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I currently receive a Int that could be anywhere between 0 and 180,000. I need to change it to fit between 0 and 3000. I know in other languages you can use something like
Map(min1,max1,min2,max2,input)
I can't seem to find something like that inside of swift.
This is what I currently have, but it always returns 0.
var newY = [(Int)(input)]
newY = newY.map _ in 0 * 3000
print(newY[0])
I think I am using the wrong function. I have never done any mapping in Swift before.
swift integer range
|
show 3 more comments
I currently receive a Int that could be anywhere between 0 and 180,000. I need to change it to fit between 0 and 3000. I know in other languages you can use something like
Map(min1,max1,min2,max2,input)
I can't seem to find something like that inside of swift.
This is what I currently have, but it always returns 0.
var newY = [(Int)(input)]
newY = newY.map _ in 0 * 3000
print(newY[0])
I think I am using the wrong function. I have never done any mapping in Swift before.
swift integer range
stackoverflow.com/questions/24132399/…
– Sahil
Mar 15 '17 at 17:50
What does "keep a number between two numbers" mean? Do you want to generate a random number?
– Andrew Brooke
Mar 15 '17 at 17:51
Your question is unclear, but documentation onmapcan be found here.
– Caleb
Mar 15 '17 at 17:55
Lets say you get a value3003: should this be mapped to3, or to an integer of the size~3000*3003/180000(~50)?
– dfri
Mar 15 '17 at 17:57
Which other language does your example come from? It might be easier to answer your question if I could see the definition of how that one behaves.
– Phillip Mills
Mar 15 '17 at 18:02
|
show 3 more comments
I currently receive a Int that could be anywhere between 0 and 180,000. I need to change it to fit between 0 and 3000. I know in other languages you can use something like
Map(min1,max1,min2,max2,input)
I can't seem to find something like that inside of swift.
This is what I currently have, but it always returns 0.
var newY = [(Int)(input)]
newY = newY.map _ in 0 * 3000
print(newY[0])
I think I am using the wrong function. I have never done any mapping in Swift before.
swift integer range
I currently receive a Int that could be anywhere between 0 and 180,000. I need to change it to fit between 0 and 3000. I know in other languages you can use something like
Map(min1,max1,min2,max2,input)
I can't seem to find something like that inside of swift.
This is what I currently have, but it always returns 0.
var newY = [(Int)(input)]
newY = newY.map _ in 0 * 3000
print(newY[0])
I think I am using the wrong function. I have never done any mapping in Swift before.
swift integer range
swift integer range
edited Mar 15 '17 at 18:28
Travis Griggs
10.8k16 gold badges63 silver badges113 bronze badges
10.8k16 gold badges63 silver badges113 bronze badges
asked Mar 15 '17 at 17:43
Jacob BashistaJacob Bashista
371 silver badge8 bronze badges
371 silver badge8 bronze badges
stackoverflow.com/questions/24132399/…
– Sahil
Mar 15 '17 at 17:50
What does "keep a number between two numbers" mean? Do you want to generate a random number?
– Andrew Brooke
Mar 15 '17 at 17:51
Your question is unclear, but documentation onmapcan be found here.
– Caleb
Mar 15 '17 at 17:55
Lets say you get a value3003: should this be mapped to3, or to an integer of the size~3000*3003/180000(~50)?
– dfri
Mar 15 '17 at 17:57
Which other language does your example come from? It might be easier to answer your question if I could see the definition of how that one behaves.
– Phillip Mills
Mar 15 '17 at 18:02
|
show 3 more comments
stackoverflow.com/questions/24132399/…
– Sahil
Mar 15 '17 at 17:50
What does "keep a number between two numbers" mean? Do you want to generate a random number?
– Andrew Brooke
Mar 15 '17 at 17:51
Your question is unclear, but documentation onmapcan be found here.
– Caleb
Mar 15 '17 at 17:55
Lets say you get a value3003: should this be mapped to3, or to an integer of the size~3000*3003/180000(~50)?
– dfri
Mar 15 '17 at 17:57
Which other language does your example come from? It might be easier to answer your question if I could see the definition of how that one behaves.
