Convert words (string) to Int
I'm sure this has been done a hundred times, but i'm hoping there is a really simple way to accomplished this. I'm wanting to change words to an int.
Like the following example
One = 1 Two = 2 Three = 3
So basically if I have the string "One" it gets converted to 1, even if I could get back a string "1" I can just convert that.
Did this for fun... there's probably many edge cases that would fail...
private static Dictionary<string,long> numberTable=
new Dictionary<string,long>
{{"zero",0},{"one",1},{"two",2},{"three",3},{"four",4},
{"five",5},{"six",6},{"seven",7},{"eight",8},{"nine",9},
{"ten",10},{"eleven",11},{"twelve",12},{"thirteen",13},
{"fourteen",14},{"fifteen",15},{"sixteen",16},
{"seventeen",17},{"eighteen",18},{"nineteen",19},{"twenty",20},
{"thirty",30},{"forty",40},{"fifty",50},{"sixty",60},
{"seventy",70},{"eighty",80},{"ninety",90},{"hundred",100},
{"thousand",1000},{"million",1000000},{"billion",1000000000},
{"trillion",1000000000000},{"quadrillion",1000000000000000},
{"quintillion",1000000000000000000}};
public static long ToLong(string numberString)
{
var numbers = Regex.Matches(numberString, @"\w+").Cast<Match>()
.Select(m => m.Value.ToLowerInvariant())
.Where(v => numberTable.ContainsKey(v))
.Select(v => numberTable[v]);
long acc = 0,total = 0L;
foreach(var n in numbers)
{
if(n >= 1000)
{
total += (acc * n);
acc = 0;
}
else if(n >= 100){
acc *= n;
}
else acc += n;
}
return (total + acc) * ( numberString.StartsWith("minus",
StringComparison.InvariantCultureIgnoreCase) ? -1 : 1);
}
Here's a method that does that. If you need a wider range, it's easily extensible; just use a long, a ulong, or even a BigInt, and add more items to the modifiers dictionary.
static int ParseEnglish(string number) {
string[] words = number.ToLower().Split(new char[] {' ', '-', ','}, StringSplitOptions.RemoveEmptyEntries);
string[] ones = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
string[] teens = {"eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
string[] tens = {"ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
Dictionary<string, int> modifiers = new Dictionary<string, int>() {
{"billion", 1000000000},
{"million", 1000000},
{"thousand", 1000},
{"hundred", 100}
};
if(number == "eleventy billion")
return int.MaxValue; // 110,000,000,000 is out of range for an int!
int result = 0;
int currentResult = 0;
int lastModifier = 1;
foreach(string word in words) {
if(modifiers.ContainsKey(word)) {
lastModifier *= modifiers[word];
} else {
int n;
if(lastModifier > 1) {
result += currentResult * lastModifier;
lastModifier = 1;
currentResult = 0;
}
if((n = Array.IndexOf(ones, word) + 1) > 0) {
currentResult += n;
} else if((n = Array.IndexOf(teens, word) + 1) > 0) {
currentResult += n + 10;
} else if((n = Array.IndexOf(tens, word) + 1) > 0) {
currentResult += n * 10;
} else if(word != "and") {
throw new ApplicationException("Unrecognized word: " + word);
}
}
}
return result + currentResult * lastModifier;
}
Here's a C# version of the algorithm above. I renamed and rewrote some code for clarity and added support for negative numbers, hyphenated numbers, zero, and combinations of text words and digits (like "100 and 5"). Thank you to Ry- for the great start.
/// <summary>
/// Convert text number strings to integer numbers. Credit to stackoverflow
/// for the main algorithm.
/// </summary>
public static int
WordNumberToInt (string number) {
// define arrays of keywords to translate text words to integer positions
// in the arrays. Thus, ordering of words in the array is important.
string[] ones = {
"one", "two", "three", "four", "five", "six",
"seven", "eight", "nine"
};
string[] teens = {
"eleven", "twelve", "thirteen", "fourteen", "fifteen",
"sixteen", "seventeen", "eighteen", "nineteen"
};
string[] tens = {
"ten", "twenty", "thirty", "forty", "fifty", "sixty",
"seventy", "eighty", "ninety"
};
var bigscales = new Dictionary<string, int> () {
{"hundred", 100}, {"hundreds", 100}, {"thousand", 1000},
{"million", 1000000}, {"billion", 1000000000},
};
string[] minusWords = {"minus", "negative"};
var splitchars = new char[] {' ', '-', ','};
// flip all words to lowercase for proper matching
var lowercase = number.ToLower ();
var inputwords = lowercase.Split (splitchars, StringSplitOptions.RemoveEmptyEntries);
// initalize loop variables and flags
int result = 0;
int currentResult = 0;
int bigMultiplierValue = 1;
bool bigMultiplierIsActive = false;
bool minusFlag = false;
foreach (string curword in inputwords) {
// input words are either bigMultipler words or little words
//
if (bigscales.ContainsKey (curword)) {
bigMultiplierValue *= bigscales[curword];
bigMultiplierIsActive = true;
}
else {
// multiply the current result by the previous word bigMultiplier
// and disable the big multiplier until next time
if (bigMultiplierIsActive) {
result += currentResult * bigMultiplierValue;
currentResult = 0;
bigMultiplierValue = 1; // reset the multiplier value
bigMultiplierIsActive = false; // turn it off until next time
}
// translate the incoming text word to an integer
int n;
if ((n = Array.IndexOf (ones, curword) + 1) > 0) {
currentResult += n;
}
else if ((n = Array.IndexOf (teens, curword) + 1) > 0) {
currentResult += n + 10;
}
else if ((n = Array.IndexOf (tens, curword) + 1) > 0) {
currentResult += n * 10;
}
// allow for negative words (like "minus")
else if (minusWords.Contains (curword)) {
minusFlag = true;
}
// allow for phrases like "zero 500" hours military time
else if (curword == "zero") {
continue;
}
// allow for text digits too, like "100 and 5"
else if (int.TryParse (curword, out int tmp)) {
currentResult += tmp;
}
else if (curword != "and") {
throw new ApplicationException ("Expected a number: " + curword);
}
}
}
var final = result + currentResult * bigMultiplierValue;
if (minusFlag)
final *= -1;
return final;
}
Here are some test cases that I ran.
-20 = minus twenty
-261 = minus two hundred sixty one
1965 = nineteen hundred and sixty five
45 = forty five
55 = fifty-five
21 = twenty-one
55 = fifty five
0 = zero
105 = one hundred 5
105 = 100 and 5