I am doing a community website that requires me to calculate the similarity between any two users. Each user is described with the following attributes:
age, skin type (oily, dry), hair type (long, short, medium), lifestyle (active outdoor lover, TV junky) and others.
Can anyone tell me how to go about this problem or point me to some resources?
Another way of computing (in R) all the pairwise dissimilarities (distances) between observations in the data set. The original variables may be of mixed types. The handling of nominal, ordinal, and (a)symmetric binary data is achieved by using the general dissimilarity coefficient of Gower (Gower, J. C. (1971) A general coefficient of similarity and some of its properties, Biometrics 27, 857–874). For more check out this on page 47. If x contains any columns of these data-types, Gower's coefficient will be used as the metric.
For example
x1 <- factor(c(10, 12, 25, 14, 29))
x2 <- factor(c("oily", "dry", "dry", "dry", "oily"))
x3 <- factor(c("medium", "short", "medium", "medium", "long"))
x4 <- factor(c("active outdoor lover", "TV junky", "TV junky", "active outdoor lover", "TV junky"))
x <- cbind(x1,x2,x3,x4)
library(cluster)
daisy(x, metric = "euclidean")
you'll get :
Dissimilarities :
1 2 3 4
2 2.000000
3 3.316625 2.236068
4 2.236068 1.732051 1.414214
5 4.242641 3.741657 1.732051 2.645751
If you are interested on a method for dimensionality reduction for categorical data (also a way to arrange variables into homogeneous clusters) check this
Give each attribute an appropriate weight, and add the differences between values.
enum SkinType
Dry, Medium, Oily
enum HairLength
Bald, Short, Medium, Long
UserDifference(user1, user2)
total := 0
total += abs(user1.Age - user2.Age) * 0.1
total += abs((int)user1.Skin - (int)user2.Skin) * 0.5
total += abs((int)user1.Hair - (int)user2.Hair) * 0.8
# etc...
return total
If you really need similarity instead of difference, use 1 / UserDifference(a, b)
You probably should take a look for
Data Mining and Data Warehousing (Essential)
Machine Learning (Extra)
Artificial Neural Networks (Especially SOM)
Pattern Recognition (Related)
These topics will let you your program recognize similarities and clusters in your users collection and try to adapt to them...
You can then know different hidden common groups of related users... (i.e users with green hair usually do not like watching TV..)
As an advice, try to use ready implemented tools for this feature instead of implementing it yourself...
Take a look at Open Directory Data Mining Projects
Three steps to achieve a simple subjective metric for difference between two datapoints that might work fine in your case:
Capture all your variables in a representative numeric variable, for example: skin type (oily=-1, dry=1), hair type (long=2, short=0, medium=1),lifestyle (active outdoor lover=1, TV junky=-1), age is a number.
Scale all numeric ranges so that they fit the relative importance you give them for indicating difference. For example: An age difference of 10 years is about as different as the difference between long and medium hair, and the difference between oily and dry skin. So 10 on the age scale is as different as 1 on the hair scale is as different as 2 on the skin scale, so scale the difference in age by 0.1, that in hair by 1 and and that in skin by 0.5
Use an appropriate distance metric to combine the differences between two people on the various scales in one overal difference. The smaller this number, the more similar they are. I'd suggest simple quadratic difference as a first attempt at your distance function.
Then the difference between two people could be calculated with (I assume Person.age, .skin, .hair, etc. have already gone through step 1 and are numeric):
double Difference(Person p1, Person p2) {
double agescale=0.1;
double skinscale=0.5;
double hairscale=1;
double lifestylescale=1;
double agediff = (p1.age-p2.age)*agescale;
double skindiff = (p1.skin-p2.skin)*skinscale;
double hairdiff = (p1.hair-p2.hair)*hairscale;
double lifestylediff = (p1.lifestyle-p2.lifestyle)*lifestylescale;
double diff = sqrt(agediff^2 + skindiff^2 + hairdiff^2 + lifestylediff^2);
return diff;
}
Note that diff in this example is not on a nice scale like (0..1). It's value can range from 0 (no difference) to something large (high difference). Also, this method is almost completely unscientific, it is just designed to quickly give you a working difference metric.
