Use 'any' to tell you true/false, if a list contains some value(s) divisible by 42
Having fun working through a Haskell tutorial...
One problem posed is as you see in the subject line... this is one I'd really like to figure out, but I'm at a loss... I've used any like so:
ghci >any (==55) [15,25,35,45,55,65,75,85,95]
True
ghci >any (==55) [15,25,35,45,54,65,75,85,95]
False
ghci >all even [2,4,6,8]
True
ghci >all even [1,3,5,7,9]
False
and it seems like checking if a list element is divisible by 42 or not, should be fairly easy...
I mean, you would check if any of the numbers in the list are n `mod` 42 == 0, right?
But how do you state that in the expression? Or do you need to write a 'helper' function?
haskell
edited Nov 13 at 3:39
duplode
22.8k44682
asked Nov 13 at 3:35
Stormy
673
add a comment |
Having fun working through a Haskell tutorial...
One problem posed is as you see in the subject line... this is one I'd really like to figure out, but I'm at a loss... I've used any like so:
ghci >any (==55) [15,25,35,45,55,65,75,85,95]
True
ghci >any (==55) [15,25,35,45,54,65,75,85,95]
False
ghci >all even [2,4,6,8]
True
ghci >all even [1,3,5,7,9]
False
and it seems like checking if a list element is divisible by 42 or not, should be fairly easy...
I mean, you would check if any of the numbers in the list are n `mod` 42 == 0, right?
But how do you state that in the expression? Or do you need to write a 'helper' function?
haskell
edited Nov 13 at 3:39
duplode
22.8k44682
asked Nov 13 at 3:35
Stormy
673
1
Without a helper definition, the most straightforward way would be with an anonymous function (a "lambda").
– duplode
Nov 13 at 3:41
Holy cow, thank you everyone... i had actually thought of the lambda way later, but couldn't figure out putting together the function for some reason... but once i see it, I'm like, "that makes total sense".
– Stormy
Nov 14 at 6:14
add a comment |
Having fun working through a Haskell tutorial...
One problem posed is as you see in the subject line... this is one I'd really like to figure out, but I'm at a loss... I've used any like so:
ghci >any (==55) [15,25,35,45,55,65,75,85,95]
True
ghci >any (==55) [15,25,35,45,54,65,75,85,95]
False
ghci >all even [2,4,6,8]
True
ghci >all even [1,3,5,7,9]
False
and it seems like checking if a list element is divisible by 42 or not, should be fairly easy...
I mean, you would check if any of the numbers in the list are n `mod` 42 == 0, right?
But how do you state that in the expression? Or do you need to write a 'helper' function?
haskell
edited Nov 13 at 3:39
duplode
22.8k44682
asked Nov 13 at 3:35
Stormy
673
Having fun working through a Haskell tutorial...
One problem posed is as you see in the subject line... this is one I'd really like to figure out, but I'm at a loss... I've used any like so:
ghci >any (==55) [15,25,35,45,55,65,75,85,95]
True
ghci >any (==55) [15,25,35,45,54,65,75,85,95]
False
ghci >all even [2,4,6,8]
True
ghci >all even [1,3,5,7,9]
False
and it seems like checking if a list element is divisible by 42 or not, should be fairly easy...
I mean, you would check if any of the numbers in the list are n `mod` 42 == 0, right?
But how do you state that in the expression? Or do you need to write a 'helper' function?
haskell
haskell
edited Nov 13 at 3:39
duplode
22.8k44682
asked Nov 13 at 3:35
Stormy
673
edited Nov 13 at 3:39
duplode
22.8k44682
asked Nov 13 at 3:35
Stormy
673
edited Nov 13 at 3:39
duplode
22.8k44682
edited Nov 13 at 3:39
duplode
22.8k44682
edited Nov 13 at 3:39
duplode
22.8k44682
22.8k44682
asked Nov 13 at 3:35
Stormy
673
asked Nov 13 at 3:35
Stormy
673
asked Nov 13 at 3:35
Stormy
673
673
1
Without a helper definition, the most straightforward way would be with an anonymous function (a "lambda").
