Re: [R] Fisher's Test 5x4 table
Dear Paul, quoting the email-footer: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. So, what exactly did you try and what was the actual problem/error message? Besides that, have you noted that two of you data rows have the same name? Have you read the online help page of fisher.test(): ?fisher.test Have you tried anything like the following? W - as.matrix( read.table( w.txt, head = T)[-1]) fisher.test( W, workspace = 1e8) # For workspace look at the help page, but it presumably # won't work because of your sample size. set.seed( 20150828) # for reproducibility fisher.test( W, simulate.p.value = TRUE, B = 1e5) # For B look at the help page. Finally: Did Minitab really report p 0.001? ;-) Hth -- Gerrit Dear all, I am trying to do a fishers test on a 5x4 table on R statistics. I have already done a chi squared test using Minitab on this data set, getting a result of (1, N = 165.953, DF 12, p0.001), yet using these results (even though they are excellent) may not be suitable for publication. I have tried numerous other statistical packages in the hope of doing this test, yet each one has just the 2x2 table. I am struggling to edit the template fishers test on R to fit my table (as according to the R book it is possible, yet i cannot get it to work). The template given on the R documentation and R book is for a 2x2 fisher test. What do i need to change to get this to work? I have attached the data with the email so one can see what i am on about. Or do i have to write my own new code to compute this. Yours Sincerely, Paul Brett __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] attribute color to a list
Hi,
attriColorValue works with one value. I would like to get the color of a
list with lappy but in input I have two variables (the value and the list).
attriColorValue - function(Value, list, colors=c(a,b,c, d,e),feet){
list - round(list, digits = 0)
Max - max(list, na.rm=TRUE)
Min - min(list, na.rm=TRUE)
my.colors - colorRampPalette(colors)
#generates Max-Min colors from the color ramp
color.df-data.frame(COLOR_VALUE=seq(Min,Max,feet),
color.name=my.colors(length(seq(Min,Max,feet
colorRef - color.df[which(color.df[,1]==Value),2]
return(colorRef)
}
list -
c(7607.2149,36.0673,26.5613,-21.094,535.1462,8460.8617,3112.3839,1810.5521,-2783.7832,-1283.5496,879.4978,-307.8481,133.6729,51.6518,-212.3436,-118.6624,912.8616,16.7501,465.6139,486.3803,1051.6673,-1529.426,198.9787,-265.013,74.0492,-52.0192,-97.655,-5963.4183,-2118.4033,5701.5644,1987.7252,1638.274,1576.775,1520.7626,1039.4264,905.7974,-966.3739,365.2626,364.8378,258.3969,-323.999,-394.7463)
works
as.character(attriColorValue(123,list, colors=c(blue,white,red) ,
feet=1))
not working??
lapply(list, function(x) attriColorValue(x,df$exprsMeanDiff ,colors, feet))
Thanks
Karim
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Piecewise regression using segmented package plotted in xyplot
I perhaps should have added a stronger warning here; note that the model
fitting in my previous post (below) uses explicit initial breakpoints for
segmented (specifically, c(30,60) at line 1 of the get.segments() ). if you
know where yours are, substitute them there. Otherwise, you'd need to use the
automated breakpoint routines documented in ?segmented, typically adjusting the
control list. To be general, you'd need a way to get a choice of at least
number of breakpoints into the model. xyplot doesn't (I think) pass extra
parameters to its panel function, but it will pick up environment parameters so
you _could_ just rely on that, or perhaps on setting something in options(), as
a quick and dirty work-round.
S Ellisaon
--
There isn't an abline method for segmented, and even if there were you'd need
segments() for a segmented line plot. You're going to have to roll your own.
That will need a function to extract the break locations and predicted values
at those points
I don't have your data, so I can't do one specifically for you. But here's a
version that works on the data in the first example in ?segmented. I've
deliberately separated segmented model fitting and line segment extraction
(get.segments) from the panel function that plots them, as that would then work
with the base graphics segments function
#Construct the data set (from ?segmented, though the z covariate is not used
here)
set.seed(12)
xx-1:100
zz-runif(100)
yy-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2)
dati-data.frame(x=xx,y=yy,z=zz)
# Function to fit a simple model using segmented
# and then get line segment data from the model
# For the latter, I've used code from plot. segmented to locate extremes
# and break points (in psi) and then just used
# predict() rather crudely to get the corresponding ordinate values
# you would have to do something more clever
# if your model is not just y~x
get.segments - function(x, y, term, ...) {
#Returns a data frame of line segment coordinates
#from a simple segmented() model
S - segmented(lm(y~x), seg.Z=~x,psi=list(x=c(30,60)),
control=seg.control(display=FALSE))
if(missing(term)) term - S$nameUV$Z[1]
x-with(S, sort(c(psi[,'Est.'], rangeZ[,term])))
#Then cheat a bit
new - data.frame(x=unname(x))
names(new) - term
y - predict(S, new=new)
L - length(x) -1
#return the line segment data in an easy-to-use format
data.frame(x0=x[1:L], y0=y[1:L], x1=x[1:L + 1], y1=y[1:L + 1])
}
#Write a panel function using that...
panel.segmented - function(x, y, ...) {
## Fit the simple model
m - get.segments( x, y )
#Plot the segments
with(m, panel.segments(x0, y0, x1, y1, ... ))
}
library(lattice)
xyplot(y~x, data=dati,
panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.segmented(x, y, ...)
}
)
#And just to see if it works for panel-grouped data:
set.seed(1023)
dati$parts - sample(gl(2, 50))
xyplot(y~x|parts, data=dati,
panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.segmented(x, y, ...)
}
)
S Ellison
***
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[R] Problem loading mvabund package
Hello, Can anybody please help me with mvabund package installation? I downloaded and installed it. It seems installed, but I can't load the package. Here is what I tried: install.packages(/Users/smitra/Documents/Soft/mvabund_3.10.4.tgz, repos = NULL, type=source) * installing *binary* package ‘mvabund’ ... * DONE (mvabund) library(mvabund) Error in loadNamespace(i, c(lib.loc, .libPaths()), versionCheck = vI[[i]]) : there is no package called ‘tweedie’ Error: package or namespace load failed for ‘mvabund’ Any help will be great. Thanks. Suparna [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Piecewise regression using segmented package plotted in xyplot
There isn't an abline method for segmented, and even if there were you'd need
segments() for a segmented line plot. You're going to have to roll your own.
