[R] R Newbie Question
Hello list,
I just started R today and tried something quite simple. I wanted to
create a colored plot and eventually after hours of fiddling around got
it working. However, my solution seems very suboptimal and I'd really
appreciate your hints on how to improve. I believe that R already offers
many functions I coded (e.g. distance between two vectors, vector
length, vector normalization and so on). I generally didn't even figure
out how to create a simple vector or how to extract a row from a matrix
so the result is a vector (to get scalars, I use the sum functions,
which is an incredibly ugly workaround).
To sum the problem up: one has a binary star system with a large star
(e.g. red giant) and a small star (e.g. white dwarf). Gravitation
between them is directly proportional to the mass and indirectly
proportional to the square of the distance. If correctly plotted, one
should be able to see the inner lagrange point L1 which is the point
where the gravitational potentials of the stars cancel out, e.g. an
object would not be attracted to any star. Well, enough background
information, here's my rookie code - please feel free to comment on
anything :-)
Kind regards,
Johannes
star1center = vector(numeric, 2)
star1center[1] = -0.5
star1center[2] = 0
star1mass = 30
star2center = vector(numeric, 2)
star2center[1] = 0.5
star2center[2] = 0
star2mass = 1
sqr = function(x) {
return(x * x)
}
distance = function(a, b) {
return(sqrt(sqr(a[1] - b[1]) + sqr(a[2] - b[2])))
}
len = function(x) {
return(sqrt(sqr(x[1]) + sqr(x[2])))
}
norm = function(x) {
return(x / len(x))
}
gravitation = function(invecx, invecy) {
invec = vector(numeric, 2)
invec[1] = invecx
invec[2] = invecy
vec1 = star1mass * norm(star1center - invec) / sqr(distance(invec,
star1center))
vec2 = star2mass * norm(star2center - invec) / sqr(distance(invec,
star2center))
return(len(vec1 + vec2))
}
vmin = -1
vmax = 1
step = 0.1
vals = ((vmax - vmin) / step) + 1
xvals = seq(vmin, vmax, step)
yvals = seq(vmin, vmax, step)
a = expand.grid(seq(vmin, vmax, step), seq(vmin, vmax, step))
b = matrix(seq(1, vals*vals), vals)
for (x in 1:vals) {
for (y in 1:vals) {
b[x, y] = gravitation(sum(a[x,][1]), sum(a[y,][1]))
}
}
filled.contour(xvals, yvals, z = b, color = heat.colors, ylim = c(-1,
1), xlim = c(-1, 1), zlim = c(0, 100), nlevels = 100)
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Re: [R] R Newbie Question
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Johannes Bauer
Sent: Monday, October 20, 2008 10:42 AM
To: r-help@r-project.org
Subject: [R] R Newbie Question
Hello list,
I just started R today and tried something quite simple. I wanted to
create a colored plot and eventually after hours of fiddling
around got
it working. However, my solution seems very suboptimal and I'd really
appreciate your hints on how to improve. I believe that R
already offers
many functions I coded (e.g. distance between two vectors, vector
length, vector normalization and so on). I generally didn't
even figure
out how to create a simple vector
See ?c (i.e. type ?c without quotes at R prompt)
star1center = c(-0.5, 0)
or how to extract a row
from a matrix
so the result is a vector (to get scalars, I use the sum functions,
which is an incredibly ugly workaround).
In your code a[] is not a matrix, it is a data frame. However, you should be
able to index it just like you would a matrix. For your code,
b[x, y] = gravitation(a[x,1], a[y,1])
Without the need for the sum() function.
