Model: AShi846/fine-tuned-embedding-model
Author: AShi846
Pipeline: sentence-similarity
Library: sentence-transformers
Tags: sentence-transformers, safetensors, bert, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:475, loss:CosineSimilarityLoss, arxiv:1908.10084, base_model:sentence-transformers/all-MiniLM-L6-v2, base_model:finetune:sentence-transformers/all-MiniLM-L6-v2, text-embeddings-inference, endpoints_compatible, region:us
Base models (finetune): id: sentence-transformers/all-MiniLM-L6-v2
Card metadata: tags: sentence-transformers, sentence-similarity, feature-extraction, generated_from_trainer, dataset_size:475, loss:CosineSimilarityLoss
base_model: sentence-transformers/all-MiniLM-L6-v2
widget: source_sentence: We will analyze the $K$-means algorithm and show that it always converge. Let us consider the $K$-means objective function: $$ \mathcal{L}(\mathbf{z}, \boldsymbol{\mu})=\sum_{n=1}^{N} \sum_{k=1}^{K} z_{n k}\left\|\mathbf{x}_{n}-\boldsymbol{\mu}_{k}\right\|_{2}^{2} $$ where $z_{n k} \in\{0,1\}$ with $\sum_{k=1}^{K} z_{n k}=1$ and $\boldsymbol{\mu}_{k} \in \mathbb{R}^{D}$ for $k=1, \ldots, K$ and $n=1, \ldots, N$. How would you choose $\left\{\boldsymbol{\mu}_{k}\right\}_{k=1}^{K}$ to minimize $\mathcal{L}(\mathbf{z}, \boldsymbol{\mu})$ for given $\left\{z_{n k}\right\}_{n, k=1}^{N, K}$ ? Compute the closed-form formula for the $\boldsymbol{\mu}_{k}$. To which step of the $K$-means algorithm does it correspond?
sentences: 1. Dynamically scheduled processors have universally more
        physical registers than the typical 32 architectural ones and
        they are used for removing WARs and WAW (name
        dependencies). In VLIW processors, the same renaming must be
        done by the compiler and all registers must be architecturally
        visible.
        2. Also, various techniques essential to improve the
        performance of VLIW processors consume more registers (e.g.,
        loop unrolling or loop fusion). , O( (f+1)n^2 )b in the binary case, or O( (f+1)n^3 )b in the non-binary case, The idea is wrong. Even if the interface remains the same since we are dealing with character strings, a decorator does not make sense because the class returning JSON cannot be used without this decorator; the logic for extracting the weather prediction naturally belongs to the weather client in question. It is therefore better to create a class containing both the download of the JSON and the extraction of the weather forecast., source_sentence: Estimate the 95% confidence intervals of the geometric mean and the arithmetic mean of pageviews using bootstrap resampling. The data is given in a pandas.DataFrame called df and the respective column is called "pageviews". You can use the scipy.stats python library.
sentences: (a) PoS tagging, but also Information Retrieval (IR), Text Classification, Information Extraction. For the later, accuracy sounds like precision (but it depends on what we actually mean by 'task' (vs. subtask)) . (b) a reference must be available, 'correct' and 'incorrect' must be clearly defined, [['break+V => breakable\xa0', 'derivational'], ['freeze+V =>\xa0frozen\xa0', 'inflectional'], ['translate+V => translation', 'derivational'], ['cat+N => cats', 'inflectional'], ['modify+V => modifies ', 'inflectional']], Dynamically scheduled out-of-order processors., source_sentence: The data contains information about submissions to a prestigious machine learning conference called ICLR. Columns:
year, paper, authors, ratings, decisions, institution, csranking, categories, authors_citations, authors_publications, authors_hindex, arxiv. The data is stored in a pandas.DataFrame format. 

