Re: [R] hypergeometric vs fisher.test
"or an incompetent" Such harsh words Peter. You're not making this a friendly environment for people to ask questions. Is there a less competent attribute mailing list so that some of us don't offend you with our questions? On Aug 13, 2010, at 2:11 PM, Peter Dalgaard wrote: Andrea Franceschini wrote: > I ask the question also because I found this line in Wikipedia: > "The test (see above) based on the hypergeometric distribution > (hypergeometric test) is identical to the corresponding one-tailed > version of Fisher's exact test". > > Is this wrong ? May I kindly ask a friendly explanation for > not-experts in statistics ? > > Thx a lot, > You never said you were a non-expert. A question like that might very well have come from a student, or an incompetent claiming to have found a bug in fisher.test... The point was that Fisher's test takes the 2x2 table a b c d and interprets the situation under the null hypothesis as taking a sample of size a+c from an urn with a+b white balls and c+d black balls. Once you read the docs for phyper properly (it _is_ tricky to get it right), you arrive at > phyper(16,17+181,449+19551,17+449, lower.tail=FALSE) [1] 3.693347e-06 -- Peter Dalgaard Center for Statistics, Copenhagen Business School Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ 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.
Re: [R] Dealing with data
But in your comment, it sounded like you were in the realm of ANOVA when you made the degrees of freedom comment. I'm not going to get into the theory of statistics with you :) I'm just trying to learn R, take it easy. Yes, I understand that in the regression problem, the degrees of freedom for regression is 1, and in ANOVA, the degrees of freedom for sprays are 5. Thanks. On Aug 13, 2010, at 12:54 PM, Greg Snow wrote: If you do as.numeric on InsectSprays and use the result as a predictor in lm, then it will only fit 1 degree of freedom, not 5, try it and see. That is why I was asking and giving an alternative that would still use 5 degrees of freedom. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message----- > From: TGS [mailto:cran.questi...@gmail.com] > Sent: Friday, August 13, 2010 1:52 PM > To: Greg Snow > Subject: Re: [R] Dealing with data > > P.S. The degrees of freedom for sprays would be 5 and not 1. > > On Aug 13, 2010, at 12:27 PM, Greg Snow wrote: > > So you want 1 degree of freedom for InsectSprays? You believe that the > difference between A and B is exactly the same as between B and C which > is exactly the same as between D and E (etc.)? that seems an odd > assumption, but you can get that by using as.numeric (as I and others > have already stated). > > If on the other hand you want InsectSprays to be treated correctly with > the correct number of degrees of freedom, but have the output on a > single line testing the overall effect, then you want to use the aov > function rather than lm (internally they do the same thing, but the > default summary output for aov is 1 line per term). > > Hope this helps, > > -- > Gregory (Greg) L. Snow Ph.D. > Statistical Data Center > Intermountain Healthcare > greg.s...@imail.org > 801.408.8111 > > >> -Original Message- >> From: TGS [mailto:cran.questi...@gmail.com] >> Sent: Friday, August 13, 2010 11:51 AM >> To: Greg Snow >> Cc: r-help@r-project.org >> Subject: Re: [R] Dealing with data >> >> # Greg, if R automatically does that then I don't know why it's >> treating each indicator >> # as a different regressor. In other words, I am interested in > treating >> 'spray' as one >> # independent variable. >> # >> # Erik, which book do you suggest I read? Thanks. >> >> data(InsectSprays) >> lm(InsectSprays$count ~ 0 + InsectSprays$spray) >> >> On Aug 13, 2010, at 10:34 AM, Greg Snow wrote: >> >> R/S does all of that automatically for you, you do not need to > manually >> create the indicator variables. >> >> If you do something like: >> >>> fit <- lm( Sepal.Width ~ Species, data=iris, x=TRUE) >> >> Then look at the matrix actually used: >> >>> fit$x >> >> Or the output: >> >>> summary(fit) >> >> You will see that Species was automatically converted into indicator >> variables and those were used in the regression. >> >> If you really need the indicator variables yourself, look at the >> model.matrix function, e.g.: >> >>> model.matrix( ~Species, data=iris ) >> >> Or >> >>> model.matrix( ~Species - 1, data=iris ) >> >> If you really want 1 for A, 2 for B, etc. then look at as.numeric on > a >> factor variable (e.g. as.numeric(iris$Species) ). >> >> Hope this helps, >> >> -- >> Gregory (Greg) L. Snow Ph.D. >> Statistical Data Center >> Intermountain Healthcare >> greg.s...