Look at algorithms for computing srting difference. Its very similar to what you need. Store your attributes as a bit string and compute the distance between the strings
You should read these two topics.
Most popular clustering algorithm k - means
And similarity matrix are essential in clustering
Related
I am trying to replicate the result of an academic paper below to practice how to apply statistical methods in R.
This is what the paper states:
Pre- and postoutbreak differences in voter intentions.
Across the 32 elections included in primary analyses, the mean voter-intention difference score was greater than zero (M=1.02%), d=0.84, t(31)=2.34, p=.026.
This result is consistent with the pre- and postelection difference in nationwide polling results for the House of Representatives elections, which indicates a general postoutbreak shift toward favoring Republican rather than Democratic candidates. (If the two outliers were
included in the analysis, the mean voter-intention difference score was not meaningfully different from zero, p=.937.) (Source screenshot)
The t-test result matched the values in the paper, however, I cannot figure out how to produce the right cohen's d estimate. I looked into the documentation for cohen.d function over and over again, googled how to do this, even watched some boring Youtube videos, but to no avail. The code itself runs, but it gives me a wrong value. Am I missing something? Can someone help me with how I should format the arguments?
# excluding outliers from the dataset
no_outliers <- study2 %>%
filter(StateSenateRace != "Rhode Island", StateSenateRace != "Hawaii")
# paired t-test
t.test(no_outliers$OctMeanVoterIntentionIndex, no_outliers$SeptMeanVoterIntentionIndex, paired = TRUE, var.equal = TRUE, na.rm = TRUE)
cohen.d(no_outliers$SeptMeanVoterIntentionIndex, no_outliers$OctMeanVoterIntentionIndex, na.rm = TRUE)
Here's the result I got.
Cohen's d
d estimate: -0.02130406 (negligible)
95 percent confidence interval:
lower upper
-0.5171041 0.4744960
Thank you in advance--I wish I could contribute to Stack Overflow as a R expert someday!
I am implementing the Skipgram model, both in Pytorch and Tensorflow2. I am having doubts about the implementation of subsampling of frequent words. Verbatim from the paper, the probability of subsampling word wi is computed as
where t is a custom threshold (usually, a small value such as 0.0001) and f is the frequency of the word in the document. Although the authors implemented it in a different, but almost equivalent way, let's stick with this definition.
When computing the P(wi), we can end up with negative values. For example, assume we have 100 words, and one of them appears extremely more often than others (as it is the case for my dataset).
import numpy as np
import seaborn as sns
np.random.seed(12345)
# generate counts in [1, 20]
counts = np.random.randint(low=1, high=20, size=99)
# add an extremely bigger count
counts = np.insert(counts, 0, 100000)
# compute frequencies
f = counts/counts.sum()
# define threshold as in paper
t = 0.0001
# compute probabilities as in paper
probs = 1 - np.sqrt(t/f)
sns.distplot(probs);
Q: What is the correct way to implement subsampling using this "probability"?
As an additional info, I have seen that in keras the function keras.preprocessing.sequence.make_sampling_table takes a different approach:
def make_sampling_table(size, sampling_factor=1e-5):
"""Generates a word rank-based probabilistic sampling table.
Used for generating the `sampling_table` argument for `skipgrams`.
`sampling_table[i]` is the probability of sampling
the i-th most common word in a dataset
(more common words should be sampled less frequently, for balance).
The sampling probabilities are generated according
to the sampling distribution used in word2vec:
```
p(word) = (min(1, sqrt(word_frequency / sampling_factor) /
(word_frequency / sampling_factor)))
```
We assume that the word frequencies follow Zipf's law (s=1) to derive
a numerical approximation of frequency(rank):
`frequency(rank) ~ 1/(rank * (log(rank) + gamma) + 1/2 - 1/(12*rank))`
where `gamma` is the Euler-Mascheroni constant.