That will need a function to extract the break locations and predicted values
at those points
I don't have your data, so I can't do one specifically for you. But here's a
version that works on the data in the first example in ?segmented. I've
deliberately separated segmented model fitting and line segment extraction
(get.segments) from the panel function that plots them, as that would then work
with the base graphics segments function
#Construct the data set (from ?segmented, though the z covariate is not used
here)
set.seed(12)
xx-1:100
zz-runif(100)
yy-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2)
dati-data.frame(x=xx,y=yy,z=zz)
# Function to fit a simple model using segmented
# and then get line segment data from the model
# For the latter, I've used code from plot. segmented to locate extremes
# and break points (in psi) and then just used
# predict() rather crudely to get the corresponding ordinate values
# you would have to do something more clever
# if your model is not just y~x
get.segments - function(x, y, term, ...) {
#Returns a data frame of line segment coordinates
#from a simple segmented() model
S - segmented(lm(y~x), seg.Z=~x,psi=list(x=c(30,60)),
control=seg.control(display=FALSE))
if(missing(term)) term - S$nameUV$Z[1]
x-with(S, sort(c(psi[,'Est.'], rangeZ[,term])))
#Then cheat a bit
new - data.frame(x=unname(x))
names(new) - term
y - predict(S, new=new)
L - length(x) -1
#return the line segment data in an easy-to-use format
data.frame(x0=x[1:L], y0=y[1:L], x1=x[1:L + 1], y1=y[1:L + 1])
}
#Write a panel function using that...
panel.segmented - function(x, y, ...) {
## Fit the simple model
m - get.segments( x, y )
#Plot the segments
with(m, panel.segments(x0, y0, x1, y1, ... ))
}
library(lattice)
xyplot(y~x, data=dati,
panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.segmented(x, y, ...)
}
)
#And just to see if it works for panel-grouped data:
set.seed(1023)
dati$parts - sample(gl(2, 50))
xyplot(y~x|parts, data=dati,
panel = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.segmented(x, y, ...)
}
)
S Ellison
***
This email and any attachments are confidential. Any use...{{dropped:8}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] attribute color to a list
Hi Karim,
I'm not sure that this is what is causing the error, but your list is
actually a vector. The following runs, but the function is obviously not
working:
sapply(list,
function(x)
as.character(attriColorValue(x,list,colors=c(blue,white,red),feet=1)))
There is probably a better name for your vector than list.
Jim
On Fri, Aug 28, 2015 at 8:16 PM, Karim Mezhoud kmezh...@gmail.com wrote:
Hi,
attriColorValue works with one value. I would like to get the color of a
list with lappy but in input I have two variables (the value and the list).
attriColorValue - function(Value, list, colors=c(a,b,c, d,e),feet){
list - round(list, digits = 0)
Max - max(list, na.rm=TRUE)
Min - min(list, na.rm=TRUE)
my.colors - colorRampPalette(colors)
#generates Max-Min colors from the color ramp
color.df-data.frame(COLOR_VALUE=seq(Min,Max,feet),
color.name=my.colors(length(seq(Min,Max,feet
colorRef - color.df[which(color.df[,1]==Value),2]
return(colorRef)
}
list -
c(7607.2149,36.0673,26.5613,-21.094,535.1462,8460.8617,3112.3839,1810.5521,-2783.7832,-1283.5496,879.4978,-307.8481,133.6729,51.6518,-212.3436,-118.6624,912.8616,16.7501,465.6139,486.3803,1051.6673,-1529.426,198.9787,-265.013,74.0492,-52.0192,-97.655,-5963.4183,-2118.4033,5701.5644,1987.7252,1638.274,1576.775,1520.7626,1039.4264,905.7974,-966.3739,365.2626,364.8378,258.3969,-323.999,-394.7463)
works
as.character(attriColorValue(123,list, colors=c(blue,white,red) ,
feet=1))
not working??
lapply(list, function(x) attriColorValue(x,df$exprsMeanDiff ,colors, feet))
Thanks
Karim
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
[[alternative HTML version deleted]]
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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] Problem loading mvabund package
Dear Suparna, See below On 28/08/2015 10:22, Suparna Mitra wrote: Hello, Can anybody please help me with mvabund package installation? I downloaded and installed it. It seems installed, but I can't load the package. Here is what I tried: install.packages(/Users/smitra/Documents/Soft/mvabund_3.10.4.tgz, repos = NULL, type=source) * installing *binary* package ‘mvabund’ ... * DONE (mvabund) library(mvabund) Error in loadNamespace(i, c(lib.loc, .libPaths()), versionCheck = vI[[i]]) : there is no package called ‘tweedie’ There are two possibilities here 1 - you have not installed tweedie 2 - you have and R cannot find it Error: package or namespace load failed for ‘mvabund’ Any help will be great. Thanks. Suparna [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. -- Michael http://www.dewey.myzen.co.uk/home.html __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] attribute color to a list
Hi,
Thank you for comments.
Yes it is a vector an not a list ;).
I need to round also the input Value (Value - round(Value, digits=0). If
not Matching is not possible. The vector is a real number and the color.df
are Integer.
Thanks for sapply is better than lapply in my case.
Karim
On Fri, Aug 28, 2015 at 11:35 AM, Jim Lemon drjimle...@gmail.com wrote:
Hi Karim,
I'm not sure that this is what is causing the error, but your list is
actually a vector. The following runs, but the function is obviously not
working:
sapply(list,
function(x)
as.character(attriColorValue(x,list,colors=c(blue,white,red),feet=1)))
There is probably a better name for your vector than list.
