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artemkorsakov's solution

to Word Count in the Go Track

Published at Feb 21 2019 · 0 comments
Instructions
Test suite
Solution

Note:

This exercise has changed since this solution was written.

Given a phrase, count the occurrences of each word in that phrase.

For example for the input "olly olly in come free"

olly: 2
in: 1
come: 1
free: 1

Running the tests

To run the tests run the command go test from within the exercise directory.

If the test suite contains benchmarks, you can run these with the --bench and --benchmem flags:

go test -v --bench . --benchmem

Keep in mind that each reviewer will run benchmarks on a different machine, with different specs, so the results from these benchmark tests may vary.

Further information

For more detailed information about the Go track, including how to get help if you're having trouble, please visit the exercism.io Go language page.

Source

This is a classic toy problem, but we were reminded of it by seeing it in the Go Tour.

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.

cases_test.go

package wordcount

// Source: exercism/problem-specifications
// Commit: 77623ec word-count: Use camel-case for property name word-count: remove newline from eof
// Problem Specifications Version: 1.2.0

var testCases = []struct {
	description string
	input       string
	output      Frequency
}{
	{
		"count one word",
		"word",
		Frequency{"word": 1},
	},
	{
		"count one of each word",
		"one of each",
		Frequency{"each": 1, "of": 1, "one": 1},
	},
	{
		"multiple occurrences of a word",
		"one fish two fish red fish blue fish",
		Frequency{"blue": 1, "fish": 4, "one": 1, "red": 1, "two": 1},
	},
	{
		"handles cramped lists",
		"one,two,three",
		Frequency{"one": 1, "three": 1, "two": 1},
	},
	{
		"handles expanded lists",
		"one,\ntwo,\nthree",
		Frequency{"one": 1, "three": 1, "two": 1},
	},
	{
		"ignore punctuation",
		"car: carpet as java: javascript!!&@$%^&",
		Frequency{"as": 1, "car": 1, "carpet": 1, "java": 1, "javascript": 1},
	},
	{
		"include numbers",
		"testing, 1, 2 testing",
		Frequency{"1": 1, "2": 1, "testing": 2},
	},
	{
		"normalize case",
		"go Go GO Stop stop",
		Frequency{"go": 3, "stop": 2},
	},
	{
		"with apostrophes",
		"First: don't laugh. Then: don't cry.",
		Frequency{"cry": 1, "don't": 2, "first": 1, "laugh": 1, "then": 1},
	},
	{
		"with quotations",
		"Joe can't tell between 'large' and large.",
		Frequency{"and": 1, "between": 1, "can't": 1, "joe": 1, "large": 2, "tell": 1},
	},
	{
		"multiple spaces not detected as a word",
		" multiple   whitespaces",
		Frequency{"multiple": 1, "whitespaces": 1},
	},
}

word_count_test.go

package wordcount

import (
	"reflect"
	"testing"
)

// wordcount API
//
// func WordCount(phrase string) Frequency  // Implement this function.
// type Frequency map[string]int            // Using this return type.

func TestWordCount(t *testing.T) {
	for _, tt := range testCases {
		expected := tt.output
		actual := WordCount(tt.input)
		if !reflect.DeepEqual(actual, expected) {
			t.Fatalf("%s\n\tExpected: %v\n\tGot: %v", tt.description, expected, actual)
		} else {
			t.Logf("PASS: %s - WordCount(%s)", tt.description, tt.input)
		}
	}
}

func BenchmarkWordCount(b *testing.B) {
	for i := 0; i < b.N; i++ {
		for _, tt := range testCases {
			WordCount(tt.input)
		}
	}
}
package wordcount

import (
	"regexp"
	"strings"
)

var reg = regexp.MustCompile(`([1-9]+|[a-z]+'?[a-z]+)`)

// Frequency is the map where the key is the word and the value is the number of the occurrences.
type Frequency map[string]int

// WordCount counts the occurrences of each word in that phrase.
func WordCount(input string) Frequency {
	result := make(map[string]int)
	words := reg.FindAllString(strings.ToLower(input), -1)
	for _, word := range words {
		result[word] = result[word] + 1
	}
	return result
}

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