我有一个我在C＃工作的项目。我有两个char数组。一个是句子，一个是单词。我必须遍历句子数组，直到找到一个与转换成单词数组的单词匹配的单词，我想知道的是一旦我找到单词，我如何向后遍历句子数组在这一点上，我发现这个词与单词数组的长度相同？
代码：
String wordString = "(Four)";
String sentenceString = "Maybe the fowl of Uruguay repeaters (Four) will be found";
char[] wordArray = wordString.ToCharArray();
List<String> words = sentenceString.Split(' ').ToList<string>();
//This would be the part where I iterate through sentence
foreach (string sentence in sentArray)
{
//Here I would need to find where the string of (Four) and trim it and see if it equals the wordString.
if (sentence.Contains(wordString)
{
//At this point I would need to go back the length of wordString which happens to be four places but I'm not sure how to do this. And for each word I go back in the sentence I need to capture that in another string array.
}
我不知道我是否对此表现得足够清楚，但如果我不是，请随意提问..先谢谢你。此外应该归还的是乌拉圭中继者的家禽＆＃34;。所以基本上用例是括号中的字母数，逻辑应该在括号中的单词之前返回相同数量的单词。
答案 0 :(得分：2)
int i ;
string outputString = (i=sentenceString.IndexOf(wordString))<0 ?
sentenceString : sentenceString.Substring(0,i) ;
答案 1 :(得分：2)
var wordString = "(Four)";
int wordStringInt = 4; // Just do switch case to convert your string to int
var sentenceString = "Maybe the fowl of Uruguay repeaters (Four) will be found";
var sentenceStringArray = sentenceString.Split(' ').ToList();
int wordStringIndexInArray = sentenceStringArray.IndexOf(wordString) - 1;
var stringOutPut = "";
if (wordStringIndexInArray > 0 && wordStringIndexInArray > wordStringInt)
{
stringOutPut = "";
while (wordStringInt > 0)
{
stringOutPut = sentenceStringArray[wordStringInt] + " " + stringOutPut;
wordStringInt--;
}
}
答案 2 :(得分：2)
我们是你。关于这个练习我几乎没有问题。 如果Word （四）刚开始它不应该返回？或者返回所有字符串？ 因为长度为4等于4 想象一下，如果该单词出现在句子的第二个单词上，它应该只返回第一个单词还是返回4个单词，甚至包括（四个）单词。
我的解决方案是最懒的，我只是看到你的问题并决定提供帮助。
再一次，这不是我最好的方法。
我看到下面的代码：
string wordString = "(Four)";
string sentenceString = "Maybe the fowl of Uruguay repeaters (Four) will be found";
//Additionally you can add splitoption to remove the empty word on split function bellow
//Just in case there are more space in sentence.
string[] splitedword = sentenceString.Split(' ');
int tempBackposition = 0;
int finalposition = 0;
for (int i = 0; i < splitedword.Length; i++)
{
if (splitedword[i].Contains(wordString))
{
finalposition = i;
break;
}
}
tempBackposition = finalposition - wordString.Replace("(","").Replace(")","").Length;
string output = "";
tempBackposition= tempBackposition<0?0:tempBackposition;
for (int i = tempBackposition; i < finalposition; i++)
{
output += splitedword[i] + " ";
}
Console.WriteLine(output);
Console.ReadLine();
如果不是您想要的，您可以回答我的问题吗？还是帮我理解这是不对的
答案 3 :(得分：2)
您所匹配的内容有点复杂，因此对于更通用的解决方案，您可以使用正则表达式。
首先我们宣布我们要搜索的内容：
string word = "(Four)";
