我确信这已经完成了一百次，但我希望有一个非常简单的方法来完成它。我想将单词更改为 int。
像下面的例子
一 = 1 二 = 2 三 = 3
所以基本上，如果我有字符串“One”，它会被转换为 1，即使我可以取回字符串“1”，我也可以转换它。
【问题讨论】：
标签： c#
这样做是为了好玩...可能有很多边缘情况会失败...
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);
}
【讨论】：
continue;？你可以使用elses。 （当然还有 +1。）
the quick brown fox 和zero 都返回0。
我对错误处理采取了稍微不同的方法……
public static bool ParseEnglishNumberPhrase(string pNumberPhrase, out int pValue) {
pValue = 0;
string[]
temporaryWords = pNumberPhrase.ToLower().Split(new char[] { '_', ' ', '-', ',' }, StringSplitOptions.RemoveEmptyEntries),
ones = { "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" },
teens = { "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" },
tens = { "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" },
minusWords = { "minus", "negative", "hyphen" };
List<string> words = temporaryWords.ToList();
bool minusFlag = false;
Dictionary<string, int> modifiers = new Dictionary<string, int>() {
{ "billion", 1000000000 },
{ "million", 1000000 },
{ "thousand", 1000 },
{ "hundred", 100 } };
int result = 0, currentResult = 0, lastModifier = 1;
if (pNumberPhrase.Equals("eleventy billion")) {
pValue = int.MaxValue; // 110,000,000,000 is out of range for an int!
return false;
}
if (words[0].Equals("zero") && (words.Count == 1)) {
pValue = 0;
return true;
}
else if (words[0].Equals("zero"))
words.RemoveAt(0);
if (pNumberPhrase.StartsWith("-"))
minusFlag = true;
foreach (string word in minusWords) {
if (pNumberPhrase.Contains(word)) {
minusFlag = true;
words.Remove(word);
}
}
if (words.Count == 1) {
if (int.TryParse(words[0], out int pOutValue)) {
pValue = pOutValue;
if (minusFlag)
pValue *= -1;
return true;
}
}
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") {
pValue = -1;
return false;
}
}
}
pValue = result + currentResult * lastModifier;
if (minusFlag)
pValue *= -1;
return true;
}
【讨论】：
这是一种方法。如果您需要更广泛的范围，它很容易扩展；只需使用long、ulong 甚至BigInt，然后将更多项目添加到modifiers 字典中。
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;
}
【讨论】：
这是上述算法的 C# 版本。为了清晰起见，我重命名并重写了一些代码，并添加了对负数、连字符数、零以及文本单词和数字组合（如“100 和 5”）的支持。感谢 Ry- 的良好开端。
/// <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;
}
这是我运行的一些测试用例。
-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
【讨论】：
我认为一个更容易理解的解决方案如下：
public static int ParseInt(string s)
{
var wordArray = s.Split(' ', '-');
int finalNumber = 0;
Dictionary<string, int> additionWords = new Dictionary<string, int>{
{"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}
};
Dictionary<string, int> multiplicationWords = new Dictionary<string, int>{
{"hundred", 100},{"thousand", 1000},{"million", 1000000}
};
int multiplier = 1;
for (int i = wordArray.Length - 1; i >= 0; i--){
if (additionWords.ContainsKey(wordArray[i])){
finalNumber += additionWords[wordArray[i]] * multiplier;
}
if (multiplicationWords.ContainsKey(wordArray[i])){
if (multiplicationWords[wordArray[i]] < multiplier){
multiplier *= multiplicationWords[wordArray[i]];
}else{
multiplier = multiplicationWords[wordArray[i]];
}
}
}
return finalNumber;
}
【讨论】：