@imail.org >> 801.408.8111 >> >> >>> -Original Message- >>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- >>> project.org] On Behalf Of TGS >>> Sent: Friday, August 13, 2010 11:22 AM >>> To: David Winsemius >>> Cc: r-help@r-project.org >>> Subject: Re: [R] Dealing with data >>> >>> To clarify, I'd like to create a column of indicators for the >>> respective letters so that I could maybe do regression on > indicators, >>> etc. >>> >>> For instance, "A" gets "1", "B" gets "2", and so on. >>> >>> On Aug 13, 2010, at 10:19 AM, David Winsemius wrote: >>> >>> >>> On Aug 13, 2010, at 1:03 PM, TGS wrote: >>> >>>> # how would I code in R to look at the letter of the alphabet >>>> # in the second column and create a indicator column for the >>>> # corresponding letter? >>
Re: [R] Dealing with data
# I wasn't trying to do ANOVA. I was simply trying to figure out how regress count on sprays (this is after I saw another poster asking an unrelated question with the InsectSprays dataset). # # Anyhow, David clarified this but also, thanks for your explanation as well. rm(list = ls()); sprays <- as.numeric(InsectSprays$spray) lm(formula = count ~ 0 + spray, data = InsectSprays) lm(formula = count ~ 0 + sprays, data = InsectSprays) # besides the point, in the ANOVA problem the degrees of freedom would be 5, not 1. On Aug 13, 2010, at 12:27 PM, Greg Snow wrote: So you want 1 degree of freedom for InsectSprays? You believe that the difference between A and B is exactly the same as between B and C which is exactly the same as between D and E (etc.)? that seems an odd assumption, but you can get that by using as.numeric (as I and others have already stated). If on the other hand you want InsectSprays to be treated correctly with the correct number of degrees of freedom, but have the output on a single line testing the overall effect, then you want to use the aov function rather than lm (internally they do the same thing, but the default summary output for aov is 1 line per term). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message----- > From: TGS [mailto:cran.questi...@gmail.com] > Sent: Friday, August 13, 2010 11:51 AM > To: Greg Snow > Cc: r-help@r-project.org > Subject: Re: [R] Dealing with data > > # Greg, if R automatically does that then I don't know why it's > treating each indicator > # as a different regressor. In other words, I am interested in treating > 'spray' as one > # independent variable. > # > # Erik, which book do you suggest I read? Thanks. > > data(InsectSprays) > lm(InsectSprays$count ~ 0 + InsectSprays$spray) > > On Aug 13, 2010, at 10:34 AM, Greg Snow wrote: > > R/S does all of that automatically for you, you do not need to manually > create the indicator variables. > > If you do something like: > >> fit <- lm( Sepal.Width ~ Species, data=iris, x=TRUE) > > Then look at the matrix actually used: > >> fit$x > > Or the output: > >> summary(fit) > > You will see that Species was automatically converted into indicator > variables and those were used in the regression. > > If you really need the indicator variables yourself, look at the > model.matrix function, e.g.: > >> model.matrix( ~Species, data=iris ) > > Or > >> model.matrix( ~Species - 1, data=iris ) > > If you really want 1 for A, 2 for B, etc. then look at as.numeric on a > factor variable (e.g. as.numeric(iris$Species) ). > > Hope this helps, > > -- > Gregory (Greg) L. Snow Ph.D. > Statistical Data Center > Intermountain Healthcare > greg.s...@imail.org > 801.408.8111 > > >> -Original Message- >> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- >> project.org] On Behalf Of TGS >> Sent: Friday, August 13, 2010 11:22 AM >> To: David Winsemius >> Cc: r-help@r-project.org >> Subject: Re: [R] Dealing with data >> >> To clarify, I'd like to create a column of indicators for the >> respective letters so that I could maybe do regression on indicators, >> etc. >> >> For instance, "A" gets "1", "B" gets "2", and so on. >> >> On Aug 13, 2010, at 10:19 AM, David Winsemius wrote: >> >> >> On Aug 13, 2010, at 1:03 PM, TGS wrote: >> >>> # how would I code in R to look at the letter of the alphabet >>> # in the second column and create a indicator column for the >>> # corresponding letter? >>> >>> data(InsectSprays) >>> InsectSprays$spray >> >> It's already what most people mean when they say "indicator column", >> i.e., a factor variable (and not a character vector) so, what > do >> _you_ mean? >>> >> >> >> -- >> >> David Winsemius, MD >> West Hartford, CT >> >> __ >> 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.
