average,regression,effect,spss,coefficients

If you test an interaction hypothesis, you have to include a number of terms in your model. In this case, you would have to include: Base effect of price Base effects of brands (dummies) Interaction effects of brand dummies * price. Since you have 5 brands, you will have to...

java,performance,ocr,correlation,coefficients

Nowadays, it's hard to find a CPU with a single core (even in mobiles). As the tasks are nicely separated, you can do it with a few lines only. So I'd go for it, though the gain is limited. In case you really mean cross-correlation, then a transform like DFT...

language-agnostic,metrics,coefficients

As originally defined by Jaccard, the similarity coefficient is the size of the intersection divided by the size of the union. Since both are sizes, a negative result obviously isn't possible. What you show in the question looks sort of like the Jaccard similarity for a bit vector. However, for...

r,table,scientific-notation,stargazer,coefficients

Here's a reproducible example: m1 <- lm(Sepal.Length ~ Petal.Length*Sepal.Width, transform(iris, Sepal.Length = Sepal.Length+1e6, Petal.Length=Petal.Length*10, Sepal.Width=Sepal.Width*100)) # Coefficients: # (Intercept) Petal.Length Sepal.Width Petal.Length:Sepal.Width # 1.000e+06 7.185e-02 8.500e-03 -7.701e-05 I don't believe stargazer has easy support for this. You could try other alternatives like xtable or any of the many options here...

python,scikit-learn,logistic-regression,coefficients

The feature names can be access from vect using the get_feature_names method. You can zip them to the coefficients like this for example: zip(vect.get_feature_names(),d.coef_[0]) This returns a tuple with (token, coefficient)...

function,transfer,coefficients

let W be the tf, I the input and O the output. in laplace IW=O, then you just need to do O/I and simplify it to get the W in the form you want,even if it would be better and more general to write it in the bode form....

python,sympy,polynomials,coefficients

I don't know why sympy truncates small coefficients when constructing a polynomial over reals, but it doesn't do this over rationals. So as a workaround, you can construct a polynomial with domain='QQ', extract coefficients, and then convert back to floats. Example using your polynomial: import sympy z_s = symbols('z_s') f...

r,expression,offset,lm,coefficients

One way of handling this would be to calculate a new column with your total offset and remove the columns used in your offset from the data set: # create copy of data withou columns used in offset dat <- df[-match(inputs_fix, names(df))] # calculate offset dat$offset <- 0 for (i...

r,logistic-regression,survival-analysis,coefficients

Another approach can be with the use of the summary function. You can see that with summary the coefficients of the model are taken as a matrix. > is.matrix(summary(B)$coefficients) [1] TRUE At this point you can store summary(B)$coefficients in an object and then subset it as you wish. summary(B)$coefficients[1,1] ...

r,logistic-regression,coefficients

Another solution I discovered is converting the results to a data frame, then extracting the row names as follows: > allpredsincld<-as.data.frame(summary(step1)$coefficients) > allpredsincld Estimate Std. Error z value Pr(>|z|) (Intercept) -7.998346 1.216048 -6.577327 4.789808e-11 i1 3.928425 0.695920 5.644939 1.652402e-08 then: > allpredsincld<-allpredsincld[-1,] > allpredsincld Estimate Std. Error z value Pr(>|z|)...

You are using glm(...) incorrectly, which IMO is a much bigger problem than offsets. The main underlying assumption in least squares regression is that the error in the response is normally distributed with constant variance. If the error in Y is normally distributed, then log(Y) most certainly is not. So,...

python,grouping,sympy,coefficients

collect is the right tool for this: Example from the above link: >>> collect(a*x**2 + b*x**2 + a*x - b*x + c, x) c + x**2*(a + b) + x*(a - b) In your case (coefficients are not normalized here): col = collect(Det, [X1, Y1], evaluate=False) A = col[X1] B...

After much trial and error I worked out how to edit the source code for the coefplot multiplot function. Creating our test datasets. library(coefplot) model1 <- lm(price ~ carat + cut, data=diamonds) model2 <- lm(price ~ carat + cut + color, data=diamonds) model3 <- lm(price ~ carat + color, data=diamonds)...