Find smallest number in dictionary Swift

Question

Im trying to find the smallest number that is also false in my dictionary. I'm also using the new Swift Language. How can I make this work?

var firstRow = [1: false, 2: false, 3: false, 4: false, 5: false, 6: false]




  for (number, bool) in firstRow {

            if bool == false {

              //  NSLog("\(number)")


                for i = number; i > 0; i-- {
                    if number <= smallest {

                        smallest = number

                        NSLog("\(smallest)")
                    }

                }
               // NSLog("\(bool, number)")
            }
      }

Well because like this my if statement will only allow the numbers that are with a false variable to pass through. Then I just want to find the smallest of those numbers and eventually the biggest — SpringN, Jun 08 '14 at 01:54
The for loop isn't necessary? Thats the part Im trying to fix — SpringN, Jun 08 '14 at 01:57
http://stackoverflow.com/questions/416266/sorted-collection-in-java — Srinath Ganesh, Jun 08 '14 at 02:17
Why are you even using NSLog? I recommend really studying The Swift Proggramming Guide before you go — Shane Hsu, Jun 08 '14 at 02:36
If you're looking for the smallest number in the dictionary, why are you running a decrement loop? That doesn't represent numbers in the dictionary, and will always end in 1 as the smallest. But the worst thing here is that you're writing it like C ... learn functional programming. — Jim Balter, Jun 08 '14 at 04:10

score 2 · Accepted Answer · answered Jun 08 '14 at 01:56

2

Here you go:

var smallest = 10000000000
for (number, bool) in firstRow {

        if bool == false && number < smallest{
             smallest = number

        }
}
println("\(smallest)")

answered Jun 08 '14 at 01:56

ASKASK

657
8
16

1

You might want to change `smallest` to `UInt8.maxValue()` – Brian Tracy Jun 08 '14 at 01:58
why UInt8? isn't its max value only 255? – ASKASK Jun 08 '14 at 02:00
Instead of 1000000000? – SpringN Jun 08 '14 at 02:00
2

@ASKASK or `typeOfNumbersInArray.maxValue()` – Brian Tracy Jun 08 '14 at 02:00
1

Well 255 would be more than enough anyway – SpringN Jun 08 '14 at 02:00
2

@BrianTracy is both right and wrong - it should be `var smallest = Int.max`, there's no need to guess about the right large value. – Nate Cook Jun 08 '14 at 02:08
if all values are greater than 255 then max value maybe an issue – Srinath Ganesh Jun 08 '14 at 02:11
@ASKASK too busy criticizing, forgot to up vote. +1, This is the right answer – Brian Tracy Jun 09 '14 at 19:00

score 0 · Answer 2 · answered Jun 08 '14 at 02:26

0

If you're doing this strictly in Swift, I'd go with ASKASK while adding the the change:

let firstRow = [1: true, 2: false, 3: true]

var smallest = Int.max

for (number, bool) in firstRow {
    if bool == false && number < smallest {
        smallest = number
    }
}

There's no guarantee about the order in which your key-value pairs will be enumerated in your Dictionary, so given this example the fact that your keys are Integers that are displayed in ascending order does not help us, you will still need to check every pair in the Dictionary.

You might benefit from using an Array of Tuples here, especially if you know beforehand that the associated integers will be added to your collection in increasing order - or even if you don't, then you could at least sort the array by the first value in each tuple and thereby break out of the for loop when appropriate; for example:

let firstRow = [(1, true), (2, false), (3: true)]
var smallest: Int!  // given that it might be possible _none_ of them are 'false'
for (number, bool) in firstRow {
    if bool == false {
        smallest = number
        break
    }
}

answered Jun 08 '14 at 02:26

fqdn

2,823
1
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You're suggesting *hand-sorting* the table? Why not just do the whole thing by hand and just say smallest = 2 ? – Jim Balter Jun 08 '14 at 04:14
not at all... I was rather suggesting that if we know that members are going to be pushed to the collection using strictly increasing integers, that we take advantage of one of the ordered data structures so that we can break from the search confident that we have found the smallest number; in any case, looks like there have been several options presented here, I'm sure one of them must have ended up serving the need! – fqdn Jun 11 '14 at 19:34
"I was rather suggesting that if we know that members are going to be pushed to the collection using strictly increasing integers" -- uh, also known as "sorted". Like I said, hand sorting ... that's the only way you would "know beforehand" that the data is sorted without having to sort it. Since the OP didn't say that's guaranteed, I would omit the break. And hopefully Apple will add functional facilities to their collection library (if they haven't already) so you can just say smallest = firstRow.min({$0.1 == false}) as in other modern languages. – Jim Balter Jun 12 '14 at 00:17

score 0 · Answer 3 · edited May 23 '17 at 12:11

0

There seems to be a built-in filter() method on arrays in Swift, so I'd go with something like:

let falseNumbers: Int[] = firstRow.keys.filter { firstRow[$0] == false }
let smallestFalseNumber: Int = sort(falseNumbers)[0]

The above assumes there is at least one false value in firstRow dictionary. To ensure that:

if find(firstRow.values, false) {
    // The above code
}

edited May 23 '17 at 12:11

Community

1
1

answered Jun 08 '14 at 09:28

Anton Strogonoff

32,294
8
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61

score 0 · Answer 4 · answered Oct 27 '20 at 20:15

You can use the built-in min(by: ) method to fetch the minimum value using key or value in dictionary.

let minmum = firstrow.min { a, b in
    return a.value < b.value
 }
 // use minimum object to print key and value 
 print(minimum.key)
 print(minimum.value)

Find smallest number in dictionary Swift

4 Answers4