Jim
On Fri, Aug 28, 2015 at 8:16 PM, Karim Mezhoud kmezh...@gmail.com wrote:
Hi,
attriColorValue works with one value. I would like to get the color of a
list with lappy but in input I have two variables (the value and the
list).
attriColorValue - function(Value, list, colors=c(a,b,c, d,e),feet){
list - round(list, digits = 0)
Max - max(list, na.rm=TRUE)
Min - min(list, na.rm=TRUE)
my.colors - colorRampPalette(colors)
#generates Max-Min colors from the color ramp
color.df-data.frame(COLOR_VALUE=seq(Min,Max,feet),
color.name=my.colors(length(seq(Min,Max,feet
colorRef - color.df[which(color.df[,1]==Value),2]
return(colorRef)
}
list -
c(7607.2149,36.0673,26.5613,-21.094,535.1462,8460.8617,3112.3839,1810.5521,-2783.7832,-1283.5496,879.4978,-307.8481,133.6729,51.6518,-212.3436,-118.6624,912.8616,16.7501,465.6139,486.3803,1051.6673,-1529.426,198.9787,-265.013,74.0492,-52.0192,-97.655,-5963.4183,-2118.4033,5701.5644,1987.7252,1638.274,1576.775,1520.7626,1039.4264,905.7974,-966.3739,365.2626,364.8378,258.3969,-323.999,-394.7463)
works
as.character(attriColorValue(123,list, colors=c(blue,white,red) ,
feet=1))
not working??
lapply(list, function(x) attriColorValue(x,df$exprsMeanDiff ,colors,
feet))
Thanks
Karim
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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.
[[alternative HTML version deleted]]
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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] heat map labeling
Hi Angela, Assuming the above data frame is named angela.df: angela.mat-as.matrix(angela.df[,2:3]) angela.mat-angela.mat[apply(angela.mat,1,function(x) all(x) 0),] will remove all of the rows that have contain at least one zero. Jim On Fri, Aug 28, 2015 at 9:00 AM, Angela via R-help r-help@r-project.org wrote: Hello, I have a dataset of 985 genes, looks something like the ones below. I want to label only those with the high intensities, since labeling all doesn't show up. Is there a way to do that? If not, is there a way to pull out the highest ones (say, highest 50, or those above X amount) and only show those in a heat map? Thanks! -Angela Z transforming gives all cells the same value, just + or - (for example, all have 0.5 and -0.5). The researchers want the actual values used. Gene var1 var2 A 800 0 B 25 30 C 75 200 D 0 0 E 400 600 E 500 70 E 100 100 F 600 600 F 70 827460 G 420930 40 H 0 0 H 100 100 I 70 60 J 0 70 K 0 0 L 20 50 L 100 300 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] lsqlin in R package pracma
I got interested in enabling the full funcionality that MATLAB's lsqlin() has, that is with equality and bound constraints. To replace an equality constraint with two inequality constraints will not work with solve.QP() because it requires positive definite matrices. I will use kernlab::ipop() instead. To handle the full MATLAB example, add the following simple linear equality constraint 3x1 + 5x2 + 7x3 + 9x4 = 4 to the example above, plus lower and upper bounds -0.1 and 2.0 for all x_i. C - matrix(c( 0.9501, 0.7620, 0.6153, 0.4057, 0.2311, 0.4564, 0.7919, 0.9354, 0.6068, 0.0185, 0.9218, 0.9169, 0.4859, 0.8214, 0.7382, 0.4102, 0.8912, 0.4447, 0.1762, 0.8936), 5, 4, byrow=TRUE) d - c(0.0578, 0.3528, 0.8131, 0.0098, 0.1388) A - matrix(c( 0.2027, 0.2721, 0.7467, 0.4659, 0.1987, 0.1988, 0.4450, 0.4186, 0.6037, 0.0152, 0.9318, 0.8462), 3, 4, byrow=TRUE) b - c(0.5251, 0.2026, 0.6721) # Add the equality constraint to matrix A Aeq - c(3, 5, 7, 9) beq - 4 A1 - rbind(A , c(3, 5, 7, 9)) b1 - c(b, 4) lb - rep(-0.1, 4) # lower and upper bounds ub - rep( 2.0, 4) r1 - c(1, 1, 1, 0) # 0 to force an equality constraint # Prepare for a quadratic solver Dmat - t(C) %*% C dvec - (t(C) %*% d) Amat - -1 * A1 bvec - -1 * b1 library(kernlab) s - ipop(-dvec, Dmat, Amat, bvec, lb, ub, r1) s # An object of class ipop # Slot primal: # [1] -0.0885 -0.0997 0.15990817 0.40895991 # ... x - s@primal # [1] -0.1000 -0.1000 0.1599 0.4090 A1 %*% x - b1 = 0 # i.e., A x = b and 3x[1] + ... + 9x[4] = 4 sum((C %*% x - d)^2)# minimum: 0.1695 And this is exactly the solution that lsqlin() in MATLAB computes. On Thu, Aug 27, 2015 at 6:06 PM, Raubertas, Richard richard_rauber...@merck.com wrote: Is it really that complicated? This looks like an ordinary quadratic programming problem, and 'solve.QP' from the 'quadprog' package seems to solve it without user-specified starting values: library(quadprog) Dmat - t(C) %*% C dvec - (t(C) %*% d) Amat - -1 * t(A) bvec - -1 * b rslt - solve.QP(Dmat, dvec, Amat, bvec) sum((C %*% rslt$solution - d)^2) [1] 0.01758538 Richard Raubertas Merck Co. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Fisher's Test 5x4 table
Paul, as the error messages of your first three attempts (see below) tell you - in an admittedly rather cryptic way - your table or its sample size, respectively, are too large, so that either the largest (hash table) key is too large, or your (i.e., R's) workspace is too small, or your hardware/os cannot allocate enough memory to calculate the p-value of Fisher Exact Test exactly by means of the implemented algorithm. One way out of this is to approximate the exact p-value through simulation, but apparently there occurred a typo in your (last) attempt to do that (Error: unexpected '' in ). So, for me the following works (and it should also for you) and gives the shown output (after a very short while): Trapz - as.matrix( read.table( w.txt, head = T, row.names = Traps)) set.seed( 20150828) # For the sake of reproducibility. fisher.test( Trapz, simulate.p.value = TRUE, + B = 1e5) Fisher's Exact Test for Count Data with simulated p-value (based on 1e+05 replicates) data: Trapz p-value = 1e-05 alternative hypothesis: two.sided Or for a higher value for B if you are patient enough (with a computing time of several seconds) : set.seed( 20150828) fisher.test( Trapz, simulate.p.value=TRUE, B = 1e7) Fisher's Exact Test for Count Data with simulated p-value (based on 1e+07 replicates) data: Trapz p-value = 1e-07 alternative hypothesis: two.sided Hth -- Gerrit (BTW, you don't have to specify arguments (in function calls) whose default values you don't want to change.) On Fri, 28 Aug 2015, paul brett wrote: Hi Gerrit, I spotted that, it was a mistake on my own part, it should read 1.trap.2.barrier. I have corrected it on the file attached. So I have done these so far: fisher.test(Trapz, workspace = 20, hybrid = FALSE, control = list(), or = 1, alternative = two.sided, conf.int = TRUE, conf.level = 0.95,simulate.p.value = FALSE, B = 2000) Error in fisher.test(Trapz, workspace = 2e+05, hybrid = FALSE, control = list(), : FEXACT error 501. The hash table key cannot be computed because the largest key is larger than the largest representable int. The algorithm cannot proceed. Reduce the workspace size or use another algorithm. fisher.test(Trapz, workspace = 2000, hybrid = FALSE, control = list(), or = 1, alternative = two.sided, conf.int = TRUE, conf.level = 0.95,simulate.p.value = FALSE, B = 2000) Error in fisher.test(Trapz, workspace = 2000, hybrid = FALSE, control = list(), : FEXACT error 40. Out of workspace. fisher.test(Trapz, workspace = 1e8, hybrid = FALSE, control = list(), or = 1, alternative = two.sided, conf.int = TRUE, conf.level = 0.95,simulate.p.value = FALSE, B = 2000) Error in fisher.test(Trapz, workspace = 1e+08, hybrid = FALSE, control = list(), : FEXACT error 501. The hash table key cannot be computed because the largest key is larger than the largest representable int. The algorithm cannot proceed. Reduce the workspace size or use another algorithm. fisher.test(Trapz, workspace = 20, hybrid = FALSE, control = list(), or = 1, alternative = two.sided, conf.int = TRUE, conf.level = 0.95,simulate.p.value = FALSE, B = 2000) Error: cannot allocate vector of size 7.5 Gb In addition: Warning messages: 1: In fisher.test(Trapz, workspace = 2e+09, hybrid = FALSE, control = list(), : Reached total allocation of 6027Mb: see help(memory.size) 2: In fisher.test(Trapz, workspace = 2e+09, hybrid = FALSE, control = list(), : Reached total allocation of 6027Mb: see help(memory.size) 3: In fisher.test(Trapz, workspace = 2e+09, hybrid = FALSE, control = list(), : Reached total allocation of 6027Mb: see help(memory.size) 4: In fisher.test(Trapz, workspace = 2e+09, hybrid = FALSE, control = list(), : Reached total allocation of 6027Mb: see help(memory.size) fisher.test(Trapz, workspace = 1e8, hybrid = FALSE, control = list(), or = 1, alternative = two.sided, conf.int = TRUE, conf.level = 0.95,simulate.p.value = TRUE, B = 1e5) Error: unexpected '' in So the issue could be perhaps that R cannot compute my sample as the workspace needed is too big? Is there a way around this? I think I have everything set out correctly. Is my only other alternative is to do a 2x2 fisher test for each of the variables? I attach on the pdf the Minitab result for the Chi squared test as proof (I know that getting very low p values are highly unlikely but sometimes it happens). Seeing is believing i suppose! Regards, Paul On Fri, Aug 28, 2015 at 8:56 AM, Gerrit Eichner gerrit.eich...@math.uni-giessen.de wrote: Dear Paul, quoting the email-footer: PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. So, what exactly did you try and what was the actual problem/error message? Besides that, have you noted that two of you data rows have the same name? Have you read the online help page of fisher.test
[R-es] Un asunto al margen
Cordial Saludo He conversado con algunos amigos y me han comentado que existe una plataforma de trabajo colaborativo, algo parecido a schoology o edmode la cual entre otras herramientas trae implementado a R y LaTeX, el problema es que no recuerdo el nombre y si bien entiendo que esta lista no es el espacio más apropiado para la consulta, quedaría muy agradecido por cualquier orientación al respecto, de antemano me excuso con todos si soy impertinente con mi pregunta y desde ya muchas gracias. [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] Rcpp, function signature
On 27 Aug 2015, at 17:40 , Dirk Eddelbuettel e...@debian.org wrote: Michael Meyer via R-help r-help at r-project.org writes: I am an (very) grateful user of Rcpp. Glad to hear that! But you are on the wrong mailing list. Please ask on rcpp-devel. But for the benefit of the rest of us: A NumericVector is a pointer, right? -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Office: A 4.23 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Piecewise regression using segmented package plotted in xyplot
Hi
Without a reproducible example I am only guessing
Try this
xyplot(threshold ~ age |frequency.a, data=subset(rage !is.na(threshold}
!is.na(age)),
groups = HL,
cex=0.5,
layout=c(7,4),
par.strip.tex=list(cex=0.8),
xlab=Age (years),
ylab=Threshold (dB SPL),
# na.rm=TRUE, # see above
type = c(p,l), # re-read ?xyplot: from the panel section:
panel.abline(lm(threshold~age))
panel = panel.superpose,
panel.groups =function(x,y,groups,...) {
panel.xyplot(x,y,...)
# panel.abline(segmented(lm(threshold~age),seg.Z =
~age, psi = NA, control = seg.control(K=1)))
},
)
I do not know what the commented line means-- you may need to look at the
subscripts section as well as panel.groups of ?xyplot
and ?panel.xyplot
Regards
Duncan
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
Email: home: mac...@northnet.com.au
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Sumitrajit
Dhar
Sent: Friday, 28 August 2015 14:12
To: r-help@r-project.org
Subject: [R] Piecewise regression using segmented package plotted in xyplot
Hi,
xyplot(threshold ~ age |frequency.a, data=rage,
groups=HL,
cex=0.5,
layout=c(7,4),
par.strip.tex=list(cex=0.8),
xlab=Age (years),
ylab=Threshold (dB SPL),
na.rm=TRUE,
panel=function(x,y,groups,...) {
panel.superpose(x,y,groups=HL,...)
# panel.abline(segmented(lm(threshold~age),seg.Z = ~age, psi = NA, control =
seg.control(K=1)))
panel.abline(lm(threshold~age))
},
)
Is there anyway to make the commented line work in lattice? I need to fit my
data in each panel using piecewise regression. Being able to use segmented
would make it easy.
