Hi
see in line
> -Original Message-
> From: R-help On Behalf Of N Meriam
> Sent: Tuesday, January 8, 2019 3:08 PM
> To: r-help@r-project.org
> Subject: [R] Warning message: NAs introduced by coercion
>
> Dear all,
>
> I have a .csv file called df4. (15752 obs. of 264 variables).
> I appl
Hi
I cannot help you with kernlab
> > pred = predict(mod, df, type = "probabilities")
> > acc = table(pred, df$cons)
> Error in table(pred, df$cons) : all arguments must have the same length
> which again is weird since mod, df and df$cons are made from the same
> dataframe.
Why not check leng
t;.
Cheers
Petr
> -Original Message-
> From: Luigi Marongiu
> Sent: Tuesday, January 8, 2019 4:40 PM
> To: PIKAL Petr
> Cc: r-help
> Subject: Re: [R] error in plotting model from kernlab
>
> Hi,
> the maintainer hasn't answered yet. The problem with &#
elaborate your results further and
it is always good to have appropriately structured results.
Cheers
Petr
From: Rachel Thompson
Sent: Tuesday, January 8, 2019 4:24 PM
To: PIKAL Petr
Cc: r-help mailing list
Subject: Re: [R] Mailinglist
Hi
Thank you for your help and suggestions!
I have tried
From: N Meriam
Sent: Tuesday, January 8, 2019 4:36 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: [R] Warning message: NAs introduced by coercion
I see...
Here's a portion of what my data looks like (csv file attached).
I run again and here are the results:
df4 <- read.
And as you use bioconductor related package you probably could get better
answers in specialised biconductor help
https://www.bioconductor.org/help/
Cheers
Petr
From: N Meriam
Sent: Tuesday, January 8, 2019 4:36 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: [R] Warning message: NAs
Hi
I do not use Rcmdr but from documentation it seems to me that it does not have
much to do with importing data from Excel.
So without some additional info from your side (at least used commands) you
hardly get any reasonable answer.
Cheers
Petr
> -Original Message-
> From: R-help O
Hi
If you want to remove rows with NA values from your data you could use
?complete.cases
or
t2 <- t1[!is.na(t1$Petal.Width),]
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Rui Barradas
> Sent: Saturday, January 12, 2019 12:55 PM
> To: Ernest Han ; r-help@r-project.org
1$Petal.Width)
[1] NA
> mean(t1$Petal.Width, na.rm=T)
[1] 1.147101
>
Cheers
Petr
> -Original Message-
> From: Ernest Han
> Sent: Wednesday, January 16, 2019 3:27 AM
> To: PIKAL Petr
> Cc: r-help@r-project.org
> Subject: Re: [R] NA rows appeared in data.frame
>
Hi
Instead of attachment which is usually removed you should use dput
Something like output from
dput(head(yourdata,30))
To remove duplicate values see
unique or duplicated
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Kevin Wamae
> Sent: Thursday, January 17, 2019 1:2
Hi
Or you could use rounding.
which(round(lut, 3)==1.8)
[1] 401
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of Ben Tupper
> Sent: Thursday, January 17, 2019 2:43 PM
> To: POLWART, Calum (COUNTY DURHAM AND DARLINGTON NHS
> FOUNDATION TRUST)
> Cc: r-help@r-project.org
> Sub
)
> Sent: Thursday, January 17, 2019 2:56 PM
> To: PIKAL Petr ; Ben Tupper
> Cc: r-help@r-project.org
> Subject: RE: [R] I can't get seq to behave how I think it should
>
> Thanks guys.
>
> I've used Petr's method and its working for me.
>
> If the data had been
Hi
what about to create all histograms in pdf device?
see
?pdf
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of akshay kulkarni
> Sent: Tuesday, January 22, 2019 2:21 PM
> To: R help Mailing list
> Subject: [R] large number of scrollable histograms
>
> dear members,
>
arni
Sent: Wednesday, January 23, 2019 8:01 AM
To: PIKAL Petr ; R help Mailing list
Subject: Re: large number of scrollable histograms
dear pikal,
Thanks for the suggestion
I have checked ?pdf. The syntax is:
> pdf("sample.pdf", 7, 5)
> hist(vector1)
Hi
Well, I tried google and this
https://www.jessesadler.com/post/network-analysis-with-r/
seems to provide some insights.