Re: [R] Dealing with data
# Greg, if R automatically does that then I don't know why it's treating each indicator # as a different regressor. In other words, I am interested in treating 'spray' as one # independent variable. # # Erik, which book do you suggest I read? Thanks. data(InsectSprays) lm(InsectSprays$count ~ 0 + InsectSprays$spray) On Aug 13, 2010, at 10:34 AM, Greg Snow wrote: R/S does all of that automatically for you, you do not need to manually create the indicator variables. If you do something like: > fit <- lm( Sepal.Width ~ Species, data=iris, x=TRUE) Then look at the matrix actually used: > fit$x Or the output: > summary(fit) You will see that Species was automatically converted into indicator variables and those were used in the regression. If you really need the indicator variables yourself, look at the model.matrix function, e.g.: > model.matrix( ~Species, data=iris ) Or > model.matrix( ~Species - 1, data=iris ) If you really want 1 for A, 2 for B, etc. then look at as.numeric on a factor variable (e.g. as.numeric(iris$Species) ). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- > project.org] On Behalf Of TGS > Sent: Friday, August 13, 2010 11:22 AM > To: David Winsemius > Cc: r-help@r-project.org > Subject: Re: [R] Dealing with data > > To clarify, I'd like to create a column of indicators for the > respective letters so that I could maybe do regression on indicators, > etc. > > For instance, "A" gets "1", "B" gets "2", and so on. > > On Aug 13, 2010, at 10:19 AM, David Winsemius wrote: > > > On Aug 13, 2010, at 1:03 PM, TGS wrote: > >> # how would I code in R to look at the letter of the alphabet >> # in the second column and create a indicator column for the >> # corresponding letter? >> >> data(InsectSprays) >> InsectSprays$spray > > It's already what most people mean when they say "indicator column", > i.e., a factor variable (and not a character vector) so, what do > _you_ mean? >> > > > -- > > David Winsemius, MD > West Hartford, CT > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting- > guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Dealing with data
To clarify, I'd like to create a column of indicators for the respective letters so that I could maybe do regression on indicators, etc. For instance, "A" gets "1", "B" gets "2", and so on. On Aug 13, 2010, at 10:19 AM, David Winsemius wrote: On Aug 13, 2010, at 1:03 PM, TGS wrote: > # how would I code in R to look at the letter of the alphabet > # in the second column and create a indicator column for the > # corresponding letter? > > data(InsectSprays) > InsectSprays$spray It's already what most people mean when they say "indicator column", i.e., a factor variable (and not a character vector) so, what do _you_ mean? > -- David Winsemius, MD West Hartford, CT __ 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] Dealing with data
# how would I code in R to look at the letter of the alphabet # in the second column and create a indicator column for the # corresponding letter? data(InsectSprays) InsectSprays$spray __ 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 64-bit and Revolution
I downloaded their Academic version and installed it on a Windows virtual machine (as there is not a Mac version available). I played around with it a little bit and wasn't overly impressed. I still like my current configuration: textmate with R bundle on Mac OSX. With textmate data entry is a breeze, executing commands is a breeze, and dealing with graphics is a breeze. Also, I can easily use textmate's Sweave bundle to produce documents. In short, from the very little that I have tried using Revolution, my current configuration beats using Revolution. I just need to continue working on obtaining the R skills! On Aug 12, 2010, at 4:12 PM, Lars Bishop wrote: Dear users, The company where I work is considering getting a license for Revolution Enterprise - Windows 64-bit. I'll appreciate for those familiar with the product if can share your experiences with it? In particular, how does it compare to the "free" version of R 64-bit? Thanks in advance. Regards, Lars. [[alternative HTML version deleted]] __ 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.
