输入1:List<string>
,例如:
&#34; hello&#34;,&#34; world&#34;,&#34; stack&#34;,&#34; overflow&#34;。
输入2:List<Foo>
(两个属性,字符串a,字符串b),例如:
Foo 1: a:&#34;你好!&#34; b:string.Empty
Foo 2: a:&#34;我喜欢Stack Overflow&#34; b:&#34;它是有史以来最好的网站!&#34;
所以我想以Dictionary<string,int>
结束。 List<Foo>
或 a
字段中的单词及其在b
中显示的次数。
当前的头版代码的第一遍/顶部,这太慢了:
var occurences = new Dictionary<string, int>();
foreach (var word in uniqueWords /* input1 */)
{
var aOccurances = foos.Count(x => !string.IsNullOrEmpty(x.a) && x.a.Contains(word));
var bOccurances = foos.Count(x => !string.IsNullOrEmpty(x.b) && x.b.Contains(word));
occurences.Add(word, aOccurances + bOccurances);
}
答案 0 :(得分:1)
大致是:
occurrences
),可选择使用不区分大小写的比较器。Foo
,使用RegEx
将a
和b
分成单词。occurrences
中是否存在密钥。如果存在,则增加并更新字典中的值。答案 1 :(得分:0)
您可以尝试将两个字符串a + b连接起来。然后做一个正则表达式将所有单词拉出到一个集合中。然后最终使用查询分组索引。
例如
void Main()
{
var a = "Hello there!";
var b = "It's the best site ever!";
var ab = a + " " + b;
var matches = Regex.Matches(ab, "[A-Za-z]+");
var occurences = from x in matches.OfType<System.Text.RegularExpressions.Match>()
let word = x.Value.ToLowerInvariant()
group word by word into g
select new { Word = g.Key, Count = g.Count() };
var result = occurences.ToDictionary(x => x.Word, x => x.Count);
Console.WriteLine(result);
}
建议进行一些更改的示例... 编辑。只是重新阅读要求....有点奇怪,但嘿......
void Main()
{
var counts = GetCount(new [] {
"Hello there!",
"It's the best site ever!"
});
Console.WriteLine(counts);
}
public IDictionary<string, int> GetCount(IEnumerable<Foo> inputs)
{
var allWords = from input in inputs
let matchesA = Regex.Matches(input.A, "[A-Za-z']+").OfType<System.Text.RegularExpressions.Match>()
let matchesB = Regex.Matches(input.B, "[A-Za-z']+").OfType<System.Text.RegularExpressions.Match>()
from x in matchesA.Concat(matchesB)
select x.Value;
var occurences = allWords.GroupBy(x => x, (x, y) => new{Key = x, Count = y.Count()}, StringComparer.OrdinalIgnoreCase);
var result = occurences.ToDictionary(x => x.Key, x => x.Count, StringComparer.OrdinalIgnoreCase);
return result;
}