# Arguments
size: Int, number of possible words to sample.
sampling_factor: The sampling factor in the word2vec formula.
# Returns
A 1D Numpy array of length `size` where the ith entry
is the probability that a word of rank i should be sampled.
"""
gamma = 0.577
rank = np.arange(size)
rank[0] = 1
inv_fq = rank * (np.log(rank) + gamma) + 0.5 - 1. / (12. * rank)
f = sampling_factor * inv_fq
return np.minimum(1., f / np.sqrt(f))
I tend to trust deployed code more than paper write-ups, especially in a case like word2vec, where the original authors' word2vec.c code released by the paper's authors has been widely used & served as the template for other implementations. If we look at its subsampling mechanism...
if (sample > 0) {
real ran = (sqrt(vocab[word].cn / (sample * train_words)) + 1) * (sample * train_words) / vocab[word].cn;
next_random = next_random * (unsigned long long)25214903917 + 11;
if (ran < (next_random & 0xFFFF) / (real)65536) continue;
}
...we see that those words with tiny counts (.cn) that could give negative values in the original formula instead here give values greater-than 1.0, and thus can never be less than the long-random-masked-and-scaled to never be more than 1.0 ((next_random & 0xFFFF) / (real)65536). So, it seems the authors' intent was for all negative-values of the original formula to mean "never discard".
As per the keras make_sampling_table() comment & implementation, they're not consulting the actual word-frequencies at all. Instead, they're assuming a Zipf-like distribution based on word-rank order to synthesize a simulated word-frequency.
If their assumptions were to hold – the related words are from a natural-language corpus with a Zipf-like frequency-distribution – then I'd expect their sampling probabilities to be close to down-sampling probabilities that would have been calculated from true frequency information. And that's probably "close enough" for most purposes.
I'm not sure why they chose this approximation. Perhaps other aspects of their usual processes have not maintained true frequencies through to this step, and they're expecting to always be working with natural-language texts, where the assumed frequencies will be generally true.
(As luck would have it, and because people often want to impute frequencies to public sets of word-vectors which have dropped the true counts but are still sorted from most- to least-frequent, just a few days ago I wrote an answer about simulating a fake-but-plausible distribution using Zipf's law – similar to what this keras code is doing.)
But, if you're working with data that doesn't match their assumptions (as with your synthetic or described datasets), their sampling-probabilities will be quite different than what you would calculate yourself, with any form of the original formula that uses true word frequencies.
In particular, imagine a distribution with one token a million times, then a hundred tokens all appearing just 10 times each. Those hundred tokens' order in the "rank" list is arbitrary – truly, they're all tied in frequency. But the simulation-based approach, by fitting a Zipfian distribution on that ordering, will in fact be sampling each of them very differently. The one 10-occurrence word lucky enough to be in the 2nd rank position will be far more downsampled, as if it were far more frequent. And the 1st-rank "tall head" value, by having its true frequency *under-*approximated, will be less down-sampled than otherwise. Neither of those effects seem beneficial, or in the spirit of the frequent-word-downsampling option - which should only "thin out" very-frequent words, and in all cases leave words of the same frequency as each other in the original corpus roughly equivalently present to each other in the down-sampled corpus.
So for your case, I would go with the original formula (probability-of-discarding-that-requires-special-handling-of-negative-values), or the word2vec.c practical/inverted implementation (probability-of-keeping-that-saturates-at-1.0), rather than the keras-style approximation.
(As a totally-separate note that nonetheless may be relevant for your dataset/purposes, if you're using negative-sampling: there's another parameter controlling the relative sampling of negative examples, often fixed at 0.75 in early implementations, that one paper has suggested can usefully vary for non-natural-language token distributions & recommendation-related end-uses. This parameter is named ns_exponent in the Python gensim implementation, but simply a fixed power value internal to a sampling-table pre-calculation in the original word2vec.c code.)
I have read an article on data leakage. In a hackathon there are two sets of data, train data on which participants train their algorithm and test set on which performance is measured.