Re: [R] Plotting one dot in a graph
# just to clean it up for my own understanding, the "difference" approach as
you had suggested would be
x <- seq(.2, .3, by = .1)
f1 <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f1(x), type = "l")
abline(h = -.1)
abline(v = x[which.min(abs(diff((f1(x) - (-.1))**2)))], lty = 'dotted')
points(x = x[which.min(abs(diff((f1(x) - (-.1))**2)))], y = -.1)
# and the uniroot approach is:
x <- seq(.2, .3, by = .01)
f1 <- function(x){
x*cos(x)-2*x**2+3*x-1
}
f2 <- function(x){
-.1
}
f3 <- function(x){
f1(x) - f2(x)
}
plot(x,f1(x), type = "l")
abline(h = -.1)
abline(v = uniroot(f = f3, interval = c(.2, .3))$root, lty = 'dotted')
points(x = uniroot(f = f3, interval = c(.2, .3))$root, y = -.1)
# Thanks David!
On Aug 12, 2010, at 1:33 PM, David Winsemius wrote:
On Aug 12, 2010, at 4:15 PM, TGS wrote:
> David, I was expecting this to work but how would I specify the vector in
> "diff()" in order for the following to work?
>
> x <- seq(.2, .3, by = .01)
> f <- function(x){
> x*cos(x)-2*x**2+3*x-1
> }
> plot(x,f(x), type = "l")
> abline(h = -.1)
> abline(v = x[which.min(abs(diff(c(-.1, f(x)], lty = 'dotted')
f2 <- function(x) -0.1
f3 <- function(x) f(x) -f2(x)
abline(v=uniroot(f3, c(0.2, 0.3) )$root)
points(x=uniroot(f3, c(0.2, 0.3) )$root, y= -0.1)
If you are going to use the differences, then you probably want to minimize
either the abs() or the square of the differences.
--
David.
>
> On Aug 12, 2010, at 1:00 PM, David Winsemius wrote:
>
>
> On Aug 12, 2010, at 3:54 PM, TGS wrote:
>
>> Actually I spoke too soon David.
>>
>> I'm looking for a function that will either tell me which point is the
>> intersection so that I'd be able to plot a point there.
>>
>> Or, if I have to solve for the roots in the ways which were demonstrated
>> yesterday, then would I be able to specify what the horizontal line is, for
>> instance in the case where y (is-not) 0?
>
> Isn't the abline h=0 represented mathematically by the equation y=0 and
> therefore you are solving just for the zeros of "f" (whaich are the same as
> for (f-0)? If it were something more interesting, like solving the
> intersection of two polynomials, you would be solving for the zeros of the
> difference of the equations. Or maybe I have not understood what you were
> requesting?
>
>
>>
>> On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
>>
>>
>> On Aug 12, 2010, at 3:43 PM, TGS wrote:
>>
>>> I'd like to plot a point at the intersection of these two curves. Thanks
>>>
>>> x <- seq(.2, .3, by = .01)
>>> f <- function(x){
>>> x*cos(x)-2*x**2+3*x-1
>>> }
>>>
>>> plot(x,f(x), type = "l")
>>> abline(h = 0)
>>
>> Would this just be the uniroot strategy applied to "f"? You then plot the x
>> and y values with points()
>>
>
>>
>
> David Winsemius, MD
> West Hartford, CT
>
>
David Winsemius, MD
West Hartford, CT
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Plotting one dot in a graph
I was meaning something like the following:
x <- seq(.2, .3, by = .01)
f <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = "l")
abline(h = -.1)
But I'm guessing "uniroot" will do this?---I haven't looked far into the
uniroot function to see if it will solve this.
On Aug 12, 2010, at 1:00 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:54 PM, TGS wrote:
> Actually I spoke too soon David.