To sum the problem up: one has a binary star system with a large star
(e.g. red giant) and a small star (e.g. white dwarf). Gravitation
between them is directly proportional to the mass and indirectly
proportional to the square of the distance. If correctly plotted, one
should be able to see the inner lagrange point L1 which is the point
where the gravitational potentials of the stars cancel out, e.g. an
object would not be attracted to any star. Well, enough background
information, here's my rookie code - please feel free to comment on
anything :-)
Kind regards,
Johannes
snip
I have edited your code to simplify some of your functions and use vectorized
computations. I'm sure others could provide even more optimized code. I will
leave it to others to respond concerning already written functions for norm,
len, etc. You could try using the various search functions with key terms
(like distance) to see what may be available.
star1center = c(-0.5, 0)
star1mass = 30
star2center = c(0.5, 0)
star2mass = 1
sqr = function(x) {
return(x * x)
}
distance = function(a, b) {
return(sum((a-b)^2)^0.5)
}
len = function(x) {
return(sum(x^2)^0.5)
}
norm = function(x) {
return(x / len(x))
}
gravitation = function(invecx, invecy) {
invec = c(invecx, invecy)
vec1 = star1mass * norm(star1center - invec) /
sqr(distance(invec,star1center))
vec2 = star2mass * norm(star2center - invec) /
sqr(distance(invec,star2center))
return(len(vec1 + vec2))
}
vmin = -1
vmax = 1
step = 0.1
vals = ((vmax - vmin) / step) + 1
xvals = seq(vmin, vmax, step)
yvals = seq(vmin, vmax, step)
a = expand.grid(seq(vmin, vmax, step), seq(vmin, vmax, step))
b = matrix(seq(1, vals*vals), vals)
for (x in 1:vals) {
for (y in 1:vals) {
b[x, y] = gravitation(a[x,1], a[y,1])
}
}
filled.contour(xvals, yvals, z = b, color = heat.colors, ylim = c(-1,
1), xlim = c(-1, 1), zlim = c(0, 100), nlevels = 100)
Hope this is helpful,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R Newbie Question
tIs a bit neater to create list objects for each of
star1 and star2 and then, since the various functions
are not used outside of gravitation stick them right
in gravitation. Also we can use distance for len
as well by giving it a default second arg of 0 and
sqr does not seem to do much so we can eliminate
that too.
Making the vec computation a function eliminates
some redundant code and using outer eliminates
the two loops.
star1 - list(mass = 30, center = c(-0.5, 0))
star2 - list(mass = 1, center = c(0.5, 0))
gravitation2 - function(v1, v2) {
distance - function(a, b = 0) sqrt(sum((a-b)^2))
norm - function(x) x/distance(x)
vec - function(x) with(x, mass * norm(center - v) / distance(v,
center)^2)
v - c(v1, v2)
distance(vec(star1) + vec(star2))
}
xvals2 - yvals2 - seq(vmin, vmax, step)
b2 - outer(xvals, yvals, Vectorize(gravitation2))
On Mon, Oct 20, 2008 at 1:41 PM, Johannes Bauer [EMAIL PROTECTED] wrote:
Hello list,
I just started R today and tried something quite simple. I wanted to
create a colored plot and eventually after hours of fiddling around got
it working. However, my solution seems very suboptimal and I'd really
appreciate your hints on how to improve. I believe that R already offers
many functions I coded (e.g. distance between two vectors, vector
length, vector normalization and so on). I generally didn't even figure
out how to create a simple vector or how to extract a row from a matrix
so the result is a vector (to get scalars, I use the sum functions,
which is an incredibly ugly workaround).
