You are using predict.segemented incorrectly. Like nearly all the predict() functions, your newdata parameter should be a data.frame, not a vector. Also, it needs to have names that match the variables used in your regression. Try predict.segmented(seg_model, data.frame(x=xtest)) instead. When using a function for the first time, be sure the...

To get the correlation between all row combinations, try cor(t(df1), t(df2)) ...

algorithm,linear-algebra,linear

In your case, since x and y only take values between 0 and 10, brute force algorithm maybe the best option as it takes less time to implement. However, if you have to find all pairs of integral solution (x, y) in a larger range, you really should apply the...

plot([150,360],[260,50],'k-') Should be simple enough to move it up or down if you need to ...

double zmat[N+1][N+1]; This is what is causing you trouble. Declaring a large matrix as a local variable in a function is not a good idea. Local variables are allocated on the stack. Many machines limit the size of the stack to a smallish amount of data. On my machine, the...

matlab,regression,numerical-methods,linear

Judging from the link you provided, and my understanding of your problem, you want to calculate the line of best fit for a set of data points. You also want to do this from first principles. This will require some basic Calculus as well as some linear algebra for solving...

python,pandas,linear,interpolate

Pandas' method='linear' interpolation will do what I call "1D" interpolation If you want to interpolate a "dependent" variable over an "independent" variable, make the "independent" variable; i.e. the Index of a Series, and use the method='index' (or method='values', they're the same) In other words: pd.Series(index=df.size, data=df.cost.values) #Make size the independent...

I've got what you want, so think in this way: you have original price you want to calculate final transferAmount that: transferAmount = originalPrice + fee but fee here depends on transferAmount fee = transferAmount * 0.029 + 2.25 now solve it: transferAmount = originalPrice + transferAmount * 0.029 +...

Looks like you've provided the arguments to polyfit in the wrong order. The first argument takes the values of the x axis which, in this case, correspond to the sampling times given by i. The second argument takes the y values. So you'd want something like: function linCoeffs = getLinearTrend(y)...

The only change you need to make is to return result(value) instead of result -- at the moment you're returning the scale object, which needs to be applied to a value to give the mapped value.

While using a separate background xml file is a good idea, i would suggest that you use a drawable instead for background. So you can design your background image with the 'sharp tip' on the left in photoshop or paint.net and set the property android:[email protected]/background_design where 'background design' is the...

I had done something similar. You will need to modify the loop for your need. Let me know if you need help with that. vars=colnames(mydata)[-1] for (i in vars) { for (j in vars) { if (i != j) { factor= paste(i,j,sep='*')} lm.fit <- lm(paste("Sales ~", factor), data=mydata) print(summary(lm.fit)) }}...

Your indentation is misleading. What you meant is: int reverse(int a[], int key) { int i = sizeof(a) - 1; while (i >= 0) { if (a[i] == key) { return i; } i--; } return -1; } What you wrote is equivalent to: int reverse(int a[], int key) {...

In the car package you can use the 'scatterplotMatrix' function. Here's an example using the Prestige dataset in that package: library(car) scatterplotMatrix(~prestige +income +education + women, data= Prestige)...

(A * x) - (y / A) = B Assume A is not zero. Solving for X: AAX - y = B * A AAX = (B * A) + y X = [(B * A) + y]/[A * A]...

r,mathematical-optimization,linear

You can use constrOptim with cost function least square and contraints defined such that ui %*% a >= ci. Suppose n=3. You want constraints such as: a1 >= 0 a2 >= 0 a3 >= 0 -a1 -a2 -a3 >= -1 Thus you have to provide constrOptim the following parameters: ui...

Your current issue is this line array = [myNumber] All that you are doing is setting array to be a list of 1 element repeatedly with this line, as opposed to adding each element to array. Use the increment operator += instead, so that you are actually adding each element...

What is it used for? It is used for open-addressed hashing, i.e. memory efficient set and map/dictionary behavior. What does it do? Defines an algorithm for deciding where to store and find members of a set, or where to store and find keys and values of a map. Which...

r,loops,dataframes,regression,linear

This is a slight modification of @BondedDust's comment. models <- sapply(unique(as.character(df$country)), function(cntry)lm(BirthRate~US.,df,subset=(country==cntry)), simplify=FALSE,USE.NAMES=TRUE) # to summarize all the models lapply(models,summary) # to run anova on all the models lapply(models,anova) This produces a named list of models, so you could extract the model for Aruba as: models[["Aruba"]] ...

gnu,linear-programming,linear,ampl,glpk

You can do it as follows: s.t. rest: x['Persones'] >= 7; ...

python,syntax,regression,linear

Your last variable, or rather pair of variables, is invalid. Syy/Sxx If that is supposed to be a single variable, know that you cannot have / in your variable name. Python identifiers and keywords Identifiers (also referred to as names) are described by the following lexical definitions: identifier ::= (letter|"_")...

c++,linear-algebra,equation,linear

You don't allocate the second dimension for the EquationHolder. Since is a 2D matrix you have to allocate the second dimension also. Change your double for loop to the following: float ** EquationHolder=new float *[3]; for (int i=0; i<NumEquations; i++) { EquationHolder[i] = new float[3]; cout<<"Please Enter The Information Of...

Remove these lines: string = array; search = a; Edit: These two lines set the reference of string to an empty String array (array) and the reference of search to an empty String (a). This means that string and search now have the same content as array and a, which...

What you are proposing to do is a massively bad idea, so much so that I'm reluctant to show you how to do it. The reason is that for OLS, assuming the residuals are normally distributed with constant variance, then the parameter estimates follow a multivariate t-distribution and we can...

Changes of scale will not change the correlation coef: > out2[,2] <- out[,2]*10 > cor(out2) [,1] [,2] [1,] 1.0 0.5 [2,] 0.5 1.0 plot(out2) > lm(out2[,2]~out2[,1]) Call: lm(formula = out2[, 2] ~ out2[, 1]) Coefficients: (Intercept) out2[, 1] -5.732e-16 5.000e+00 ...

math,integer,equation,linear,diophantine

existence Bezout's identity tells you indeed that with the greatest common divisor (gcd) d of a and b, we have : i) d is the smallest positive integer that can be written as ax + by, and ii) every integer of the form ax + by is a multiple of...

You could try using apply() # data df = data.frame(a,b,c,d,e,f) # apply function out = t(apply(df, 1, function(x){ coefs = matrix( c(x['a'], x['d'], x['b'], x['e']), 2, 2); ys = array(c(x['c'],x['f']),2); out = solve(coefs, ys); names(out) = c('P','Q'); out})) Or else using sapply() out = t(sapply(seq(100), function(i){ coefs = matrix(c(a[i], d[i],...

python,arrays,numpy,python-imaging-library,linear

It does pretty much what you'd expect: it tests the array img to see if it's greater than 0. But since it's a NumPy array, this is an elementwise comparison: each element is compared to 0. This creates an array of Booleans with the result of the comparison for each...