>
> I'm looking for a function that will either tell me which point is the
> intersection so that I'd be able to plot a point there.
>
> Or, if I have to solve for the roots in the ways which were demonstrated
> yesterday, then would I be able to specify what the horizontal line is, for
> instance in the case where y (is-not) 0?
Isn't the abline h=0 represented mathematically by the equation y=0 and
therefore you are solving just for the zeros of "f" (whaich are the same as for
(f-0)? If it were something more interesting, like solving the intersection of
two polynomials, you would be solving for the zeros of the difference of the
equations. Or maybe I have not understood what you were requesting?
>
> On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
>
>
> On Aug 12, 2010, at 3:43 PM, TGS wrote:
>
>> I'd like to plot a point at the intersection of these two curves. Thanks
>>
>> x <- seq(.2, .3, by = .01)
>> f <- function(x){
>> x*cos(x)-2*x**2+3*x-1
>> }
>>
>> plot(x,f(x), type = "l")
>> abline(h = 0)
>
> Would this just be the uniroot strategy applied to "f"? You then plot the x
> and y values with points()
>
>
David Winsemius, MD
West Hartford, CT
__
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] Plotting one dot in a graph
Actually I spoke too soon David.
I'm looking for a function that will either tell me which point is the
intersection so that I'd be able to plot a point there.
Or, if I have to solve for the roots in the ways which were demonstrated
yesterday, then would I be able to specify what the horizontal line is, for
instance in the case where y (is-not) 0?
On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:43 PM, TGS wrote:
> I'd like to plot a point at the intersection of these two curves. Thanks
>
> x <- seq(.2, .3, by = .01)
> f <- function(x){
> x*cos(x)-2*x**2+3*x-1
> }
>
> plot(x,f(x), type = "l")
> abline(h = 0)
Would this just be the uniroot strategy applied to "f"? You then plot the x and
y values with points()
--
David Winsemius, MD
West Hartford, CT
__
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] Plotting one dot in a graph
Yes, I'm playing around with other things but the "points()" function is what I
was looking for. Thanks
On Aug 12, 2010, at 12:47 PM, David Winsemius wrote:
On Aug 12, 2010, at 3:43 PM, TGS wrote:
> I'd like to plot a point at the intersection of these two curves. Thanks
>
> x <- seq(.2, .3, by = .01)
> f <- function(x){
> x*cos(x)-2*x**2+3*x-1
> }
>
> plot(x,f(x), type = "l")
> abline(h = 0)
Would this just be the uniroot strategy applied to "f"? You then plot the x and
y values with points()
--
David Winsemius, MD
West Hartford, CT
__
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] Plotting one dot in a graph
I'd like to plot a point at the intersection of these two curves. Thanks
x <- seq(.2, .3, by = .01)
f <- function(x){
x*cos(x)-2*x**2+3*x-1
}
plot(x,f(x), type = "l")
abline(h = 0)
__
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] Error in rowSums
Yes, please do as Erik said in the future but here's one way to do it. (A <- matrix(data = rnorm(n = 9, mean = 0, sd = 1), nrow = 3, ncol = 3, byrow = FALSE, dimnames = NULL)) matrix(rowSums(A)) On Aug 12, 2010, at 11:28 AM, Amit Patel wrote: Hi I am trying to calculate the row sums of a matrix i have created The matrix ( FeaturePresenceMatrix) has been created by 1) Read csv 2) Removing unnecesarry data using [-1:4,] command 3) replacing all the NA values with as.numeric(0) and all others with as.numeric (1) When I carry out the command TotalFeature <- rowrowSums(FeaturePresenceMatrix, na.rm = TRUE) I get the following error. Error in rowSums(FeaturePresenceMatrix, na.rm = TRUE) : 'x' must be numeric Any tips onhow I can get round this? Thanks in Advance Amit __ 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.
