Hi Pijush,
Without access to the data, we can only guess. However, in such cases
the problem is often a factor variable where you expect a numeric one.
Try this:
is.factor(x)
and if the answer is TRUE, you have found your problem.
Jim
On Wed, Jun 3, 2015 at 4:13 PM, Pijush Das wrote:
> Dear S
Dear Sir,
I am facing an error when I am trying to use svm and found similar kind of
problem faced by other. But I unable to solve the problem. The problem is
given below.
> rm(list=ls(all=TRUE))
> CombinedGeneList <- read.xlsx(file.choose(), sheet = 1, colNames =
TRUE,rowNames = TRUE)
> Noemalis
On Jun 2, 2015, at 3:31 PM, t.k. t.k. wrote:
> Hi everyone
> This is a general question.
> I have imported a "Dates" variable as character. I used the "as.Date"
> command and convert it to Date,format..
>
> I want to change the display of the date to weekday (e.g., I want my time
> series to be
Hi tk.tk.,
Once you have the date as a Date object, leave it that way for your
time series analysis. If you want to get labels (as opposed to date
values) in the format you request, do this:
date1<-"3/6/2015"
Date1<-as.Date(date1,"%d/%m/%Y")
[1] "2015-06-03"
# get a character date in the -- format
All,
I am using gglpot to produce combination density and histogram plots, which are
actually kinda cool, everything works well and the plots look nice. Â However
after each plot run I receive the following message:
stat_bin: binwidth defaulted to range/30. Use 'bin width = x' to adjust this.
Hi everyone
This is a general question.
I have imported a "Dates" variable as character. I used the "as.Date"
command and convert it to Date,format..
I want to change the display of the date to weekday (e.g., I want my time
series to be instead of 15-04-2010 as Friday-04-2010). I used the "format"
Hi Ben,
While Jean's answer looks correct, I think that there is something
amiss with your specification of the problem. You have eight
combinations in your "possibilities". So if you draw samples "x"
where:
If p(x = possibilities[1,] | possibilities[5,]) = 0.5 AND
p(x = possibilities[2,] | possi
Hi Dan,
With a bit of arm-twisting, you can get pretty close with pyramid.plot
(plotrix).
lx<-c(0,0,0,0,0,0,0,0,0,0,0,0,0,20,30,55,60,60,70,80,100,90)
rx<-c(90,95,95,80,70,65,75,54,52,50,40,20,0,0,0,0,0,0,0,0,0,0)
library(plotrix)
x11(width=10)
pyramid.plot(lx,rx,labels=rep("",22),gap=0,top.labels
Hi Hanna,
That is because "dotplot" is a lattice graphics function and "legend"
is a base graphics function. There are two things you can do to fix
this. One is to use the "dotchart" function in base graphics. The
other is to use simpleKey in the latticeExtra package for the legend.
Jim
On Wed,
Hi all,
I wanted to add the legend to a dotplot using legend funciton. If
does not seem to be working? Anyone have any suggestions?
Thanks!
Hanna
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Hi Erica,
The problem may be that you are specifying a grouping factor (mdl) in
which the group sizes are unequal. If one case in group "tot" is
missing, is it possible to identify the corresponding cases in the
other factor levels and delete them?
Jim
On Tue, Jun 2, 2015 at 11:59 PM, Erica Csek
Hi valerio,
This is a guess, but try running your code with "htmlize" (prettyR).
Change the "pdf" call to:
png("C:/Users/Administrator/Desktop/treemap1.png")
map.market(id = data1$Id, area = data1$size, group = data1$Storage,
color = data1$col, lab=c(TRUE,TRUE), main="Test Map")
dev.off()
png("C:
On 2015-06-02 Tue 14:20, Jacob Wegelin wrote:
I want to use a specialized function to compute knots on the fly to pass to
splines::bs within a call to nlme::lme.
The specialized function will compute knots from the x-axis variable (the x
argument in a call to splines::bs).
