[R] Remove rows that have repeated items in a particular column
Dear R Users, I apology for not being able to provide an adequately informative subject. Let me describe my problem with an example. For a matrix *(mat - matrix(c(1,1,2,2,2,3,3, 5,9,1,3,7,4,8), ncol = 2))* my desired output is *(desired - matrix(c(1,2,3, 5,1,4), ncol = 2))* That is, the first column is numerically grouped and only the first item in each group is wanted. The second column is in increasing order within each group. My actual data will be of size 10^6 by 100 so I am hoping to solve this by a simple function. Thank you very much for your help. Best, Zhongyi Yuan [[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.
Re: [R] Remove rows that have repeated items in a particular column
Worked like a charm. Thanks. Zhongyi On Wed, Jul 25, 2012 at 11:58 PM, MK mkao006rm...@gmail.com wrote: Very simple mat[!duplicated(mat[, 1]), ] M [[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] color of math annotation in legend
Dear useRs,
Can someone help me to adjust the color of math annotation in a legend? The
following code gives me a black alpha = 2.
x=y=1:100
z=seq(0.5,50,by=0.5)
plot(x,y,type='l',col='black')
lines(x,z,col='red')
legend('topleft',c(expression(paste(alpha, = , 1)),
expression(paste(alpha, = , 2))),col=c('black','red'))
Is it possible to make it red? Thanks in advance for your help.
Best,
Zhongyi Yuan
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Re: [R] color of math annotation in legend
Thank you Tyler.
I thought the problem was due to the use of expression(...).
Best,
Zhongyi
On Thu, Jul 28, 2011 at 2:49 AM, Tyler Rinker tyler_rin...@hotmail.comwrote:
Use the text.col argument as below
?legend
x=y=1:100
z=seq(0.5,50,by=0.5)
plot(x,y,type='l',col='black')
lines(x,z,col='red')
legend('topleft',c(expression(paste(alpha, = , 1)),
expression(paste(alpha, = , 2))),text.col=c(black,red))
Date: Thu, 28 Jul 2011 02:32:04 -0500
From: zhongyi-y...@uiowa.edu
To: r-help@r-project.org
Subject: [R] color of math annotation in legend
Dear useRs,
Can someone help me to adjust the color of math annotation in a legend?
The
following code gives me a black alpha = 2.
x=y=1:100
z=seq(0.5,50,by=0.5)
plot(x,y,type='l',col='black')
lines(x,z,col='red')
legend('topleft',c(expression(paste(alpha, = , 1)),
expression(paste(alpha, = , 2))),col=c('black','red'))
Is it possible to make it red? Thanks in advance for your help.
Best,
Zhongyi Yuan
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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[R] roundoff error in recursion
Dear R users,
I run into the following problem and hope someone can help me out. Thank you
in advance for your time.
I have a function defined recursively as follows:
recFun = function(k,x){ #i is less than 10. x can be any real number. I need
the value of recFun(k=1,2,3,4, x=0.12).
if(k==1)
integrand = function(z) (1 + pi*exp(z+0.9*x+0.012))^(-1.5) * dnorm(z)
else
integrand = function(z) (1 + pi*exp(z+0.9*x+0.012))^(-1.5) *
recFun(k-1,z+0.9*x+0.012) * dnorm(z)
integrate(integrand, -Inf, Inf)
}
The goal of the function is to evaluate a quantity satisfying the recursive
relation that
$$I(1,x)=\int_{-\infty }^{\infty }\left( 1+\pi
e^{z+0.9x+0.012}\right)^{-1.5}\Phi (dz),$$ and
$$I(k,x)=\int_{-\infty }^{\infty }\left( 1+\pi e^{z+0.9x+0.012}\right)
^{-1.5} I(k-1,z+0.9x+0.012)\Phi (dz),$$
where $\Phi$ is the distribution function of the standard normal
distribution.
But for k=2, a roundoff error occurs:
recFun(2,0.12)
Error in integrate(integrand, -Inf, Inf) : roundoff error was detected
Can someone suggest a solution to this problem? Thanks a lot.
Best,
Zhongyi
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[R] sample exponential r.v. by MCMC
Dear R users, I am leaning MCMC sampling, and have a problem while trying to sample exponential r.v.'s via the following code: samp - MCMCmetrop1R(dexp, theta.init=1, rate=2, mcmc=5000, burnin=500, thin=10, verbose=500, logfun=FALSE) I tried other distribtions such as Normal, Gamma with shape1, it works perfectly fine. Can someon tell me why? Thank you for your help. Best, Zhongyi [[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.
Re: [R] sample exponential r.v. by MCMC
Hi Dennis,
Sorry for not being considerate. I should have at least mentioned I was
using MCMCpack.
I have no idea about traceback() though.
I appreciate your suggestion.
Best,
Zhongyi
On Wed, Sep 29, 2010 at 5:00 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
It might be helpful to inform the list which package you're using -
evidently it's MCMCpack, but I had to search to learn that.
