我之前发现过类似的帖子，但没有真正回答这个问题。
在我的指纹识别中，我生成了一个包含 5 个整数的记录集。例如：
33,42,88,121,194
这些对应于特定音乐样本的最高幅度的频率。
例如：对于 30ms 的音频样本，我有以下频率的桶：
0-40
40-80
80-120
120-180
180-250
我试图生成一个哈希（一个宽容的哈希），它可能会生成相同的哈希
一样
33,42,88,121,194，就像33,43,88,122,195
，其中形成类似散列的频率存在微小差异。
第一关是LSH吗？据我所知，这最适合音频指纹识别。
如果没有，任何人都可以提供一些伪代码或 c# 来实现我正在寻找的功能吗？我已经阅读了 LSH 和 matlab 和 perl 实现，但我不理解它们，所以发布它们的链接并不能真正帮助我太多。
再次感谢！
Ive found simlar posts before about this but nothing really answers the question.
In my fingerprinting, i produce a recordset which has 5 integers. For example:
33,42,88,121,194
These correspond to the frequencies which have the highest magnitude for a particular sample of music.
Eg: for 30ms of audio sample i have buckets of the following frequencies:
0-40
40-80
80-120
120-180
180-250
Im trying to produce a hash (a forgiving one) which will perhaps produce the same hash for
33,42,88,121,194 as it would for say
33,43,88,122,195
where there are minor differences in the frequencies a similar hash would be formed.
1st off is this LSH? as i have read that this is best for Audio Fingerprinting.
If not, could anyone provide some psuedocode or c# for a function that might do what im looking for? i have read up on LSH and matlab and perl implementations but i dont understand them so posting a link to them won't really help me too much.
thanks again!
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这可能与此重复：比较两个频谱图来查找它们匹配算法的偏移，看来您正在尝试做的是为样本中峰值的粗略分布生成直方图。有多种方法可以做到这一点，另一个“示例”在这里： 比较两个频谱图以找到它们匹配算法的偏移
一种方法是使用峰值数据及其分布（随时间变化）的快速傅里叶变换来生成粗略的结果蒸馏形式样品的当量。为此，您要做的事情大致类似于：
。比对指纹，你对第二个样本运行相同的过程，然后使用 Diff 算法来比较两者，使用一些“模糊”来确定它们的接近程度，您将需要在二维上比较指纹，离散指纹的顺序，以及每个示例的总体差异。
这篇关于制作与 Shazaam 大致相同的 Java 的文章不久前发布：http://www.redcode.nl/blog/2010/06/creating-shazam-in-java/ 可能对您有一些帮助。
This might be a duplicate of this: Compare two spectogram to find the offset where they match algorithm, what it appears you are trying to do is produce a histogram for the rough distribution of the peaks in the sample. There are several methods to do this, another "example" is here: Compare two spectogram to find the offset where they match algorithm
One method of doing this is to use a Fast-Fourier-Transform of the peak data and its distribution (over time) to produce a rough equivalence of the sample in a distilled form. To do this you do something roughly similar to:
To compare the fingerprint, you run the same process over the second sample, and then use a Diff algorithm to compare the two, using some "fuzz" to decide how close they are. You will need to compare the fingerprints on two dimensions, the order of the discrete fingerprints, as well as the overall difference in each sample.
This article on making a rough Java equivalent to Shazaam was posted a while ago: http://www.redcode.nl/blog/2010/06/creating-shazam-in-java/ and may be of some help to you.