If it is not appropriate to your situation, you should be more specific.
Cheers
Petr
BTW, what is factor in continuous scale? I am almost sure that factors are
discrete
do not know R basics and want to do highly sophisticated
analysis. Spending few hours reading R intro could save you much time and
headache.
Sorry, I could not help you with network analysis as I do not know anything
about it.
Cheers
Petr
From: Bhubaneswor Dhakal
Sent: Tuesday, January 29, 201
Dear all
I enclose 3 electron microscope images in which I would like to evaluate plane
spacing.
Before I start to dig deeper and use trial and error in trying to find some
packages/functions for such pattern evaluation in electron microscopy pictures
I would like to ask if anybody could point
: S Ellison
> Sent: Monday, February 11, 2019 12:54 PM
> To: PIKAL Petr ; r-help@r-project.org
> Subject: RE: pattern evaluation in electron microscopy images
>
> Not really my field, but would you not approach this using FFT on selected
> regions?
>
> I think IMageJ ha
Hi
and if you consider stringsAsFactor option a nuisance you probably could use
as.character
df1$x2[match(df2$x1, df1$x1)] <- as.character(df2$x2)
Slightly different approach is merge
merge(df1, df2, by=c("x1"), all=T)
which gives you additional column from df2 with NA values together with mer
Hi
Could you show what is your intention with your data? What do you mean by sort
data to have the same number of leaves? Do you want to trim excessive rows in
both data.frames to meet such condition?
I would suggest using merge.
merge(test1, test2, by.x=c("plot", "plant"), by.y=c("plot_lai",
Dear all
Sorry, this is probably the most off-topic mail I have ever sent to this help
list. However maybe somebody could point me to right direction or give some
advice.
In microscopy particle counting you have finite viewing field and some
particles could be partly outside of this field. My
for bigger particle.
OTOH, if I take your first reasoning I get quite satisfactory values.
> 1-(10-c(0.1, 1))* (10-c(0.1,1))/(10^2)
[1] 0.0199 0.1900
Cheers.
Petr
> -Original Message-
> From: Jim Lemon
> Sent: Thursday, February 21, 2019 12:24 AM
> To: Rolf Turner
> Cc
OK. I got it.
Thanks.
Petr
> -Original Message-
> From: Jim Lemon
> Sent: Thursday, February 21, 2019 11:36 AM
> To: PIKAL Petr
> Cc: Rolf Turner ; r-help@r-project.org
> Subject: Re: [R] particle count probability
>
> Hi Petr,
> My second message was t
Hi.
Without going too deep in your messy code:
"title" is function to make titles in plots.
therefore
titles <- c(title, extractTitle(data.combined[i, "name"]))
put in your titles object in the first place this "title" function.
And I believe, that instead of multiple if/else there is better v
Hi
aggregate is another option.
aggregate(df[, 2:4], list(df$Year), sum)
Cheers
Petr
> -Original Message-
> From: R-help On Behalf Of
> p...@philipsmith.ca
> Sent: Monday, March 4, 2019 4:15 AM
> To: r-help@r-project.org
> Subject: [R] Tidyverse data frame conversion from monthly to an
Hi
Do you want something like this?
> x <- c(1,2,NA, 3, 4, 5, NA, 6,7,8, NA, NA, 9,10)
> y <- c(1,2,NA, NA, 3, 4, 5, 6, NA, 7,8, NA, NA, 9,10)
> identical(x[which(!is.na(x))], y[which(!is.na(y))])
[1] TRUE
If I expect NA and want to extract or compare something, I tend to use which to
select onl
Hi
You shouldn't use HTML formating unless you want to surprise us with weird and
messy email.
function dcast is in data.table and/or in reshape2 packages. For both
install.packages("data.table")
install.packages("reshape2")
followed by
library(data.table)
library(reshape2)
should be enough t
Hallo Bernard
I did not follow all emails in this thread but it seems to me that your request
is similar to Bioconductor packages dealing with Flow Cytometry data.
Especially flowViz package is designed to visualise such data.
Cheers
Petr
> -Original Message-
> From: R-help On Behalf
Hi
If your problem is not bigger than example (regarding columns following maybe
can help (I shortened target.column to tc)
dt$target.value <- dt[,3]*(dt$tc==3)+dt[,2]*(dt$tc==2)+dt[,1]*(dt$tc==1)
and changed dt to data frame as I do not use data table. But it shall work with
data table too.
