英文:
Equivalent Map.Compute in C#
问题
Sure, here's the translated code:
Dictionary<int, int> map = new Dictionary<int, int>();
foreach (int i in elements)
map[i] = map.ContainsKey(i) ? map[i] + 1 : 1;
Let me know if you need further assistance!
英文:
I am trying to convert some code from java to C# but I´m stuck with this:
Map<Integer, Integer> map = new HashMap<>(); // converted to Dictionary<int, int>
for (int i : elements)//converted to foreach (int i in elements)
map.compute(i, (k, v) -> (v == null) ? 1 : v+1)
What is the equivalent If I'm doing it with a dictionary in C#?
答案1
得分: 1
我不知道存在确切的等同物，所以可能需要在某种程度上自己实现它。如果你正在寻找一个看起来和感觉相同的解决方案，你可以为 Dictionary<K,V> 添加一个扩展。
这些语言之间有足够的差异，需要一些手工操作，但要抓住要点，这应该足够作为一个起点。
在C#的Dictionary上不存在Compute，所以你可以添加它。理想情况下，对于一个“等效”的版本，你会希望确保它涵盖了所有原始版本的情况，而不仅仅是这个特定的情况（依我之见）。我不敢断言以下代码完全等同，但你可以通过调查你的Java版本中compute的源代码来确保你的版本与原始版本相符（这是我在查看HashMap时得出的，可能不准确）。
// 你可以轻松地添加到Dictionary<K, V>的实现中；太棒了！
public static class DictExtensions
{
public static V Compute<K, V>(this Dictionary<K, V> dict, K key, Func<K, V, V> func)
{
// 如果没有提供func，抛出异常。
if (func == null) throw new ArgumentNullException(nameof(func));
// 如果没有映射，返回null。
if (!dict.TryGetValue(key, out var value)) return default;
// 从func获取新值。
var result = func(key, value);
if (result == null)
{
// 如果映射存在但func => null，
// 删除映射并返回null。
dict.Remove(key);
return default;
}
// 映射存在，func返回非null值。
// 设置新值并返回。
dict[key] = result;
return result;
}
}
然后，在将正确的using语句添加到你想要在其中使用Compute的代码之后，可以像你期望的那样使用它：map.Compute(i, (k, v) => (v == null) ? 1 : v + 1);。
英文:
I don't know of an exact equivalent that exists so it may come down to implementing it yourself to some degree. If you're looking for a solution that looks and feels the same, you can add an extension to Dictionary<K,V>.
There's sufficient differences between these languages to require a bit of hand waiving but, to get to the gist of it, this should suffice as a starting point.
Compute doesn't exist on the C# Dictionary so you can add it. Ideally, for an "equivalent", you'd want to ensure it covers all the cases as the original, not just this particular case (IMHO). I don't claim that the following does but you can ensure yours does by investigating the source code for compute in your Java version. (this is what I came up with looking over HashMap; I could be off).
// You can easily add to the Dictionary<K, V> implementation; SWEET!
public static class DictExtensions
{
public static V Compute<K, V>(this Dictionary<K, V> dict, K key, Func<K, V, V> func)
{
// if no func given, throw.
if (func == null) throw new ArgumentNullException(nameof(func));
// if no mapping, return null.
if (!dict.TryGetValue(key, out var value)) return default;
// get the new value from func.
var result = func(key, value);
if (result == null)
{
// if the mapping exists but func => null,
// remove the mapping and return null.
dict.Remove(key);
return default;
}
// mapping exists and func returned a non-null value.
// set and return the new value
dict[key] = result;
return result;
}
}
Then, after adding a correct using statement to the code you'd like to use Compute in, use it like you expect: map.Compute(i, (k, v) => (v == null) ? 1 : v + 1);
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