Re: [R] Creating vectors
I think I understand your question and the following would produce the result you've posted. (x <- matrix(c(1, 2, 2, 3, 1, 2, 1, 2, 3, 4), nrow=5, byrow=TRUE)) On Aug 12, 2010, at 5:41 AM, clips10 wrote: Thanks for the help, I tried to apply this to a vector with two columns, well I suppose it is not a vector but for instance like this: [,1] [,2] [1,]1 2 [2,]2 3 [3,]1 2 [4,]1 2 [5,]3 4 and return a vector : 1,2,1,1,3, so that it recognises both columns together. I tried match(x, unique(x)) as earlier suggested but this returns a vector of length 10 as opposed to 5, even though unique(x) does remove the repeated rows. Sorry if this is confusing, I am trying to do as originally posted but with 2 columns Thanks -- View this message in context: http://r.789695.n4.nabble.com/Creating-vectors-tp2321440p2322646.html Sent from the R help mailing list archive at Nabble.com. __ 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.
Re: [R] Derivative
This following works for me but I still favor the quick and dirty method
suggested originally by David.
options(scipen = 10)
x <- seq(0,2, by = .01)
f <- expression(5*cos(2*x)-2*x*sin(2*x))
D(f, 'x')
f.prime <- function(x){
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) * 2)))
}
curve(expr = f.prime, from = 1, to = 2, n = 101)
uniroot(f = f.prime, interval = c(1,2), tol = .1)
On Aug 11, 2010, at 10:56 PM, David Winsemius wrote:
On Aug 12, 2010, at 12:49 AM, Dennis Murphy wrote:
> Hi:
>
> Try the following:
>
> f <- function(x) 5*cos(2*x)-2*x*sin(2*x)
> curve(f, -5, 5)
> abline(0, 0, lty = 'dotted')
>
> This shows rather clearly that your function has multiple roots, which isn't
> surprising given that it's a linear combination of sines and cosines. To
> find a specific root numerically, use function uniroot on f, as follows:
>
>> uniroot(f, c(0, 2))
Except he was asking for the root of the derivative. If the classroom
assignment allows use of R's limited symbolic differentiation you could try:
> df.dx <- D(expression(5*cos(2*x)-2*x*sin(2*x)), "x")
> df.dx
-(5 * (sin(2 * x) * 2) + (2 * sin(2 * x) + 2 * x * (cos(2 * x) *
2)))
(Which as one of the examples in the deriv help page notes is not the most
simple form.)
I was assuming that the OP wanted a solution to:
d( abs(f(x)) )/dt = 0 in the domain [1,2]
So:
f.df.dx <- function (x) {
eval(parse(text=D(expression(5*cos(2*x)-2*x*sin(2*x)), "x") ) )
}
# no abs() but we should be satisfied with either a minimum or a maximum
uniroot(f.df.dx, c(1,2) )
$root
[1] 1.958218267
$f.root
[1] 1.138013788e-05
$iter
[1] 4
$estim.prec
[1] 6.103515625e-05
It doesn't agree with my earlier method and I think this one has a greater
probablity of being correct. I don't think I needed to take second differences.
--
David.
> $root
> [1] 0.6569286
>
> $f.root
> [1] -0.0001196119
>
> $iter
> [1] 6
>
> $estim.prec
> [1] 6.103516e-05
>
> This catches the root that lies between x = 0 and x = 2. If you want to find
> a set of roots, you can try a loop. Fortunately, since the function is even,
> you really only need to find the roots on one side of zero, since the ones
> on the other side are the same with opposite sign.
>
> lo <- seq(0, 4.5, by = 1.5)
> hi <- seq(1.5, 6, by = 1.5)
> roots <- numeric(length(lo))
>
> for(i in seq_along(lo)) roots[i] <- uniroot(f, c(lo[i], hi[i]))$root
> roots
>
> See ?uniroot for other options and tolerance settings.
>
> HTH,
> Dennis
>
> On Wed, Aug 11, 2010 at 6:21 PM, TGS wrote:
>
>> How would I numerically find the x value where the derivative of the
>> function below is zero?
>>
>> x <- seq(1,2, by = .01)
>> y <- 5*cos(2*x)-2*x*sin(2*x)
>> plot(x,abs(y), type = "l", ylab = "|y|")
>
--
David Winsemius, MD
West Hartford, CT
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