Problem of the Day -- Character Frequency

Character Frequency

Another great Monday is upon us! Today's Problem of the Day comes from user D. Thanks for the submission!

For some input string, for example "Feel free to include attribution details such as a Twitter handle so you can get credit for submitting the problem.", determine the character frequencies and display a simple visualization:

      ...................
    e ............
    t ............
    i ........
    r ......
    o ......
    a ......
    l .....
    n .....
    u .....
    s .....
    c ....
    d ....
    b ...
    h ...
    f ..
    g ..
    m ..
    F .
    T .
    w .
    y .
    p .
    . .

Permalink: http://problemotd.com/problem/character-frequency/

Comments:

Emre Ozdemir - 10 years, 7 months ago

#include <iostream>
#include <string>
#include <algorithm>
using namespace std;

int main()
{
    int  k, i;
    int arr[123][2]= {0};
    string str;
    getline(cin,str);

    for(k=0;k<=122;k++)
        arr[k][0]=k;


    for(i=65;i<=122;i++)            //Letter counter
    {
        for(k=0;k<str.length();k++)
        {
            if(str[k]==i)
                arr[i][1]++;
        }
    }

    for(i=65;i<=122;i++)        //Sorts by number of letters and swaps
    {
        for(k=i+1;k<=122;k++)
        {
            if(arr[i][1]<arr[k][1])
            {
            swap(arr[i][0],arr[k][0]);
            swap(arr[i][1],arr[k][1]);
            }

        }
    }

    for(i=65;i<=122;i++)        //Dot and letter printer
    {
        if(arr[i][1]>0)
        {
            printf("%c", arr[i][0]);
            for(k=1;k<=arr[i][1];k++)
                cout<<'.';

            puts("\n");
        }
    }

    getchar();
    getchar();
    return 0;
}

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Pyro - 10 years, 7 months ago

My code in python. It's not really optimised but works well!

def print_dots(n):
    print " ",
    for i in range(n):
        print ".",
    print " "

lst = [[i,0] for i in xrange(127)]
text=raw_input("text?\n")

for i in range(len(text)):
    lst[ord(text[i])][1]=lst[ord(text[i])][1]+1
lst.sort(key=lambda x: x[1], reverse=True)
not_finished=True

while not_finished:
    for i in xrange(127):
        if lst[i][1]>0:
            print chr(lst[i][0]),
            print_dots(lst[i][1])
    not_finished=False

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Pearce Keesling - 10 years, 7 months ago

Getting them to display in descending was a little ugly, but the output is quite elegant IMHO opinion. Alphabetical order was a side-effect :). Wrote this in C.

#include <stdio.h>
#include <string.h>

int compare (const void * a, const void * b)
{
    return(*(int*)a - *(int*)b);
}

int main(int argc, char* argv[])
{
    int frequency[256] = {0};
    for(int i = 0; i < strlen(argv[1]);i++)
    {
        frequency[argv[1][i]] ++;
    }
    int max_i = 0;
    int max = 0;
    do
    {
        max =0;
        for(int i = 0; i < 256;i++)
        {
            if(frequency[i]>max)
            {
                max = frequency[i];
                max_i = i;
            }
        }
        if(frequency[max_i]==0)continue;
        printf("%c ",max_i);
        while(frequency[max_i]>0)
        {
            printf(".");
            frequency[max_i]--;
        }
        printf("\n");
    }while(max>0);

    return 0;
}

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Tim - 10 years, 7 months ago
Happy to add another python submission to the pile. How does one enable pretty markup?

def counter(input_string): dictionary = {} for l in input_string: if l != " ": dictionary[l] = dictionary.get(l, 0) + 1 return dictionary

def graph(input_string): data = counter(input_string) tdata = data.items() tdata.sort(key=lambda tup: tup[1]) for tup in reversed(tdata): print tup[0], "."*tup[1]

graph("Feel free to include attribution details such as a Twitter handle so you can get credit for submitting the problem.")

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Pyro - 10 years, 7 months ago
You have to put ``` before and after the code!

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Tim - 10 years, 7 months ago

Thanks!

def counter(input_string):
    dictionary = {}
    for l in input_string:
        if l != " ":
            dictionary[l] = dictionary.get(l, 0) + 1
    return dictionary

def graph(input_string):
    data = counter(input_string)
    tdata = data.items()
    tdata.sort(key=lambda tup: tup[1])
    for tup in reversed(tdata):
        print tup[0], "."*tup[1]

graph("Feel free to include attribution details such as a Twitter handle so you can get credit for submitting the problem.")