You may also want to include the following information:
samp - MCMCmetrop1R(dexp, theta.init=1, rate=2,
+ mcmc=5000, burnin=500,
+ thin=10, verbose=500, logfun=FALSE)
Error in optim(theta.init, maxfun, control = optim.control, lower =
optim.lower, :
non-finite finite-difference value [1]
traceback()
2: optim(theta.init, maxfun, control = optim.control, lower = optim.lower,
upper = optim.upper, method = optim.method, hessian = TRUE,
...)
1: MCMCmetrop1R(dexp, theta.init = 1, rate = 2, mcmc = 5000, burnin = 500,
thin = 10, verbose = 500, logfun = FALSE)
along with the results of sessionInfo(). You may also want to cc the
package maintainer in case this happens to be a bug:
maintainer('MCMCpack')
[1] Andrew D. Martin admar...@wustl.edu
I don't know enough about the package to help you, but providing this
information to the list will increase the probability of a successful
resolution in a shorter period of time.
Cheers,
Dennis
On Wed, Sep 29, 2010 at 2:19 PM, Zhongyi Yuan zhongyi-y...@uiowa.eduwrote:
Dear R users,
I am leaning MCMC sampling, and have a problem while trying to sample
exponential r.v.'s via the following code:
samp - MCMCmetrop1R(dexp, theta.init=1, rate=2,
mcmc=5000, burnin=500,
thin=10, verbose=500, logfun=FALSE)
I tried other distribtions such as Normal, Gamma with shape1, it works
perfectly fine. Can someon tell me why?
Thank you for your help.
Best,
Zhongyi
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Re: [R] labels gone
Hi Milton, Thanks for your help. I must confess that I don't understand what your code means at this moment. But you really helped me out and I'll figure it out later. Thanks. Cheers. Zhongyi On Tue, Mar 16, 2010 at 12:26 AM, milton ruser milton.ru...@gmail.comwrote: Hi Zhongyi, I must confess I not understood completely what you need, but... Tau-seq(0.05,0.95,0.05); Pi - seq(0.19,0.01,-0.01); par(cex.axis=0.8,ps=9,mar=c(1.5,1,0.5,1), oma=c(1,1,0.2,1) ,tck=-0.01); plot(Tau,Pi, type='l', xlab=Tau,ylab=Pi,col=4, xaxt=n, yaxt=n); axis(1,labels=F) axis(1,line=-1, lwd=0) axis(2,labels=F) axis(2,line=-0.5, lwd=0) mtext(x_txt,side=1, line=0.5) mtext(y_txt,side=2, line=1.2) cheers milton On Tue, Mar 16, 2010 at 12:28 AM, Zhongyi Yuan zhongyi-y...@uiowa.eduwrote: Dear R users: I am drawing a graph with the following code: Tau-seq(0.05,0.95,0.05); Pi - seq(0.19,0.01,-0.01); par(cex.axis=0.8,ps=9,mar=c(1.5,1,0.5,1), oma=c(1,1,0.2,1) ,tck=-0.01); plot(Tau,Pi, type='l', xlab=Tau,ylab=Pi,col=4); I want to make the graph take as little space as possible. Here I run into two problems. One is that the labels are gone, the other is the scale numbers of axises are a little too far from my axises. Can anyone help me with this? I am new to graphics and have spent a lot of time on par but still not yet got it. Thanks. best, Zhongyi [[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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[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] labels gone
Dear R users: I am drawing a graph with the following code: Tau-seq(0.05,0.95,0.05); Pi - seq(0.19,0.01,-0.01); par(cex.axis=0.8,ps=9,mar=c(1.5,1,0.5,1), oma=c(1,1,0.2,1) ,tck=-0.01); plot(Tau,Pi, type='l', xlab=Tau,ylab=Pi,col=4); I want to make the graph take as little space as possible. Here I run into two problems. One is that the labels are gone, the other is the scale numbers of axises are a little too far from my axises. Can anyone help me with this? I am new to graphics and have spent a lot of time on par but still not yet got it. Thanks. best, Zhongyi [[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.
Re: [R] construct a list that consists of lists
Hi Jim,
Dennis Murphy solved my problem by the following code.