Data leakage helps in getting a perfect score in test data, with out viewing train data by exploiting the leak.
I have read the article, but I am missing the crux how the leakage is exploited.
Steps as shown in article are following:
Let's load the test data.
Note, that we don't have any training data here, just test data. Moreover, we will not even use any features of test objects. All we need to solve this task is the file with the indices for the pairs, that we need to compare.
Let's load the data with test indices.
test = pd.read_csv('../test_pairs.csv')
test.head(10)
pairId FirstId SecondId
0 0 1427 8053
1 1 17044 7681
2 2 19237 20966
3 3 8005 20765
4 4 16837 599
5 5 3657 12504
6 6 2836 7582
7 7 6136 6111
8 8 23295 9817
9 9 6621 7672
test.shape[0]
368550
For example, we can think that there is a test dataset of images, and each image is assigned a unique Id from 0 to N−1 (N -- is the number of images). In the dataframe from above FirstId and SecondId point to these Id's and define pairs, that we should compare: e.g. do both images in the pair belong to the same class or not. So, for example for the first row: if images with Id=1427 and Id=8053 belong to the same class, we should predict 1, and 0 otherwise.
But in our case we don't really care about the images, and how exactly we compare the images (as long as comparator is binary).
print(test['FirstId'].nunique())
print(test['SecondId'].nunique())
26325
26310
So the number of pairs we are given to classify is very very small compared to the total number of pairs.
To exploit the leak we need to assume (or prove), that the total number of positive pairs is small, compared to the total number of pairs. For example: think about an image dataset with 1000 classes, N images per class. Then if the task was to tell whether a pair of images belongs to the same class or not, we would have 1000*N*(N−1)/2 positive pairs, while total number of pairs was 1000*N(1000N−1)/2.
Another example: in Quora competitition the task was to classify whether a pair of qustions are duplicates of each other or not. Of course, total number of question pairs is very huge, while number of duplicates (positive pairs) is much much smaller.
Finally, let's get a fraction of pairs of class 1. We just need to submit a constant prediction "all ones" and check the returned accuracy. Create a dataframe with columns pairId and Prediction, fill it and export it to .csv file. Then submit
test['Prediction'] = np.ones(test.shape[0])
sub=pd.DataFrame(test[['pairId','Prediction']])
sub.to_csv('sub.csv',index=False)
All ones have accuracy score is 0.500000.
So, we assumed the total number of pairs is much higher than the number of positive pairs, but it is not the case for the test set. It means that the test set is constructed not by sampling random pairs, but with a specific sampling algorithm. Pairs of class 1 are oversampled.
Now think, how we can exploit this fact? What is the leak here? If you get it now, you may try to get to the final answer yourself, othewise you can follow the instructions below.
Building a magic feature
In this section we will build a magic feature, that will solve the problem almost perfectly. The instructions will lead you to the correct solution, but please, try to explain the purpose of the steps we do to yourself -- it is very important.
Incidence matrix
First, we need to build an incidence matrix. You can think of pairs (FirstId, SecondId) as of edges in an undirected graph.
The incidence matrix is a matrix of size (maxId + 1, maxId + 1), where each row (column) i corresponds i-th Id. In this matrix we put the value 1to the position [i, j], if and only if a pair (i, j) or (j, i) is present in a given set of pais (FirstId, SecondId). All the other elements in the incidence matrix are zeros.
Important! The incidence matrices are typically very very sparse (small number of non-zero values). At the same time incidence matrices are usually huge in terms of total number of elements, and it is impossible to store them in memory in dense format. But due to their sparsity incidence matrices can be easily represented as sparse matrices. If you are not familiar with sparse matrices, please see wiki and scipy.sparse reference. Please, use any of scipy.sparseconstructors to build incidence matrix.
For example, you can use this constructor: scipy.sparse.coo_matrix((data, (i, j))). We highly recommend to learn to use different scipy.sparseconstuctors, and matrices types, but if you feel you don't want to use them, you can always build this matrix with a simple for loop. You will need first to create a matrix using scipy.sparse.coo_matrix((M, N), [dtype]) with an appropriate shape (M, N) and then iterate through (FirstId, SecondId) pairs and fill corresponding elements in matrix with ones.
Note, that the matrix should be symmetric and consist only of zeros and ones. It is a way to check yourself.
import networkx as nx
import numpy as np
import pandas as pd
import scipy.sparse
import matplotlib.pyplot as plt
test = pd.read_csv('../test_pairs.csv')
x = test[['FirstId','SecondId']].rename(columns={'FirstId':'col1', 'SecondId':'col2'})
y = test[['SecondId','FirstId']].rename(columns={'SecondId':'col1', 'FirstId':'col2'})
comb = pd.concat([x,y],ignore_index=True).drop_duplicates(keep='first')
comb.head()
col1 col2
0 1427 8053
1 17044 7681
2 19237 20966
3 8005 20765
4 16837 599
data = np.ones(comb.col1.shape, dtype=int)
inc_mat = scipy.sparse.coo_matrix((data,(comb.col1,comb.col2)), shape=(comb.col1.max() + 1, comb.col1.max() + 1))
rows_FirstId = inc_mat[test.FirstId.values,:]
rows_SecondId = inc_mat[test.SecondId.values,:]
f = rows_FirstId.multiply(rows_SecondId)
f = np.asarray(f.sum(axis=1))
f.shape
(368550, 1)
f = f.sum(axis=1)
f = np.squeeze(np.asarray(f))
print (f.shape)
Now build the magic feature
Why did we build the incidence matrix? We can think of the rows in this matix as of representations for the objects. i-th row is a representation for an object with Id = i. Then, to measure similarity between two objects we can measure similarity between their representations. And we will see, that such representations are very good.
Now select the rows from the incidence matrix, that correspond to test.FirstId's, and test.SecondId's.
So do not forget to convert pd.series to np.array
These lines should normally run very quickly
rows_FirstId = inc_mat[test.FirstId.values,:]
rows_SecondId = inc_mat[test.SecondId.values,:]
Our magic feature will be the dot product between representations of a pair of objects. Dot product can be regarded as similarity measure -- for our non-negative representations the dot product is close to 0 when the representations are different, and is huge, when representations are similar.
Now compute dot product between corresponding rows in rows_FirstId and rows_SecondId matrices.
From magic feature to binary predictions
But how do we convert this feature into binary predictions? We do not have a train set to learn a model, but we have a piece of information about test set: the baseline accuracy score that you got, when submitting constant. And we also have a very strong considerations about the data generative process, so probably we will be fine even without a training set.
We may try to choose a thresold, and set the predictions to 1, if the feature value f is higer than the threshold, and 0 otherwise. What threshold would you choose?
How do we find a right threshold? Let's first examine this feature: print frequencies (or counts) of each value in the feature f.
For example use np.unique function, check for flags
Function to count frequency of each element
from scipy.stats import itemfreq
itemfreq(f)
array([[ 14, 183279],
[ 15, 852],
[ 19, 546],
[ 20, 183799],
[ 21, 6],
[ 28, 54],
[ 35, 14]])
Do you see how this feature clusters the pairs? Maybe you can guess a good threshold by looking at the values?
In fact, in other situations it can be not that obvious, but in general to pick a threshold you only need to remember the score of your baseline submission and use this information.
Choose a threshold below:
pred = f > 14 # SET THRESHOLD HERE
pred
array([ True, False, True, ..., False, False, False], dtype=bool)
submission = test.loc[:,['pairId']]
submission['Prediction'] = pred.astype(int)
submission.to_csv('submission.csv', index=False)
I want to understand the idea behind this. How we are exploiting the leak from the test data only.
There's a hint in the article. The number of positive pairs should be 1000*N*(N−1)/2, while the number of all pairs is 1000*N(1000N−1)/2. Of course, the number of all pairs is much, much larger if the test set was sampled at random.
As the author mentions, after you evaluate your constant prediction of 1s on the test set, you can tell that the sampling was not done at random. The accuracy you obtain is 50%. Had the sampling been done correctly, this value should've been much lower.
Thus, they construct the incidence matrix and calculate the dot product (the measure of similarity) between the representations of our ID features. They then reuse the information about the accuracy obtained with constant predictions (at 50%) to obtain the corresponding threshold (f > 14). It's set to be greater than 14 because that constitutes roughly half of our test set, which in turn maps back to the 50% accuracy.
The "magic" value didn't have to be greater than 14. It could have been equal to 14. You could have adjusted this value after some leader board probing (as long as you're capturing half of the test set).
It was observed that the test data was not sampled properly; same-class pairs were oversampled. Thus there is a much higher probability of each pair in the training set to have target=1 than any random pair. This led to the belief that one could construct a similarity measure based only on the pairs that are present in the test, i.e., whether a pair made it to the test is itself a strong indicator of similarity.
Using this insight one can calculate an incidence matrix and represent each id j as a binary array (the i-th element representing the presence of i-j pair in test, and thus representing the strong probability of similarity between them). This is a pretty accurate measure, allowing one to find the "similarity" between two rows just by taking their dot product.
The cutoff arrived at is purely by the knowledge of target-distribution found by leaderboard probing.
Given 1 Billion records containing following information:
ID x1 x2 x3 ... x100
1 0.1 0.12 1.3 ... -2.00
2 -1 1.2 2 ... 3
...
For each ID above, I want to find the top 10 closest IDs, based on Euclidean distance of their vectors (x1, x2, ..., x100).
What's the best way to compute this?
As it happens, I have a solution to this, involving combining sklearn with Spark: https://adventuresindatascience.wordpress.com/2016/04/02/integrating-spark-with-scikit-learn-visualizing-eigenvectors-and-fun/
The gist of it is:
Use sklearn’s k-NN fit() method centrally
But then use sklearn’s k-NN kneighbors() method distributedly
Performing a brute-force comparison of all records against all records is a losing battle. My suggestion would be to go for a ready-made implementation of k-Nearest Neighbor algorithm such as the one provided by scikit-learn then broadcast the resulting arrays of indices and distances and go further.
Steps in this case would be:
1- vectorize the features as Bryce suggested and let your vectorizing method return a list (or numpy array) of floats with as many elements as your features
2- fit your scikit-learn nn to your data:
nbrs = NearestNeighbors(n_neighbors=10, algorithm='auto').fit(vectorized_data)
3- run the trained algorithm on your vectorized data (training and query data are the same in your case)
distances, indices = nbrs.kneighbors(qpa)
Steps 2 and 3 will run on your pyspark node and are not parallelizable in this case. You will need to have enough memory on this node. In my case with 1.5 Million records and 4 features, it took a second or two.
Until we get a good implementation of NN for spark I guess we would have to stick to these workarounds. If you'd rather like to try something new, then go for http://spark-packages.org/package/saurfang/spark-knn
You haven't provided a lot of detail, but the general approach I would take to this problem would be to:
Convert the records to a data structure like like a LabeledPoint with (ID, x1..x100) as label and features
Map over each record and compare that record to all the other records (lots of room for optimization here)
Create some cutoff logic so that once you start comparing ID = 5 with ID = 1 you interrupt the computation because you have already compared ID = 1 with ID = 5
Some reduce step to get a data structure like {id_pair: [1,5], distance: 123}
Another map step to find the 10 closest neighbors of each record
You've identified pyspark and I generally do this type of work using scala, but some pseudo code for each step might look like:
# 1. vectorize the features
def vectorize_raw_data(record)
arr_of_features = record[1..99]
LabeledPoint( record[0] , arr_of_features)
# 2,3 + 4 map over each record for comparison
broadcast_var = []
def calc_distance(record, comparison)
# here you want to keep a broadcast variable with a list or dictionary of
# already compared IDs and break if the key pair already exists
# then, calc the euclidean distance by mapping over the features of
# the record and subtracting the values then squaring the result, keeping
# a running sum of those squares and square rooting that sum
return {"id_pair" : [1,5], "distance" : 123}
for record in allRecords:
for comparison in allRecords:
broadcast_var.append( calc_distance(record, comparison) )
# 5. map for 10 closest neighbors
def closest_neighbors(record, n=10)
broadcast_var.filter(x => x.id_pair.include?(record.id) ).takeOrdered(n, distance)
The psuedocode is terrible, but I think it communicates the intent. There will be a lot of shuffling and sorting here as you are comparing all records with all other records. IMHO, you want to store the keypair/distance in a central place (like a broadcast variable that gets updated though this is dangerous) to reduce the total euclidean distance calculations you perform.
