
let avg = numbers.reduce(0) { return $0 + $1 / numbers.count }
print(avg) // 1
Which is obviously incorrect. However, if I remove the division to the outside of the closure:
let numbers = [1,2,3,4,5]
let avg = numbers.reduce(0) { return $0 + $1 } / numbers.count
print(avg) // 3
Bingo! I think I remember reading somewhere (can't recall if it was in relation to Swift, JavaScript or programming math in general) that this has something to do with the fact that dividing the sum by the length yields a float / double e.g. (1 + 2) / 5 = 0.6 which will be rounded down within the sum to 0. However I would expect ((1 + 2) + 3) / 5 = 1.2 to return 1, however it too seems to return 0.
With doubles, the calculation works as expected whichever way it's calculated, as long as I box the count integer to a double:
let numbers = [1.0,2.0,3.0,4.0,5.0]
let avg = numbers.reduce(0) { return $0 + $1 / Double(numbers.count) }
print(avg) // 3
I think I understand the why (maybe not?). But I can't come up with a solid example to prove it.
Any help and / or explanation is very much appreciated. Thanks.
The division does not yield a double; you're doing integer division.
You're not getting ((1 + 2) + 3 etc.) / 5.
In the first case, you're getting (((((0 + (1/5 = 0)) + (2/5 = 0)) + (3/5 = 0)) + (4/5 = 0)) + (5/5 = 1)) = 0 + 0 + 0 + 0 + 0 + 1 = 1.
In the second case, you're getting ((((((0 + 1) + 2) + 3) + 4) + 5) / 5) = 15 / 5 = 3.
In the third case, double precision loss is much smaller than the integer, and you get something like (((((0 + (1/5.0 = 0.2)) + (2/5.0 = 0.4)) + (3/5.0 = 0.6)) + (4/5.0 = 0.8)) + (5/5.0 = 1.0)).
The problem is that what you are attempting with the first piece of code does not make sense mathematically.
The average of a sequence is the sum of the entire sequence divided by the number of elements.
reduce calls the lambda function for every member of the collection it is being called on. Thus you are summing and dividing all the way through.
For people finding it hard to understand the original answer.
Consider.
let x = 4
let y = 3
let answer = x/y
You expect the answer to be a Double, but no, it is an Int. For you to get an answer which is not a rounded down Int. You must explicitly state the values to be Double. See below
let doubleAnswer = Double(x)/Double(y)
Hope this helped.
Related
Having an infinite sequence s = 1234567891011...
Let's find the number at the n position (n <= 10^18)
EX: n = 12 => 1; n = 15 => 2
import Foundation
func findNumber(n: Int) -> Character {
var i = 1
var z = ""
while i < n + 1 {
z.append(String(i))
i += 1
}
print(z)
return z[z.index(z.startIndex, offsetBy: n-1)]
}
print(findNumber(n: 12))
That's my code but when I find the number at 100.000th position, it returns an error, I thought I appended too many i to z string.
Can anyone help me, in swift language?
The problem we have here looks fairly straight forward. Take a list of all the number 1-infinity and concatenate them into a string. Then find the nth digit. Straight forward problem to understand. The issue that you are seeing though is that we do not have an infinite amount of memory nor time to be able to do this reasonably in a computer program. So we must find an alternative way around this that does not just add the numbers onto a string and then find the nth digit.
The first thing we can say is that we know what the entire list is. It will always be the same. So can we use any properties of this list to help us?
Let's call the input number n. This is the position of the digit that we want to find. Let's call the output digit d.
Well, first off, let's look at some examples.
We know all the single digit numbers are just in the same position as the number itself.
So, for n<10 ... d = n
What about for two digit numbers?
Well, we know that 10 starts at position 10. (Because there are 9 single digit numbers before it). 9 + 1 = 10
11 starts at position 12. Again, 9 single digits + one 2 digit number before it. 9 + 2 + 1 = 12
So how about, say... 25? Well that has 9 single digit numbers and 15 two digit numbers before it. So 25 starts at 9*1 + 15*2 + 1 = 40 (+ 1 as the sum gets us to the end of 24 not the start of 25).
So... 99 starts at? 9*1 + 89*2 + 1 = 188.
The we do the same for the three digit numbers...