Hi
It is not exactly clear what do you want from your function.
foo <- function(x, y, option = TRUE)
in that cese you want to return list
x,y,z or x,y,z,w with w=NA
the same apply with
foo <- function(x, y, option = FALSE)
return is
x,y,w or x,y,z=NA,w?
anyway I would use option like that
{
Hi
I like to use logical values directly in computations if possible.
yourData[,10] <- yourData[,9]/(yourData[,8]+(yourData[,8]==0))
Logical values are automagicaly considered FALSE=0 and TRUE=1 and can be used
in computations. If you really want to change 0 to 1 in column 8 you can use
yourDa
Hi
see in line
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Pavneet Arora
> Sent: Monday, July 28, 2014 11:08 AM
> To: r-help@r-project.org
> Subject: [R] Differencing between 2 previous values
>
> Hello All,
>
> I am tryi
Hi
Above what David said, there is chance that you do not need cycle for your
computation.
From what you describe about your csv files there seems to be some mismatch in
your write.csv statement.
Make a small example code together with data set, preferably as an output from
dput, and try agai
Hi
Maybe
?aggregate
Use dput for data presentation and no HTML as everything gets scrambled with it.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Abhinaba Roy
> Sent: Wednesday, July 30, 2014 2:46 PM
> To:
Hi
Maybe others will disagree but I find for cycle for this type of task better
than sapply.
for(i in 1:length(ind)) {
if (there are more than 3 date items*) {
postscript(ind[i])
do all plotting
dev.off()
}}
If you want to plot with gaps you need to add all relevant YEARs for x axis
with mis
Hi
I do not see any solution without loops but maybe others find it.
I think that you can do it in one loop. Best structure for loop will be list.
In each cycle you will compute matrix with diagonal NA
mat<-matrix(1,nrow=number of items, ncol=number of items)
diag(mat) <- NA
apply(price chunk
Hi
If your function works for you why to bother with apply? If it does not give
you required results, please post some data and show us what is expected.
I would remove unlist from your function and declare list for storing data,
which seems to me more natural for storing results.
t.test.list=
3057 0.9007913 Numeric,2 Numeric,2 0 "two.sided"
methoddata.name
[1,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
[2,] "Welch Two Sample t-test" "rnorm(20) and rnorm(20)"
Regards
Petr
From: my1stbox [mai
you are interested, the source code is available.
Regards
Petr
From: my1stbox [mailto:my1st...@163.com]
Sent: Tuesday, August 26, 2014 12:24 PM
To: PIKAL Petr
Cc: R Help
Subject: Re: RE: RE: [R] How to do t.test to rows of a dataframe using apply
family function?
Thank you so much!
Why do do.c
Hi
Please be more specific and try to post some example to simplify elaborating
answer.
How big is your matrix. Double loop seems to be the first which come to my mind
however it can be unpractical when dealing with big matrix or if you want to do
this task repeatedly.
Petr
> -Original
That is bad, especially if it is password to your bank account. Maybe you shall
write it down somewhere next time.
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Ernesto Villarino
> Sent: Friday, August 29, 2014 9:32 A
Hi
Can you be more specific or show some code? I am completely lost in your
question.
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of PO SU
> Sent: Sunday, August 31, 2014 7:04 AM
> To: R. Help
> Subject: [R] w
Hi
I am puzzled about what do you want?
You was discouraged using $ operator due to partial matching, however you still
use it.
In your example you do not have named element being NULL. Just check it
yourself.
> a<-list()
> a$ress<-1
> a$res<-NULL
> a
$ress
[1] 1
> str(a)
List of 1
$ ress:
Hi
> Now confused by "Choices"! :) What is my error please?
Your main error is that you do not read help which was offered by Wiliam.
Choices is not a function it is name of a variable. You can call it e.g.
ctsanddogs if you wish.
see in line
> -Original Message-
> From: r-help-boun..
Hi
You still do not disclose important info about details of your functions.
However, when you want to perform indexing like you show, you maybe can get rid
of NULL and use zero instead.