I would like to do an algebraic curve fit of 2D data points, but for various reasons - it isn't really possible to have much of the sample data in memory at once, and iterating through all of it is an expensive process.
(The reason for this is that actually I need to fit thousands of curves simultaneously based on gigabytes of data which I'm reading off disk, and which is therefore sloooooow).
Note that the number of polynomial coefficients will be limited (perhaps 5-10), so an exact fit will be extremely unlikely, but this is ok as I'm trying to find an underlying pattern in data with a lot of random noise.
I understand how one can use a genetic algorithm to fit a curve to a dataset, but this requires many passes through the sample data, and thus isn't practical for my application.
Is there a way to fit a curve with a single pass of the data, where the state that must be maintained from sample to sample is minimal?
I should add that the nature of the data is that the points may lie anywhere on the X axis between 0.0 and 1.0, but the Y values will always be either 1.0 or 0.0.
So, in Java, I'm looking for a class with the following interface:
public interface CurveFit {
public void addData(double x, double y);
public List<Double> getBestFit(); // Returns the polynomial coefficients
}
The class that implements this must not need to keep much data in its instance fields, no more than a kilobyte even for millions of data points. This means that you can't just store the data as you get it to do multiple passes through it later.
edit: Some have suggested that finding an optimal curve in a single pass may be impossible, however an optimal fit is not required, just as close as we can get it in a single pass.
The bare bones of an approach might be if we have a way to start with a curve, and then a way to modify it to get it slightly closer to new data points as they come in - effectively a form of gradient descent. It is hoped that with sufficient data (and the data will be plentiful), we get a pretty good curve. Perhaps this inspires someone to a solution.
Yes, it is a projection. For
y = X beta + error
where lowercased terms are vectors, and X is a matrix, you have the solution vector
\hat{beta} = inverse(X'X) X' y
as per the OLS page. You almost never want to compute this directly but rather use LR, QR or SVD decompositions. References are plentiful in the statistics literature.
If your problem has only one parameter (and x is hence a vector as well) then this reduces to just summation of cross-products between y and x.
If you don't mind that you'll get a straight line "curve", then you only need six variables for any amount of data. Here's the source code that's going into my upcoming book; I'm sure that you can figure out how the DataPoint class works:
Interpolation.h:
#ifndef __INTERPOLATION_H
#define __INTERPOLATION_H
#include "DataPoint.h"
class Interpolation
{
private:
int m_count;
double m_sumX;
double m_sumXX; /* sum of X*X */
double m_sumXY; /* sum of X*Y */
double m_sumY;
double m_sumYY; /* sum of Y*Y */
public:
Interpolation();
void addData(const DataPoint& dp);
double slope() const;
double intercept() const;
double interpolate(double x) const;
double correlate() const;
};
#endif // __INTERPOLATION_H
Interpolation.cpp:
#include <cmath>
#include "Interpolation.h"
Interpolation::Interpolation()
{
m_count = 0;
m_sumX = 0.0;
m_sumXX = 0.0;
m_sumXY = 0.0;
m_sumY = 0.0;
m_sumYY = 0.0;
}
void Interpolation::addData(const DataPoint& dp)
{
m_count++;
m_sumX += dp.getX();
m_sumXX += dp.getX() * dp.getX();
m_sumXY += dp.getX() * dp.getY();
m_sumY += dp.getY();
m_sumYY += dp.getY() * dp.getY();
}
double Interpolation::slope() const
{
return (m_sumXY - (m_sumX * m_sumY / m_count)) /
(m_sumXX - (m_sumX * m_sumX / m_count));
}
double Interpolation::intercept() const
{
return (m_sumY / m_count) - slope() * (m_sumX / m_count);
}
double Interpolation::interpolate(double X) const
{
return intercept() + slope() * X;
}
double Interpolation::correlate() const
{
return m_sumXY / sqrt(m_sumXX * m_sumYY);
}
Why not use a ring buffer of some fixed size (say, the last 1000 points) and do a standard QR decomposition-based least squares fit to the buffered data? Once the buffer fills, each time you get a new point you replace the oldest and re-fit. That way you have a bounded working set that still has some data locality, without all the challenges of live stream (memoryless) processing.
