**dict1** **dict2**
Id value Id value
24379 348 24379 270451
24368 348 24368 270451
24377 90 24377 270450
24366 90 24366 270450
24369 10 24369 270450
24300 25
Result:
24379 696
24368 696
24377 190
24366 190
24369 190
我有以下逻辑,并希望优化此解决方案:
Dictionary<int, int> result = new Dictionary<int, int>();
foreach (int itemKey in dict1.keys)
{
result.add (itemKey, dict1.Where(a => dict2.ContainsKey(a.key)
&& dict2.ContiansKey(itemKey)
&& dict2[a.key] == dict2[itemKey])
.Sum(a => a.value);
}
答案 0 :(得分:2)
您可以分两步完成:
dict2
的值dict1
,并从查找词典中插入值以下是如何做到这一点:
var lookup = dict1
.Where(p => dict2.ContainsKey(p.Key))
.GroupBy(p => dict2[p.Key])
.ToDictionary(g => g.Key, g => g.Sum(p => p.Value));
var res = dict1.Keys
.Where(k => dict2.ContainsKey(k))
.ToDictionary(k => k, k => lookup[dict2[k]]);
答案 1 :(得分:1)
public static void DicAddTest()
{
Dictionary<int, int> dic1 = new Dictionary<int, int>() { {24379,348}, { 24368, 348 }, { 24377, 90 }, { 24366, 90 } };
Dictionary<int, int> dic2 = new Dictionary<int, int>() { { 24379, 270451 }, { 24368, 270451 }, { 24377, 270450 }, { 24366, 270450 } };
Dictionary<int, int> dicResult = DicAdd(dic1, dic2);
foreach (KeyValuePair<int, int> kvp in dicResult)
Debug.WriteLine("{0} {1}", kvp.Key, kvp.Value);
Debug.WriteLine("");
}
public static Dictionary<int, int> DicAdd(Dictionary<int, int> dic1, Dictionary<int, int> dic2)
{
Dictionary<int, int> dicResult = new Dictionary<int, int>(dic1);
foreach (int k in dic1.Keys.Where(x => dic2.Keys.Contains(x)))
dicResult[k] = dicResult[k] + dicResult[k];
return dicResult;
}
问题不明确
public static Dictionary<int, int> DicAdd2(Dictionary<int, int> dic1, Dictionary<int, int> dic2)
{
Dictionary<int, int> dicResult = new Dictionary<int, int>();
foreach (KeyValuePair<int, int> kvp in dic1.Where(x => dic2.Keys.Contains(x.Key)))
dicResult.Add(kvp.Key, 2 * kvp.Value);
return dicResult;
}
答案 2 :(得分:0)
如果您不确定dict1
和dict2
是否具有相同的密钥,也许这样做会更容易:
var result = new Dictionary<int, int>();
foreach(var kvp in dict1)
{
int value;
if(dict2.TryGetValue(kvp.Key, out value))
{
result[kvp.Key] = kvp.Value * 2;
}
}
这只会添加两个词典中的值。如果您的词典非常大,您可以使用Parallel For,或者考虑使用Hashtable代替。