string sentence = "Maybe the fowl of Uruguay repeaters (Four) will be found";
然后我们将使用正则表达式搜索此字符串中的单词。由于我们不想匹配空格，我们需要知道每个匹配实际开始的位置和我们需要知道括号内的单词，我们告诉它我们可选择要打开和结束括号，但我们也希望这些内容匹配：
var words = Regex.Matches(sentence, @"[\p{Ps}]*(?<Content>[\w]+)[\p{Pe}]*").Cast<Match>().ToList();
[\p{Ps}]表示我们需要打开标点符号([{等，而*表示零或更多。
Followed是一个名为Content（由?<Content>指定）的子捕获，带有一个或多个单词字符。
最后，我们指定我们想要零或更多结束标点符号。
然后我们需要在匹配列表中找到该词：
var item = words.Single(x => x.Value == word);
然后我们需要找到这个项目的索引：
int index = words.IndexOf(item);
此时我们只需要知道内容的长度：
var length = item.Groups["Content"].Length;
这个长度我们用来回到字符串4个单词
var start = words.Skip(index - length).First();
现在我们拥有了所需的一切：
var result = sentence.Substring(start.Index, item.Index - start.Index);
结果应包含fowl of Uruguay repeaters。
编辑：从单词而不是内容中找出计数可能要简单得多。在这种情况下，完整的代码应如下：
string word = "(Four)";
string sentence = "Maybe the fowl of Uruguay repeaters (Four) will be found";
var wordMatch = Regex.Match(word, @"[\p{Ps}]*(?<Content>[\w]+)[\p{Pe}]*");
var length = wordMatch.Groups["Content"].Length;
var words = Regex.Matches(sentence, @"\S+").Cast<Match>().ToList();
var item = words.Single(x => x.Value == word);
int index = words.IndexOf(item);
var start = words.Skip(index - length).First();
var result = sentence.Substring(start.Index, item.Index - start.Index);
\S+表示＆＃34;匹配一个或多个非空白字符＆＃34;。
答案 4 :(得分：2)
您应该尝试以下内容，在找到数字后使用Array.Copy。您仍然必须正确实现ConvertToNum函数（现在它已经硬编码），但这应该是一个快速简单的解决方案。
string[] GetWords()
{
string sentenceString = "Maybe the fowl of Uruguay repeaters (Four) will be found";
string[] words = sentenceString.Split();
int num = 0;
int i; // scope of i should remain outside the for loop
for (i = 0; i < words.Length; i++)
{
string word = words[i];
if (word.StartsWith("(") && word.EndsWith(")"))
{
num = ConvertToNum(word.Substring(1, word.Length - 1));
// converted the number word we found, so we break
break;
}
}
if (num == 0)
{
// no number word was found in the string - return empty array
return new string[0];
}
// do some extra checking if number word exceeds number of previous words
int startIndex = i - num;
// if it does - just start from index 0
startIndex = startIndex < 0 ? 0 : startIndex;
int length = i - startIndex;
string[] output = new string[length];
Array.Copy(words, startIndex, output, 0, length);
return output;
}
// Convert the number word to an integer
int ConvertToNum(string numberStr)
{
return 4; // you should implement this method correctly
}
有关实施ConvertToNum解决方案的帮助，请参阅 - Convert words (string) to Int。显然，它可以根据您期望处理的数字范围进行简化。
答案 5 :(得分：2)
这是我的解决方案，我根本不使用正则表达式，以便于理解：
static void Main() {
var wordString = "(Four)";
int wordStringLength = wordString.Replace("(","").Replace(")","").Length;
//4, because i'm assuming '(' and ')' doesn't count.
