我正在寻找一个关于哈希表如何工作的解释-用简单的英语为我这样的傻瓜!
例如,我知道它接受键,计算哈希(我正在寻找一个解释),然后执行某种模运算来计算出它在存储值的数组中的位置,但我的知识到此为止。
谁能解释一下过程吗?
编辑:我并不是特别问哈希码是如何计算的,而是对哈希表如何工作的一般概述。
我正在寻找一个关于哈希表如何工作的解释-用简单的英语为我这样的傻瓜!
例如,我知道它接受键,计算哈希(我正在寻找一个解释),然后执行某种模运算来计算出它在存储值的数组中的位置,但我的知识到此为止。
谁能解释一下过程吗?
编辑:我并不是特别问哈希码是如何计算的,而是对哈希表如何工作的一般概述。
当前回答
其实比这更简单。
哈希表不过是一个包含键/值对的向量数组(通常是稀疏数组)。此数组的最大大小通常小于哈希表中存储的数据类型的可能值集中的项数。
哈希算法用于根据将存储在数组中的项的值生成该数组的索引。
This is where storing vectors of key/value pairs in the array come in. Because the set of values that can be indexes in the array is typically smaller than the number of all possible values that the type can have, it is possible that your hash algorithm is going to generate the same value for two separate keys. A good hash algorithm will prevent this as much as possible (which is why it is relegated to the type usually because it has specific information which a general hash algorithm can't possibly know), but it's impossible to prevent.
因此,您可以使用多个键来生成相同的散列代码。当这种情况发生时,将遍历向量中的项,并在向量中的键和正在查找的键之间进行直接比较。如果找到,则返回与该键关联的值,否则不返回任何值。
其他回答
简短而甜蜜:
哈希表封装了一个数组,我们称之为internalArray。将项以如下方式插入数组:
let insert key value =
internalArray[hash(key) % internalArray.Length] <- (key, value)
//oversimplified for educational purposes
有时两个键会散列到数组中的同一个索引,而您希望保留这两个值。我喜欢把两个值都存储在同一个索引中,通过将internalArray作为一个链表数组来编码很简单:
let insert key value =
internalArray[hash(key) % internalArray.Length].AddLast(key, value)
所以,如果我想从哈希表中检索一个项,我可以这样写:
let get key =
let linkedList = internalArray[hash(key) % internalArray.Length]
for (testKey, value) in linkedList
if (testKey = key) then return value
return null
删除操作写起来也很简单。正如你所知道的,从我们的链表数组中插入、查找和删除几乎是O(1)。
当我们的internalArray太满时,可能在85%左右的容量,我们可以调整内部数组的大小,并将所有项目从旧数组移动到新数组中。
你们已经很接近完整地解释了这个问题,但是遗漏了一些东西。哈希表只是一个数组。数组本身将在每个槽中包含一些内容。至少要将哈希值或值本身存储在这个插槽中。除此之外,您还可以存储在此插槽上碰撞的值的链接/链表,或者您可以使用开放寻址方法。您还可以存储一个或多个指针,这些指针指向您希望从该槽中检索的其他数据。
It's important to note that the hashvalue itself generally does not indicate the slot into which to place the value. For example, a hashvalue might be a negative integer value. Obviously a negative number cannot point to an array location. Additionally, hash values will tend to many times be larger numbers than the slots available. Thus another calculation needs to be performed by the hashtable itself to figure out which slot the value should go into. This is done with a modulus math operation like:
uint slotIndex = hashValue % hashTableSize;
这个值是该值将要进入的槽。在开放寻址中,如果槽位已经被另一个哈希值和/或其他数据填充,将再次运行模运算来查找下一个槽:
slotIndex = (remainder + 1) % hashTableSize;
我想可能还有其他更高级的方法来确定槽索引,但这是我见过的最常见的方法……会对其他表现更好的公司感兴趣。
With the modulus method, if you have a table of say size 1000, any hashvalue that is between 1 and 1000 will go into the corresponding slot. Any Negative values, and any values greater than 1000 will be potentially colliding slot values. The chances of that happening depend both on your hashing method, as well as how many total items you add to the hash table. Generally, it's best practice to make the size of the hashtable such that the total number of values added to it is only equal to about 70% of its size. If your hash function does a good job of even distribution, you will generally encounter very few to no bucket/slot collisions and it will perform very quickly for both lookup and write operations. If the total number of values to add is not known in advance, make a good guesstimate using whatever means, and then resize your hashtable once the number of elements added to it reaches 70% of capacity.
我希望这对你有所帮助。
PS - In C# the GetHashCode() method is pretty slow and results in actual value collisions under a lot of conditions I've tested. For some real fun, build your own hashfunction and try to get it to NEVER collide on the specific data you are hashing, run faster than GetHashCode, and have a fairly even distribution. I've done this using long instead of int size hashcode values and it's worked quite well on up to 32 million entires hashvalues in the hashtable with 0 collisions. Unfortunately I can't share the code as it belongs to my employer... but I can reveal it is possible for certain data domains. When you can achieve this, the hashtable is VERY fast. :)
这是另一种看待它的方式。
我假设你理解数组A的概念,它支持索引操作,你可以一步找到第I个元素,A[I],不管A有多大。
因此,例如,如果您想存储一组恰好年龄不同的人的信息,一个简单的方法是有一个足够大的数组,并使用每个人的年龄作为数组的索引。这样,你就可以一步获取任何人的信息。
But of course there could be more than one person with the same age, so what you put in the array at each entry is a list of all the people who have that age. So you can get to an individual person's information in one step plus a little bit of search in that list (called a "bucket"). It only slows down if there are so many people that the buckets get big. Then you need a larger array, and some other way to get more identifying information about the person, like the first few letters of their surname, instead of using age.
这是基本思想。不使用年龄,可以使用任何能产生良好价值观传播的人的函数。这就是哈希函数。比如你可以把这个人名字的ASCII表示的每三分之一,按某种顺序打乱。重要的是,您不希望太多人散列到同一个存储桶,因为速度取决于存储桶保持较小。
Hashtable inside contains cans in which it stores the key sets. The Hashtable uses the hashcode to decide to which the key pair should plan. The capacity to get the container area from Key's hashcode is known as hash work. In principle, a hash work is a capacity which when given a key, creates an address in the table. A hash work consistently returns a number for an item. Two equivalent items will consistently have a similar number while two inconsistent objects may not generally have various numbers. When we put objects into a hashtable then it is conceivable that various objects may have equal/ same hashcode. This is known as a collision. To determine collision, hashtable utilizes a variety of lists. The sets mapped to a single array index are stored in a list and then the list reference is stored in the index.
你取一堆东西,和一个数组。
对于每一个东西,你为它建立一个索引,称为哈希。关于哈希的重要事情是它“分散”了很多;你不希望两个相似的东西有相似的哈希值。
你把东西放到数组中哈希值表示的位置。在一个给定的哈希中可以有多个对象,所以你可以将这些对象存储在数组或其他合适的东西中,我们通常称之为bucket。
当你在哈希中查找东西时,你会经历相同的步骤,计算哈希值,然后查看那个位置的bucket中有什么,并检查它是否是你要寻找的东西。
当你的哈希工作得很好并且你的数组足够大时,在数组的任何特定下标处最多只会有很少的东西,所以你不需要看太多。
额外的好处是,当你的哈希表被访问时,它会把找到的东西(如果有的话)移动到桶的开头,这样下次它就会是第一个被检查的东西。