How can I sum the middle elements of an expanding list in Haskell?
So far I know how to expand a list from its ends, but they end up getting doubled because of the first condition, which is to double a singleton. Would it make sense for the code to be like this:
sumExpand :: [Integer] -> [Integer]
sumExpand l = expand l
where
expand a = a
expand (x:) a = x: expand (x:a)
expand (x:xs) a = expand (x:a) (expand xs a)
And for me to work on its output:
[1,1,2,2,3,3] from [1,2,3]
instead of [1,3,5,3]
The latter being my desire? Here's how I got to a temporary solution for a list of two elements:
expand (x:xs) a = x: tail (expand (map (x+) xs) (last xs:a))
Output:
*Main> sumExpand [1,2]
[1,3,2]
*Main> sumExpand [1,2,3]
[1,7,4,3]
EDIT: basically, I want the algorithm to work like this: [a, b, c] => [a, a+b, b+c, c]
list haskell sum expand
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
add a comment |
So far I know how to expand a list from its ends, but they end up getting doubled because of the first condition, which is to double a singleton. Would it make sense for the code to be like this:
sumExpand :: [Integer] -> [Integer]
sumExpand l = expand l
where
expand a = a
expand (x:) a = x: expand (x:a)
expand (x:xs) a = expand (x:a) (expand xs a)
And for me to work on its output:
[1,1,2,2,3,3] from [1,2,3]
instead of [1,3,5,3]
The latter being my desire? Here's how I got to a temporary solution for a list of two elements:
expand (x:xs) a = x: tail (expand (map (x+) xs) (last xs:a))
Output:
*Main> sumExpand [1,2]
[1,3,2]
*Main> sumExpand [1,2,3]
[1,7,4,3]
EDIT: basically, I want the algorithm to work like this: [a, b, c] => [a, a+b, b+c, c]
list haskell sum expand
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
1
What precisely is your algorithm supposed to do? What are you having trouble with?
– jberryman
Nov 20 '18 at 17:01
You want to insert the sum of each adjacent pair between the pair?[a, b, c] => [a, a+b, b, b + c, c]?
– chepner
Nov 20 '18 at 17:17
What I want to do is the following:[a, b, c] => [a, a+b, b+c, c]. It's basically the same as what @chepner suggested, but without b repeating itself.
– Marco_O
Nov 20 '18 at 17:30
1
How about more than 3 elements in list? say, [a, b, c, d] => [a, a+b, b+c, c+d, d]?
– assembly.jc
Nov 20 '18 at 17:44
I have the solution, but could you give me any pointers on how to improve my research or problem solving skills, for when stuff like this happens? I just got down voted and I want to improve. Thanks in advance.
– Marco_O
Nov 20 '18 at 18:07
add a comment |
So far I know how to expand a list from its ends, but they end up getting doubled because of the first condition, which is to double a singleton. Would it make sense for the code to be like this:
sumExpand :: [Integer] -> [Integer]
sumExpand l = expand l
where
expand a = a
expand (x:) a = x: expand (x:a)
expand (x:xs) a = expand (x:a) (expand xs a)
And for me to work on its output:
[1,1,2,2,3,3] from [1,2,3]
instead of [1,3,5,3]
The latter being my desire? Here's how I got to a temporary solution for a list of two elements:
expand (x:xs) a = x: tail (expand (map (x+) xs) (last xs:a))
Output:
*Main> sumExpand [1,2]
[1,3,2]
*Main> sumExpand [1,2,3]
[1,7,4,3]
EDIT: basically, I want the algorithm to work like this: [a, b, c] => [a, a+b, b+c, c]
list haskell sum expand
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
So far I know how to expand a list from its ends, but they end up getting doubled because of the first condition, which is to double a singleton. Would it make sense for the code to be like this:
sumExpand :: [Integer] -> [Integer]
sumExpand l = expand l
where
expand a = a
expand (x:) a = x: expand (x:a)
expand (x:xs) a = expand (x:a) (expand xs a)
And for me to work on its output:
[1,1,2,2,3,3] from [1,2,3]
instead of [1,3,5,3]
The latter being my desire? Here's how I got to a temporary solution for a list of two elements:
expand (x:xs) a = x: tail (expand (map (x+) xs) (last xs:a))
Output:
*Main> sumExpand [1,2]
[1,3,2]
*Main> sumExpand [1,2,3]
[1,7,4,3]
EDIT: basically, I want the algorithm to work like this: [a, b, c] => [a, a+b, b+c, c]
list haskell sum expand
list haskell sum expand
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
edited Nov 20 '18 at 17:34
Marco_O
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
asked Nov 20 '18 at 16:52
Marco_OMarco_O
167
167
1
What precisely is your algorithm supposed to do? What are you having trouble with?
– jberryman
Nov 20 '18 at 17:01
You want to insert the sum of each adjacent pair between the pair?[a, b, c] => [a, a+b, b, b + c, c]?
– chepner
Nov 20 '18 at 17:17
What I want to do is the following:[a, b, c] => [a, a+b, b+c, c]. It's basically the same as what @chepner suggested, but without b repeating itself.
– Marco_O
Nov 20 '18 at 17:30
1
How about more than 3 elements in list? say, [a, b, c, d] => [a, a+b, b+c, c+d, d]?
– assembly.jc
Nov 20 '18 at 17:44
I have the solution, but could you give me any pointers on how to improve my research or problem solving skills, for when stuff like this happens? I just got down voted and I want to improve. Thanks in advance.
– Marco_O
Nov 20 '18 at 18:07
add a comment |
1
What precisely is your algorithm supposed to do? What are you having trouble with?