The syntax works w
Easy to write - here's a quick piece of code (without bells and whistles
though...)
dat1 <- rnorm(1000)
dat2 <- rgamma(3000, shape=3)
biHist <- function(d1, d2) {
col1 <- "#FFCCBB"
col2 <- "#BBCCFF"
hAll <- hist(c(d1, d2))
hD1 <- hist(d1, breaks=hAll$breaks)
On Jun 2, 2015, at 11:13 AM, Dan Abner wrote:
> Hi all,
>
> Does anyone know how to create a graph of propensity scores like the
> one on the left in the attachment? I can easily generate the one on
> the right: How does one force the respective histograms to share the
> same rotated x-axis? Is
You model is reaching the error threshold otherwise you would be receiving
an 'actual' error message. You model is just converging very quickly. If
you want to see the error reducing, change lifesign="minimal" to
lifesign="full" and set the lifesign.step=1 to make it very verbose for
this model.
Hi,
Developing a Neural Network in R to predict the quality of wine.Attached is
the Wine Data Set.Tried twice once by selecting specific features and once
by using all features to predict quality of wine
Each time I run the Neural Network I am getting a huge error.Tried using
feature selection a
I want to use a specialized function to compute knots on the fly to pass to
splines::bs within a call to nlme::lme.
The specialized function will compute knots from the x-axis variable (the x
argument in a call to splines::bs).
The syntax works with lm. But when I try it with lme, the followin
Points later in the input vectors may be obscuring earlier
points. If that is the problem then use pch="." or cex=.2
(or some other small number) to make the plot symbols
smaller so they don't overlap as much. Sometimes using
transparency helps also - try using adjustcolor(Color(n), alpha.f=0.5)
HI All,
i need to call the tree map function in R to display my multiple disks usage,
using ‘portfolio' library. I need furthermore to generate multiple page each
showing the treemap of each disk. It works fine if use pdf file , but my boss
wants to save the result into an html file. Any help?
Dear listers,
I'm performing a PERMANOVA (adonis{vegan}) to compare the results (ROC,
TSS) of models based on two factors (model, algo). I was not able to find a
pairwise test for adonis, on PRIMER it would be a Tukey test. Though, I
chose to perform a pairwise.t.test what would be quite simple. H
Ben,
Perhaps I am missing something, but couldn't you simply reduce your
possibilities to:
possibilities[c(1, 5, 2, 4), ]
Var1 Var2 Var3
[1,] "A" "A" "C"
[2,] "A" "A" "T"
[3,] "C" "A" "C"
[4,] "C" "G" "C"
If you sample from these four rows you will have a 50% chance that Var1 and
Va
Try this.
Jean
D <- structure(list(
id = structure(1:6, .Label = c("O13297", "O13329", "O13525",
"O13539", "O13541", "O13547"), class = "factor"),
X = c(44.44, 31.272085, 6.865672, 14.176245, 73.275862,
28.991597),
Y = c(21.6122, 4.0159, 2.43884, 7.81217, 3.59012, 258.999)),
.
Package nlmrt (function nlxb) tries to use symbolic derivatives. In
fact, Duncan Murdoch and I have a very slowly developing nls14 package
to substitute for nls that should advance this even further.
nlxb also allows "masked" (i.e., fixed) parameters, which would let you
combine your runs, fixing
I wrote Rcgmin to do bounds constraints only. Linear constraints are
much more complicated to include.
If your constraints are equality ones, you could "solve", but that could
make it awkward to evaluate the gradient.
For inequality constraints, especially if there are only a couple, I
think I'd
> plot(1:33292, 1:33292,col=Color(33292))
Error in plot.xy(xy, type, ...) : could not find function "Color"
Please tell us what you're trying to accomplish. "not working" is rather vague.
Without a reproducible example that includes some sample data (fake is
fine), the code you used, and some cle
Dear R-List,
I have a set of possibilities I want to sample from:
bases <- list(c('A', 'C'), c('A', 'G'), c('C', 'T'))
possibilities <- as.matrix(expand.grid(bases))
>possibilities
Var1 Var2 Var3
[1,] "A" "A" "C"
[2,] "C" "A" "C"
[3,] "A" "G" "C"
[4,] "C" "G" "C"
[5,] "A" "A" "T"
[6,]
sir i done this plot(1:33292, 1:33292,col=Color(33292)) command then it gives
the coloured line but again it is not working in my data why?
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Hello everyone,
I have a data frame D with 4 columns id,X,Y,C.
I want to plot a simple scatter plot of D$X vs. D$Y and using D$C values as a
color. (id is just a text string not used for the plot)
But actually, I don't want to use the raw values of D$C, I would prefer to
calculate the average v
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