[英]RemoveAll from map in F#
C#: C#:
In C# I have something like this: 在C#我有这样的事情:
IImmutableDictionary<string, string> map = new Dictionary<string, string>
{
{"K1", "V1"},
{"K2", "V2"},
{"K3", "V3"},
}.ToImmutableDictionary();
IEnumerable<string> keys = new[] {"K1,K3"};
map = map.RemoveRange(keys);
I assume that the method ImmutableDictionary<K,V>.RemoveRange Method (IEnumerable<K>)
was introduced since it is much more efficient than series of Remove(K)
calls. 我假设引入了
ImmutableDictionary<K,V>.RemoveRange Method (IEnumerable<K>)
,因为它比一系列Remove(K)
调用更有效。 It creates the resulting immutable object only once instead of once for every element from keys
to remove. 它仅为要删除的
keys
每个元素创建一次生成的不可变对象,而不是一次。
F#: F#:
What is the best way to achieve the same in F# . 在F#中实现相同目标的最佳方法是什么? I came up with this recursive solution:
我想出了这个递归解决方案:
let rec removeAll (map:Map<string, string>, keys:list<string>) =
match keys with
| [] -> map
| _ -> removeAll(map.Remove(keys |> Seq.head), keys.Tail)
but I doubt it is as efficient as the RemoveRange
from above. 但我怀疑它和上面的
RemoveRange
一样有效。
Questions: 问题:
RemoveAll
in F#? RemoveAll
最有效的等价物是什么? Map.filter
will be useful here and will presumably prevent creating many intermediate maps, though to use it efficiently for many keys you'll want to put the keys into a set first. Map.filter
在这里很有用,可能会阻止创建许多中间贴图,但是要有效地使用它,你需要先将键放入一组中。
let removeAll keys map =
let keySet = set keys
map |> Map.filter (fun k _ -> k |> keySet.Contains |> not)
[ 1, 2
3, 4
5, 6
7, 8 ]
|> Map
|> removeAll [1; 5]
// map [(3, 4); (7, 8)]
This is small enough that it's potentially not worth breaking out into a function. 这足够小,可能不值得突破功能。 For example in a case where you have an array of no more than 10 keys, then it may be less efficient to make a set out of them first.
例如,如果您的数组不超过10个键,那么首先设置它们的效率可能会低一些。
Your current function is tail-recursive so it should be optimised in terms of not creating multiple stack frames, however you can write it a bit more simply by using pattern matching on the list: 您当前的函数是尾递归的,因此应该在不创建多个堆栈帧方面进行优化,但是您可以通过在列表上使用模式匹配来更简单地编写它:
let rec removeAll (map:Map<_,_>, keys:list<_>) =
match keys with
| [] -> map
| key :: rest -> removeAll(map.Remove(key), rest)
Also note that it can automatically be made generic by either removing type annotations or replacing parts of them with _
, telling the compiler to infer the type. 另请注意,它可以通过删除类型注释或用
_
替换它们的一部分来自动变为通用,告诉编译器推断类型。
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