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SithEngineer - 10 years, 7 months ago

a C# solution...

    public struct CharCount {
        public ulong Count;
        public char Value;
    }

    public static LinkedList<CharCount> CountChars(string toCount) {
        char[] charArray = toCount.ToCharArray();
        int charCounterLength = (int) Math.Pow(2, sizeof(char) * 8);
        ulong[] charCounter = new ulong[charCounterLength];

        // count chars
        for(int i = 0 ; i < charArray.Length ; ++i) {
            charCounter[charArray[i]]++;
        }

        // render response
        LinkedList<CharCount> result = new LinkedList<CharCount>(); 
        for(int i = 0 ; i < charCounterLength ; ++i) {
            if(charCounter[i] > 0) {
                result.AddFirst( new CharCount { Count = charCounter[i], Value = (char)i } );
            }
        }
        return result;
    }

and test code...

public static void Main() 
{
    var data = "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Suspendisse commodo est eu massa scelerisque vehicula. Integer vehicula dolor eu aliquet.";
    Console.WriteLine ("counting chars frequency of: {0}", data);

    var result = Algorithms.CountChars(data);
    foreach(var countResult in result) {
        Console.WriteLine ("{0} apeared {1} times", countResult.Value, countResult.Count);    
    }

    // Keep the console window open in debug mode.
    Console.WriteLine ("Press any key to exit.");
    Console.ReadKey();
}

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Nick Krichevsky - 10 years, 7 months ago

Python Code!

d={}
for char in string:
    if char not in d:
        d[char]=1
    else:
        d[char]+=1
for item in d:
    print item,
    print " ",
    for i in range(d[item]):
        print '.',
    print ""

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Akshay Bist - 10 years, 7 months ago

My Python solution

from collections import Counter
import operator


def get_sorted_freq(text):
    freq = Counter(text)
    return sorted(freq.iteritems(), key=operator.itemgetter(1), reverse=True)


def print_result(sorted_freq):
    for char, freq in sorted_freq:
        print "{0}:\t{1}".format(char, "." * freq)


def main():
    text = raw_input("Enter text: ")
    sorted_freq = get_sorted_freq(text)
    print_result(sorted_freq)


if __name__ == "__main__":
    main()

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rav - 10 years, 7 months ago

Python:


import sys
import operator

phrase = "Feel free to include attribution details such as a Twitter handle so you can get credit for submitting the problem".lower()
frequency = {}

for letter in phrase:
  try:
    frequency[letter]
  except KeyError:
    frequency[letter] = 1
  else:
    frequency[letter] = frequency[letter] + 1

frequency.pop(" ", None)

sortFreq = sorted(frequency.iteritems(), key=operator.itemgetter(1),reverse=True)

for key,value in sortFreq:
  dots = ''
  for i in range(0,value):
    dots = dots + '.'
  print key + ": " + dots

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ujjl - 10 years, 7 months ago

Haskell solution

import Data.Map
charFreq input = putStr $ formatFreq ( toList $ fromListWith (+) [(c, 1) | c <- input] )

formatFreq [] = []
formatFreq ((a,b):xs) = a:' ':(take b (repeat '.'))++"\n"++(formatFreq xs)

Output:

*Main> charFreq "abcdefgabc deabcdabcaba"
  .
a ......
b .....
c ....
d ...
e ..
f .
g .

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James - 10 years, 7 months ago

Python solution. Anyone give feedback?

def charfreq(string):
    l = list(string)
    s = list(set(l))
    p = []
    for i in s:
        p.append((i, l.count(i)))
    p = sorted(p, key=lambda x: x[1], reverse=True)
    for i in p:
        print('{}{}'.format(i[0], i[1]*'.'))

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Ben - 10 years, 7 months ago

A C# solution using LINQ:

public static void Main()
{
    var input = Console.ReadLine();
    var inputArray = input.ToCharArray();
    var distinctChars = inputArray.GroupBy(i => i);

    foreach (var character in distinctChars.OrderByDescending(d => d.Count()))
    {
        Console.Write(character.Key + " ");
        for (var i = 0; i < character.Count(); i++)
        {
            Console.Write('.');
        }

        Console.WriteLine();
    }

    Console.ReadKey();
}

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Hueho - 10 years, 7 months ago

def frequencies(phrase)
  multiset = Hash.new { |hash, key| hash[key] = 0 }
  phrase.chars.each { |ch| multiset[ch] += 1 }
  multiset
end

def sort(frequencies)
  frequencies.to_a.sort_by { |arr| arr[1] }
end

def output(ordered_freq)
  ordered_freq.each do |ch, count|
    puts "#{ch} " + ('.' * count)
  end
end

output sort(frequencies(ARGF.read)).reverse

Reads the phrase from the standard input or loading any file passed to script as a argument

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Anonymous - 10 years, 6 months ago

#python 2.7
import string
from operator import itemgetter

def charfreq(b):

    mydict = {}

    for letters in b:
        mydict.update({letters: 0})

    for letters in b:
        mydict[letters] = mydict[letters] + 1

    sortedletters = sorted(mydict.items(), key=itemgetter(1))[::-1]

    i = 0
    while i < len(sortedletters):
        print sortedletters[i][0], ":", "*"*sortedletters[i][1]
        i = i + 1

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Problem of the Day A new programming or logic puzzle every Mon-Fri

Character Frequency

Comments:

Problem of the Day
A new programming or logic puzzle every Mon-Fri