> a<-1:5
> a[-c(1,3)]
[1] 2 4 5
> a[-c(0,1,3)]
[1] 2 4 5
> a[-c(1,0,3)]
[1] 2 4 5
> a[-c(0,1,0,3,0)]
[1] 2 4
behaviour
would change almost any R code I am afraid that you need to reprogram it
yourself.
Regards
Petr
> -Original Message-
> From: PO SU [mailto:rhelpmaill...@163.com]
> Sent: Thursday, September 11, 2014 9:27 AM
> To: PIKAL Petr
> Cc: R. Help
> Subject: Re
Hi
look at str(aggr)
I bet you will find that your first column is factor and lines you see are from
making a boxplot with your plot command.
You can change it by various ways but I would prefer to change factor to
numeric, which requires some regular expression.
Here is one but I believe that
Hi
> Selection: 1
> Error in if (menu1 == 1) menu1a <- menu(c("1a", "2a", "3a", "4a"),
> graphics = FALSE, :
> argument is of length zero
>
> How to correct this error please, for this zero length argument?
Instead of writing functions which you are unable to debug and resolve look
what e
ot;)
>
So if result of menu is 0 (you did not choose anything) you can either stay
with 0, then switch does not return anything or add 1 and let evaluate
something meaningful specified in second and following positions of switch
command.
Regards
Petr
> -Original Message-
> From: r...@op
Hi
> -Original Message-
> From: r...@openmailbox.org [mailto:r...@openmailbox.org]
> Sent: Thursday, September 18, 2014 4:35 PM
> To: PIKAL Petr
> Cc: r-help@r-project.org
> Subject: RE: [R] apply block of if statements with menu function
>
> On 2014-09-16
Hi.
Partly. You got response from David regarding HTML formating. I polished
scrammbled text but I got
> AsciiGridImpute(type.rf, xfile, outfile)
Error in file(xfiles[[i]], open = "rt") : cannot open the connection
In addition: Warning message:
In file(xfiles[[i]], open = "rt") :
cannot open f
Hi
I will second Jeff. If you want several plots on one page it could be more
convenient to use standard plot.
see
?layout or ?split.screen
for complex figures
or ?par mfrow
for simple layout.
In all cases it is difficult to combine base and grid graphics though.
Regards
Petr
> -Origina
Hi
Please, be more clear in what do you want. I get many errors trying your code
and your explanation does not help much.
> for(i in length(1:(2*nrow(X{
+ Y$IID1new <- ifelse((as.character(Y[,2]) == as.characterXl[,i]) &
X$IID1new != '') , as.character(as.matrix(X[,(2*nrow(X)+i)])),'')
e(1:3, .Label = c("samas4", "samas5", "samas6"
), class = "factor")), .Names = c("FID", "IID"), class = "data.frame",
row.names = c(NA,
-3L))
> -Original Message-
> From: Kate Ignatius [mailto:kate.ignat...@g
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Frank S.
> Sent: Monday, September 29, 2014 9:17 PM
> To: r-help@r-project.org
> Subject: [R] Loop does not work: Error in else statement
>
> Hi to all members of R list,
>
>
Hi
Slightly better but still html scrambled.
see in line
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Frank S.
> Sent: Tuesday, September 30, 2014 2:55 PM
> To: r-help@r-project.org
> Subject: [R] Loop does not work: Erro
Hi
Everything seems to be OK. Please check your if your objects are as required by.
str(objects) (and post result if you are in doubt :)
from your attempts I presume that datetime was not created properly
Maybe this can do it, without actual data it is hard to tell exactly.
datetime=as.POSIXct(
Hi
in this case correct syntax is not to use if at all
vectorx*(vectorx>vectory)
1] 0 0 0 70
if you insist you can use
?ifelse
which is "vectorised" if
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of r..
Hi
when trying your code I got
pval = summary(model)$coeff[2,4]
Error in summary(model)$coeff[2, 4] : subscript out of bounds
> str(summary(model)$coeff)
num [1, 1:4] 1.73e-17 7.07e-01 2.44e-17 1.00
- attr(*, "dimnames")=List of 2
..$ : chr "(Intercept)"
..$ : chr [1:4] "Estimate" "Std. Er
Hi
maybe
which(abs(data)>0.3, arr.ind=T)
row col
Loss_EV_Amygdala_SF_left_hemisphere 15 2
Loss_EV_Amygdala_SF_left_hemisphere 15 3
Loss_PE_Amygdala_SF_right_hemisphere 5 7
Loss_PE_Amygdala_SF_left_hemisphere 13 9
Gives you what you want.