100... 9*1 + 90*2 + 1 = 190
300... 9*1 + 90*2 + 199*3 + 1 = 787
1000...? 9*1 + 90*2 + 900*3 + 1 = 2890
OK... so now I'm seeing a pattern here that seems to need to know the number of digits in each number. Well... I can get the number of digits in a number by rounding up the log(base 10) of that number.
rounding up log base 10 of 5 = 1
rounding up log base 10 of 23 = 2
rounding up log base 10 of 99 = 2
rounding up log base 10 of 627 = 3
OK... so I think I need something like...
// in pseudo code
let lengthOfNumber = getLengthOfNumber(n)
var result = 0
for each i from 0 to lengthOfNumber - 1 {
result += 9 * 10^i * (i + 1) // this give 9*1 + 90*2 + 900*3 + ...
}
let remainder = n - 10^(lengthOfNumber - 1) // these have not been added in the loop above
result += remainder * lengthOfNumber
So, in the above pseudo code you can give it any number and it will return the position in the list that that number starts on.
This isn't the exact same as the problem you are trying to solve. And I don't want to solve it for you.
This is just a leg up on how I would go about solving it. Hopefully, this will give you some guidance on how you can take this further and solve the problem that you are trying to solve.
I am writing code in c++ to detect if an input number is a Palindrome Number, which means its reverse is the same as the origin. I have problems computing the reverse int.
e.g.
121 returns true;
123 returns false;
12321 returns true;
10 returns false;
I input 123 and the sum should be 321. However, my code keeps returning 386. I stepped into the function with xcode. Still, I have no idea why reverse += (3 * 10) + 2 turns to be 35 or why the final reverse number to be 386.
int origin = x;
int reverse = 0;
while (x != 0) {
int digit = x % 10;
reverse += ((reverse * 10) + digit);
x /= 10;
}
why reverse += (3 * 10) + 2 turns to be 35
Because += adds what is on the right to the existing value of what’s on the left. (3 * 10) + 2 is 32, but reverse was already 3 and so you are adding your 32 to the existing 3, which is 35.
You don’t want to add to the value of reverse; you want to replace it.
Change
reverse += ((reverse * 10) + digit)
To
reverse = ((reverse * 10) + digit)
I am trying to code an RSI (which has been a good way for me to learn API data fetching and algorithms already).
The API I am fetching data from comes from a reputable exchange so I know the values my algorithm is analyzing are correct, that's a good start.
The issue I'm having is that the result of my calculations are completely off from what I can read on that particular exchange and which also provides an RSI indicator (I assume they analyze their own data, so the same data as I have).
I used the exact same API to translate the Ichimoku indicator into code and this time everything is correct! I believe my RSI calculations might be wrong somehow but I've checked and re-checked many times.
I also have a "literal" version of the code where every step is calculated like an excel sheet. It's pretty stupid in code but it validates the logic of the calculation and the results are the same as the following code.
Here is my code to calculate the RSI :
let period = 14
// Upward Movements and Downward Movements
var upwardMovements : [Double] = []
var downwardMovements : [Double] = []
for idx in 0..<15 {
let diff = items[idx + 1].close - items[idx].close
upwardMovements.append(max(diff, 0))
downwardMovements.append(max(-diff, 0))
}
// Average Upward Movements and Average Downward Movements
let averageUpwardMovement1 = upwardMovements[0..<period].reduce(0, +) / Double(period)
let averageDownwardMovement1 = downwardMovements[0..<period].reduce(0, +) / Double(period)
let averageUpwardMovement2 = (averageUpwardMovement1 * Double(period - 1) + upwardMovements[period]) / Double(period)
let averageDownwardMovement2 = (averageDownwardMovement1 * Double(period - 1) + downwardMovements[period]) / Double(period)
// Relative Strength
let relativeStrength1 = averageUpwardMovement1 / averageDownwardMovement1
let relativeStrength2 = averageUpwardMovement2 / averageDownwardMovement2
// Relative Strength Index
let rSI1 = 100 - (100 / (relativeStrength1 + 1))
let rSI2 = 100 - (100 / (relativeStrength2 + 1))
// Relative Strength Index Average
let relativeStrengthAverage = (rSI1 + rSI2) / 2
BitcoinRelativeStrengthIndex.bitcoinRSI = relativeStrengthAverage
Readings at 3:23pm this afternoon give 73.93 for my algorithm and 18.74 on the exchange. As the markets are crashing right now and I have access to different RSIs on different exchanges, they all display an RSI below 20 so my calculations are off.
