3 ms·

Fun with a Functional Programming Interview Question

11 points by amscotti 6y ago

badhabit 6y ago
newlisp: (set 'L '()) (dolist (s (explode "aaaabbbcca")) (if (!= (ref-all s L) '((0 0))) (push (list s 1) L) (inc (L 0 1)))) (print (reverse L))
deleted 6y ago
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al2o3cr 6y ago
Surprising that F# doesn't have a "chunkBy" function for sequences; it makes the solution in Elixir vastly more readable than the version in the article: "aaaabbbcca" |> String.codepoints() |> Enum.chunk_by(& &1) |> Enum.map(&{hd(&1), length(&1)})
- DanielBMarkham 6y ago
  There is a chunkBy, it's just not this nice. This would be a great project for the F#+ guys.
deleted 6y ago
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riwsky 6y ago
meanwhile python be out here like: [(m.group(1), len(m.group())) for m in re.finditer(r"([a-z])\1*", “aaaabbbcca")
- eesmith 6y ago
  import itertools [(c, len(list(it))) for (c, it) in itertools.groupby("aaaabbbcca")] It's not the first time I've wanted def ilen(it): return sum(1 for _ in it) as part of itertools. Which would make the second line be: [(c, itertools.ilen(it)) for (c, it) in itertools.groupby("aaaabbbcca")] Rejected many times, eg, from 2010: https://bugs.python.org/issue9547 https://bugs.python.org/issue9547 .
d_t_w 6y ago
#(mapv (juxt (comp str first) count) (partition-by identity %))
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homic 6y ago
I attempted this as a Haskell beginner. Positive it can be improved upon. runLengthEncode :: String -> [(Char, Int)] runLengthEncode [] = [] runLengthEncode input = let chFirst = head input zipped = zip input $ chFirst : input in rle zipped (chFirst, 0) where rle :: [(Char, Char)] -> (Char, Int) -> [(Char, Int)] rle [] (ch, count) = [(ch, count)] rle ((ch1, ch2) : cs) (ch, count) = if ch1 == ch2 then rle cs (ch1, count + 1) else (ch2, count) : rle cs (ch1, 1)
dirkt 6y ago
Haskell in REPL: Prelude> :m Data.List Prelude Data.List> let f xs = [(head ys, length ys) | ys <- group xs] Prelude Data.List> f "aaaabbbcca" [('a',4),('b',3),('c',2),('a',1)] Of course using the "group" function instead of programming it myself is cheating a bit ...
zzo38computer 6y ago
I did it this way: map (head &&& length) . group (Although in an actual full program, there would be other considerations (what considerations those are depends on the program being written), and in a full program I probably would not be using Haskell, but rather C.)

11 points by amscotti 6y ago

newlisp: (set 'L '()) (dolist (s (explode "aaaabbbcca")) (if (!= (ref-all s L) '((0 0))) (push (list s 1) L) (inc (L 0 1)))) (print (reverse L))

deleted 6y ago

[deleted]

al2o3cr 6y ago

Surprising that F# doesn't have a "chunkBy" function for sequences; it makes the solution in Elixir vastly more readable than the version in the article: "aaaabbbcca" |> String.codepoints() |> Enum.chunk_by(& &1) |> Enum.map(&{hd(&1), length(&1)})

DanielBMarkham 6y ago
There is a chunkBy, it's just not this nice. This would be a great project for the F#+ guys.

deleted 6y ago

[deleted]

riwsky 6y ago

meanwhile python be out here like: [(m.group(1), len(m.group())) for m in re.finditer(r"([a-z])\1*", “aaaabbbcca")

eesmith 6y ago
import itertools [(c, len(list(it))) for (c, it) in itertools.groupby("aaaabbbcca")] It's not the first time I've wanted def ilen(it): return sum(1 for _ in it) as part of itertools. Which would make the second line be: [(c, itertools.ilen(it)) for (c, it) in itertools.groupby("aaaabbbcca")] Rejected many times, eg, from 2010: https://bugs.python.org/issue9547 https://bugs.python.org/issue9547 .

d_t_w 6y ago

#(mapv (juxt (comp str first) count) (partition-by identity %))

deleted 6y ago

[deleted]

homic 6y ago

I attempted this as a Haskell beginner. Positive it can be improved upon. runLengthEncode :: String -> [(Char, Int)] runLengthEncode [] = [] runLengthEncode input = let chFirst = head input zipped = zip input $ chFirst : input in rle zipped (chFirst, 0) where rle :: [(Char, Char)] -> (Char, Int) -> [(Char, Int)] rle [] (ch, count) = [(ch, count)] rle ((ch1, ch2) : cs) (ch, count) = if ch1 == ch2 then rle cs (ch1, count + 1) else (ch2, count) : rle cs (ch1, 1)

dirkt 6y ago

Haskell in REPL: Prelude> :m Data.List Prelude Data.List> let f xs = [(head ys, length ys) | ys <- group xs] Prelude Data.List> f "aaaabbbcca" [('a',4),('b',3),('c',2),('a',1)] Of course using the "group" function instead of programming it myself is cheating a bit ...

zzo38computer 6y ago

I did it this way: map (head &&& length) . group (Although in an actual full program, there would be other considerations (what considerations those are depends on the program being written), and in a full program I probably would not be using Haskell, but rather C.)