Question

我使用以下代码将字符串数组拆分为列表。

private List<string> GenerateTerms(string[] docs)
    {
        return docs.SelectMany(doc => ProcessDocument(doc)).Distinct().ToList();
    }

    private IEnumerable<string> ProcessDocument(string doc)
    {
        return doc.Split(' ')
                  .GroupBy(word => word)
                  .OrderByDescending(g => g.Count())
                  .Select(g => g.Key)
                  .Take(1000);
    }

我想要做的是用

替换返回的列表

Dictionary <string, int>

即。而不是返回列表，我想返回字典

有人可以帮忙吗？提前致谢。

Answer 1

string doc = "This is a test sentence with some words with some words repeating like: is a test";
var result = doc.Split(' ')
                   .GroupBy(word => word)
                   .OrderByDescending(g=> g.Count())
                   .Take(1000)
                   .ToDictionary(r => r.Key ,r=> r.Count());

修改

我相信您希望从字符串数组中获取最终字典，基于单词作为键并将其最终计数作为值。由于字典不能包含重复值，因此您不需要使用Distict。您必须将您的方法重写为：

private Dictionary<string,int> GenerateTerms(string[] docs) { List<Dictionary<string, int>> combinedDictionaryList = new List<Dictionary<string, int>>(); foreach (string str in docs) { //Add returned dictionaries to a list combinedDictionaryList.Add(ProcessDocument(str)); } //return a single dictionary from list od dictionaries return combinedDictionaryList .SelectMany(dict=> dict) .ToLookup(pair => pair.Key, pair => pair.Value) .ToDictionary(group => group.Key, group => group.Sum(value => value)); } private Dictionary<string,int> ProcessDocument(string doc) { return doc.Split(' ') .GroupBy(word => word) .OrderByDescending(g => g.Count()) .Take(1000) .ToDictionary(r => r.Key, r => r.Count()); }

然后你可以这样称呼它：

string[] docs = new[] { "This is a test sentence with some words with some words repeating like: is a test", "This is a test sentence with some words with some words repeating like: is a test", "This is a test sentence with some words", "This is a test sentence with some words", }; Dictionary<string, int> finalDictionary = GenerateTerms(docs);

Answer 2

试试这个：

string[] docs = {"aaa bbb", "aaa ccc", "sss, ccc"};        

var result = docs.SelectMany(doc => doc.Split())
                 .GroupBy(word => word)
                 .OrderByDescending(g => g.Count())
                 .ToDictionary(g => g.Key, g => g.Count())
                 .Take(1000);

修改

var result = docs.SelectMany( doc => doc.Split() .GroupBy(word => word) .OrderByDescending(g => g.Count()) .Take(1000)) .Select(g => new {Word = g.Key, Cnt = g.Count()}) .GroupBy(t => t.Word) .ToDictionary(g => g.Key, g => g.Sum(t => t.Cnt));

Answer 3

如果没有任何额外的瑕疵，以下情况应该有效。

return doc.Split(' ')
          .GroupBy(word => word)
          .ToDictionary(g => g.Key, g => g.Count());

根据您的具体情况，通过Take，OrderBy等方式对其进行定制。

Answer 4

尝试这样的事情：

    var keys = new List<string>();
    var values = new List<string>();
    var dictionary = keys.ToDictionary(x => x, x => values[keys.IndexOf(x)]);

ienumerable <string> to dictionary <string，int =“”> </string，> </string>

4 个答案: