return a list from a complex string
up vote
-2
down vote
favorite
I start with haskell.
I was wondering if I can convert a string to a list of different element's type like that : "1*30%4" -> ['1', '*', '30', '%', '4'] without Parsec
I have already found answers but none really help me...
like map (:) "1*30%4"with GHCI
or with the help of intersperse.
but I do not know how to keep the right formats, I can not, for example, have numbers or float/double in my list because everything is cut one by one : "1*30%4" -> ['1', '*', '3', '0', '%', '4'] or "1*30.4%4" -> ['1', '*', '3', '0', '.', '4', '%', '4']
Someone can help me ?
parsing haskell operators
asked Nov 8 at 11:29
astrocurieux
115
add a comment |
up vote
-2
down vote
favorite
I start with haskell.
I was wondering if I can convert a string to a list of different element's type like that : "1*30%4" -> ['1', '*', '30', '%', '4'] without Parsec
I have already found answers but none really help me...
like map (:) "1*30%4"with GHCI
or with the help of intersperse.
but I do not know how to keep the right formats, I can not, for example, have numbers or float/double in my list because everything is cut one by one : "1*30%4" -> ['1', '*', '3', '0', '%', '4'] or "1*30.4%4" -> ['1', '*', '3', '0', '.', '4', '%', '4']
Someone can help me ?
parsing haskell operators
asked Nov 8 at 11:29
astrocurieux
115
Would you like to do it with a parsing library like Parsec or just do it with vanilla Haskell?
– cmdv
Nov 8 at 11:34
I actually forgot to mention that I do not want to use Parsec
– astrocurieux
Nov 8 at 11:39
3
List in Haskell cannot hold different type of element. like '30' is not a type of Char, so, it must be broken into '3' and '0' in order to store them in list, say ['1', ...'3', '0' ...]. The thing you need may be list of String, say ["1", ... "30", "%"..] right?
– assembly.jc
Nov 8 at 11:51
yes it's exactly that, with the 'Words' function my program works well, but I have to separate everything with spaces, it's very annoying ...
– astrocurieux
Nov 8 at 12:04
3
What's wrong with parser libraries, it's the right tool for the job?
– Li-yao Xia
Nov 8 at 12:19
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I start with haskell.
I was wondering if I can convert a string to a list of different element's type like that : "1*30%4" -> ['1', '*', '30', '%', '4'] without Parsec
I have already found answers but none really help me...
like map (:) "1*30%4"with GHCI
or with the help of intersperse.
but I do not know how to keep the right formats, I can not, for example, have numbers or float/double in my list because everything is cut one by one : "1*30%4" -> ['1', '*', '3', '0', '%', '4'] or "1*30.4%4" -> ['1', '*', '3', '0', '.', '4', '%', '4']
Someone can help me ?
parsing haskell operators
asked Nov 8 at 11:29
astrocurieux
115
I start with haskell.
I was wondering if I can convert a string to a list of different element's type like that : "1*30%4" -> ['1', '*', '30', '%', '4'] without Parsec
I have already found answers but none really help me...
like map (:) "1*30%4"with GHCI
or with the help of intersperse.
but I do not know how to keep the right formats, I can not, for example, have numbers or float/double in my list because everything is cut one by one : "1*30%4" -> ['1', '*', '3', '0', '%', '4'] or "1*30.4%4" -> ['1', '*', '3', '0', '.', '4', '%', '4']
Someone can help me ?
parsing haskell operators
parsing haskell operators
asked Nov 8 at 11:29
astrocurieux
115
asked Nov 8 at 11:29
astrocurieux
115
edited Nov 8 at 11:40
asked Nov 8 at 11:29
astrocurieux
115
asked Nov 8 at 11:29
astrocurieux
115
asked Nov 8 at 11:29
astrocurieux
115
115
Would you like to do it with a parsing library like Parsec or just do it with vanilla Haskell?