– Phillip Mills
Mar 15 '17 at 18:02
stackoverflow.com/questions/24132399/…
– Sahil
Mar 15 '17 at 17:50
stackoverflow.com/questions/24132399/…
– Sahil
Mar 15 '17 at 17:50
What does "keep a number between two numbers" mean? Do you want to generate a random number?
– Andrew Brooke
Mar 15 '17 at 17:51
What does "keep a number between two numbers" mean? Do you want to generate a random number?
– Andrew Brooke
Mar 15 '17 at 17:51
Your question is unclear, but documentation on
map can be found here.– Caleb
Mar 15 '17 at 17:55
Your question is unclear, but documentation on
map can be found here.– Caleb
Mar 15 '17 at 17:55
Lets say you get a value
3003: should this be mapped to 3, or to an integer of the size ~3000*3003/180000 (~50)?– dfri
Mar 15 '17 at 17:57
Lets say you get a value
3003: should this be mapped to 3, or to an integer of the size ~3000*3003/180000 (~50)?– dfri
Mar 15 '17 at 17:57
Which other language does your example come from? It might be easier to answer your question if I could see the definition of how that one behaves.
– Phillip Mills
Mar 15 '17 at 18:02
Which other language does your example come from? It might be easier to answer your question if I could see the definition of how that one behaves.
– Phillip Mills
Mar 15 '17 at 18:02
|
show 3 more comments
4 Answers
4
active
oldest
votes
The map function on collections is going to do something very different. It applies a mapping function to each element of a collection and returns a new collection based on the results.
What you're looking for would be:
func map(minRange:Int, maxRange:Int, minDomain:Int, maxDomain:Int, value:Int) -> Int
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
print(map(minRange: 0, maxRange: 1800000, minDomain: 0, maxDomain: 3000, value: 200000))
With only a little more work you can make it generic over all integer types:
func map<T:IntegerArithmetic>(minRange:T, maxRange:T, minDomain:T, maxDomain:T, value:T) -> T
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
Another option would be to take advantage of the Swift Range type to make calling more succinct:
func map<T:IntegerArithmetic>(range:Range<T>, domain:Range<T>, value:T) -> T
return domain.lowerBound + (domain.upperBound - domain.lowerBound) * (value - range.lowerBound) / (range.upperBound - range.lowerBound)
map(range:0..<3000, domain:0..<180000, value: 1500)
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
1
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existingmap?
– Travis Griggs
Mar 15 '17 at 18:25
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
add a comment |
It's not clear whether the OP simply wants to clamp (title seems to say that) or wants to interpolate from one bounding range to another. The title makes me think one, but the dual max/min values make me think they're after an interpolation. I answered the former. David Berry has a good answer for the latter.
What you may want is min/max functions. Swift has these. To "clamp" a value between a low and high value, you usually combine the two:
var bottom = 13
var top = 42
var tooLow = 7
var clamped = min(top, max(bottom, tooLow)) -> 13
var justRight = 23
clamped = min(top, max(bottom, justRight)) --> 23
var tooHigh = 99
clamped = min(top, max(bottom, tooHigh)) --> 42
This is usually the route most people go, and is probably good enough for most. I personally hate writing that again and again, and I get tired of having to think about which side to feed into the max and the min. And I don't like that it uses what looks like a free function, I'm an object oriented message sending sort of guy, so I do the following:
precedencegroup MinMaxPrecedence
associativity: left
higherThan: NilCoalescingPrecedence, AdditionPrecedence, MultiplicationPrecedence
infix operator <> : MinMaxPrecedence
func <><T:Comparable>(a:T, b:T) -> T
return a < b ? a : b
infix operator >< : MinMaxPrecedence
func ><<T:Comparable>(a:T, b:T) -> T
return a < b ? b : a
Basically, this defines two new operators (<> and ><) that can be used between any type that adopts Comparable. They're easy for me to remember, the one that tucks in smaller wants the smaller value, and the one that opens up bigger returns the bigger value. What's nice is that you can then put them in simpler expressions:
var bottom = 13
var top = 42
var tooLow = 7
var justRight = 23
var tooHigh = 99
bottom >< tooLow <> top --> 13
bottom >< justRight <> top --> 23
bottom >< tooHigh <> top --> 42
Although I agree it's not absolutely clear, hismapfunction takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen amapfunction somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.
