具有多个键的字典不能按预期工作
本文关键字:不能按 工作 字典 | 更新日期: 2023-09-27 18:15:26
我正在制作一个控制台程序,其中我有多个值映射到dictionary
keyLookup
。我正在使用if命令,使用键输出一些console.writeline = ("stuff");
,但只有当我的值和键相同(在字典中)时才有效。我不知道这是为什么。我一直在玩弄list
和foreach
和一些变量试图找出我做错了什么,但即使它继续工作,它现在如何工作,它仍然不工作我想要的方式。
同样,如果我的console.readline();
中有一个词不在我的字典中,整个事情就会崩溃。这是我不想要的,我也不确定它为什么要这样做,因为在某种程度上它没有这样做。此外,我的mathFunction
字典工作就像我想我的keyLookup
字典工作。虽然我认为不同之处在于我如何使用列表来交叉引用keyLookup
。
class MainClass
{
public static string Line;
static string foundKey;
public static void Main (string[] args)
{
while (true)
{
if (Line == null)
{Console.WriteLine ("Enter Input"); }
WordChecker ();
}
}
public static void WordChecker()
{
string inputString = Console.ReadLine ();
inputString = inputString.ToLower();
string[] stripChars = { ";", ",", ".", "-", "_", "^", "(", ")", "[", "]",
"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "'n", "'t", "'r" };
foreach (string character in stripChars)
{
inputString = inputString.Replace(character, "");
}
// Split on spaces into a List of strings
List<string> wordList = inputString.Split(' ').ToList();
// Define and remove stopwords
string[] stopwords = new string[] { "and", "the", "she", "for", "this", "you", "but" };
foreach (string word in stopwords)
{
// While there's still an instance of a stopword in the wordList, remove it.
// If we don't use a while loop on this each call to Remove simply removes a single
// instance of the stopword from our wordList, and we can't call Replace on the
// entire string (as opposed to the individual words in the string) as it's
// too indiscriminate (i.e. removing 'and' will turn words like 'bandage' into 'bdage'!)
while ( wordList.Contains(word) )
{
wordList.Remove(word);
}
}
// Create a new Dictionary object
Dictionary<string, int> dictionary = new Dictionary<string, int>();
// Loop over all over the words in our wordList...
foreach (string word in wordList)
{
// If the length of the word is at least three letters...
if (word.Length >= 3)
{
// ...check if the dictionary already has the word.
if ( dictionary.ContainsKey(word) )
{
// If we already have the word in the dictionary, increment the count of how many times it appears
dictionary[word]++;
}
else
{
// Otherwise, if it's a new word then add it to the dictionary with an initial count of 1
dictionary[word] = 1;
}
}
List<string> dicList = new List<string>();
dicList = dictionary.Keys.ToList ();
Dictionary<string, string> keyLookup = new Dictionary<string, string>();
keyLookup["hey"] = "greeting";
keyLookup["hi"] = "greeting";
keyLookup["greeting"] = "greeting";
keyLookup["math"] = "math";
keyLookup["calculate"] = "math";
keyLookup["equation"] = "math";
foundKey = keyLookup[word];
List<string> keyList = new List<string>();
foreach (string keyWord in dicList)
{
if(keyWord == foundKey)
{keyList.Add (keyWord); }
}
foreach (string mKey in keyList)
{
if(mKey == "greeting")
{Greetings ();}
if (mKey == "math")
{Math ();}
}
}
}
public static void Math()
{
Console.WriteLine ("What do you want me to math?");
Console.WriteLine ("input a number");
string input = Console.ReadLine ();
decimal a = Convert.ToDecimal (input);
Console.WriteLine("Tell me math function");
string mFunction = Console.ReadLine();
Console.WriteLine ("tell me another number");
string inputB = Console.ReadLine();
decimal b = Convert.ToDecimal (inputB);
Dictionary<string, string> mathFunction = new Dictionary<string, string>();
mathFunction["multiply"] = "multiply";
mathFunction["times"] = "multiply";
mathFunction["x"] = "multiply";
mathFunction["*"] = "multiply";
mathFunction["divide"] = "divide";
mathFunction["/"] = "divide";
mathFunction["subtract"] = "subtract";
mathFunction["minus"] = "subtract";
mathFunction["-"] = "subtract";
mathFunction["add"] = "add";
mathFunction["+"] = "add";
mathFunction["plus"] = "add";
string foundKey = mathFunction[mFunction];
if (foundKey == "add")
{
Console.WriteLine (a + b);
}
else if (foundKey == "subtract")
{
Console.WriteLine (a - b);
}
else if (foundKey == "multiply")
{
Console.WriteLine (a * b);
}
else if (foundKey == "divide")
{
Console.WriteLine (a / b);
}
else
{
Console.WriteLine ("not a math");
}
}
public static void Greetings()
{
Console.WriteLine("You said hello");
}
}'
你应该以不同的方式遍历字典(不要使用tollist - function)。
试试这个:
foreach (KeyValuePair kvp (Of String, String) In testDictionary)
{
Debug.WriteLine("Key:" + kvp.Key + " Value:" + kvp.Value);
}
如果单词不匹配,你的应用程序会崩溃,因为这段代码(你没有这样创建一个新条目):
// Otherwise, if it's a new word then add it to the dictionary with an initial count of 1
dictionary[word] = 1;
EDIT:我错了,dictionary[word] = 1
不会创建一个新的元素。这样很好。
foundKey = keyLookup[word];
如果keyLookup中不存在word
,那么它将崩溃。
string foundKey = mathFunction[mFunction];
如果mathFunction中不存在mFunction
,那么它将崩溃。
如果你试图使它成为一个"会话"程序,那么单词查找是最重要的部分。您不使用谓词或LINQ,它们都可以使字符串函数非常简单。当前您使用的是字典。为什么不为每个关键字使用列表呢?
List<string> GreetingKeywords;
GreetingKeywords.Add("hello"); // ...
List<string> MathKeywords;
MathKeywords.Add("math"); // ...
foreach (var word in dicList)
{
if (GreetingKeywords.Contains(word))
{ Greetings(); }
if (MathKeywords.Contains(word))
{ Maths(); }
}
我建议你仔细阅读谓词和列表/字典函数,如Find, IndexOf等。