linear-regression

I fit a linear regression model on 75% of my data set that includes ~11000 observations and 143 variables: gl.fit <- lm(y[1:ceiling(length(y)*(3/4))] ~ ., data= x[1:ceiling(length(y)*(3/4)),]) #3/4 for training , and I got an R^2 of 0.43. I then tried predicting on my test data using the rest of the data: ytest=y[(ceiling(length(y)*(3/4))+1):length(y)] x.test <- cbind(1,x[(ceiling(length(y)*(3/4))+1):length(y),]) #The rest for test yhat <- as.matrix(x.test)%*%gl.fit$coefficients #Calculate the predicted values I now would like to calculate the R^2 value on my test data. Is there any easy way

My goal is to solve: Kc=y with the pseudo-inverse (i.e. minimum norm solution ): c=K^{+}y such that the model is (hopefully) high degree polynomial model f(x) = sum_i c_i x^i . I am specially interested in the underdetermined case where we have more polynomial features than data (few equation too many variables/unknowns) columns = deg+1 > N = rows . Note K is the vandermode matrix of polynomial features. I was initially using the python function np.linalg.pinv but then I noticed something funky was going on as I noted here: Why do different methods for solving Xc=y in python give different

I have the following R code library(forecast) value <- c(1.2, 1.7, 1.6, 1.2, 1.6, 1.3, 1.5, 1.9, 5.4, 4.2, 5.5, 6, 5.6, 6.2, 6.8, 7.1, 7.1, 5.8, 0, 5.2, 4.6, 3.6, 3, 3.8, 3.1, 3.4, 2, 3.1, 3.2, 1.6, 0.6, 3.3, 4.9, 6.5, 5.3, 3.5, 5.3, 7.2, 7.4, 7.3, 7.2, 4, 6.1, 4.3, 4, 2.4, 0.4, 2.4) sensor<-ts(value,frequency=24) fit <- auto.arima(sensor) LH.pred<-predict(fit,n.ahead=24) plot(sensor,ylim=c(0,10),xlim=c(0,5),type="o", lwd="1") lines(LH.pred$pred,col="red",type="o",lwd="1") grid() The resulting graph is But I am not satisfied with the prediction. Is there any way to make the prediction look

I'm trying to write out a bit of code for the gradient descent algorithm explained in the Stanford Machine Learning lecture ( lecture 2 at around 25:00 ). Below is the implementation I used at first, and I think it's properly copied over from the lecture, but it doesn't converge when I add large numbers ( >8 ) to the training set. I'm inputting a number X , and the point (X,X) is added to the training set, so at the moment, I'm only trying to get it to converge to y=ax+b where a=1=theta\[1\] and b=0=theta\[0\] . The training set is the array x and y , where (x[i],y[i]) is a point. void train()

Can anyone explain to me the difference between ols in statsmodel.formula.api versus ols in statsmodel.api? Using the Advertising data from the ISLR text, I ran an ols using both, and got different results. I then compared with scikit-learn's LinearRegression. import numpy as np import pandas as pd import statsmodels.formula.api as smf import statsmodels.api as sm from sklearn.linear_model import LinearRegression df = pd.read_csv("C:\...\Advertising.csv") x1 = df.loc[:,['TV']] y1 = df.loc[:,['Sales']] print "Statsmodel.Formula.Api Method" model1 = smf.ols(formula='Sales ~ TV', data=df).fit()

I'm new to Python and trying to perform linear regression using sklearn on a pandas dataframe. This is what I did: data = pd.read_csv('xxxx.csv') After that I got a DataFrame of two columns, let's call them 'c1', 'c2'. Now I want to do linear regression on the set of (c1,c2) so I entered X=data['c1'].values Y=data['c2'].values linear_model.LinearRegression().fit(X,Y) which resulted in the following error IndexError: tuple index out of range What's wrong here? Also, I'd like to know visualize the result make predictions based on the result? I've searched and browsed a large number of sites but

I need to colour datapoints that are outside of the the confidence bands on the plot below differently from those within the bands. Should I add a separate column to my dataset to record whether the data points are within the confidence bands? Can you provide an example please? Example dataset: ## Dataset from http://www.apsnet.org/education/advancedplantpath/topics/RModules/doc1/04_Linear_regression.html ## Disease severity as a function of temperature # Response variable, disease severity diseasesev<-c(1.9,3.1,3.3,4.8,5.3,6.1,6.4,7.6,9.8,12.4) # Predictor variable, (Centigrade) temperature<