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I've noticed that AI community refers to various tensors as 512-d, meaning 512 dimensional tensor, where the term 'dimension' seems to mean 512 different float values in the representation for a single datapoint. e.g. in 512-d word-embeddings means 512 length vector of floats used to represent 1 english-word e.g. https://medium.com/#jonathan_hui/nlp-word-embedding-glove-5e7f523999f6
But it isn't 512 different dimensions, it's only 1 dimensional vector? Why is the term dimension used in such a different manner than usual?
When we use the term conv1d or conv2d which are convolutions over 1-dimension and 2-dimensions, a dimension is used in the typical way it's used in math/sciences but in the word-embedding context, a 1-d vector is said to be a 512-d vector, or am I missing something?
Why is this overloaded use of the term dimension? What context determines what dimension means in machine-learning as the term seems overloaded?
In the context of word embeddings in neural networks, dimensionality reduction, and many other machine learning areas, it is indeed correct to call the vector (which is typically, an 1D array or tensor) as n-dimensional where n is usually greater than 2. This is because we usually work in the Euclidean space where a (data) point in a certain dimensional (Euclidean) space is represented as an n-tuple of real numbers (i.e. real n-space ℝn).
Below is an exampleref of a (data) point in a 3D (Euclidean) space. To represent any point in this space, say d1, we need a tuple of three real numbers (x1, y1, z1).
Now, your confusion arises why this point d1 is called as 3 dimensional instead of 1 dimensional array. The reason is because it lies or lives in this 3D space. The same argument can be extended to all points in any n-dimensional real space, as it is done in the case of embeddings with 300d, 512d, 1024d vector etc.
However, in all nD array compute frameworks such as NumPy, PyTorch, TensorFlow etc, these are still 1D arrays because the length of the above said vectors can be represented using a single number.
But, what if you have more than 1 data point? Then, you have to stack them in some (unique) way. And this is where the need for a second dimension arises. So, let's say you stack 4 of these 512d vectors vertically, then you'd end up with a 2D array/tensor of shape (4, 512). Note that here we call the array as 2D because two integer numbers are required to represent the extent/length along each axis.
To understand this better, please refer my other answer on axis parameter visualization for nD arrays, the visual representation of which I will include it below.
ref: Euclidean space wiki
It is not overloading, but standard usage. What are the elements of a 512-dimensional vector space? They are 512 dimensional vectors. Each of which can be represented by 512 floating point number as in your equation. Each such vector spans a 1-dimensional subspace of the 512-dimensional space.
When you talk of the dimension of a tensor, a tensor is a linear map (roughly speaking, I am omitting the duals) from the product of N vector spaces to the reals. The dimension of a TENSOR is the N.
If you want to be more specific, you need to be clear on the terms dimension, rank, and shape.
The dimensionality of a tensor means the rank, which has a specific definition: the rank is the number of indices. When you see "3-dimensional tensor", you can take that to mean that the tensor has 3 indices, namely T[i][j][k]. So a vector has rank 1, a matrix has rank 2, a cube has rank 3, etc.
When you want to specify the size of each dimension, you should prefer to use the term shape. A 3-dimensional (aka rank 3) tensor can have shape [10, 20, 30] if the 0th dimension has 10 values, the 1st dimension has 20 values, and the 2nd dimension has 30 values. (This shape might represent, say, a batch of 10 images, each of shape 20x30.)
Note, though, that when talking about vectors, it is common to say "512-D vector". As you mentioned, this terminology comes up a lot with word embeddings (e.g. "we used 512-D word embeddings"). Since "vector" by definition means rank 1, then people will interpret that statement to mean "a structure of rank 1 with 512 values".
You might encounter someone saying "I have a 5-d vector", in which case you'd need to follow up with "wait, do you mean a 5-d tensor or a 1-d vector with 5 values?".
I am not a mathematician, by the way.