The code above works to give me a linear fit.
Thanks for your help in advance.
Regards,
Sumit
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Re: [R] lsqlin in R package pracma
Nice and interesting! Something to remember. Lesson (for me): Always first look in Task Views on CRAN. Choose Optimization and look at Mathematical Programming Solvers. Berend On 28 Aug 2015, at 12:56, Hans W Borchers hwborch...@gmail.com wrote: I got interested in enabling the full funcionality that MATLAB's lsqlin() has, that is with equality and bound constraints. To replace an equality constraint with two inequality constraints will not work with solve.QP() because it requires positive definite matrices. I will use kernlab::ipop() instead. To handle the full MATLAB example, add the following simple linear equality constraint 3x1 + 5x2 + 7x3 + 9x4 = 4 to the example above, plus lower and upper bounds -0.1 and 2.0 for all x_i. C - matrix(c( 0.9501, 0.7620, 0.6153, 0.4057, 0.2311, 0.4564, 0.7919, 0.9354, 0.6068, 0.0185, 0.9218, 0.9169, 0.4859, 0.8214, 0.7382, 0.4102, 0.8912, 0.4447, 0.1762, 0.8936), 5, 4, byrow=TRUE) d - c(0.0578, 0.3528, 0.8131, 0.0098, 0.1388) A - matrix(c( 0.2027, 0.2721, 0.7467, 0.4659, 0.1987, 0.1988, 0.4450, 0.4186, 0.6037, 0.0152, 0.9318, 0.8462), 3, 4, byrow=TRUE) b - c(0.5251, 0.2026, 0.6721) # Add the equality constraint to matrix A Aeq - c(3, 5, 7, 9) beq - 4 A1 - rbind(A , c(3, 5, 7, 9)) b1 - c(b, 4) lb - rep(-0.1, 4) # lower and upper bounds ub - rep( 2.0, 4) r1 - c(1, 1, 1, 0) # 0 to force an equality constraint # Prepare for a quadratic solver Dmat - t(C) %*% C dvec - (t(C) %*% d) Amat - -1 * A1 bvec - -1 * b1 library(kernlab) s - ipop(-dvec, Dmat, Amat, bvec, lb, ub, r1) s # An object of class ipop # Slot primal: # [1] -0.0885 -0.0997 0.15990817 0.40895991 # ... x - s@primal # [1] -0.1000 -0.1000 0.1599 0.4090 A1 %*% x - b1 = 0 # i.e., A x = b and 3x[1] + ... + 9x[4] = 4 sum((C %*% x - d)^2)# minimum: 0.1695 And this is exactly the solution that lsqlin() in MATLAB computes. On Thu, Aug 27, 2015 at 6:06 PM, Raubertas, Richard richard_rauber...@merck.com wrote: Is it really that complicated? This looks like an ordinary quadratic programming problem, and 'solve.QP' from the 'quadprog' package seems to solve it without user-specified starting values: library(quadprog) Dmat - t(C) %*% C dvec - (t(C) %*% d) Amat - -1 * t(A) bvec - -1 * b rslt - solve.QP(Dmat, dvec, Amat, bvec) sum((C %*% rslt$solution - d)^2) [1] 0.01758538 Richard Raubertas Merck Co. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Frequency count of terms only in a given column in R
Dear Agrima As well as Sarah's seven possibilities an eighth occurs to me: you have not yet read it into R in the first place. If that is the case you may be able to use read.table to get it into a data frame with columns corresponding to your words. ?read.table may be your friend here. On 28/08/2015 15:42, Sarah Goslee wrote: Hi, On Fri, Aug 28, 2015 at 7:49 AM, agrima seth sethagr...@gmail.com wrote: i have a text file with data of the given format: white snow lived snow in snow lived place in place a place called place as place That doesn't specify the format. I can think of at least seven things that could be: a character vector a two-column matrix a one-column matrix a two-column data frame, with or without values correctly specified as character a one-column data frame, with or without values correctly specified as character The correct answer depends on what the format actually is; you need to use dput() or some other unambiguous way of providing sample data. Here are some suggestions for creating a good reproducible example: http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example Some combination of strsplit() and table(), possibly with apply(), will be the answer, though. Sarah here i have to find the frequency of the terms only in the first column (i.e.) white - 1 lived- 2 in -2 a-1 called - 1 as -1 Could you please guide me how to do the above in R. [[alternative HTML version deleted]] and please don't post in HTML, as it makes figuring out what you meant even more difficult. Sarah -- Michael http://www.dewey.myzen.co.uk/home.html __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Frequency count of terms only in a given column in R
i have a text file with data of the given format: white snow lived snow in snow lived place in place a place called place as place here i have to find the frequency of the terms only in the first column (i.e.) white - 1 lived- 2 in -2 a-1 called - 1 as -1 Could you please guide me how to do the above in R. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Frequency count of terms only in a given column in R
Hi, On Fri, Aug 28, 2015 at 7:49 AM, agrima seth sethagr...@gmail.com wrote: i have a text file with data of the given format: white snow lived snow in snow lived place in place a place called place as place That doesn't specify the format. I can think of at least seven things that could be: a character vector a two-column matrix a one-column matrix a two-column data frame, with or without values correctly specified as character a one-column data frame, with or without values correctly specified as character The correct answer depends on what the format actually is; you need to use dput() or some other unambiguous way of providing sample data. Here are some suggestions for creating a good reproducible example: http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example Some combination of strsplit() and table(), possibly with apply(), will be the answer, though. Sarah here i have to find the frequency of the terms only in the first column (i.e.) white - 1 lived- 2 in -2 a-1 called - 1 as -1 Could you please guide me how to do the above in R. [[alternative HTML version deleted]] and please don't post in HTML, as it makes figuring out what you meant even more difficult. Sarah -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Rcpp, function signature
peter dalgaard pdalgd at gmail.com writes: But for the benefit of the rest of us: A NumericVector is a pointer, right? Effectively even though it is not treated as one by the users. But you know what P in SEXP stands for, and Rcpp objects really are what we call proxy objects for the respective underlying SEXP objects. Searching the rcpp-devel list archives for the clone() function will bring a number of preceding discussions. As I said in the previous email, this list is not the best place to discuss Rcpp matters --- rcpp-devel is. Please feel free to bring follow-up questions there. Dirk __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Referencing a component of a large object (Error: slot NULL)
s - paste('chain:,j,' ,sep=)
You don't want the single quotes around the string you are constructing.