Do you guys have any idea?
I am answering this 2 years later, but hopefully it helps someone.
RSI gets more precise the more data points you feed into it. For a default RSI period of 14, you should have at least 200 previous data points. The more, the better!
Let's suppose you have an array of close candle prices for a given market. The following function will return RSI values for each candle. You should always ignore the first data points, since they are not precise enough or the number of candles is not the 14 (or whatever your periods number is).
func computeRSI(on prices: [Double], periods: Int = 14, minimumPoints: Int = 200) -> [Double] {
precondition(periods > 1 && minimumPoints > periods && prices.count >= minimumPoints)
return Array(unsafeUninitializedCapacity: prices.count) { (buffer, count) in
buffer.initialize(repeating: 50)
var (previousPrice, gain, loss) = (prices[0], 0.0, 0.0)
for p in stride(from: 1, through: periods, by: 1) {
let price = prices[p]
let value = price - previousPrice
if value > 0 {
gain += value
} else {
loss -= value
}
previousPrice = price
}
let (numPeriods, numPeriodsMinusOne) = (Double(periods), Double(periods &- 1))
var avg = (gain: gain / numPeriods, loss: loss /numPeriods)
buffer[periods] = (avg.loss > .zero) ? 100 - 100 / (1 + avg.gain/avg.loss) : 100
for p in stride(from: periods &+ 1, to: prices.count, by: 1) {
let price = prices[p]
avg.gain *= numPeriodsMinusOne
avg.loss *= numPeriodsMinusOne
let value = price - previousPrice
if value > 0 {
avg.gain += value
} else {
avg.loss -= value
}
avg.gain /= numPeriods
avg.loss /= numPeriods
if avgLoss > .zero {
buffer[p] = 100 - 100 / (1 + avg.gain/avg.loss)
} else {
buffer[p] = 100
}
previousPrice = price
}
count = prices.count
}
}
Please note that the code is very imperative to reduce the amount of operations/loops and get the maximum compiler optimizations. You might be able to squeeze more performance using the Accelerate framework, though. We are also handling the edge case where you might get all gains or losses in a periods range.
If you want to have a running RSI calculation. Just store the last RSI value and perform the RSI equation for the new price.
Like in many languages, Swift's division operator defaults to integer division, so:
let n = 1 / 2
print(n) // 0
If you want floating point division, you have to do
let n1 = 1.0 / 2
let n2 = 1 / 2.0
let n3 = Double(1) / 2
let n4 = 1 / Double(2)
print(n1) // 0.5
print(n2) // 0.5
print(n3) // 0.5
print(n4) // 0.5
Again, like most other languages, you can't cast the whole operation:
let n5 = Double(1 / 2)
print(n5) // 0.0
Which happens because swift performs the integer division of 1 and 2 (1 / 2) and gets 0, which it then tries to cast to a Double, effectively giving you 0.0.
I am curious as to why the following works:
let n6 = (1 / 2) as Double
print(n6) // 0.5
I feel like this should produce the same results as Double(1 / 2). Why doesn't it?
1 and 2 are literals. They have no type unless you give them a type from context.
let n6 = (1 / 2) as Double
is essentially the same as
let n6: Double = 1 / 2
that means, you tell the compiler that the result is a Double. That means the compiler searches for operator / with a Double result, and that means it will find the operator / on two Double operands and therefore considers both literals as of type Double.
On the other hand,
let n5 = Double(1 / 2)
is a cast (or better said, initialization of a Double). That means the expression 1 / 2 gets evaluated first and then converted to Double.
I'm looking to create a function that returns a solve for x math equation that can be preformed in ones head (Clearly thats a bit subjective but I'm not sure how else to phrase it).
Example problem: (x - 15)/10 = 6
Note: Only 1 x in the equation
I want to use the operations +, -, *, /, sqrt (Only applied to X -> sqrt(x))
I know that let mathExpression = NSExpression(format: question) converts strings into math equations but when solving for x I'm not sure how to go about doing this.
I previously asked Generating random doable math problems swift for non solving for x problems but I'm not sure how to convert that answer into solving for x
Edit: Goal is to generate an equation and have the user solve for the variable.