Are you limiting the number of polynomial coefficients (i.e. fitting to a max power of x in your polynomial)?
If not, then you don't need a "best fit" algorithm - you can always fit N data points EXACTLY to a polynomial of N coefficients.
Just use matrices to solve N simultaneous equations for N unknowns (the N coefficients of the polynomial).
If you are limiting to a max number of coefficients, what is your max?
Following your comments and edit:
What you want is a low-pass filter to filter out noise, not fit a polynomial to the noise.
Given the nature of your data:
the points may lie anywhere on the X axis between 0.0 and 1.0, but the Y values will always be either 1.0 or 0.0.
Then you don't need even a single pass, as these two lines will pass exactly through every point:
X = [0.0 ... 1.0], Y = 0.0
X = [0.0 ... 1.0], Y = 1.0
Two short line segments, unit length, and every point falls on one line or the other.
Admittedly, an algorithm to find a good curve fit for arbitrary points in a single pass is interesting, but (based on your question), that's not what you need.
Assuming that you don't know which point should belong to which curve, something like a Hough Transform might provide what you need.
The Hough Transform is a technique that allows you to identify structure within a data set. One use is for computer vision, where it allows easy identification of lines and borders within the field of sight.
Advantages for this situation:
Each point need be considered only once
You don't need to keep a data structure for each candidate line, just one (complex, multi-dimensional) structure
Processing of each line is simple
You can stop at any point and output a set of good matches
You never discard any data, so it's not reliant on any accidental locality of references
You can trade off between accuracy and memory requirements
Isn't limited to exact matches, but will highlight partial matches too.
An approach
To find cubic fits, you'd construct a 4-dimensional Hough space, into which you'd project each of your data-points. Hotspots within Hough space would give you the parameters for the cubic through those points.
You need the solution to an overdetermined linear system. The popular methods are Normal Equations (not usually recommended), QR factorization, and singular value decomposition (SVD). Wikipedia has decent explanations, Trefethen and Bau is very good. Your options:
Out-of-core implementation via the normal equations. This requires the product A'A where A has many more rows than columns (so the result is very small). The matrix A is completely defined by the sample locations so you don't have to store it, thus computing A'A is reasonably cheap (very cheap if you don't need to hit memory for the node locations). Once A'A is computed, you get the solution in one pass through your input data, but the method can be unstable.
Implement an out-of-core QR factorization. Classical Gram-Schmidt will be fastest, but you have to be careful about stability.
Do it in-core with distributed memory (if you have the hardware available). Libraries like PLAPACK and SCALAPACK can do this, the performance should be much better than 1. The parallel scalability is not fantastic, but will be fine if it's a problem size that you would even think about doing in serial.
Use iterative methods to compute an SVD. Depending on the spectral properties of your system (maybe after preconditioning) this could converge very fast and does not require storage for the matrix (which in your case has 5-10 columns each of which are the size of your input data. A good library for this is SLEPc, you only have to find a the product of the Vandermonde matrix with a vector (so you only need to store the sample locations). This is very scalable in parallel.
I believe I found the answer to my own question based on a modified version of this code. For those interested, my Java code is here.