比较字典中值之间的差异
本文关键字:之间 字典 比较 | 更新日期: 2023-09-27 18:07:58
我有一个Dictionary<string, float>
,并希望从中选择KeyValuePairs,其中浮动值之间的差异小于某个阈值。
这是字典:
Dictionary<string, float> heights = new Dictionary<string, float> ();
示例条目:
"first", 61.456
"second", 80.567
"third", 62.456
"4", 59.988
"5", 90.34
"6", 82.123
我需要这些元素,它们的值有很小的不同,例如:"第一","第三"answers"4"在一个列表或类似的东西。差值是给定的浮点数,比如3.5
有可能通过Linq实现吗?
我试着用循环来做这个,但不知怎么的,它真的很乱…
您可以创建查找接近值的自定义方法,并接受两个参数:保存值的dictionary和用于保存最大查找范围的float:
static Dictionary<string, float> FindRange(Dictionary<string, float> dict, float precision)
{
Dictionary<string, float> temp = new Dictionary<string, float>();
List<int> counter = new int[dict.Count].ToList(); float[] values = dict.Values.ToArray();
for (int i = 0; i < values.Length; i++)
for (int i2 = 0; i2 < values.Length; i2++)
if (i2 != i && Math.Abs(values[i] - values[i2]) < precision) counter[i]++;
for (int i = 0; i < values.Length; i++)
if (Math.Abs(values[i] - values[counter.IndexOf(counter.Max())]) < precision)
temp.Add(dict.FirstOrDefault(kv => kv.Value == values[i]).Key, values[i]);
return temp;
}
用法示例:
static void Main()
{
Dictionary<string, float> heights = new Dictionary<string, float>()
{
{"first", 61.456f},
{"second", 80.567f},
{"third", 62.456f},
{"4", 59.988f},
{"5", 90.34f},
{"6", 82.123f}
};
// returns max sequence of elements with difference less than 3f
var newDict = FindRange(heights, 3f);
foreach (var item in newDict)
{
Console.WriteLine(item.Key + " " + item.Value);
}
}
输出:
first 61,456
third 62,456
4 59,988
您可以先按值对条目进行排序,然后比较附近的条目会更容易。
var list = heights.ToList();
list.Sort((a,b) => {return a.Value.CompareTo(b.Value);});
bool first = true;
for (int i = 1; i < list.Count; ++i)
{
if (Math.Abs(list[i-1].Value - list[i].Value) < threshold)
{
if (first)
{
first = false;
Console.WriteLine(list[i-1]);
}
Console.WriteLine(list[i]);
}
}
这里是dotnetfiddle的一个工作示例
类似于
List<KeyValuePair<string,float>> matching = new List<KeyValuePair<string,float>>();
int i = 0;
var all = _dict.Select(kvp => kvp).ToList().OrderBy(kvp => kvp.Value);
all.ForEach(kvp => {
if(i < all.Count() - 1 && Math.Abs(all[i+1].Value - kvp.Value) < threshhold)
{
matching.Add(kvp);
if(i == all.Count() - 1) matching.add(all[i+1]); // Need to manually add the final entry if it's a match
}
i++;
});