Re: [R] Variable name as string
So here is the next version.
Why does the intercept needs lower.tail=TRUE to give the same result as
summary() for value=0?
# See Verzani, simpleR (pdf), p. 80
coeff.test - function(lm.result, idx, value) {
# idx = 1 is the intercept, idx1 the other coefficients
# null hypothesis: coeff = value
# alternative hypothesis: coeff != value
coeff - coefficients(lm.result)[idx]
SE - coefficients(summary(lm.result))[idx,Std. Error]
n - df.residual(lm.result)
t - (coeff - value )/SE
if (idx == 1) {
2 * pt(t,n,lower.tail=TRUE) # times two because problem is two-sided
} else {
2 * pt(t,n,lower.tail=FALSE)
}
}
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[R] Variable name as string
Hello,
from Verzani, simpleR (pdf), p. 80, I created the following function to
test the coefficient of lm() against an arbitrary value.
coeff.test - function(lm.result, var, coeffname, value) {
# null hypothesis: coeff = value
# alternative hypothesis: coeff != value
es - resid(lm.result)
coeff - (coefficients(lm.result))[[coeffname]]
# degrees of freedom = length(var) - number of coefficients?
n - df.residual(lm.result)
s - sqrt( sum( es^2 ) / n )
SE - s/sqrt(sum((var - mean(var))^2))
t - (coeff - value )/SE
2 * pt(t,n,lower.tail=FALSE) # times two because problem is two-sided
}
E.g. it needs to be called coeff.test(lm(N ~ D), D, D, 70) if I want
to test the probability slope of the linear model being 70.
Is there a more elegant way to avoid passing D and D both as
parameters?
Also, as a non-professional, I would like to know whether the function
is valid for all coefficients of lm(), e.g. coeff.test(lm(N ~ D + H), H,
H, 70). I am aware that Verzani gives a different formula for testing
the intercept.
Thanks!
Jan Rheinländer
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Re: [R] Variable name as string
coeffname - deparse(substitute(var)) Is that what you wanted? Yes! That gets rid of the extra parameter. Thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Programming: loop versus vector oriented
Hello Petr, thank you for your ideas. The split() looks most realistic. What about this idea: 1. Define three functions Refun1, Refun2, Refun3 for the three different sections of the calculations (same as you suggested) 2. lambda = (Re = 2320) * Refun1(Re) + ((Re 2320) (Re 65 * dk)) * Refun2(Re) etc. But my thought is that probably the values of RefunXYZ will be calculated for every value of Re, even if the condition (Re = 2320) is FALSE (= 0). So that would give a lot of unnecessary function evaluations. Regards, Jan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Programming: loop versus vector oriented
Hello Petr, but I think this is how your code really works. Did you try it? it does, but the R documentation says somewhere: Warning: for() loops are used in R code much less often than in compiled languages. Code that takes a `whole object' view is likely to be both clearer and faster in R. So I am wondering in what way the whole object view could be applied to my function. Best regards, Jan The function should work on the nth elements of the two input vectors and put the result into the nth element of the output vector. So it would work like c - a + b, only instead of '+' there are more complex calculations. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Programming: loop versus vector oriented
Hello Petr, If you want to get results of your function for a vector of reynolds and dk you can use function outer and probably get rid of for cycle in the function. outer(c(100, 530,2410), c(10, 150,200),lambda_wall) [,1] [,2] [,3] [1,] 0.640 0.6400 0.6400 [2,] 0.1207547 0.12075472 0.12075472 [3,] 0.1081338 0.04515774 0.04515774 that gives me an array as an answer (and does more calculations than necessary, in this case 9 instead of 3). The result should be a vector. The function should work on the nth elements of the two input vectors and put the result into the nth element of the output vector. So it would work like c - a + b, only instead of '+' there are more complex calculations. Best regards, Jan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Programming: loop versus vector oriented
Dear all,
I am new to R and to it's programming philosophy. The following function
is supposed to work on a vector, but I can't figure out how to do that
without looping through every element of it. Is there a more elegant
way?
Note: I have shortened it, so it is NOT correct from the pipe hydraulics
point of view
# Calculate wall friction factor
# reynolds: Reynolds number
# dk: relative roughness of pipe
lambda_wall - function (reynolds, dk) {
z - 1
result - 0
for (Re in reynolds) {
if (Re = 2320) {
# Laminar flow
lambda - 64/Re
} else if (Re 65 * dk[z]) {
# Turbulent flow
if (Re 1e+5) {
lambda - 0.3164 / sqrt(sqrt(Re))
} else {
lambda - 0.309/(log10(Re/7))^2
}
} else {
# Intermediate area
lambdanew - 1 / (2 * log10(3.71 * dk[z]))^2 # Start value
iter - 0
repeat {
lambda - lambdanew
lambdanew - 1 / (2 * log10(2.51/(Re * sqrt(lambda)) + 0.27/dk[z]))^2
iter - iter + 1
if ((abs(lambdanew - lambda) 0.001) || (iter 100)) break
}
lambda = lambdanew
}
result[z] - lambda
z - z + 1
}
result
} # lambda_wall()
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Re: [R] Calculating with tolerances (error propagation)
Hello Bernardo, - If I understood your problem this script solve your problem: q-0.15 + c(-.1,0,.1) h-10 + c(-.1,0,.1) 5*q*h [1] 2.475 7.500 12.625 - OK, this solves the simple example. But what if the example is not that simple. E.g. P = 5 * q/h Here, to get the maximum tolerances for P, we need to divide the maximum value for q by the minimum value for h, and vice versa. Is there any way to do this automatically, without thinking about every single step? There is a thing called interval arithmetic (I saw it as an Octave package) which would do something like this. I would have thought that tracking how a (measuring) error propagates through a complex calculation would be a standard problem of statistics?? In other words, I am looking for a data type which is a number with a deviation +- somehow attached to it, with binary operators that automatically knows how to handle the deviation. Thank you, Jan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Calculating with tolerances
Dear list, I am from an engineering background, accustomed to work with tolerances. For example, I have measured Q = 0.15 +- 0.01 m^3/s H = 10 +- 0.1 m and now I want to calculate P = 5 * Q * H and get a value with a tolerance +- What is the elegant way of doing this in R? Thank you, Jan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