– duplode
Nov 13 at 3:41
Holy cow, thank you everyone... i had actually thought of the lambda way later, but couldn't figure out putting together the function for some reason... but once i see it, I'm like, "that makes total sense".
– Stormy
Nov 14 at 6:14
add a comment |
1
Without a helper definition, the most straightforward way would be with an anonymous function (a "lambda").
– duplode
Nov 13 at 3:41
Holy cow, thank you everyone... i had actually thought of the lambda way later, but couldn't figure out putting together the function for some reason... but once i see it, I'm like, "that makes total sense".
– Stormy
Nov 14 at 6:14
1
1
Without a helper definition, the most straightforward way would be with an anonymous function (a "lambda").
– duplode
Nov 13 at 3:41
Without a helper definition, the most straightforward way would be with an anonymous function (a "lambda").
– duplode
Nov 13 at 3:41
Holy cow, thank you everyone... i had actually thought of the lambda way later, but couldn't figure out putting together the function for some reason... but once i see it, I'm like, "that makes total sense".
– Stormy
Nov 14 at 6:14
Holy cow, thank you everyone... i had actually thought of the lambda way later, but couldn't figure out putting together the function for some reason... but once i see it, I'm like, "that makes total sense".
– Stormy
Nov 14 at 6:14
add a comment |
2 Answers
2
active
oldest
votes
Composing (0==) and (`mod 42`):
f :: [Integer] -> Bool
f = any ((0==).(`mod` 42))
futher reducing parenthesis noise:
f :: [Integer] -> Bool
f = any $ (0==).(`mod` 42)
Helper function:
f :: [Integer] -> Bool
f = any div42
where
div42 n = n `mod` 42 == 0
Stylistically, for this function, either way seems fine. However, say you wanted to check if any values are divisible by 42 OR 52, then utilizing composition may become more obfuscated/complex/futile. Whereas using a helper function keeps things readable: div42Or52 n = n `mod` 42 == 0 || n `mod` 52 == 0. Note, I've used a where clause above, but a let in expression or a lambda are possible alternative ways to structure helper functions.
Ultimately, its up to the developer to balance concise code with understandable code.
answered Nov 13 at 4:07
DavOS
6441621
2
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might writedivisibleBy d = (== 0) . (`mod` d)(“to be divisible by d means to be equal to zero modulo d”) and then useany (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in awhereclause, though.
– Jon Purdy
Nov 13 at 8:17
1
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.isDivisibleBy x ysends a different message thanx `isDivisibleBy` y.
– dfeuer
Nov 13 at 15:35
1
@dfeuer: Yeah, same problem with other names really;divideswith the same definition but used asany (42 `divides`)makes just as much sense, butdivides 42is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)
– Jon Purdy
Nov 13 at 20:27
add a comment |
You can define function inplace:
any (n -> n `mod` 42 == 0) [1, 2, 42]
answered Nov 13 at 4:12
talex
9,1031546
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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active
oldest
votes
active
oldest
votes
Composing (0==) and (`mod 42`):
f :: [Integer] -> Bool
f = any ((0==).(`mod` 42))
futher reducing parenthesis noise:
f :: [Integer] -> Bool
f = any $ (0==).(`mod` 42)
Helper function:
f :: [Integer] -> Bool
f = any div42
where
div42 n = n `mod` 42 == 0
Stylistically, for this function, either way seems fine. However, say you wanted to check if any values are divisible by 42 OR 52, then utilizing composition may become more obfuscated/complex/futile. Whereas using a helper function keeps things readable: div42Or52 n = n `mod` 42 == 0 || n `mod` 52 == 0. Note, I've used a where clause above, but a let in expression or a lambda are possible alternative ways to structure helper functions.
