text,data-mining,information-retrieval,euclidean-distance,cosine-similarity

If your data is normalized to unit length, then it is very easy to prove that Euclidean(A,B) = 2 - Cos(A,B) This does hold if ||A||=||B||=1. It does not hold in the general case, and it depends on the exact order in which you perform your normalization steps. I.e. if...

python,pandas,dataframes,cosine-similarity

One of the issue in addition to my main goal that I have at this point of the code is my dataframe still has NaN. That's beacause df.fillna does not modify DataFrame, but returns a new one. Fix it and your result will be fine....

java,c,performance,cosine-similarity

Math.pow(a, b) does math.exp( math.log (a)*b) it's going to a very expensive way to square a number. I suggest you write the Java code similar to the way you wrote the C code to get a closer result. Note: the JVM can take a couple of seconds to warm up...

wordnet,cosine-similarity,word2vec,sentence-similarity

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)....

c++,image,opencv,vector,cosine-similarity

oh, not too difficult. step 1: fill your feature(Mat) with numbers, one after the other: Mat feature; // you could use a std::vector, too, but cv::Mat has the // handy dot-product used below already built in. feature.push_back(aspect_ratio); feature.push_back(area); feature.push_back(center.x); feature.push_back(center.y); feature.push_back(more_stuff); ... step 2: to compare those features, use the...

python,vectorization,text-processing,cosine-similarity

Is this what you wanted to do? text_file = ['hello','world','testing'] term_dict = {'some':0, 'word':0, 'world':0} for word in text_file: if word in term_dict: term_dict[word] = 1 If you've tokenized your file (.split() method in Python), then they will be available in a list. Assuming that you've normalized each term (lowered,...

machine-learning,recommendation-engine,user-profile,cosine-similarity

Similarity measures between object in clustering analysis is a broad subject. What I would suggest for You is to consider approach of 'divide and conquer'. Treat similarity between two user profiles as weighted average from all attributes similarity. Just remember to user normalized values for Your attributes similarity before doing...

python,nlp,nltk,wordnet,cosine-similarity

There's no easy way to get similarity between words that are not nouns/verbs. As noted, nouns/verbs similarity are easily extracted from >>> from nltk.corpus import wordnet as wn >>> dog = wn.synset('dog.n.1') >>> cat = wn.synset('cat.n.1') >>> car = wn.synset('car.n.1') >>> wn.path_similarity(dog, cat) 0.2 >>> wn.path_similarity(dog, car) 0.07692307692307693 >>> wn.wup_similarity(dog,...

You want to compute the similarities between the given row and each row in the Matrix. Hence, inner product and norms must be computed getRowDimension times. But the initializations are in the wrong place - move them into the loop over all rows. And you want to use += and...

python,numpy,scipy,euclidean-distance,cosine-similarity

You can use scipy.spatial.distance.squareform to convert between a full m x n distance matrix and the upper triangle: import numpy as np from scipy.spatial import distance m = 100 n = 200 X = np.random.randn(m, n) d = distance.pdist(X, metric='jaccard') print(d.shape) # (4950,) D = distance.squareform(d) print D.shape # (100,...

python,out-of-memory,fork,scikit-learn,cosine-similarity

What version of scikit-learn are you using? And does it run with n_jobs=1? The result should fit in memory, it is 8 * 42588 ** 2 / 1024 ** 3 = 13 Gb. But the data is about 2gb, and will be replicated to each core. So if you have...

python,dictionary,typeerror,cosine-similarity

Could you try changing them to lists via list(v1.values())? dict_values is a type, so converting it to a list may solve the issue. return up / (np.sqrt(np.dot(list(v1.values()), list(v1.values()))) * np.sqrt(np.dot(list(v2.values()), list(v2.values())))) Reference: Python: simplest way to get list of values from dict?...

cosine-similarity,word2vec,sentence-similarity

Cosine measures the angle between two vectors and does not take the length of either vector into account. When you divide by the length of the phrase, you are just shortening the vector, not changing its angular position. So your results look correct to me.

c++,arrays,opencv,mat,cosine-similarity

The correct definition of cosine similarity is : Your code does not compute the denominator, hence the values are wrong. double cosine_similarity(double *A, double *B, unsigned int Vector_Length) { double dot = 0.0, denom_a = 0.0, denom_b = 0.0 ; for(unsigned int i = 0u; i < Vector_Length; ++i) {...