I have task of sentence similarity where i calculate the cosine of two sentence to decide how similar they are . It seems that for sentence with digits the similarity is not affected no matter how "far" the numbers are . For an example:
a = generate_embedding('issue 845')
b = generate_embedding('issue 11')
cosine_sim(a,b) = 0.9307
is there a way to distance the hashing of numbers or any other hack to handle that issue?
If your sentence embedding are produced using the embeddings of individual words (or tokens), then a hack could be the following:
to add dimensions to the word embedding. These dimensions would be set to zero for all non-numeric tokens, and for numeric tokens these dimensions would contain values reflecting the magnitude of the numeric value. It would get a bit mathematical because cosine similarity uses angles, so the extra dimensions added to the embedding would have to reflect the magnitude of the numeric values through larger or smaller angles.
An easier (workaround) hack would be to extract the numeric values from the sentences using regular expressions and compute their distance and combine that information with the similarity score in order to obtain a new similarity score.
I am using WEKA for performing text collection. Suppose i have n documents with text, i calculated TFID as feature vector for each document and than calculated cosine similarity between each of each of the document.it generated nXn matrix. Now i wonder how to use this nxn matrix in k-mean algorithm . i know i can apply some dimension reduction such as MDS or PCA. What I am confused here is that after applying dimension reduction how will i identify that document itself, for example if i have 3 documents d1,d2 d3 than cosine will give me distances between d11,d12,d13
d21,d22,d23
d31,d32,d33
now i am not sure what will be output after PCA or MDS and how i will identify the documents after kmean. Please suggest. I hope i have put my question clearly
PCA is used on the raw data, not on distances, i.e. PCA(X).
MDS uses a distance function, i.e. MDS(X, cosine).
You appear to believe you need to run PCA(cosine(X))? That doesn't work.
You want to run MDS(X, cosine).
Assuming that I have a word similarity score for each pair of words in two sentences, what is a decent approach to determining the overall sentence similarity from those scores?
The word scores are calculated using cosine similarity from vectors representing each word.
Now that I have individual word scores, is it too naive to sum the individual word scores and divide by the total word count of both sentences to get a score for the two sentences?
I've read about further constructing vectors to represent the sentences, using the word scores, and then again using cosine similarity to compare the sentences. But I'm not familiar with how to construct sentence vectors from the existing word scores. Nor am I aware of what the tradeoffs are compared with the naive approach described above, which at the very least, I can easily comprehend. :).
Any insights are greatly appreciated.
Thanks.
What I ended up doing, was taking the mean of each set of vectors, and then applying cosine-similarity to the two means, resulting in a score for the sentences.
I'm not sure how mathematically sound this approach is, but I've seen it done in other places (like python's gensim).
It would be better to use contextual word embeddings(vector representations) for words.
Here is an approach to sentence similarities by pairwise word similarities: BERTScore.
You can check the math here.
I have a set of vectors in multidimensional space (may be several thousands of dimensions). In this space, I can calculate distance between 2 vectors (as a cosine of the angle between them, if it matters). What I want is to visualize these vectors keeping the distance. That is, if vector a is closer to vector b than to vector c in multidimensional space, it also must be closer to it on 2-dimensional plot. Is there any kind of diagram that can clearly depict it?
I don't think so. Imagine any twodimensional picture of a tetrahedron. There is no way of depicting the four vertices in two dimensions with equal distances from each other. So you will have a hard time trying to depict more than three n-dimensional vectors in 2 dimensions conserving their mutual distances.
(But right now I can't think of a rigorous proof.)
Update:
Ok, second idea, maybe it's dumb: If you try and find clusters of closer associated objects/texts, then calculate the center or mean vector of each cluster. Then you can reduce the problem space. At first find a 2D composition of the clusters that preserves their relative distances. Then insert the primary vectors, only accounting for their relative distances within a cluster and their distance to the center of to two or three closest clusters.
This approach will be ok for a large number of vectors. But it will not be accurate in that there always will be somewhat similar vectors ending up at distant places.