linear-regression

问题 While looking for a R related solution I found some inconsistency between R and SPSS (ver. 24) in computing standardized residuals in a simple linear model. It appears that what SPSS calls standarized residuals matches R studentized residuals I'm far for assuming there is a software bug somewhere, but clearly things differ between those two programs. Have a look at this example #generate data in R set.seed(111) y = rnorm(20, 0, 1) x = rnorm(20, 1, 1) #calculate and standarized residuals

问题 I have this code: library(biglm) library(ff) myData <- read.csv.ffdf(file = "myFile.csv") testData <- read.csv(file = "test.csv") form <- dependent ~ . model <- biglm(form, data=myData) predictedData <- predict(model, newdata=testData) the model is created without problems, but when I make the prediction... it runs out of memory: unable to allocate a vector of size xx.x MB some hints? or how to use ff to reserve memory for predictedData variable? 回答1: I have not used biglm package before.

问题 Hi I have a linear regression model that i am trying to optimise. I am optimising the span of an exponential moving average and the number of lagged variables that I use in the regression. However I keep finding that the results and the calculated mse keep coming up with different final results. No idea why can anyone help? Process after starting loop: 1. Create new dataframe with three variables 2. Remove nil values 3. Create ewma's for each variable 4. Create lags for each variable 5. Drop

问题 In a recent homework, I was asked to perform a Bayesian fit over a set of data a and b using a Metropolis algorithm. The relationship between a and b is given: e(t) = e_0*cos(w*t) w = 2 * pi The Metropolis algorithm is (it works fine with other fit): def metropolis(logP, args, v0, Nsteps, stepSize): vCur = v0 logPcur = logP(vCur, *args) v = [] Nattempts = 0 for i in range(Nsteps): while(True): #Propose step: vNext = vCur + stepSize*np.random.randn(*vCur.shape) logPnext = logP(vNext, *args)

问题 I have data and I expect several linear correlations of the form y_i = a_i + b_i * t_i, i = 1 .. N where N is a priori unknown. The short version of the question is: Given a fit how can I extract N ? how can I extract the equations? In the reproducible example below, I have data (t,y) with corresponding parameters p1 (levels p1_1 , p1_2 ) and p2 (levels p2_1 , p2_2 , p2_3 ). Thus the data looks like (t, y, p1, p2) which has at most 2*3 different best-fit lines and the linear fit from has then

问题 Ignore the red fitted curve first. I'd like to get a curve to the blue datapoints. I know the first part (up to y~200 in this case) is linear, then a different curve (combination of two logarithmic curves but could also be approximated differently) and then it saturates at about 250 or 255. I tried it like this: func = fittype('(x>=0 & x<=xTrans1).*(A*x+B)+(x>=xTrans1 & x<=xTrans2).*(C*x+D)+(x>=xTrans2).*E*255'); freg = fit(foundData(:,1), foundData(:,2), func); plot(freg, foundData(:,1),

问题 I have found some odd behavior with predict and the svyglm object from the survey package. If your newdata in predict has a factor/character with one level it spits out error: Error in `contrasts<-`(`*tmp*`, value = contr.funs[1 + isOF[nn]]) : contrasts can be applied only to factors with 2 or more levels This error makes sense if I was putting a one level variable as the predictor for a model, but for newdata I don't see the problem. With regular glm this works fine. MRE: library(survey)

问题 I am suppose to use the predict function to predict when fjbjor is 5.5 and I always get this warning message and I have tried many ways but it always comes so is there anyone who can see what I am doing wrong here This is my code fit.lm <- lm(fjbjor~amagn, data=bjor) summary(fit.lm) new.bjor<- data.frame(fjbjor=5.5) predict(fit.lm,new.bjor) and this comes out 1 2 3 4 5 6 7 8 9 10 11 5.981287 2.864521 9.988559 5.758661 4.645530 2.419269 4.645530 5.313409 6.871792 3.309773 4.200278 12 13 14 15

问题 features = [tf.contrib.layers.real_valued_column("x", dimension=1)] estimator = tf.contrib.learn.LinearRegressor(feature_columns=features) y = np.array([0., -1., -2., -3.]) input_fn = tf.contrib.learn.io.numpy_input_fn({"x":x}, y, batch_size=4, num_epochs=1000) estimator.fit(input_fn=input_fn, steps=1000) For example, do these "steps=1000" and "num_epochs=1000" mean exactly the same thing? If yes, why does it need to be duplicated? If not, can i set these two parameters differently? 回答1: Here

问题 I am running the example multiple linear regression for Flink (0.10-SNAPSHOT). I can't figure out how to extract the weights (e.g. slope and intercept, beta0-beta1, what ever you want to call them). I'm not super seasoned in Scala, that is probably half my problem. Thanks for any help any one can give. object Job { def main(args: Array[String]) { // set up the execution environment val env = ExecutionEnvironment.getExecutionEnvironment val survival = env.readCsvFile[(String, String, String,