Try
s - paste0(chain:, j)
which will give you, e.g.,
chain:1
instead of
'chain:1'
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Aug 28, 2015 at 6:08 PM, Andrea Lamont alamont...@gmail.com wrote:
Hm.
It still doesn't seem to be working and has me completely stumped.
This works:
mi.control.i@data[['chain:1']]@variables$y.obs.tx@parameters[30,]
But this doesn't:
mi.control.i@data[[s]]@variables$y.obs.tx@parameters[30,]
Error: trying to get slot variables from an object of a basic class
(NULL) with no slots
s
[1] 'chain:1'
On Fri, Aug 28, 2015 at 5:52 PM, William Dunlap wdun...@tibco.com wrote:
tf - vector(list, numberofchains)
for (j in 1:numberofchains){
s - paste('chain:,j,' ,sep=)
tf[[j]] = mi.control.i@data$s@variables$y.obs.tx@parameters[30,]
}
If you want the component whose name is the value of the variable 's'
use data[[s]].
The syntax 'data$s' means to get the component of 'data' called s,
usually the same as 'data[[s]]'.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Aug 28, 2015 at 12:08 PM, Andrea Lamont alamont...@gmail.com
wrote:
I am running a simulation and need to refer to a matrix of parameters
from
a large object. Here is a snippet of the object structure itself:
Formal class 'mi' [package mi] with 3 slots
..@ call : language .local(y = y, n.chains = ..2, max.minutes =
2)
..@ data :List of 100
.. ..$ chain:1 :'data.frame':1 obs. of 76 variables:
Formal class 'missing_data.frame' [package mi] with 17 slots
.. .. .. ..@ .Data : list()
.. .. .. ..@ variables :List of 76
.. .. .. .. ..$ y.obs.tx:Formal class 'binary' [package mi] with 27
slots...
The parameter list I need can be referenced by:
mi.control.i@data$'chain:1'@variables$y.obs.tx@parameters[30,]
Since this is a simulation, I would like to grab the parameters from each
chain and join them together in a matrix. I have this:
tf - vector(list, numberofchains)for (j in 1:numberofchains){
s - paste('chain:,j,' ,sep=)
tf[[j]] = mi.control.i@data$s@variables$y.obs.tx@parameters[30,]}
Within the loop, however, the reference to the object (tf[[j]]) does not
work. I get an error that reads
Error: trying to get slot variables from an object of a basic class
(NULL) with no slots
I am happy to send reproducible code, however, I obtain this object
through
the package 'mi' and is computationally intensive. I'm not sure if it
makes
it easier for folks. Let me know.
Any ideas?
--
Andrea Lamont, PhD
Post-Doctoral Fellow
University of South Carolina
Columbia, SC 29208
*Please consider the environment before printing this email.*
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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.
--
Andrea Lamont, PhD
Post-Doctoral Fellow
University of South Carolina
Columbia, SC 29208
*Please consider the environment before printing this email.*
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and provide commented, minimal, self-contained, reproducible code.
[R] Change the maximum likelihood of multinomial logic model in R
Dear all, Can anyone help me to change the maximum likelihood of multinomial logic model in R? Thanks __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Change the maximum likelihood of multinomial logic model in R
Dear All, can anybody help me with an example on how to use mlogit.optim? Regards On Aug 28, 2015, at 9:00 PM, Alaa Sindi alaasi...@gmail.com wrote: Dear all, Can anyone help me to change the maximum likelihood of multinomial logic model in R? Thanks __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] heat map labeling
Hi Angela, It depends upon what you want to illustrate. If you are just interested in the relative values, you can suppress the labels. One solution is to create a very high PDF to look at the colors, which you could then expand and scroll around to see the labels. Jim On Sat, Aug 29, 2015 at 6:06 AM, Angela via R-help r-help@r-project.org wrote: Hi Jim, Thank you, that definitely reduced it but there are still about 600 genes, so too many to label. It does make the heat map itself look cleaner. Maybe labeling isn't necessary for the heat map? -Angela [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Multiple Integrals
Hello all,
For a study I want to find E|X1/3+X2/3+X3/3-X4| for variables following
lognormal distribution. To get the value we need to use four integrals. This is
the code which I is used
fx-function(x){
dlnorm(x,meanlog=2.185,sdlog=0.562)
}
U31-integrate(function(y1) { sapply(y1, function(y1) {
+ integrate(function(y2){ sapply(y2, function(y2) {
+ integrate(function(x1){ sapply(x1, function(x1) {
+ integrate(function(x2)
+ abs(y1/3+y2/3+x1/3-x2)*fx(y1)*fx(y2)*fx(x1)*fx(x2),0, Inf)$value
+ })},0, Inf)$value })},0, Inf)$value})},0,Inf)$value
The error I received is the following:
Error in integrate(function(y2) { :
maximum number of subdivisions reached
I can understand the problem, but I am unable to figure out what can be done..
It would be great if you can let me know a solution to the problem so as to
find a value for the integral.
Shant
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and provide commented, minimal, self-contained, reproducible code.
