You can have a look at scipy.stats: from pydoc import help from scipy.stats.stats import pearsonr help(pearsonr) >>> Help on function pearsonr in module scipy.stats.stats: pearsonr(x, y) Calculates a Pearson correlation coefficient and the p-value for testing non-correlation. The Pearson correlation coefficient measures the linear relationship between two datasets. Strictly speaking, Pearson’s correlation requires that each … Read more
While the sum of squares algorithm works fine most of the time, it can cause big trouble if you are dealing with very large numbers. You basically may end up with a negative variance… Plus, don’t never, ever, ever, compute a^2 as pow(a,2), a * a is almost certainly faster. By far the best way … Read more
This is kind of overkill but let’s give it a go. First lets use statsmodel to find out what the p-values should be import pandas as pd import numpy as np from sklearn import datasets, linear_model from sklearn.linear_model import LinearRegression import statsmodels.api as sm from scipy import stats diabetes = datasets.load_diabetes() X = diabetes.data y … Read more
See help(Sys.sleep). For example, from ?Sys.sleep testit <- function(x) { p1 <- proc.time() Sys.sleep(x) proc.time() – p1 # The cpu usage should be negligible } testit(3.7) Yielding > testit(3.7) user system elapsed 0.000 0.000 3.704
The answer is to use Welford’s algorithm, which is very clearly defined after the “naive methods” in: Wikipedia: Algorithms for calculating variance It’s more numerically stable than either the two-pass or online simple sum of squares collectors suggested in other responses. The stability only really matters when you have lots of values that are close … Read more
sklearn.linear_model.LinearRegression will do it: from sklearn import linear_model clf = linear_model.LinearRegression() clf.fit([[getattr(t, ‘x%d’ % i) for i in range(1, 8)] for t in texts], [t.y for t in texts]) Then clf.coef_ will have the regression coefficients. sklearn.linear_model also has similar interfaces to do various kinds of regularizations on the regression.
If you’re using scikit-learn you can use sklearn.preprocessing.normalize: import numpy as np from sklearn.preprocessing import normalize x = np.random.rand(1000)*10 norm1 = x / np.linalg.norm(x) norm2 = normalize(x[:,np.newaxis], axis=0).ravel() print np.all(norm1 == norm2) # True
I would use a recursive function. Here’s a (working) example with comments. Hope this works for you! function sampling($chars, $size, $combinations = array()) { # if it’s the first iteration, the first set # of combinations is the same as the set of characters if (empty($combinations)) { $combinations = $chars; } # we’re done if … Read more
Give the basic info, with a clear route for how to go further – update your browser! I think sacrificing functionality for 99% of users to accommodate 1% is sheer bloody mindedness. Sure, it is possible to allow for non-javascript enabled content for every aspect of a website, plus provide the optimal experience – but … Read more
import numpy as np import scipy.stats def mean_confidence_interval(data, confidence=0.95): a = 1.0 * np.array(data) n = len(a) m, se = np.mean(a), scipy.stats.sem(a) h = se * scipy.stats.t.ppf((1 + confidence) / 2., n-1) return m, m-h, m+h You can calculate like this.