Number each list element and format text in haskell
I want to give each one a number from 1 to length(x:xs), like a book's page number. Unfortunately it only works backwards.
numberL :: [String] -> [String]
numberL =
numberL (x:xs) = ([show (length(x:xs)) ++ ": " ++ x] ++ numberL (xs))
Also how do I remove any new line and tab from the text and replace it with the actual new line and tabulator?
haskell numbers element
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
add a comment |
I want to give each one a number from 1 to length(x:xs), like a book's page number. Unfortunately it only works backwards.
numberL :: [String] -> [String]
numberL =
numberL (x:xs) = ([show (length(x:xs)) ++ ": " ++ x] ++ numberL (xs))
Also how do I remove any new line and tab from the text and replace it with the actual new line and tabulator?
haskell numbers element
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
add a comment |
I want to give each one a number from 1 to length(x:xs), like a book's page number. Unfortunately it only works backwards.
numberL :: [String] -> [String]
numberL =
numberL (x:xs) = ([show (length(x:xs)) ++ ": " ++ x] ++ numberL (xs))
Also how do I remove any new line and tab from the text and replace it with the actual new line and tabulator?
haskell numbers element
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
I want to give each one a number from 1 to length(x:xs), like a book's page number. Unfortunately it only works backwards.
numberL :: [String] -> [String]
numberL =
numberL (x:xs) = ([show (length(x:xs)) ++ ": " ++ x] ++ numberL (xs))
Also how do I remove any new line and tab from the text and replace it with the actual new line and tabulator?
haskell numbers element
haskell numbers element
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
asked Nov 23 '18 at 16:02
Akos FajsziAkos Fajszi
445
445
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
There are multiple built-in Haskell functions in Prelude that are good to learn and use them. zip and zipWith are two of them, when you think about something to be done using two different lists into one result list:
[1..] will generate the list of indices for you, it's an infinite list
appendIndex :: String -> Int -> String
appendIndex s i = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = zipWith appendIndex l [1..]
if you wanted to use zip, which is more basic but a little more verbose:
appendIndex :: (String,Int) -> String
appendIndex (s,i) = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = fmap appendIndex $ zip l [1..]
-- if you dont know about Functors yet, `fmap` is the generic way of doing `map`
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
add a comment |
To get it right, it's important to understand why you're thinking wrong. Your recursion looks like this:
numberL (x:xs) = ... ++ numberL xs
So you calculate numberL xs and then put something in front of it. If numberL xs were correct, then then it would be numbered from 1 onwards, like: 1:..., 2:..., 3:.... So you could never build numberL (x:xs) from numberL xs just by adding new elements at the front. The whole numbering would be wrong. Instead you'd have to change the whole numbering of numberL xs.
The problem therefore is that it's not very useful to know numberL xs in order to calculate numberL (x:xs), due to the fact numberL always starts numbering from 1.
So lift that restriction. Build a function that numbers starting at n,
numberLFrom :: Int -> [String] -> [String]
numberLFrom n = ...
numberLFrom n (x:xs) = ...
Now the question you have to ask yourself is, in order to number (x:xs) starting at n you need to number xs starting at which number? And then how do you introduced the numbered x to that result?
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
There are multiple built-in Haskell functions in Prelude that are good to learn and use them. zip and zipWith are two of them, when you think about something to be done using two different lists into one result list:
[1..] will generate the list of indices for you, it's an infinite list
appendIndex :: String -> Int -> String
appendIndex s i = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = zipWith appendIndex l [1..]
if you wanted to use zip, which is more basic but a little more verbose:
appendIndex :: (String,Int) -> String
appendIndex (s,i) = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = fmap appendIndex $ zip l [1..]
-- if you dont know about Functors yet, `fmap` is the generic way of doing `map`
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
add a comment |
There are multiple built-in Haskell functions in Prelude that are good to learn and use them. zip and zipWith are two of them, when you think about something to be done using two different lists into one result list:
[1..] will generate the list of indices for you, it's an infinite list
appendIndex :: String -> Int -> String
appendIndex s i = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = zipWith appendIndex l [1..]
if you wanted to use zip, which is more basic but a little more verbose:
appendIndex :: (String,Int) -> String
appendIndex (s,i) = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = fmap appendIndex $ zip l [1..]
-- if you dont know about Functors yet, `fmap` is the generic way of doing `map`
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
add a comment |
There are multiple built-in Haskell functions in Prelude that are good to learn and use them. zip and zipWith are two of them, when you think about something to be done using two different lists into one result list:
[1..] will generate the list of indices for you, it's an infinite list
appendIndex :: String -> Int -> String
appendIndex s i = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = zipWith appendIndex l [1..]
if you wanted to use zip, which is more basic but a little more verbose:
appendIndex :: (String,Int) -> String
appendIndex (s,i) = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = fmap appendIndex $ zip l [1..]