– jberryman
Nov 20 '18 at 17:01
You want to insert the sum of each adjacent pair between the pair?[a, b, c] => [a, a+b, b, b + c, c]?
– chepner
Nov 20 '18 at 17:17
What I want to do is the following:[a, b, c] => [a, a+b, b+c, c]. It's basically the same as what @chepner suggested, but without b repeating itself.
– Marco_O
Nov 20 '18 at 17:30
1
How about more than 3 elements in list? say, [a, b, c, d] => [a, a+b, b+c, c+d, d]?
– assembly.jc
Nov 20 '18 at 17:44
I have the solution, but could you give me any pointers on how to improve my research or problem solving skills, for when stuff like this happens? I just got down voted and I want to improve. Thanks in advance.
– Marco_O
Nov 20 '18 at 18:07
1
1
What precisely is your algorithm supposed to do? What are you having trouble with?
– jberryman
Nov 20 '18 at 17:01
What precisely is your algorithm supposed to do? What are you having trouble with?
– jberryman
Nov 20 '18 at 17:01
You want to insert the sum of each adjacent pair between the pair?
[a, b, c] => [a, a+b, b, b + c, c]?– chepner
Nov 20 '18 at 17:17
You want to insert the sum of each adjacent pair between the pair?
[a, b, c] => [a, a+b, b, b + c, c]?– chepner
Nov 20 '18 at 17:17
What I want to do is the following:
[a, b, c] => [a, a+b, b+c, c]. It's basically the same as what @chepner suggested, but without b repeating itself.– Marco_O
Nov 20 '18 at 17:30
What I want to do is the following:
[a, b, c] => [a, a+b, b+c, c]. It's basically the same as what @chepner suggested, but without b repeating itself.– Marco_O
Nov 20 '18 at 17:30
1
1
How about more than 3 elements in list? say, [a, b, c, d] => [a, a+b, b+c, c+d, d]?
– assembly.jc
Nov 20 '18 at 17:44
How about more than 3 elements in list? say, [a, b, c, d] => [a, a+b, b+c, c+d, d]?
– assembly.jc
Nov 20 '18 at 17:44
I have the solution, but could you give me any pointers on how to improve my research or problem solving skills, for when stuff like this happens? I just got down voted and I want to improve. Thanks in advance.
– Marco_O
Nov 20 '18 at 18:07
I have the solution, but could you give me any pointers on how to improve my research or problem solving skills, for when stuff like this happens? I just got down voted and I want to improve. Thanks in advance.
– Marco_O
Nov 20 '18 at 18:07
add a comment |
1 Answer
1
active
oldest
votes
Basically, all you want to compute component-wise sums between your input list and a shifted version of it:
a b c d e
a b c d e
---------------------------
a a+b b+c c+d d+e e
Fill each empty slot with a 0 (0:x and x++[0]), and you just need zipWith
> x -> zipWith (+) (0:x) (x++[0]) $ [1,2,3]
[1,3,5,3]
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
add a comment |
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Basically, all you want to compute component-wise sums between your input list and a shifted version of it:
a b c d e
a b c d e
---------------------------
a a+b b+c c+d d+e e
Fill each empty slot with a 0 (0:x and x++[0]), and you just need zipWith
> x -> zipWith (+) (0:x) (x++[0]) $ [1,2,3]
[1,3,5,3]
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
add a comment |
Basically, all you want to compute component-wise sums between your input list and a shifted version of it:
a b c d e
a b c d e
---------------------------
a a+b b+c c+d d+e e
Fill each empty slot with a 0 (0:x and x++[0]), and you just need zipWith
> x -> zipWith (+) (0:x) (x++[0]) $ [1,2,3]
[1,3,5,3]
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
add a comment |
Basically, all you want to compute component-wise sums between your input list and a shifted version of it:
a b c d e
a b c d e
---------------------------
a a+b b+c c+d d+e e
Fill each empty slot with a 0 (0:x and x++[0]), and you just need zipWith
> x -> zipWith (+) (0:x) (x++[0]) $ [1,2,3]
[1,3,5,3]
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
Basically, all you want to compute component-wise sums between your input list and a shifted version of it:
a b c d e
a b c d e
---------------------------
a a+b b+c c+d d+e e
Fill each empty slot with a 0 (0:x and x++[0]), and you just need zipWith
> x -> zipWith (+) (0:x) (x++[0]) $ [1,2,3]
[1,3,5,3]
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
answered Nov 20 '18 at 17:48
chepnerchepner
246k32233324
246k32233324
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
add a comment |
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
Thank you! That's very useful. Now I have something to study its meaning.
– Marco_O
Nov 20 '18 at 17:50
add a comment |
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1
What precisely is your algorithm supposed to do? What are you having trouble with?
– jberryman
Nov 20 '18 at 17:01
You want to insert the sum of each adjacent pair between the pair?
[a, b, c] => [a, a+b, b, b + c, c]?– chepner
Nov 20 '18 at 17:17
What I want to do is the following:
[a, b, c] => [a, a+b, b+c, c]. It's basically the same as what @chepner suggested, but without b repeating itself.– Marco_O
Nov 20 '18 at 17:30
1
How about more than 3 elements in list? say, [a, b, c, d] => [a, a+b, b+c, c+d, d]?
– assembly.jc
Nov 20 '18 at 17:44
I have the solution, but could you give me any pointers on how to improve my research or problem solving skills, for when stuff like this happens? I just got down voted and I want to improve. Thanks in advance.
– Marco_O
Nov 20 '18 at 18:07