– cmdv
Nov 8 at 11:34
I actually forgot to mention that I do not want to use Parsec
– astrocurieux
Nov 8 at 11:39
3
List in Haskell cannot hold different type of element. like '30' is not a type of Char, so, it must be broken into '3' and '0' in order to store them in list, say ['1', ...'3', '0' ...]. The thing you need may be list of String, say ["1", ... "30", "%"..] right?
– assembly.jc
Nov 8 at 11:51
yes it's exactly that, with the 'Words' function my program works well, but I have to separate everything with spaces, it's very annoying ...
– astrocurieux
Nov 8 at 12:04
3
What's wrong with parser libraries, it's the right tool for the job?
– Li-yao Xia
Nov 8 at 12:19
add a comment |
Would you like to do it with a parsing library like Parsec or just do it with vanilla Haskell?
– cmdv
Nov 8 at 11:34
I actually forgot to mention that I do not want to use Parsec
– astrocurieux
Nov 8 at 11:39
3
List in Haskell cannot hold different type of element. like '30' is not a type of Char, so, it must be broken into '3' and '0' in order to store them in list, say ['1', ...'3', '0' ...]. The thing you need may be list of String, say ["1", ... "30", "%"..] right?
– assembly.jc
Nov 8 at 11:51
yes it's exactly that, with the 'Words' function my program works well, but I have to separate everything with spaces, it's very annoying ...
– astrocurieux
Nov 8 at 12:04
3
What's wrong with parser libraries, it's the right tool for the job?
– Li-yao Xia
Nov 8 at 12:19
Would you like to do it with a parsing library like Parsec or just do it with vanilla Haskell?
– cmdv
Nov 8 at 11:34
Would you like to do it with a parsing library like Parsec or just do it with vanilla Haskell?
– cmdv
Nov 8 at 11:34
I actually forgot to mention that I do not want to use Parsec
– astrocurieux
Nov 8 at 11:39
I actually forgot to mention that I do not want to use Parsec
– astrocurieux
Nov 8 at 11:39
3
3
List in Haskell cannot hold different type of element. like '30' is not a type of Char, so, it must be broken into '3' and '0' in order to store them in list, say ['1', ...'3', '0' ...]. The thing you need may be list of String, say ["1", ... "30", "%"..] right?
– assembly.jc
Nov 8 at 11:51
List in Haskell cannot hold different type of element. like '30' is not a type of Char, so, it must be broken into '3' and '0' in order to store them in list, say ['1', ...'3', '0' ...]. The thing you need may be list of String, say ["1", ... "30", "%"..] right?
– assembly.jc
Nov 8 at 11:51
yes it's exactly that, with the 'Words' function my program works well, but I have to separate everything with spaces, it's very annoying ...
– astrocurieux
Nov 8 at 12:04
yes it's exactly that, with the 'Words' function my program works well, but I have to separate everything with spaces, it's very annoying ...
– astrocurieux
Nov 8 at 12:04
3
3
What's wrong with parser libraries, it's the right tool for the job?
– Li-yao Xia
Nov 8 at 12:19
What's wrong with parser libraries, it's the right tool for the job?
– Li-yao Xia
Nov 8 at 12:19
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
As some users have pointed out, your returning type is [String] instead of [Char]. You can easily achieve this by the following:
import Data.Char
import Data.List
expresionToList :: String -> [String]
expresionToList = groupBy readAsNumber
where readAsNumber c d = pred c && pred d
pred x = isDigit x || x == '.'
predfunction returns True when its input is a digit or a dot, False otherwisereadAsNumbertakes two returns True if both are a digit or a dot- finally you group your string by
readAsNumber
answered Nov 8 at 12:47
Luis Morillo
70512
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
@astrocurieuxpredfunction is defined in the where clause. It is not built-in function.