-- if you dont know about Functors yet, `fmap` is the generic way of doing `map`
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
There are multiple built-in Haskell functions in Prelude that are good to learn and use them. zip and zipWith are two of them, when you think about something to be done using two different lists into one result list:
[1..] will generate the list of indices for you, it's an infinite list
appendIndex :: String -> Int -> String
appendIndex s i = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = zipWith appendIndex l [1..]
if you wanted to use zip, which is more basic but a little more verbose:
appendIndex :: (String,Int) -> String
appendIndex (s,i) = (show i) ++ " :" ++ s
indexThem :: [String] -> [String]
indexThem l = fmap appendIndex $ zip l [1..]
-- if you dont know about Functors yet, `fmap` is the generic way of doing `map`
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
edited Nov 24 '18 at 5:48
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
answered Nov 23 '18 at 16:30
Stephane RollandStephane Rolland
19.1k2698146
19.1k2698146
add a comment |
add a comment |
To get it right, it's important to understand why you're thinking wrong. Your recursion looks like this:
numberL (x:xs) = ... ++ numberL xs
So you calculate numberL xs and then put something in front of it. If numberL xs were correct, then then it would be numbered from 1 onwards, like: 1:..., 2:..., 3:.... So you could never build numberL (x:xs) from numberL xs just by adding new elements at the front. The whole numbering would be wrong. Instead you'd have to change the whole numbering of numberL xs.
The problem therefore is that it's not very useful to know numberL xs in order to calculate numberL (x:xs), due to the fact numberL always starts numbering from 1.
So lift that restriction. Build a function that numbers starting at n,
numberLFrom :: Int -> [String] -> [String]
numberLFrom n = ...
numberLFrom n (x:xs) = ...
Now the question you have to ask yourself is, in order to number (x:xs) starting at n you need to number xs starting at which number? And then how do you introduced the numbered x to that result?
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
add a comment |
To get it right, it's important to understand why you're thinking wrong. Your recursion looks like this:
numberL (x:xs) = ... ++ numberL xs
So you calculate numberL xs and then put something in front of it. If numberL xs were correct, then then it would be numbered from 1 onwards, like: 1:..., 2:..., 3:.... So you could never build numberL (x:xs) from numberL xs just by adding new elements at the front. The whole numbering would be wrong. Instead you'd have to change the whole numbering of numberL xs.
The problem therefore is that it's not very useful to know numberL xs in order to calculate numberL (x:xs), due to the fact numberL always starts numbering from 1.
So lift that restriction. Build a function that numbers starting at n,
numberLFrom :: Int -> [String] -> [String]
numberLFrom n = ...
numberLFrom n (x:xs) = ...
Now the question you have to ask yourself is, in order to number (x:xs) starting at n you need to number xs starting at which number? And then how do you introduced the numbered x to that result?
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
add a comment |
To get it right, it's important to understand why you're thinking wrong. Your recursion looks like this:
numberL (x:xs) = ... ++ numberL xs
So you calculate numberL xs and then put something in front of it. If numberL xs were correct, then then it would be numbered from 1 onwards, like: 1:..., 2:..., 3:.... So you could never build numberL (x:xs) from numberL xs just by adding new elements at the front. The whole numbering would be wrong. Instead you'd have to change the whole numbering of numberL xs.
The problem therefore is that it's not very useful to know numberL xs in order to calculate numberL (x:xs), due to the fact numberL always starts numbering from 1.
So lift that restriction. Build a function that numbers starting at n,
numberLFrom :: Int -> [String] -> [String]
numberLFrom n = ...
numberLFrom n (x:xs) = ...
Now the question you have to ask yourself is, in order to number (x:xs) starting at n you need to number xs starting at which number? And then how do you introduced the numbered x to that result?
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
To get it right, it's important to understand why you're thinking wrong. Your recursion looks like this:
numberL (x:xs) = ... ++ numberL xs
So you calculate numberL xs and then put something in front of it. If numberL xs were correct, then then it would be numbered from 1 onwards, like: 1:..., 2:..., 3:.... So you could never build numberL (x:xs) from numberL xs just by adding new elements at the front. The whole numbering would be wrong. Instead you'd have to change the whole numbering of numberL xs.
The problem therefore is that it's not very useful to know numberL xs in order to calculate numberL (x:xs), due to the fact numberL always starts numbering from 1.
So lift that restriction. Build a function that numbers starting at n,
numberLFrom :: Int -> [String] -> [String]
numberLFrom n = ...
numberLFrom n (x:xs) = ...
Now the question you have to ask yourself is, in order to number (x:xs) starting at n you need to number xs starting at which number? And then how do you introduced the numbered x to that result?
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
edited Nov 23 '18 at 16:45
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
answered Nov 23 '18 at 16:36
Jorge AdrianoJorge Adriano
2,210918
2,210918
add a comment |
add a comment |
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