– David Berry
Mar 15 '17 at 18:26
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
add a comment |
Given that you are looking for mapping the relative position of a value in a given range to another range, you could do something along the lines:
// dummy function name
func transform(_ number: Int, fromRange: (Int, Int), toRange: (Int, Int)) -> Int?
guard number >= fromRange.0 && number <= fromRange.1,
toRange.0 <= toRange.1 else return nil
return toRange.0 + (number-fromRange.0)*(toRange.1-toRange.0)/(fromRange.1-fromRange.0)
// ex1
let numberA = 3001
let transformedNumberA = transform(numberA, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberA ?? -1) // 50
// ex2
let numberB = 134_000
let transformedNumberB = transform(numberB, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberB ?? -1) // 2233
// ex3
let numberC = 200_000
let transformedNumberC = transform(numberC, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberC ?? -1) // -1 (nil return)
Just take care to notice that (number-fromRange.0)*(toRange.1-toRange.0) (left associativity of / and * operators) may overflow.
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
add a comment |
//Map function
func mapy(n:Double, start1:Double, stop1:Double, start2:Double, stop2:Double) -> Double
return ((n-start1)/(stop1-start1))*(stop2-start2)+start2;
;
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
The map function on collections is going to do something very different. It applies a mapping function to each element of a collection and returns a new collection based on the results.
What you're looking for would be:
func map(minRange:Int, maxRange:Int, minDomain:Int, maxDomain:Int, value:Int) -> Int
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
print(map(minRange: 0, maxRange: 1800000, minDomain: 0, maxDomain: 3000, value: 200000))
With only a little more work you can make it generic over all integer types:
func map<T:IntegerArithmetic>(minRange:T, maxRange:T, minDomain:T, maxDomain:T, value:T) -> T
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
Another option would be to take advantage of the Swift Range type to make calling more succinct:
func map<T:IntegerArithmetic>(range:Range<T>, domain:Range<T>, value:T) -> T
return domain.lowerBound + (domain.upperBound - domain.lowerBound) * (value - range.lowerBound) / (range.upperBound - range.lowerBound)
map(range:0..<3000, domain:0..<180000, value: 1500)
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
1
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existingmap?
– Travis Griggs
Mar 15 '17 at 18:25
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
add a comment |
The map function on collections is going to do something very different. It applies a mapping function to each element of a collection and returns a new collection based on the results.
What you're looking for would be:
func map(minRange:Int, maxRange:Int, minDomain:Int, maxDomain:Int, value:Int) -> Int
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
print(map(minRange: 0, maxRange: 1800000, minDomain: 0, maxDomain: 3000, value: 200000))
With only a little more work you can make it generic over all integer types:
func map<T:IntegerArithmetic>(minRange:T, maxRange:T, minDomain:T, maxDomain:T, value:T) -> T
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
Another option would be to take advantage of the Swift Range type to make calling more succinct:
func map<T:IntegerArithmetic>(range:Range<T>, domain:Range<T>, value:T) -> T
return domain.lowerBound + (domain.upperBound - domain.lowerBound) * (value - range.lowerBound) / (range.upperBound - range.lowerBound)
map(range:0..<3000, domain:0..<180000, value: 1500)
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
1
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existingmap?
– Travis Griggs
Mar 15 '17 at 18:25
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
add a comment |
The map function on collections is going to do something very different. It applies a mapping function to each element of a collection and returns a new collection based on the results.
What you're looking for would be:
func map(minRange:Int, maxRange:Int, minDomain:Int, maxDomain:Int, value:Int) -> Int
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
print(map(minRange: 0, maxRange: 1800000, minDomain: 0, maxDomain: 3000, value: 200000))
With only a little more work you can make it generic over all integer types:
func map<T:IntegerArithmetic>(minRange:T, maxRange:T, minDomain:T, maxDomain:T, value:T) -> T
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
Another option would be to take advantage of the Swift Range type to make calling more succinct:
func map<T:IntegerArithmetic>(range:Range<T>, domain:Range<T>, value:T) -> T
return domain.lowerBound + (domain.upperBound - domain.lowerBound) * (value - range.lowerBound) / (range.upperBound - range.lowerBound)
map(range:0..<3000, domain:0..<180000, value: 1500)
The map function on collections is going to do something very different. It applies a mapping function to each element of a collection and returns a new collection based on the results.
