linear-regression
predict.lm() with an unknown factor level in test data
You have to remove the extra levels before any calculation, like: > id <- which(!(foo.new$predictor %in% levels(foo$predictor))) > foo.new$predictor[id] <- NA > predict(model,newdata=foo.new) 1 2 3 4 -0.1676941 -0.6454521 0.4524391 NA This is a more general way of doing it, it will set all levels that do not occur in the original data to NA. … Read more
Linear Regression with a known fixed intercept in R
You could subtract the explicit intercept from the regressand and then fit the intercept-free model: > intercept <- 1.0 > fit <- lm(I(x – intercept) ~ 0 + y, lin) > summary(fit) The 0 + suppresses the fitting of the intercept by lm. edit To plot the fit, use > abline(intercept, coef(fit)) P.S. The variables … Read more
How to overplot a line on a scatter plot in python?
import numpy as np from numpy.polynomial.polynomial import polyfit import matplotlib.pyplot as plt # Sample data x = np.arange(10) y = 5 * x + 10 # Fit with polyfit b, m = polyfit(x, y, 1) plt.plot(x, y, ‘.’) plt.plot(x, b + m * x, ‘-‘) plt.show()
Linear regression with matplotlib / numpy
arange generates lists (well, numpy arrays); type help(np.arange) for the details. You don’t need to call it on existing lists. >>> x = [1,2,3,4] >>> y = [3,5,7,9] >>> >>> m,b = np.polyfit(x, y, 1) >>> m 2.0000000000000009 >>> b 0.99999999999999833 I should add that I tend to use poly1d here rather than write out … Read more
lm function in R does not give coefficients for all factor levels in categorical data
GE is dropped, alphabetically, as the intercept term. As eipi10 stated, you can interpret the coefficients for the other levels in states with GE as the baseline (statesLA = 0.1 meaning LA is, on average, 0.1x more than GE). EDIT: To respond to your updated question: If you include all of the levels in a … Read more
How to Loop/Repeat a Linear Regression in R
You want to run 22,000 linear regressions and extract the coefficients? That’s simple to do from a coding standpoint. set.seed(1) # number of columns in the Lung and Blood data.frames. 22,000 for you? n <- 5 # dummy data obs <- 50 # observations Lung <- data.frame(matrix(rnorm(obs*n), ncol=n)) Blood <- data.frame(matrix(rnorm(obs*n), ncol=n)) Age <- sample(20:80, … Read more
Is there a better alternative than string manipulation to programmatically build formulas?
reformulate will do what you want. reformulate(termlabels = c(‘x’,’z’), response=”y”) ## y ~ x + z Or without an intercept reformulate(termlabels = c(‘x’,’z’), response=”y”, intercept = FALSE) ## y ~ x + z – 1 Note that you cannot construct formulae with multiple reponses such as x+y ~z+b reformulate(termlabels = c(‘x’,’y’), response = c(‘z’,’b’)) z … Read more