Ultimately, its up to the developer to balance concise code with understandable code.
answered Nov 13 at 4:07
DavOS
6441621
2
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might writedivisibleBy d = (== 0) . (`mod` d)(“to be divisible by d means to be equal to zero modulo d”) and then useany (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in awhereclause, though.
– Jon Purdy
Nov 13 at 8:17
1
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.isDivisibleBy x ysends a different message thanx `isDivisibleBy` y.
– dfeuer
Nov 13 at 15:35
1
@dfeuer: Yeah, same problem with other names really;divideswith the same definition but used asany (42 `divides`)makes just as much sense, butdivides 42is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)
– Jon Purdy
Nov 13 at 20:27
add a comment |
Composing (0==) and (`mod 42`):
f :: [Integer] -> Bool
f = any ((0==).(`mod` 42))
futher reducing parenthesis noise:
f :: [Integer] -> Bool
f = any $ (0==).(`mod` 42)
Helper function:
f :: [Integer] -> Bool
f = any div42
where
div42 n = n `mod` 42 == 0
Stylistically, for this function, either way seems fine. However, say you wanted to check if any values are divisible by 42 OR 52, then utilizing composition may become more obfuscated/complex/futile. Whereas using a helper function keeps things readable: div42Or52 n = n `mod` 42 == 0 || n `mod` 52 == 0. Note, I've used a where clause above, but a let in expression or a lambda are possible alternative ways to structure helper functions.
Ultimately, its up to the developer to balance concise code with understandable code.
answered Nov 13 at 4:07
DavOS
6441621
2
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might writedivisibleBy d = (== 0) . (`mod` d)(“to be divisible by d means to be equal to zero modulo d”) and then useany (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in awhereclause, though.
– Jon Purdy
Nov 13 at 8:17
1
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.isDivisibleBy x ysends a different message thanx `isDivisibleBy` y.
– dfeuer
Nov 13 at 15:35
1
@dfeuer: Yeah, same problem with other names really;divideswith the same definition but used asany (42 `divides`)makes just as much sense, butdivides 42is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)
– Jon Purdy
Nov 13 at 20:27
add a comment |
Composing (0==) and (`mod 42`):
f :: [Integer] -> Bool
f = any ((0==).(`mod` 42))
futher reducing parenthesis noise:
f :: [Integer] -> Bool
f = any $ (0==).(`mod` 42)
Helper function:
f :: [Integer] -> Bool
f = any div42
where
div42 n = n `mod` 42 == 0
Stylistically, for this function, either way seems fine. However, say you wanted to check if any values are divisible by 42 OR 52, then utilizing composition may become more obfuscated/complex/futile. Whereas using a helper function keeps things readable: div42Or52 n = n `mod` 42 == 0 || n `mod` 52 == 0. Note, I've used a where clause above, but a let in expression or a lambda are possible alternative ways to structure helper functions.
Ultimately, its up to the developer to balance concise code with understandable code.
answered Nov 13 at 4:07
DavOS
6441621
Composing (0==) and (`mod 42`):
f :: [Integer] -> Bool
f = any ((0==).(`mod` 42))
futher reducing parenthesis noise:
f :: [Integer] -> Bool
f = any $ (0==).(`mod` 42)
Helper function:
f :: [Integer] -> Bool
f = any div42
where
div42 n = n `mod` 42 == 0
Stylistically, for this function, either way seems fine. However, say you wanted to check if any values are divisible by 42 OR 52, then utilizing composition may become more obfuscated/complex/futile. Whereas using a helper function keeps things readable: div42Or52 n = n `mod` 42 == 0 || n `mod` 52 == 0. Note, I've used a where clause above, but a let in expression or a lambda are possible alternative ways to structure helper functions.
Ultimately, its up to the developer to balance concise code with understandable code.
answered Nov 13 at 4:07
DavOS
6441621
edited Nov 13 at 5:06
answered Nov 13 at 4:07
DavOS
6441621
answered Nov 13 at 4:07
DavOS
6441621
answered Nov 13 at 4:07
DavOS
6441621
6441621
2
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might writedivisibleBy d = (== 0) . (`mod` d)(“to be divisible by d means to be equal to zero modulo d”) and then useany (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in awhereclause, though.
