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How can I use the last result from a scala map as input to the next function?
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I'm working through some project euler questions to practice my scala. For problem 7 I have to find the 10001st prime. I have a working solution, but dont feel its as functional as it could be.
def first_n_primes(n: Long) : List[Long] =
var last_prime = 1L
(1L to n).map(x => last_prime = get_next_prime(x, last_prime); last_prime).toList
Specifically, I feel there might be a way to get rid of the var last_prime, but I dont know how to use the result of the nth map evaluation as the input to the n+1 evaluation. How can I do this more functionally?
scala
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
add a comment |
I'm working through some project euler questions to practice my scala. For problem 7 I have to find the 10001st prime. I have a working solution, but dont feel its as functional as it could be.
def first_n_primes(n: Long) : List[Long] =
var last_prime = 1L
(1L to n).map(x => last_prime = get_next_prime(x, last_prime); last_prime).toList
Specifically, I feel there might be a way to get rid of the var last_prime, but I dont know how to use the result of the nth map evaluation as the input to the n+1 evaluation. How can I do this more functionally?
scala
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
What is your question exactly ? Is there a way to use result of map or how to solve the problem 7?
– Shantiswarup Tunga
Mar 22 at 12:26
I've solved problem 7. get_next prime starts from the last prime so its not rediscovering old primes - I just want to know if theres a better way to get the result from the nth evalauation to be an input to the n+1 evaluation.
– Andrew Bucknell
Mar 22 at 23:08
You can usescan,scanLeft,fold,foldLeft,reduce, reduceLefthigher order function for getting the result from nth evaluation
– Shantiswarup Tunga
Mar 23 at 4:13
add a comment |
I'm working through some project euler questions to practice my scala. For problem 7 I have to find the 10001st prime. I have a working solution, but dont feel its as functional as it could be.
def first_n_primes(n: Long) : List[Long] =
var last_prime = 1L
(1L to n).map(x => last_prime = get_next_prime(x, last_prime); last_prime).toList
Specifically, I feel there might be a way to get rid of the var last_prime, but I dont know how to use the result of the nth map evaluation as the input to the n+1 evaluation. How can I do this more functionally?
scala
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
I'm working through some project euler questions to practice my scala. For problem 7 I have to find the 10001st prime. I have a working solution, but dont feel its as functional as it could be.
def first_n_primes(n: Long) : List[Long] =
var last_prime = 1L
(1L to n).map(x => last_prime = get_next_prime(x, last_prime); last_prime).toList
Specifically, I feel there might be a way to get rid of the var last_prime, but I dont know how to use the result of the nth map evaluation as the input to the n+1 evaluation. How can I do this more functionally?
scala
scala
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
asked Mar 22 at 11:47
Andrew BucknellAndrew Bucknell
7683927
7683927
What is your question exactly ? Is there a way to use result of map or how to solve the problem 7?
– Shantiswarup Tunga
Mar 22 at 12:26
I've solved problem 7. get_next prime starts from the last prime so its not rediscovering old primes - I just want to know if theres a better way to get the result from the nth evalauation to be an input to the n+1 evaluation.
– Andrew Bucknell
Mar 22 at 23:08
You can usescan,scanLeft,fold,foldLeft,reduce, reduceLefthigher order function for getting the result from nth evaluation
– Shantiswarup Tunga
Mar 23 at 4:13
add a comment |
What is your question exactly ? Is there a way to use result of map or how to solve the problem 7?
– Shantiswarup Tunga
Mar 22 at 12:26
I've solved problem 7. get_next prime starts from the last prime so its not rediscovering old primes - I just want to know if theres a better way to get the result from the nth evalauation to be an input to the n+1 evaluation.
– Andrew Bucknell
Mar 22 at 23:08
You can usescan,scanLeft,fold,foldLeft,reduce, reduceLefthigher order function for getting the result from nth evaluation
– Shantiswarup Tunga
Mar 23 at 4:13
What is your question exactly ? Is there a way to use result of map or how to solve the problem 7?
– Shantiswarup Tunga
Mar 22 at 12:26
What is your question exactly ? Is there a way to use result of map or how to solve the problem 7?
– Shantiswarup Tunga
Mar 22 at 12:26
I've solved problem 7. get_next prime starts from the last prime so its not rediscovering old primes - I just want to know if theres a better way to get the result from the nth evalauation to be an input to the n+1 evaluation.
– Andrew Bucknell
Mar 22 at 23:08
I've solved problem 7. get_next prime starts from the last prime so its not rediscovering old primes - I just want to know if theres a better way to get the result from the nth evalauation to be an input to the n+1 evaluation.