Thank you for you suggestion. Will check out listBuilder function too.
best,
Zhongyi
# (2) A little more general: prespecify the number of list components
# and run a one-line loop to populate the list
l - vector('list', 6)
for(i in seq_along(l)) l[[i]] - v
l
[[1]]
[[1]]$para1
[1] 1 2 3 4 5
[[1]]$para2
[1] 5 6 7 8 9
[[2]]
[[2]]$para1
[1] 1 2 3 4 5
[[2]]$para2
[1] 5 6 7 8 9
[[3]]
[[3]]$para1
[1] 1 2 3 4 5
[[3]]$para2
[1] 5 6 7 8 9
[[4]]
[[4]]$para1
[1] 1 2 3 4 5
[[4]]$para2
[1] 5 6 7 8 9
[[5]]
[[5]]$para1
[1] 1 2 3 4 5
[[5]]$para2
[1] 5 6 7 8 9
[[6]]
[[6]]$para1
[1] 1 2 3 4 5
[[6]]$para2
[1] 5 6 7 8 9
On Fri, Mar 12, 2010 at 3:06 AM, Jim Lemon j...@bitwrit.com.au wrote:
On 03/12/2010 05:13 PM, Zhongyi Yuan wrote:
Dear R users:
I am hoping that someone can help with constructing a list that consists
of
list with the number of lists variable.
i.e. to find a convenient express(or loop sentences) to realize the
following:
list( list(para1=p1, para2=p2), list(para1=p1, para2=p2), ,
list(para1=p1,para2=p2) )
Hi Zhongyi,
Have a look at the listBuilder function in the crank package.
Jim
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] construct a list that consists of lists
Hi Henrique,
Great.
I tried a similar thing but used
list( replicate(6, list(para1 = 1:5, para2 = 5:9), simplify = FALSE) )
and didn't work. Seems I don't need list(...).
Thanks.
Zhongyi
On Fri, Mar 12, 2010 at 11:56 AM, Henrique Dallazuanna www...@gmail.comwrote:
You can do this also, using replicate:
replicate(6, list(para1 = 1:5, para2 = 5:9), simplify = FALSE)
On Fri, Mar 12, 2010 at 2:51 PM, Zhongyi Yuan zhongyi-y...@uiowa.edu
wrote:
Hi Jim,
Dennis Murphy solved my problem by the following code.
Thank you for you suggestion. Will check out listBuilder function too.
best,
Zhongyi
# (2) A little more general: prespecify the number of list components
# and run a one-line loop to populate the list
l - vector('list', 6)
for(i in seq_along(l)) l[[i]] - v
l
[[1]]
[[1]]$para1
[1] 1 2 3 4 5
[[1]]$para2
[1] 5 6 7 8 9
[[2]]
[[2]]$para1
[1] 1 2 3 4 5
[[2]]$para2
[1] 5 6 7 8 9
[[3]]
[[3]]$para1
[1] 1 2 3 4 5
[[3]]$para2
[1] 5 6 7 8 9
[[4]]
[[4]]$para1
[1] 1 2 3 4 5
[[4]]$para2
[1] 5 6 7 8 9
[[5]]
[[5]]$para1
[1] 1 2 3 4 5
[[5]]$para2
[1] 5 6 7 8 9
[[6]]
[[6]]$para1
[1] 1 2 3 4 5
[[6]]$para2
[1] 5 6 7 8 9
On Fri, Mar 12, 2010 at 3:06 AM, Jim Lemon j...@bitwrit.com.au wrote:
On 03/12/2010 05:13 PM, Zhongyi Yuan wrote:
Dear R users:
I am hoping that someone can help with constructing a list that
consists
of
list with the number of lists variable.
i.e. to find a convenient express(or loop sentences) to realize the
following:
list( list(para1=p1, para2=p2), list(para1=p1, para2=p2), ,
list(para1=p1,para2=p2) )
Hi Zhongyi,
Have a look at the listBuilder function in the crank package.
Jim
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--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
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[R] construct a list that consists of lists
Dear R users: I am hoping that someone can help with constructing a list that consists of list with the number of lists variable. i.e. to find a convenient express(or loop sentences) to realize the following: list( list(para1=p1, para2=p2), list(para1=p1, para2=p2), , list(para1=p1,para2=p2) ) I appreciate your help. best, Zhongyi [[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] append something to an xls file
Hi everyone, I run into this problem and hope somebody can help. Thank you in advance. I have an excel file with the first column being a list of names. What I want is to add values that I compute for each person to the right of the corresponding names. Can someone help me with this? Thanks. Best, Zhongyi [[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] format
Hi,
Could anybody please help me with this? In the following function, I try to
return a good format of x, i.e. if my x[1] is 800, I don't want it to give
me 800.0.
I tried format(x, trim = TRUE), but didn't work.
If my x has only first three components, which are all integers, I don't run
into this problem. So I guess fractional numbers cause this problem.
The comment line works, but I wondering if there are better ways of doing
this.
Thank you.
problem1-function(){
x - round(runif(1,0.5,9.5))*100;
if (x=500){
x-c(x,x+50+round(runif(1,0.5,19.5))*5,x+50+round(runif(1,0.5,19.5))*5,
round(runif(1,0.5,4.5))+0.5, round(runif(1,4,9),1));}
else{
x-c(x,x+100+round(runif(1,0.5,19.5))*5,x+100+round(runif(1,0.5,19.5))*5,
round(runif(1,0.5,4.5))+0.5, round(runif(1,4,9),1));}
c(format(x))
# c(format(x[1]),format(x[2]),format(x[3]))
}
Best,
Zhongyi
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