– Jon Purdy
Nov 13 at 8:17
1
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.isDivisibleBy x ysends a different message thanx `isDivisibleBy` y.
– dfeuer
Nov 13 at 15:35
1
@dfeuer: Yeah, same problem with other names really;divideswith the same definition but used asany (42 `divides`)makes just as much sense, butdivides 42is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)
– Jon Purdy
Nov 13 at 20:27
add a comment |
2
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might writedivisibleBy d = (== 0) . (`mod` d)(“to be divisible by d means to be equal to zero modulo d”) and then useany (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in awhereclause, though.
– Jon Purdy
Nov 13 at 8:17
1
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.isDivisibleBy x ysends a different message thanx `isDivisibleBy` y.
– dfeuer
Nov 13 at 15:35
1
@dfeuer: Yeah, same problem with other names really;divideswith the same definition but used asany (42 `divides`)makes just as much sense, butdivides 42is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)
– Jon Purdy
Nov 13 at 20:27
2
2
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might write
divisibleBy d = (== 0) . (`mod` d) (“to be divisible by d means to be equal to zero modulo d”) and then use any (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in a where clause, though.– Jon Purdy
Nov 13 at 8:17
For point-free code, it often helps readability to factor things into many small, generic, reusable definitions. For example, in this case I might write
divisibleBy d = (== 0) . (`mod` d) (“to be divisible by d means to be equal to zero modulo d”) and then use any (divisibleBy 42). Nothing wrong with using a lambda or auxiliary definition in a where clause, though.– Jon Purdy
Nov 13 at 8:17
1
1
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.
isDivisibleBy x y sends a different message than x `isDivisibleBy` y.– dfeuer
Nov 13 at 15:35
@JonPurdy, I completely agree. The only challenge is figuring out the order of the arguments.
isDivisibleBy x y sends a different message than x `isDivisibleBy` y.– dfeuer
Nov 13 at 15:35
1
1
@dfeuer: Yeah, same problem with other names really;
divides with the same definition but used as any (42 `divides`) makes just as much sense, but divides 42 is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)– Jon Purdy
Nov 13 at 20:27
@dfeuer: Yeah, same problem with other names really;
divides with the same definition but used as any (42 `divides`) makes just as much sense, but divides 42 is the wrong way around. This is why I prefer functions not be named such that they’re intended to be used infix, because it’s in conflict with using them in prefix; if you do want infix, an operator is preferable. (Besides, I have a personal objection to grave accents being used as quotation characters in a programming language, haha)– Jon Purdy
Nov 13 at 20:27
add a comment |
You can define function inplace:
any (n -> n `mod` 42 == 0) [1, 2, 42]
answered Nov 13 at 4:12
talex
9,1031546
add a comment |
You can define function inplace:
any (n -> n `mod` 42 == 0) [1, 2, 42]
answered Nov 13 at 4:12
talex
9,1031546
add a comment |
You can define function inplace:
any (n -> n `mod` 42 == 0) [1, 2, 42]
answered Nov 13 at 4:12
talex
9,1031546
You can define function inplace:
any (n -> n `mod` 42 == 0) [1, 2, 42]
answered Nov 13 at 4:12
talex
9,1031546
answered Nov 13 at 4:12
talex
9,1031546
answered Nov 13 at 4:12
talex
9,1031546
answered Nov 13 at 4:12
talex
9,1031546
9,1031546
add a comment |
add a comment |
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Without a helper definition, the most straightforward way would be with an anonymous function (a "lambda").
– duplode
Nov 13 at 3:41
Holy cow, thank you everyone... i had actually thought of the lambda way later, but couldn't figure out putting together the function for some reason... but once i see it, I'm like, "that makes total sense".
– Stormy
Nov 14 at 6:14