[R] kknn::predict and kknn$fitted.values
I am noticing that there is a difference between the fitted.values returned by train.kknn, and the values returned using predict with the same model and dataset. For example: data (glass) tmp - train.kknn(Type ~ ., glass, kmax=1, kernel=rectangular, distance=1) tmp$fitted.values [[1]] [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 1 [62] 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 1 2 2 1 2 2 5 2 2 2 6 2 2 2 2 2 2 2 2 2 2 2 2 [123] 2 2 2 2 3 2 2 2 5 5 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 2 3 3 3 3 3 3 2 3 7 5 5 5 5 5 5 5 5 5 5 2 5 6 6 6 6 6 6 6 [184] 2 6 7 7 2 6 7 7 7 7 7 7 7 7 7 7 7 7 5 7 7 7 7 7 7 7 7 7 7 7 7 attr(,kernel) [1] rectangular attr(,k) [1] 1 Levels: 1 2 3 5 6 7 predict (tmp,glass) [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 [62] 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 [123] 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 [184] 6 6 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 Levels: 1 2 3 5 6 7 When I check the confusion matricies for these I see that fitted.values is giving some confusion, that is, like it is a true fit, whereas predict is returning the exact answers. table (tmp$fitted.values[[1]],glass$Type) 1 2 3 5 6 7 1 69 4 0 0 0 0 2 1 67 2 1 1 1 3 0 1 15 0 0 0 5 0 3 0 11 0 1 6 0 1 0 0 8 1 7 0 0 0 1 0 26 table (predict(tmp,glass),glass$Type) 1 2 3 5 6 7 1 70 0 0 0 0 0 2 0 76 0 0 0 0 3 0 0 17 0 0 0 5 0 0 0 13 0 0 6 0 0 0 0 9 0 7 0 0 0 0 0 29 Can anyone clarify what fitted.values and predict actually do? I would have expected they would give the same output. Thanks... Jonathan -- View this message in context: http://r.789695.n4.nabble.com/kknn-predict-and-kknn-fitted-values-tp4711625.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Referencing a component of a large object (Error: slot NULL)
I am running a simulation and need to refer to a matrix of parameters from
a large object. Here is a snippet of the object structure itself:
Formal class 'mi' [package mi] with 3 slots
..@ call : language .local(y = y, n.chains = ..2, max.minutes = 2)
..@ data :List of 100
.. ..$ chain:1 :'data.frame':1 obs. of 76 variables:
Formal class 'missing_data.frame' [package mi] with 17 slots
.. .. .. ..@ .Data : list()
.. .. .. ..@ variables :List of 76
.. .. .. .. ..$ y.obs.tx:Formal class 'binary' [package mi] with 27 slots...
The parameter list I need can be referenced by:
mi.control.i@data$'chain:1'@variables$y.obs.tx@parameters[30,]
Since this is a simulation, I would like to grab the parameters from each
chain and join them together in a matrix. I have this:
tf - vector(list, numberofchains)for (j in 1:numberofchains){
s - paste('chain:,j,' ,sep=)
tf[[j]] = mi.control.i@data$s@variables$y.obs.tx@parameters[30,]}
Within the loop, however, the reference to the object (tf[[j]]) does not
work. I get an error that reads
Error: trying to get slot variables from an object of a basic class
(NULL) with no slots
I am happy to send reproducible code, however, I obtain this object through
the package 'mi' and is computationally intensive. I'm not sure if it makes
it easier for folks. Let me know.
Any ideas?
--
Andrea Lamont, PhD
Post-Doctoral Fellow
University of South Carolina
Columbia, SC 29208
*Please consider the environment before printing this email.*
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change the maximum likelihood of multinomial logic model in R
Have you looked at the help for mlogit.optim? At the minimum you need a likelihood function and starting value(s). If you don’t understand the function syntax you may have difficulty interpreting any output that you do get. On Aug 28, 2015, at 11:26 AM, Alaa Sindi alaasi...@gmail.com wrote: Dear All, can anybody help me with an example on how to use mlogit.optim? Regards On Aug 28, 2015, at 9:00 PM, Alaa Sindi alaasi...@gmail.com wrote: Dear all, Can anyone help me to change the maximum likelihood of multinomial logic model in R? Thanks __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Separate point sizes in rgl.points()?
Hi,
DrawDensity3D-function in package VecStatGraphs3D utilizes
rgl.points-function {rgl}:
function (vectors, Div = 40, Layers = 3, DrawAxes = FALSE)
{
open3d(windowRect = c(100, 100, 800, 800))
bg3d(white)
Cx = vectors[, 1]
Cy = vectors[, 2]
Cz = vectors[, 3]
Cr - kde3d(x = Cx, y = Cy, z = Cz, n = Div)
th - seq(min(Cr$d), max(Cr$d), len = Layers + 2)
ramp - colorRamp(c(white, yellow, red))
colo - rgb(ramp(seq(0, 1, length = Layers)), maxColorValue = 255)
al - seq(0.1, 0.6, len = Layers)
module = sqrt(Cx * Cx + Cy * Cy + Cz * Cz)
spheres3d(0, 0, 0, radius = max(module), color = black,
front = line, back = line, lwd = 1, smooth = TRUE,
lit = TRUE, line_antialias = FALSE, alpha = 0.2)
x - c(0, max(module), 0, 0)
y - c(0, 0, max(module), 0)
z - c(0, 0, 0, max(module))
labels - c(, X, Y, Z)
i - c(1, 2, 1, 3, 1, 4)
text3d(x, y, z, labels, adj = 0.8, cex = 1.5, font = 2, color =
black)
segments3d(x[i], y[i], z[i], lwd = 3)
rgl.points(x = Cx, y = Cy, z = Cz, size = 3, color = black)
contour3d(Cr$d, level = th[c(-1, -(Layers + 2))], x = Cr$x,
y = Cr$y, z = Cr$z, alpha = al, color = colo, add = TRUE,
engine = rgl, fill = TRUE, smooth = 2, material = shiny)
if (DrawAxes == TRUE) {
axes3d()
}
}
Is it somehow possible to define the sizes of the points all separately?
I tried by adding ”Psize” to function arguments and changing
rgl.points(x = Cx, y = Cy, z = Cz, size = Psize, color = black”),
then giving individual point size to each point but this does not work.
This does’t work either:
for(i in 1:length(Cx))
{
rgl.points(x=Cx[i], y=Cz[i], z=Cz[i], size=PSize[i], col= Colors[i])
}
Atte Tenkanen
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] heat map labeling
Hi Jim, Thank you, that definitely reduced it but there are still about 600 genes, so too many to label. It does make the heat map itself look cleaner. Maybe labeling isn't necessary for the heat map? -Angela On Fri, 8/28/15, Jim Lemon drjimle...@gmail.com wrote: Subject: Re: [R] heat map labeling Cc: r-help mailing list r-help@r-project.org Date: Friday, August 28, 2015, 6:16 AM Hi Angela,Assuming the above data frame is named angela.df: angela.mat-as.matrix(angela.df[,2:3])angela.mat-angela.mat[apply(angela.mat,1,function(x) all(x) 0),] will remove all of the rows that have contain at least one zero. Jim On Fri, Aug 28, 2015 at 9:00 AM, Angela via R-help r-help@r-project.org wrote: Hello, I have a dataset of 985 genes, looks something like the ones below. I want to label only those with the high intensities, since labeling all doesn't show up. Is there a way to do that? If not, is there a way to pull out the highest ones (say, highest 50, or those above X amount) and only show [[elided Yahoo spam]] -Angela Z transforming gives all cells the same value, just + or - (for example, all have 0.5 and -0.5). The researchers want the actual values used. Gene var1 var2 A 800 0 B 25 30 C 75 200 D 0 0 E 400 600 E 500 70 E 100 100 F 600 600 F 70 827460 G 420930 40 H 0 0 H 100 100 I 70 60 J 0 70 K 0 0 L 20 50 L 100 300 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Referencing a component of a large object (Error: slot NULL)
tf - vector(list, numberofchains)
for (j in 1:numberofchains){
s - paste('chain:,j,' ,sep=)
tf[[j]] = mi.control.i@data$s@variables$y.obs.tx@parameters[30,]
}
If you want the component whose name is the value of the variable 's'
use data[[s]].