What you're looking for would be:
func map(minRange:Int, maxRange:Int, minDomain:Int, maxDomain:Int, value:Int) -> Int
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
print(map(minRange: 0, maxRange: 1800000, minDomain: 0, maxDomain: 3000, value: 200000))
With only a little more work you can make it generic over all integer types:
func map<T:IntegerArithmetic>(minRange:T, maxRange:T, minDomain:T, maxDomain:T, value:T) -> T
return minDomain + (maxDomain - minDomain) * (value - minRange) / (maxRange - minRange)
Another option would be to take advantage of the Swift Range type to make calling more succinct:
func map<T:IntegerArithmetic>(range:Range<T>, domain:Range<T>, value:T) -> T
return domain.lowerBound + (domain.upperBound - domain.lowerBound) * (value - range.lowerBound) / (range.upperBound - range.lowerBound)
map(range:0..<3000, domain:0..<180000, value: 1500)
edited Mar 15 '17 at 18:36
answered Mar 15 '17 at 18:09
David BerryDavid Berry
32.7k11 gold badges71 silver badges83 bronze badges
32.7k11 gold badges71 silver badges83 bronze badges
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
1
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existingmap?
– Travis Griggs
Mar 15 '17 at 18:25
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
add a comment |
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
1
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existingmap?
– Travis Griggs
Mar 15 '17 at 18:25
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
There's a substantial difference between dfri's answer and mine in that due to the order of operations mine will have more even distribution while his is less likely to result in overflows. All depends on what you're looking for.
– David Berry
Mar 15 '17 at 18:16
1
1
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existing
map?– Travis Griggs
Mar 15 '17 at 18:25
Wouldn't it be better to suggest a different name of the new function, to avoid confusion with the existing
map?– Travis Griggs
Mar 15 '17 at 18:25
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Probably, I was just keeping his name intact.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
Combined with explicit parameter naming, it seems clear enough.
– David Berry
Mar 15 '17 at 18:28
add a comment |
It's not clear whether the OP simply wants to clamp (title seems to say that) or wants to interpolate from one bounding range to another. The title makes me think one, but the dual max/min values make me think they're after an interpolation. I answered the former. David Berry has a good answer for the latter.
What you may want is min/max functions. Swift has these. To "clamp" a value between a low and high value, you usually combine the two:
var bottom = 13
var top = 42
var tooLow = 7
var clamped = min(top, max(bottom, tooLow)) -> 13
var justRight = 23
clamped = min(top, max(bottom, justRight)) --> 23
var tooHigh = 99
clamped = min(top, max(bottom, tooHigh)) --> 42
This is usually the route most people go, and is probably good enough for most. I personally hate writing that again and again, and I get tired of having to think about which side to feed into the max and the min. And I don't like that it uses what looks like a free function, I'm an object oriented message sending sort of guy, so I do the following:
precedencegroup MinMaxPrecedence
associativity: left
higherThan: NilCoalescingPrecedence, AdditionPrecedence, MultiplicationPrecedence
infix operator <> : MinMaxPrecedence
func <><T:Comparable>(a:T, b:T) -> T
return a < b ? a : b
infix operator >< : MinMaxPrecedence
func ><<T:Comparable>(a:T, b:T) -> T
return a < b ? b : a
Basically, this defines two new operators (<> and ><) that can be used between any type that adopts Comparable. They're easy for me to remember, the one that tucks in smaller wants the smaller value, and the one that opens up bigger returns the bigger value. What's nice is that you can then put them in simpler expressions:
var bottom = 13
var top = 42
var tooLow = 7
var justRight = 23
var tooHigh = 99
bottom >< tooLow <> top --> 13
bottom >< justRight <> top --> 23
bottom >< tooHigh <> top --> 42
Although I agree it's not absolutely clear, hismapfunction takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen amapfunction somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.
