Search in dictionary on a variety of conditions Swift - swift

Question

I have simple dictionary
var countOfR = ["R0": 0, "R1": 0, "R2": 0, "R3": 0, "R4": 0, "R5": 0, "R6": 0]
I need to check this dictionary by multiple conditions. For example, next statement works perfectly:
for index in countOfR {
if index == ("R0",2) || index == ("R1",2) || index == ("R2",2) || index == ("R3",2) || index == ("R4",2) || index == ("R5",2) || index == ("R6",2) {
type = "P"
This will find one "pair". But next I need to check for "two pairs" - "PP". It's terrible to write something like this:
if index == ("R0",2) && index == ("R1",2) || index == ("R0",2) && index == ("R2",2) || index == ("R0",2) && index == ("R3",2) || index == ("R0",2) && index == ("R4",2) || index == ("R0",2) && index == ("R5",2) || index == ("R0",2) && index == ("R6",2) || ...
and so on... I also need to search for "pair and trine", "three pairs" and many others.
For better understanding:
["R0": 1, "R1": 2, "R2": 1, "R3": 1, "R4": 0, "R5": 1, "R6": 0] is "P",
["R0": 1, "R1": 0, "R2": 0, "R3": 1, "R4": 0, "R5": 2, "R6": 0] is "P" too,
["R0": 1, "R1": 0, "R2": 2, "R3": 1, "R4": 0, "R5": 2, "R6": 0] is "PP"
How can I solve this task? Please, give me some advice!

Answer 1

Your something like this makes no sense as it never gets true. Your index (or what name it might be) cannot be equal to two different things at the same time.
I guess you just need to count the number of entries whose value is 2.
You can write something like this:
func getType(_ countOfR: [String: Int]) -> String {
let pairs = countOfR.filter{$0.value == 2}.count
let trines = countOfR.filter{$0.value == 3}.count
let type = String(repeating: "P", count: pairs) + String(repeating: "T", count: trines)
return type
}
print(getType(["R0": 1, "R1": 2, "R2": 1, "R3": 1, "R4": 0, "R5": 1, "R6": 0]))
//->P
print(getType(["R0": 1, "R1": 0, "R2": 0, "R3": 1, "R4": 0, "R5": 2, "R6": 0]))
//->P
print(getType(["R0": 1, "R1": 0, "R2": 2, "R3": 1, "R4": 0, "R5": 2, "R6": 0]))
//->PP

Answer 2

Create a dictionary with your conditions, and filter the main dictionary by comparing the condition dictionary. Type will be create based on the filtered result count.
var condition = ["R0":2,"R1":2,"R2":2,"R3":2,"R4":2,"R5":2,"R6":2]
let test1 = ["R0": 1, "R1": 2, "R2": 1, "R3": 1, "R4": 0, "R5": 1, "R6": 0]// is "P",
let count1 = test1.filter { index in condition.contains(where: { $0 == index }) }.count
let type1 = String(repeating: "P", count: count1)
print(type1)//P
let test2 = ["R0": 1, "R1": 0, "R2": 0, "R3": 1, "R4": 0, "R5": 2, "R6": 0]// is "P" too,
let count2 = test2.filter { index in condition.contains(where: { $0 == index }) }.count
let type2 = String(repeating: "P", count: count2)
print(type2)//P
let test3 = ["R0": 1, "R1": 0, "R2": 2, "R3": 1, "R4": 0, "R5": 2, "R6": 0]// is "PP"
let count3 = test3.filter { index in condition.contains(where: { $0 == index }) }.count
let type3 = String(repeating: "P", count: count3)
print(type3)//PP