Create 3 new fields in the dataframe corresponding to the median value of the number of citations per author, the number of publications per author, and the h-index per author. So for instance, for the row authors_publications, you will create an additional column, e.g. authors_publications_median, containing the median number of publications per author in each paper.
sentences: Consider an deterministic online algorithm $\Alg$ and set $x_1 = W$. There are two cases depending on whether \Alg trades the $1$ Euro the  first day or not.  Suppose first that $\Alg$ trades the Euro at day $1$. Then we set $x_2 = W^2$ and so the algorithm is only $W/W^2 = 1/W$ competitive. For the other case when  \Alg waits for the second day,  we set $x_2 = 1$. Then \Alg gets $1$ Swiss franc whereas optimum would get $W$ and so the algorithm is only $1/W$ competitive again., This is an abstraction leak: the notion of JavaScript and even a browser is a completely different level of abstraction than users, so this method will likely lead to bugs., We could consider at least two approaches here: either binomial confidence interval or t-test. • binomial confidence interval: evaluation of a binary classifier (success or not) follow a binomial law with parameters (perror,T), where T is the test-set size (157 in the above question; is it big enough?). Using normal approximation of the binomial law, the width of the confidence interval around estimated error probability is q(α)*sqrt(pb*(1-pb)/T),where q(α) is the 1-α quantile (for a 1 - α confidence level) and pb is the estimation of perror. We here want this confidence interval width to be 0.02, and have pb = 0.118 (and 'know' that q(0.05) = 1.96 from normal distribution quantile charts); thus we have to solve: (0.02)^2 = (1.96)^2*(0.118*(1-0.118))/T Thus T ≃ 1000. • t-test approach: let's consider estimating their relative behaviour on each of the test cases (i.e. each test estimation subset is of size 1). If the new system as an error of 0.098 (= 0.118 - 0.02), it can vary from system 3 between 0.02 of the test cases (both systems almost always agree but where the new system improves the results) and 0.216 of the test cases (the two systems never make their errors on the same test case, so they disagree on 0.118 + 0.098 of the cases). Thus μ of the t-test is between 0.02 and 0.216. And s = 0.004 (by assumption, same variance). Thus t is between 5*sqrt(T) and 54*sqrt(T) which is already bigger than 1.645 for any T bigger than 1. So this doesn't help much. So all we can say is that if we want to have a (lowest possible) difference of 0.02 we should have at least 1/0.02 = 50 test cases ;-) And if we consider that we have 0.216 difference, then we have at least 5 test cases... The reason why these numbers are so low is simply because we here make strong assumptions about the test setup: that it is a paired evaluation. In such a case, having a difference (0.02) that is 5 times bigger than the standard deviation is always statistically significant at a 95% level., source_sentence: In order to summarize the degree distribution in a single number, would you recommend using the average degree? Why, or why not? If not, what alternatives can you think of? Please elaborate!
sentences: inflectional morphology: no change in the grammatical category (e.g. give, given, gave, gives ) derivational morphology: change in category (e.g. process, processing, processable, processor, processabilty), $ 	ext{Var}[\wv^	op \xx] = rac1N \sum_{n=1}^N (\wv^	op \xx_n)^2$ %
, - $E(X) = 0.5$, $Var(X) = 1/12$
- $E(Y) = 0.5$, $Var(Y) = 1/12$
- $E(Z) = 0.6$, $Var(Z) = 1/24$
- $E(K) = 0.6$, $Var(K) = 1/12$, source_sentence:  The [t-statistic](https://en.wikipedia.org/wiki/T-statistic) is the ratio of the departure of the estimated value of a parameter from its hypothesized value to its standard error. In a t-test, the higher the t-statistic, the more confidently we can reject the null hypothesis. Use `numpy.random` to create four samples, each of size 30:
- $X \sim Uniform(0,1)$
- $Y \sim Uniform(0,1)$
- $Z = X/2 + Y/2 + 0.1$
- $K = Y + 0.1$
sentences: The simplest solution to produce the indexing set associated with a document is to use a stemmer associated with stop lists allowing to ignore specific non content bearing terms. In this case, the indexing set associated with D might be:

$I(D)=\{2006$, export, increas, Switzerland, USA $\}$

A more sophisticated approach would consist in using a lemmatizer in which case, the indexing set might be:

$I(D)=\left\{2006 \_N U M\right.$, export\_Noun, increase\_Verb, Switzerland\_ProperNoun, USA\_ProperNoun\}, Including a major bugfix in a minor release instead of a bugfix release will cause an incoherent changelog and an inconvenience for users who wish to only apply the patch without any other changes. The bugfix could be as well an urgent security fix and should not wait to the next minor release date., def get_vocabulary_frequency(documents):     """     It parses the input documents and creates a dictionary with the terms and term frequencies.          INPUT:     Doc1: hello hello world     Doc2: hello friend          OUTPUT:     {'hello': 3,     'world': 1,     'friend': 1}      :param documents: list of list of str, with the tokenized documents.     :return: dict, with keys the words and values the frequency of each word.     """     vocabulary = dict()          for document in documents:         for word in document:             if word in vocabulary:                 vocabulary[word] += 1             else:                 vocabulary[word] = 1      return vocabulary
pipeline_tag: sentence-similarity
library_name: sentence-transformers
Likes: 0 | All-time downloads: 11
Created: 2025-06-10 20:11:37