To sum the problem up: one has a binary star system with a large star
(e.g. red giant) and a small star (e.g. white dwarf). Gravitation
between them is directly proportional to the mass and indirectly
proportional to the square of the distance. If correctly plotted, one
should be able to see the inner lagrange point L1 which is the point
where the gravitational potentials of the stars cancel out, e.g. an
object would not be attracted to any star. Well, enough background
information, here's my rookie code - please feel free to comment on
anything :-)
Kind regards,
Johannes
star1center = vector(numeric, 2)
star1center[1] = -0.5
star1center[2] = 0
star1mass = 30
star2center = vector(numeric, 2)
star2center[1] = 0.5
star2center[2] = 0
star2mass = 1
sqr = function(x) {
return(x * x)
}
distance = function(a, b) {
return(sqrt(sqr(a[1] - b[1]) + sqr(a[2] - b[2])))
}
len = function(x) {
return(sqrt(sqr(x[1]) + sqr(x[2])))
}
norm = function(x) {
return(x / len(x))
}
gravitation = function(invecx, invecy) {
invec = vector(numeric, 2)
invec[1] = invecx
invec[2] = invecy
vec1 = star1mass * norm(star1center - invec) / sqr(distance(invec,
star1center))
vec2 = star2mass * norm(star2center - invec) / sqr(distance(invec,
star2center))
return(len(vec1 + vec2))
}
vmin = -1
vmax = 1
step = 0.1
vals = ((vmax - vmin) / step) + 1
xvals = seq(vmin, vmax, step)
yvals = seq(vmin, vmax, step)
a = expand.grid(seq(vmin, vmax, step), seq(vmin, vmax, step))
b = matrix(seq(1, vals*vals), vals)
for (x in 1:vals) {
for (y in 1:vals) {
b[x, y] = gravitation(sum(a[x,][1]), sum(a[y,][1]))
}
}
filled.contour(xvals, yvals, z = b, color = heat.colors, ylim = c(-1,
1), xlim = c(-1, 1), zlim = c(0, 100), nlevels = 100)
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R Newbie Question/Data Model
[EMAIL PROTECTED] wrote: Given a data set and a set of predictors and a response in the data, we would like to find a model that fits the data set best. Suppose that we do not know what kind of model (linear, polynomial regression,... ) might be good, we are wondering if there is R-package(s) can auctomatically do this. Otherwise, can you direct me, or point out reference(s), basic steps to do this. Thanks. -james The best-fitting model for any data is a model with a lot of parameters, so maybe the best fitting model for any data is a model with an infinite number of parameters. However, any model with more parameters than data will have a negative number of degrees of freedom, and you do not want that. The best-fitting model for any data subject to the constraint that the number of degrees of freedom is non-negative, is the data itself, with zero degrees of freedom. The AIC tells you this too. The AIC for the model formed by the data itsel is 2n, whereas the AIC for any model with negative degrees of freedom is 2n. But I guess you want to make inference from sample to population. If that is indeed the case, then you should consider changing your focus from finding a model that fits the data set best to a model that best summarizes the information contained in your sample about the population the sample comes from. To do that, start by defining the nature of your response variable. What is the nature of the natural process generating this response variable? Is it continuous or discrete? Is it univariate or multivariate? Can it take negative and positive values? Can it take values of zero? After you have clarified the probabilistic model for the response variable, then you can start thinking about the mathematical relation between the response variable and the predictors. Is it linear or nonlinear? Are the predictors categorical or continuous? Read the posting guide, formulate a clear question, and maybe you will be given more specific help. Rubén __ 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] R Newbie Question/Data Model
Given a data set and a set of predictors and a response in the data, we would like to find a model that fits the data set best. Suppose that we do not know what kind of model (linear, polynomial regression,... ) might be good, we are wondering if there is R-package(s) can auctomatically do this. Otherwise, can you direct me, or point out reference(s), basic steps to do this. Thanks. -james __ 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] R Newbie Question/Data Model
First off there are multiple definitions of best, you need to decide which best is best for you. Second, for reasonable definitions of best, deciding between, linear, polynomial, and ... Requires backgroud knowledge and real thought. R can fit many different models and give you numerical and graphical summaries of those models, but determining the best model requires at least 1 person, some actual knowledge, and some real thought. Basic steps: Take a bunch more statistics classes and/or hire a statistician. If you can come back with a bit more of an example of what you are trying to accomplish along with background, what you mean by best, and show what your thought process has been so far, then we may be able to direct you better. You should also read through the posting guide that is linked to at the bottom of most posts to the group. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] (801) 408-8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Thursday, April 24, 2008 10:58 AM To: r-help@r-project.org Subject: [R] R Newbie Question/Data Model Given a data set and a set of predictors and a response in the data, we would like to find a model that fits the data set best. Suppose that we do not know what kind of model (linear, polynomial regression,... ) might be good, we are wondering if there is R-package(s) can auctomatically do this. Otherwise, can you direct me, or point out reference(s), basic steps to do this. Thanks. -james __ 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-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.