var sentenceString = "Maybe the fowl of Uruguay repeaters (Four) will be found";
//Transform into a list of words, ToList() to future use of IndexOf Method
var sentenceStringWords = sentenceString.Split(' ').ToList();
//Find the index of the word in the list of words
int wordIndex = sentenceStringWords.IndexOf(wordString);
//Get a subrange from the original list of words, going back x Times the legnth of word (in this case 4),
var wordsToConcat = sentenceStringWords.GetRange(wordIndex-wordStringLength, wordStringLength);
//Finally concat the output;
var outPut = string.Join(" ", wordsToConcat);
//Output: fowl of Uruguay repeaters
}
答案 6 :(得分：0)
我有一个解决方案：
string wordToMatch = "(Four)";
string sentence = "Maybe the fowl of Uruguay repeaters (Four) will be found";
if (sentence.Contains(wordToMatch))
{
int length = wordToMatch.Trim(new[] { '(', ')' }).Length;
int indexOfMatchedWord = sentence.IndexOf(wordToMatch);
string subString1 = sentence.Substring(0, indexOfMatchedWord);
string[] words = subString1.Split(new[] { ' ' }, StringSplitOptions.RemoveEmptyEntries);
var reversed = words.Reverse().Take(length);
string result = string.Join(" ", reversed.Reverse());
Console.WriteLine(result);
Console.ReadLine();
}
可能这可以改善表现，但我有一种感觉你不关心这一点。确保您使用的是System.Linq＆＃39;
答案 7 :(得分：0)
当输入不完整时，我假设空的回报，随时纠正我。你的帖子中没有100％清楚如何处理这个问题。
private string getPartialSentence(string sentence, string word)
{
if (string.IsNullOrEmpty(sentence) || string.IsNullOrEmpty(word))
return string.Empty;
int locationInSentence = sentence.IndexOf(word, StringComparison.Ordinal);
if (locationInSentence == -1)
return string.Empty;
string partialSentence = sentence.Substring(0, locationInSentence);
string[] words = partialSentence.Split(new[] {' '}, StringSplitOptions.RemoveEmptyEntries);
int nbWordsRequired = word.Replace("(", "").Replace(")", "").Length;
if (words.Count() >= nbWordsRequired)
return String.Join(" ", words.Skip(words.Count() - nbWordsRequired));
return String.Join(" ", words);
}
答案 8 :(得分：0)
我使用枚举和关联字典将“（四）”类型字符串与其整数值配对。您可以使用
轻松（也可能更容易）使用switch语句case "(Four)": { currentNumber = 4; };
我觉得enum允许更多的灵活性。
public enum NumberVerb
{
one = 1,
two = 2,
three = 3,
four = 4,
five = 5,
six = 6,
seven = 7,
eight = 8,
nine = 9,
ten = 10,
};
public static Dictionary<string, NumberVerb> m_Dictionary
{
get
{
Dictionary<string, NumberVerb> temp = new Dictionary<string, NumberVerb>();
temp.Add("(one)", NumberVerb.one);
temp.Add("(two)", NumberVerb.two);
temp.Add("(three)", NumberVerb.three);
temp.Add("(four)", NumberVerb.four);
temp.Add("(five)", NumberVerb.five);
temp.Add("(six)", NumberVerb.six);
temp.Add("(seven)", NumberVerb.seven);
temp.Add("(eight)", NumberVerb.eight);
temp.Add("(nine)", NumberVerb.nine);
temp.Add("(ten)", NumberVerb.ten);
return temp;
}
}
static void Main(string[] args)
{
string resultPhrase = "";
// Get the sentance that will be searched.
Console.WriteLine("Please enter the starting sentance:");
Console.WriteLine("(don't forget your keyword: ie '(Four)')");
string sentance = Console.ReadLine();
// Get the search word.
Console.WriteLine("Please enter the search keyword:");
string keyword = Console.ReadLine();
// Set the associated number of words to backwards-iterate.
int currentNumber = -1;
try
{
currentNumber = (int)m_Dictionary[keyword.ToLower()];
}
catch(KeyNotFoundException ex)
{
Console.WriteLine("The provided keyword was not found in the dictionary.");
}
// Search the sentance string for the keyword, and get the starting index.
Console.WriteLine("Searching for phrase...");
string[] words = sentance.Split(' ');
int searchResultIndex = -1;
for (int i = 0; (searchResultIndex == -1 && i < words.Length); i++)
{
if (words[i].Equals(keyword))
{
searchResultIndex = i;
}
}
// Handle the search results.
if (searchResultIndex == -1)
{
resultPhrase = "The keyword was not found.";
}
else if (searchResultIndex < currentNumber)
{
// Check the array boundaries with the given indexes.
resultPhrase = "Error: Out of bounds!";
}
else
{
// Get the preceding words.
for (int i = 0; i < currentNumber; i++)
{
resultPhrase = string.Format(" {0}{1}", words[searchResultIndex - 1 - i], resultPhrase);
}
}
// Display the preceding words.
Console.WriteLine(resultPhrase.Trim());
// Exit.
Console.ReadLine();
}