– Andrew Bucknell
Mar 22 at 23:08
You can use
scan,scanLeft,fold,foldLeft,reduce, reduceLeft higher order function for getting the result from nth evaluation– Shantiswarup Tunga
Mar 23 at 4:13
You can use
scan,scanLeft,fold,foldLeft,reduce, reduceLeft higher order function for getting the result from nth evaluation– Shantiswarup Tunga
Mar 23 at 4:13
add a comment |
1 Answer
1
active
oldest
votes
You are looking for scanLeft:
(1l to n).scanLeft(1) case (x, last) => get_next_prime(x, last)
Or just (1l to n).scanLeft(1)(get_next_prime)
Note however that this is not a very good algorithm looking for the primes, because there is a lot of repetitive work that could be saved (to find the next prime, you need to re-discover all the previous ones).
This sort of task is better done in scala with recursive streams:
lazy val primes: Stream[Long] = 2 #:: Stream.iterate(3l)(_+1).filter n =>
val stop = math.sqrt(n)
primes.takeWhile _ <= stop .forall k => n % k != 0
primes.take(n).toList
answered Mar 22 at 11:55
DimaDima
26.7k32640
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You are looking for scanLeft:
(1l to n).scanLeft(1) case (x, last) => get_next_prime(x, last)
Or just (1l to n).scanLeft(1)(get_next_prime)
Note however that this is not a very good algorithm looking for the primes, because there is a lot of repetitive work that could be saved (to find the next prime, you need to re-discover all the previous ones).
This sort of task is better done in scala with recursive streams:
lazy val primes: Stream[Long] = 2 #:: Stream.iterate(3l)(_+1).filter n =>
val stop = math.sqrt(n)
primes.takeWhile _ <= stop .forall k => n % k != 0
primes.take(n).toList
answered Mar 22 at 11:55
DimaDima
26.7k32640
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
add a comment |
You are looking for scanLeft:
(1l to n).scanLeft(1) case (x, last) => get_next_prime(x, last)
Or just (1l to n).scanLeft(1)(get_next_prime)
Note however that this is not a very good algorithm looking for the primes, because there is a lot of repetitive work that could be saved (to find the next prime, you need to re-discover all the previous ones).
This sort of task is better done in scala with recursive streams:
lazy val primes: Stream[Long] = 2 #:: Stream.iterate(3l)(_+1).filter n =>
val stop = math.sqrt(n)
primes.takeWhile _ <= stop .forall k => n % k != 0
primes.take(n).toList
answered Mar 22 at 11:55
DimaDima
26.7k32640
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
add a comment |
You are looking for scanLeft:
(1l to n).scanLeft(1) case (x, last) => get_next_prime(x, last)
Or just (1l to n).scanLeft(1)(get_next_prime)
Note however that this is not a very good algorithm looking for the primes, because there is a lot of repetitive work that could be saved (to find the next prime, you need to re-discover all the previous ones).
This sort of task is better done in scala with recursive streams:
lazy val primes: Stream[Long] = 2 #:: Stream.iterate(3l)(_+1).filter n =>
val stop = math.sqrt(n)
primes.takeWhile _ <= stop .forall k => n % k != 0
primes.take(n).toList
answered Mar 22 at 11:55
DimaDima
26.7k32640
You are looking for scanLeft:
(1l to n).scanLeft(1) case (x, last) => get_next_prime(x, last)
Or just (1l to n).scanLeft(1)(get_next_prime)
Note however that this is not a very good algorithm looking for the primes, because there is a lot of repetitive work that could be saved (to find the next prime, you need to re-discover all the previous ones).
This sort of task is better done in scala with recursive streams:
lazy val primes: Stream[Long] = 2 #:: Stream.iterate(3l)(_+1).filter n =>
val stop = math.sqrt(n)
primes.takeWhile _ <= stop .forall k => n % k != 0
primes.take(n).toList
answered Mar 22 at 11:55
DimaDima
26.7k32640
edited Mar 23 at 12:48
answered Mar 22 at 11:55
DimaDima
26.7k32640
answered Mar 22 at 11:55
DimaDima
26.7k32640
answered Mar 22 at 11:55
DimaDima
26.7k32640
26.7k32640
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
add a comment |
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
scanLeft is perfect! Thankyou! The one change I made is last is actually the first parameter as it holds the result of the previous operation. So I changed the signature of get_next_prime to reflect this. The use of scanleft lets me avoid recalculating previous primes because now I always start from the last prime calculated.
– Andrew Bucknell
Mar 22 at 23:36
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
It's not just where you start that matters. When you check whether a given number is prime, you don't have to try to divide it by every smaller number, it is enough to verify only primes.
– Dima
Mar 23 at 12:24
add a comment |
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What is your question exactly ? Is there a way to use result of map or how to solve the problem 7?
– Shantiswarup Tunga
Mar 22 at 12:26
I've solved problem 7. get_next prime starts from the last prime so its not rediscovering old primes - I just want to know if theres a better way to get the result from the nth evalauation to be an input to the n+1 evaluation.
– Andrew Bucknell
Mar 22 at 23:08
You can use
scan,scanLeft,fold,foldLeft,reduce, reduceLefthigher order function for getting the result from nth evaluation– Shantiswarup Tunga
Mar 23 at 4:13