– Luis Morillo
Nov 12 at 7:41
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
As some users have pointed out, your returning type is [String] instead of [Char]. You can easily achieve this by the following:
import Data.Char
import Data.List
expresionToList :: String -> [String]
expresionToList = groupBy readAsNumber
where readAsNumber c d = pred c && pred d
pred x = isDigit x || x == '.'
predfunction returns True when its input is a digit or a dot, False otherwisereadAsNumbertakes two returns True if both are a digit or a dot- finally you group your string by
readAsNumber
answered Nov 8 at 12:47
Luis Morillo
70512
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
@astrocurieuxpredfunction is defined in the where clause. It is not built-in function.
– Luis Morillo
Nov 12 at 7:41
add a comment |
up vote
3
down vote
accepted
As some users have pointed out, your returning type is [String] instead of [Char]. You can easily achieve this by the following:
import Data.Char
import Data.List
expresionToList :: String -> [String]
expresionToList = groupBy readAsNumber
where readAsNumber c d = pred c && pred d
pred x = isDigit x || x == '.'
predfunction returns True when its input is a digit or a dot, False otherwisereadAsNumbertakes two returns True if both are a digit or a dot- finally you group your string by
readAsNumber
answered Nov 8 at 12:47
Luis Morillo
70512
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
@astrocurieuxpredfunction is defined in the where clause. It is not built-in function.
– Luis Morillo
Nov 12 at 7:41
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
As some users have pointed out, your returning type is [String] instead of [Char]. You can easily achieve this by the following:
import Data.Char
import Data.List
expresionToList :: String -> [String]
expresionToList = groupBy readAsNumber
where readAsNumber c d = pred c && pred d
pred x = isDigit x || x == '.'
predfunction returns True when its input is a digit or a dot, False otherwisereadAsNumbertakes two returns True if both are a digit or a dot- finally you group your string by
readAsNumber
answered Nov 8 at 12:47
Luis Morillo
70512
As some users have pointed out, your returning type is [String] instead of [Char]. You can easily achieve this by the following:
import Data.Char
import Data.List
expresionToList :: String -> [String]
expresionToList = groupBy readAsNumber
where readAsNumber c d = pred c && pred d
pred x = isDigit x || x == '.'
predfunction returns True when its input is a digit or a dot, False otherwisereadAsNumbertakes two returns True if both are a digit or a dot- finally you group your string by
readAsNumber
answered Nov 8 at 12:47
Luis Morillo
70512
edited Nov 8 at 13:02
answered Nov 8 at 12:47
Luis Morillo
70512
answered Nov 8 at 12:47
Luis Morillo
70512
answered Nov 8 at 12:47
Luis Morillo
70512
70512
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
@astrocurieuxpredfunction is defined in the where clause. It is not built-in function.
– Luis Morillo
Nov 12 at 7:41
add a comment |
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
@astrocurieuxpredfunction is defined in the where clause. It is not built-in function.
– Luis Morillo
Nov 12 at 7:41
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
first time I hear about 'pred'.
– astrocurieux
Nov 8 at 14:22
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
thank you for your help, with the expresionToList function my problem is solved.
– astrocurieux
Nov 8 at 14:29
@astrocurieux
pred function is defined in the where clause. It is not built-in function.– Luis Morillo
Nov 12 at 7:41
@astrocurieux
pred function is defined in the where clause. It is not built-in function.– Luis Morillo
Nov 12 at 7:41
add a comment |
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Would you like to do it with a parsing library like Parsec or just do it with vanilla Haskell?
– cmdv
Nov 8 at 11:34
I actually forgot to mention that I do not want to use Parsec
– astrocurieux
Nov 8 at 11:39
3
List in Haskell cannot hold different type of element. like '30' is not a type of Char, so, it must be broken into '3' and '0' in order to store them in list, say ['1', ...'3', '0' ...]. The thing you need may be list of String, say ["1", ... "30", "%"..] right?
– assembly.jc
Nov 8 at 11:51
yes it's exactly that, with the 'Words' function my program works well, but I have to separate everything with spaces, it's very annoying ...
– astrocurieux
Nov 8 at 12:04
3
What's wrong with parser libraries, it's the right tool for the job?
– Li-yao Xia
Nov 8 at 12:19