– David Berry
Mar 15 '17 at 18:26
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
add a comment |
It's not clear whether the OP simply wants to clamp (title seems to say that) or wants to interpolate from one bounding range to another. The title makes me think one, but the dual max/min values make me think they're after an interpolation. I answered the former. David Berry has a good answer for the latter.
What you may want is min/max functions. Swift has these. To "clamp" a value between a low and high value, you usually combine the two:
var bottom = 13
var top = 42
var tooLow = 7
var clamped = min(top, max(bottom, tooLow)) -> 13
var justRight = 23
clamped = min(top, max(bottom, justRight)) --> 23
var tooHigh = 99
clamped = min(top, max(bottom, tooHigh)) --> 42
This is usually the route most people go, and is probably good enough for most. I personally hate writing that again and again, and I get tired of having to think about which side to feed into the max and the min. And I don't like that it uses what looks like a free function, I'm an object oriented message sending sort of guy, so I do the following:
precedencegroup MinMaxPrecedence
associativity: left
higherThan: NilCoalescingPrecedence, AdditionPrecedence, MultiplicationPrecedence
infix operator <> : MinMaxPrecedence
func <><T:Comparable>(a:T, b:T) -> T
return a < b ? a : b
infix operator >< : MinMaxPrecedence
func ><<T:Comparable>(a:T, b:T) -> T
return a < b ? b : a
Basically, this defines two new operators (<> and ><) that can be used between any type that adopts Comparable. They're easy for me to remember, the one that tucks in smaller wants the smaller value, and the one that opens up bigger returns the bigger value. What's nice is that you can then put them in simpler expressions:
var bottom = 13
var top = 42
var tooLow = 7
var justRight = 23
var tooHigh = 99
bottom >< tooLow <> top --> 13
bottom >< justRight <> top --> 23
bottom >< tooHigh <> top --> 42
Although I agree it's not absolutely clear, hismapfunction takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen amapfunction somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.
– David Berry
Mar 15 '17 at 18:26
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
add a comment |
It's not clear whether the OP simply wants to clamp (title seems to say that) or wants to interpolate from one bounding range to another. The title makes me think one, but the dual max/min values make me think they're after an interpolation. I answered the former. David Berry has a good answer for the latter.
What you may want is min/max functions. Swift has these. To "clamp" a value between a low and high value, you usually combine the two:
var bottom = 13
var top = 42
var tooLow = 7
var clamped = min(top, max(bottom, tooLow)) -> 13
var justRight = 23
clamped = min(top, max(bottom, justRight)) --> 23
var tooHigh = 99
clamped = min(top, max(bottom, tooHigh)) --> 42
This is usually the route most people go, and is probably good enough for most. I personally hate writing that again and again, and I get tired of having to think about which side to feed into the max and the min. And I don't like that it uses what looks like a free function, I'm an object oriented message sending sort of guy, so I do the following:
precedencegroup MinMaxPrecedence
associativity: left
higherThan: NilCoalescingPrecedence, AdditionPrecedence, MultiplicationPrecedence
infix operator <> : MinMaxPrecedence
func <><T:Comparable>(a:T, b:T) -> T
return a < b ? a : b
infix operator >< : MinMaxPrecedence
func ><<T:Comparable>(a:T, b:T) -> T
return a < b ? b : a
Basically, this defines two new operators (<> and ><) that can be used between any type that adopts Comparable. They're easy for me to remember, the one that tucks in smaller wants the smaller value, and the one that opens up bigger returns the bigger value. What's nice is that you can then put them in simpler expressions:
var bottom = 13
var top = 42
var tooLow = 7
var justRight = 23
var tooHigh = 99
bottom >< tooLow <> top --> 13
bottom >< justRight <> top --> 23
bottom >< tooHigh <> top --> 42
It's not clear whether the OP simply wants to clamp (title seems to say that) or wants to interpolate from one bounding range to another. The title makes me think one, but the dual max/min values make me think they're after an interpolation. I answered the former. David Berry has a good answer for the latter.
