Andres

Andres

Count the number of non-numeric characters in a string and then sum their values

Hello.
Trying to solve this problem I wrote the following algorithm:

string = "3340c9571y40m47ku49t9315mrvzqo667k36e"

# %{
#  "c" => 1,
#  "e" => 1,
#  "k" => 2,
#  "m" => 2,
#  "o" => 1,
#  "q" => 1,
#  "r" => 1,
#  "t" => 1,
#  "u" => 1,
#  "v" => 1,
#  "y" => 1,
#  "z" => 1
#  }
#  [1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1]
# Expected result => 14

def is_numeric?(char) do
    Regex.match?(~r/^\d+$/, char)
end

def total_non_numeric_chars(string) do
    String.codepoints(string)
    |> Enum.reduce(%{}, fn char, acc ->
      cond do
        !is_numeric?(char) ->
          if Map.has_key?(acc, char),
            do: Map.update!(acc, char, &(&1 + 1)),
            else: Map.put(acc, char, 1)

        true ->
          acc
      end
    end)
    |> Map.values()
    |> Enum.sum()
end

I would like to know if is possible to avoid the nested cond and if with some patter matching or other technique.
Any suggestions to improve the algorithm is welcome.

Thanks.

Marked As Solved

michalmuskala

michalmuskala

I just thought of using the new reduce comprehensions. With that it can be a one-liner:

for << <<char>> <- string >>, char not in ?0..?9, reduce: 0, do: (acc -> acc + 1)

Also Liked

michalmuskala

michalmuskala

Since you’re summing them up at the end, I guess you don’t really need to keep the frequency map - you can just keep the sum (unless you also output the map somewhere).

With that in mind, this is how I’d write it:

def total_non_numeric_chars(string), do: loop(string, 0)

defp loop(<<char, rest::binary>>, count) when char in ?0..?9, do: loop(rest, count)
defp loop(<<_, rest::binary>>, count), do: loop(rest, count + 1)
defp loop(<<>>, count), do: count
NobbZ

NobbZ

Just be aware of the fact that is_numeric? is an unidiomatic name for a function, either we use the prefix is_ when we have guardsave predicates or we use the suffix ? when the predicate is not guardsafe.

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