Adjacency Matrix
A square grid recording which pairs of nodes in a graph are connected, with one row and one column per node.
Related: Matrix, Sparse Matrix, Matrix Rank
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Transformers, LLMs, RAG, Agents, and the generative AI stack
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Data engineering, pipelines, and data-centric AI
Linear algebra, calculus, probability for ML
A square grid recording which pairs of nodes in a graph are connected, with one row and one column per node.
Related: Matrix, Sparse Matrix, Matrix Rank
Read the full conceptRepresenting a document as counts over a fixed vocabulary, discarding word order entirely.
Related: TF-IDF, Tokenization, Sparse Matrix
Read the full conceptThe rule for updating a belief after seeing evidence, by combining how well the evidence fits with how plausible the belief was beforehand.
Related: Likelihood, Random Variable
Read the full conceptThe number of distinct values a column contains — two for a yes/no flag, fifty thousand for a merchant ID.
Related: One-Hot Encoding, Hashing Trick, Feature
Read the full conceptA classification problem where one class is far rarer than another, so accuracy stops describing anything useful.
Related: SMOTE, Stratified Sampling, Ground Truth
Read the full conceptA change in the relationship between inputs and the correct answer, so the same input now means something different.
Related: Data Drift, Covariate Shift, Ground Truth
Read the full conceptEvery narrow-format announcement leads with a throughput multiple. The number is real. What it leaves out is that you are buying it with dynamic range, and the bill arrives as a NaN four hundred steps into a run.
Model launches report a single benchmark score with no spread, and readers compare those scores as if they were measurements. Two systems at 91% can be entirely different claims, and the missing number is the one that would tell you.
Quick answer
Use Welford's algorithm, which tracks the running mean and the sum of squared deviations from it. The textbook one-pass formula E[x²] − E[x]² loses all precision on large means and can return a negative variance.
Quick answer
Compare it against a central finite difference at h around 1e-5 in float64, using a relative tolerance. Making h smaller makes the check worse, not better.
Quick answer
Dot product divided by both L2 norms. Guard the zero vector, use np.linalg.norm rather than sqrt of sum of squares, and note that pre-normalised vectors make the dot product alone sufficient.
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