What you may want is min/max functions. Swift has these. To "clamp" a value between a low and high value, you usually combine the two:
var bottom = 13
var top = 42
var tooLow = 7
var clamped = min(top, max(bottom, tooLow)) -> 13
var justRight = 23
clamped = min(top, max(bottom, justRight)) --> 23
var tooHigh = 99
clamped = min(top, max(bottom, tooHigh)) --> 42
This is usually the route most people go, and is probably good enough for most. I personally hate writing that again and again, and I get tired of having to think about which side to feed into the max and the min. And I don't like that it uses what looks like a free function, I'm an object oriented message sending sort of guy, so I do the following:
precedencegroup MinMaxPrecedence
associativity: left
higherThan: NilCoalescingPrecedence, AdditionPrecedence, MultiplicationPrecedence
infix operator <> : MinMaxPrecedence
func <><T:Comparable>(a:T, b:T) -> T
return a < b ? a : b
infix operator >< : MinMaxPrecedence
func ><<T:Comparable>(a:T, b:T) -> T
return a < b ? b : a
Basically, this defines two new operators (<> and ><) that can be used between any type that adopts Comparable. They're easy for me to remember, the one that tucks in smaller wants the smaller value, and the one that opens up bigger returns the bigger value. What's nice is that you can then put them in simpler expressions:
var bottom = 13
var top = 42
var tooLow = 7
var justRight = 23
var tooHigh = 99
bottom >< tooLow <> top --> 13
bottom >< justRight <> top --> 23
bottom >< tooHigh <> top --> 42
edited Mar 15 '17 at 18:26
answered Mar 15 '17 at 18:20
Travis GriggsTravis Griggs
10.8k16 gold badges63 silver badges113 bronze badges
10.8k16 gold badges63 silver badges113 bronze badges
Although I agree it's not absolutely clear, hismapfunction takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen amapfunction somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.
– David Berry
Mar 15 '17 at 18:26
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
add a comment |
Although I agree it's not absolutely clear, hismapfunction takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen amapfunction somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.
– David Berry
Mar 15 '17 at 18:26
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
Although I agree it's not absolutely clear, his
map function takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen a map function somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.– David Berry
Mar 15 '17 at 18:26
Although I agree it's not absolutely clear, his
map function takes two ranges which implies he's actually wanting to interpolate. Combine that with having seen a map function somewhere somewhen that does exactly that, it seems likely that's what they're actually wanting.– David Berry
Mar 15 '17 at 18:26
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
Yup, realized that was likely after writing all of the above out. :(
– Travis Griggs
Mar 15 '17 at 18:27
add a comment |
Given that you are looking for mapping the relative position of a value in a given range to another range, you could do something along the lines:
// dummy function name
func transform(_ number: Int, fromRange: (Int, Int), toRange: (Int, Int)) -> Int?
guard number >= fromRange.0 && number <= fromRange.1,
toRange.0 <= toRange.1 else return nil
return toRange.0 + (number-fromRange.0)*(toRange.1-toRange.0)/(fromRange.1-fromRange.0)
// ex1
let numberA = 3001
let transformedNumberA = transform(numberA, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberA ?? -1) // 50
// ex2
let numberB = 134_000
let transformedNumberB = transform(numberB, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberB ?? -1) // 2233
// ex3
let numberC = 200_000
let transformedNumberC = transform(numberC, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberC ?? -1) // -1 (nil return)
Just take care to notice that (number-fromRange.0)*(toRange.1-toRange.0) (left associativity of / and * operators) may overflow.
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
add a comment |
Given that you are looking for mapping the relative position of a value in a given range to another range, you could do something along the lines:
// dummy function name
func transform(_ number: Int, fromRange: (Int, Int), toRange: (Int, Int)) -> Int?
guard number >= fromRange.0 && number <= fromRange.1,
toRange.0 <= toRange.1 else return nil
return toRange.0 + (number-fromRange.0)*(toRange.1-toRange.0)/(fromRange.1-fromRange.0)
// ex1
let numberA = 3001
let transformedNumberA = transform(numberA, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberA ?? -1) // 50
// ex2
let numberB = 134_000
let transformedNumberB = transform(numberB, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberB ?? -1) // 2233
// ex3
let numberC = 200_000
let transformedNumberC = transform(numberC, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberC ?? -1) // -1 (nil return)
Just take care to notice that (number-fromRange.0)*(toRange.1-toRange.0) (left associativity of / and * operators) may overflow.