The syntax 'data$s' means to get the component of 'data' called s,
usually the same as 'data[[s]]'.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Aug 28, 2015 at 12:08 PM, Andrea Lamont alamont...@gmail.com
wrote:
I am running a simulation and need to refer to a matrix of parameters from
a large object. Here is a snippet of the object structure itself:
Formal class 'mi' [package mi] with 3 slots
..@ call : language .local(y = y, n.chains = ..2, max.minutes =
2)
..@ data :List of 100
.. ..$ chain:1 :'data.frame':1 obs. of 76 variables:
Formal class 'missing_data.frame' [package mi] with 17 slots
.. .. .. ..@ .Data : list()
.. .. .. ..@ variables :List of 76
.. .. .. .. ..$ y.obs.tx:Formal class 'binary' [package mi] with 27
slots...
The parameter list I need can be referenced by:
mi.control.i@data$'chain:1'@variables$y.obs.tx@parameters[30,]
Since this is a simulation, I would like to grab the parameters from each
chain and join them together in a matrix. I have this:
tf - vector(list, numberofchains)for (j in 1:numberofchains){
s - paste('chain:,j,' ,sep=)
tf[[j]] = mi.control.i@data$s@variables$y.obs.tx@parameters[30,]}
Within the loop, however, the reference to the object (tf[[j]]) does not
work. I get an error that reads
Error: trying to get slot variables from an object of a basic class
(NULL) with no slots
I am happy to send reproducible code, however, I obtain this object through
the package 'mi' and is computationally intensive. I'm not sure if it makes
it easier for folks. Let me know.
Any ideas?
--
Andrea Lamont, PhD
Post-Doctoral Fellow
University of South Carolina
Columbia, SC 29208
*Please consider the environment before printing this email.*
[[alternative HTML version deleted]]
__
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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.
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem loading mvabund package
Thank you Michael :) S On 28 August 2015 at 12:09, Michael Dewey li...@dewey.myzen.co.uk wrote: Dear Suparna, See below On 28/08/2015 10:22, Suparna Mitra wrote: Hello, Can anybody please help me with mvabund package installation? I downloaded and installed it. It seems installed, but I can't load the package. Here is what I tried: install.packages(/Users/smitra/Documents/Soft/mvabund_3.10.4.tgz, repos = NULL, type=source) * installing *binary* package ‘mvabund’ ... * DONE (mvabund) library(mvabund) Error in loadNamespace(i, c(lib.loc, .libPaths()), versionCheck = vI[[i]]) : there is no package called ‘tweedie’ There are two possibilities here 1 - you have not installed tweedie 2 - you have and R cannot find it Error: package or namespace load failed for ‘mvabund’ Any help will be great. Thanks. Suparna [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. -- Michael http://www.dewey.myzen.co.uk/home.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Change the maximum likelihood of multinomial logic model in R
On Aug 28, 2015, at 11:26 AM, Alaa Sindi wrote: Dear All, can anybody help me with an example on how to use mlogit.optim? That is just a helper function. The examples are all on the mogit help page. On Aug 28, 2015, at 9:00 PM, Alaa Sindi alaasi...@gmail.com wrote: Dear all, Can anyone help me to change the maximum likelihood of multinomial logic model in R? Thanks __ David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Referencing a component of a large object (Error: slot NULL)
I am running a simulation and need to refer to a matrix of parameters from
a large object. Here is a snippet of the object structure itself:
Formal class 'mi' [package mi] with 3 slots
..@ call : language .local(y = y, n.chains = ..2, max.minutes = 2)
..@ data :List of 100
.. ..$ chain:1 :'data.frame':1 obs. of 76 variables:
Formal class 'missing_data.frame' [package mi] with 17 slots
.. .. .. ..@ .Data : list()
.. .. .. ..@ variables :List of 76
.. .. .. .. ..$ y.obs.tx:Formal class 'binary' [package mi] with 27 slots...
The parameter list I need can be referenced by:
mi.control.i@data$'chain:1'@variables$y.obs.tx@parameters[30,]
Since this is a simulation, I would like to grab the parameters from each
chain and join them together in a matrix. I have this:
tf - vector(list, numberofchains)for (j in 1:numberofchains){
s - paste('chain:,j,' ,sep=)
tf[[j]] = mi.control.i@data$s@variables$y.obs.tx@parameters[30,]}
Within the loop, however, the reference to the object (tf[[j]]) does not
work. I get an error that reads
Error: trying to get slot variables from an object of a basic class
(NULL) with no slots
I am happy to send reproducible code, however, I obtain this object through
the package 'mi' and is computationally intensive. I'm not sure if it makes
it easier for folks. Let me know.
Any ideas?
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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-es] Una inquietud
Estoy elaborando un pequeño script en el que pido a los usuarios introducir un conjunto de valores, esto lo estoy haciendo con la función cat() el asunto es que no se como hacer para que luego de leídos, ellos sean asignados a un vector x-c(???...??) agradezco cualquier orientación y desde ya gracias. [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R-es] una ayudita
Estoy elaborando un pequeño script en el que pido a los usuarios introducir un conjunto de valores, esto lo estoy haciendo con la función cat() el asunto es que no se como hacer para que luego de leídos, ellos sean asignados a un vector x-c(???...??) agradezco cualquier orientación y desde ya gracias. [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