Since all you want is a string representing an equation and a value for x, you don't need to do any solving. Just start with x and transform it until you have a nice equation. Here's a sample: (copy and paste it into a Playground to try it out)
import UIKit
enum Operation: String {
case addition = "+"
case subtraction = "-"
case multiplication = "*"
case division = "/"
static func all() -> [Operation] {
return [.addition, .subtraction, .multiplication, .division]
}
static func random() -> Operation {
let all = Operation.all()
let selection = Int(arc4random_uniform(UInt32(all.count)))
return all[selection]
}
}
func addNewTerm(formula: String, result: Int) -> (formula: String, result: Int) {
// choose a random number and operation
let operation = Operation.random()
let number = chooseRandomNumberFor(operation: operation, on: result)
// apply to the left side
let newFormula = applyTermTo(formula: formula, number: number, operation: operation)
// apply to the right side
let newResult = applyTermTo(result: result, number: number, operation: operation)
return (newFormula, newResult)
}
func applyTermTo(formula: String, number:Int, operation:Operation) -> String {
return "\(formula) \(operation.rawValue) \(number)"
}
func applyTermTo(result: Int, number:Int, operation:Operation) -> Int {
switch(operation) {
case .addition: return result + number
case .subtraction: return result - number
case .multiplication: return result * number
case .division: return result / number
}
}
func chooseRandomNumberFor(operation: Operation, on number: Int) -> Int {
switch(operation) {
case .addition, .subtraction, .multiplication:
return Int(arc4random_uniform(10) + 1)
case .division:
// add code here to find integer factors
return 1
}
}
func generateFormula(_ numTerms:Int = 1) -> (String, Int) {
let x = Int(arc4random_uniform(10))
var leftSide = "x"
var result = x
for i in 1...numTerms {
(leftSide, result) = addNewTerm(formula: leftSide, result: result)
if i < numTerms {
leftSide = "(" + leftSide + ")"
}
}
let formula = "\(leftSide) = \(result)"
return (formula, x)
}
func printFormula(_ numTerms:Int = 1) {
let (formula, x) = generateFormula(numTerms)
print(formula, " x = ", x)
}
for i in 1...30 {
printFormula(Int(arc4random_uniform(3)) + 1)
}
There are some things missing. The sqrt() function will have to be implemented separately. And for division to be useful, you'll have to add in a system to find factors (since you presumably want the results to be integers). Depending on what sort of output you want, there's a lot more work to do, but this should get you started.
Here's sample output:
(x + 10) - 5 = 11 x = 6
((x + 6) + 6) - 1 = 20 x = 9
x - 2 = 5 x = 7
((x + 3) * 5) - 6 = 39 x = 6
(x / 1) + 6 = 11 x = 5
(x * 6) * 3 = 54 x = 3
x * 9 = 54 x = 6
((x / 1) - 6) + 8 = 11 x = 9
Okay, let’s assume from you saying “Note: Only 1 x in the equation” that what you want is a linear equation of the form y = 0 = β1*x + β0, where β0 and β1 are the slope and intercept coefficients, respectively.
The inverse of (or solution to) any linear equation is given by x = -β0/β1. So what you really need to do is generate random integers β0 and β1 to create your equation. But since it should be “solvable” in someone’s head, you probably want β0 to be divisible by β1, and furthermore, for β1 and β0/β1 to be less than or equal to 12, since this is the upper limit of the commonly known multiplication tables. In this case, just generate a random integer β1 ≤ 12, and β0 equal to β1 times some integer n, 0 ≤ n ≤ 12.
If you want to allow simple fractional solutions like 2/3, just multiply the denominator and the numerator into β0 and β1, respectively, taking care to prevent the numerator or denominator from getting too large (12 is again a good limit).
Since you probably want to make y non-zero, just generate a third random integer y between -12 and 12, and change your output equation to y = β1*x + β0 + y.
Since you mentioned √ could occur over the x variable only, that is pretty easy to add; the solution (to 0 = β1*sqrt(x) + β0) is just x = (β0/β1)**2.
Here is some very simple (and very problematic) code for generating random integers to get you started:
import func Glibc.srand
import func Glibc.rand
import func Glibc.time
srand(UInt32(time(nil)))
print(rand() % 12)
There are a great many answers on this website that deal with better ways to generate random integers.