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
add a comment |
Given that you are looking for mapping the relative position of a value in a given range to another range, you could do something along the lines:
// dummy function name
func transform(_ number: Int, fromRange: (Int, Int), toRange: (Int, Int)) -> Int?
guard number >= fromRange.0 && number <= fromRange.1,
toRange.0 <= toRange.1 else return nil
return toRange.0 + (number-fromRange.0)*(toRange.1-toRange.0)/(fromRange.1-fromRange.0)
// ex1
let numberA = 3001
let transformedNumberA = transform(numberA, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberA ?? -1) // 50
// ex2
let numberB = 134_000
let transformedNumberB = transform(numberB, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberB ?? -1) // 2233
// ex3
let numberC = 200_000
let transformedNumberC = transform(numberC, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberC ?? -1) // -1 (nil return)
Just take care to notice that (number-fromRange.0)*(toRange.1-toRange.0) (left associativity of / and * operators) may overflow.
Given that you are looking for mapping the relative position of a value in a given range to another range, you could do something along the lines:
// dummy function name
func transform(_ number: Int, fromRange: (Int, Int), toRange: (Int, Int)) -> Int?
guard number >= fromRange.0 && number <= fromRange.1,
toRange.0 <= toRange.1 else return nil
return toRange.0 + (number-fromRange.0)*(toRange.1-toRange.0)/(fromRange.1-fromRange.0)
// ex1
let numberA = 3001
let transformedNumberA = transform(numberA, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberA ?? -1) // 50
// ex2
let numberB = 134_000
let transformedNumberB = transform(numberB, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberB ?? -1) // 2233
// ex3
let numberC = 200_000
let transformedNumberC = transform(numberC, fromRange: (0, 180_000), toRange: (0,3000))
print(transformedNumberC ?? -1) // -1 (nil return)
Just take care to notice that (number-fromRange.0)*(toRange.1-toRange.0) (left associativity of / and * operators) may overflow.
edited Mar 15 '17 at 18:58
answered Mar 15 '17 at 17:54
dfridfri
37.4k4 gold badges66 silver badges106 bronze badges
37.4k4 gold badges66 silver badges106 bronze badges
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
add a comment |
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
You need to subtract the minimum input range value before math, otherwise it only works if the input range starts at zero.
– David Berry
Mar 15 '17 at 18:15
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
@DavidBerry: you are right, thank you!
– dfri
Mar 15 '17 at 18:22
add a comment |
//Map function
func mapy(n:Double, start1:Double, stop1:Double, start2:Double, stop2:Double) -> Double
return ((n-start1)/(stop1-start1))*(stop2-start2)+start2;
;
add a comment |
//Map function
func mapy(n:Double, start1:Double, stop1:Double, start2:Double, stop2:Double) -> Double
return ((n-start1)/(stop1-start1))*(stop2-start2)+start2;
;
add a comment |
//Map function
func mapy(n:Double, start1:Double, stop1:Double, start2:Double, stop2:Double) -> Double
return ((n-start1)/(stop1-start1))*(stop2-start2)+start2;
;
//Map function
func mapy(n:Double, start1:Double, stop1:Double, start2:Double, stop2:Double) -> Double
return ((n-start1)/(stop1-start1))*(stop2-start2)+start2;
;
answered Mar 26 at 1:02
omarojoomarojo
5616 silver badges22 bronze badges
5616 silver badges22 bronze badges
add a comment |
add a comment |
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stackoverflow.com/questions/24132399/…
– Sahil
Mar 15 '17 at 17:50
What does "keep a number between two numbers" mean? Do you want to generate a random number?
– Andrew Brooke
Mar 15 '17 at 17:51
Your question is unclear, but documentation on
mapcan be found here.– Caleb
Mar 15 '17 at 17:55
Lets say you get a value
3003: should this be mapped to3, or to an integer of the size~3000*3003/180000(~50)?– dfri
Mar 15 '17 at 17:57
Which other language does your example come from? It might be easier to answer your question if I could see the definition of how that one behaves.
– Phillip Mills
Mar 15 '17 at 18:02