> se
Hi
If Jean's guess is correct, after simple changing Timestamp to real date
see ?strptime and ?as.POSIXct
you can use
result <- merge(mydata, myframe, all=TRUE)
use function ?na.locf from zoo package to fill NAs in Event column and get rid
of all rows with NA in location e.g. by
?complete.ca
Hi
So if I understand correctly, you want to spread value "high" to times 5
minutes before its occurrence and 5 minutes after its occurrence.
If your dates are not extremely big you can prepare its expanded version and
use code suggestions I sent previously
myframe <- data.frame (Timestamp=c("
: Monday, October 06, 2014 9:25 PM
To: PIKAL Petr
Cc: R-help@r-project.org
Subject: Re: [R] Identifying Values in Dataframe
Thanks Pikal -
I'd like to actually get the names of the variables for those indices both
above and below .3/-.3 with the intention of generating a grid of scatterplots
(
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Frederic Ntirenganya
> Sent: Thursday, October 09, 2014 1:32 PM
> To: Jim Lemon
> Cc: r-help@r-project.org
> Subject: Re: [R] Changing date format
>
> This idea substract 1 fo
ome data. I do not believe that
anybody violates calendar in such a way that all years have 366 days.
I may be mistaken but isn’t ?merge function what you really want?
Petr
From: Frederic Ntirenganya [mailto:ntfr...@gmail.com]
Sent: Friday, October 10, 2014 8:33 AM
To: Duncan Murdoch
Cc: PIK
Hi
twodays <- rowSums(embed(Samaru56$Rain,2))
gives you sum of rain in 2 cosecutive days
sel <- which(twodays>20)
gives you vector of row numbers in which above condition results in TRUE value
Samaru56[sel,]
selects these rows
Regards
Petr
> -Original Message-
> From: r-help-boun...
Hi
is
> plot(11:20, ylim=c(0,40))
> abline(h=40)
>
what you want?
Regards
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of PO SU
> Sent: Tuesday, October 14, 2014 11:33 AM
> To: R. Help
> Subject: [R] how to ajust y-a
Hi
First of all you shall transfer your data to R.
Maybe it can be solved by apply but in your case I would use for cycle
Let say your data frame is named doc
lll<-vector(nrow(doc), mode="list")
for(i in 1:nrow(doc)) lll[[i]]<-colnames(doc)[(which(doc[i,]>0))]
Cheers
Petr
> -Original Mess
Hi
It will be even worse with age, try to contact optician :-)
If you want to get better answer you need to provide more info about your file,
what you did and how it failed.
Cheers
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org]
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Vikash Kumar
> Sent: Thursday, October 16, 2014 7:06 AM
> To: adam.n.jenkin...@gmail.com
> Cc: r-help@r-project.org
> Subject: Re: [R] Retrieving lists of colnames
>
> Hi Adam
Hi
Your .R attachment did not come through. Included txt file did not have
headers. Maybe you just want
cor(whatever is the name of your numeric object)
Cheers
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Swapnil
Dear all.
Before I start fishing in (for me) murky regular expression waters I try to ask
community about changing format of negative numbers.
For some reason I get a file with negative numbers formatted with negative sign
at end of number.
something like
0.123-
It is imported as factors and
at negative numbers
>
> Is it what you want?
>
> > st <- "0.123-"
> > gsub("(.+)(-)", "\\2\\1", st)
> [1] "-0.123"
> > st <- "0.123"
> > gsub("(.+)(-)", "\\2\\1", st)
> [1] "
ginal Message-
> From: David Winsemius [mailto:dwinsem...@comcast.net]
> Sent: Monday, October 20, 2014 11:04 PM
> To: PIKAL Petr
> Cc: Marc Girondot; r-help@r-project.org
> Subject: Re: [R] format negative numbers
>
>
> On Oct 20, 2014, at 4:32 AM, PIKAL Petr wrote:
>
&g
Hi Patricia
You are somewhat circling around solution.
Is this what you wanted?
for (i in 5:7) {
plotname = paste("Graph", names(scores)[i], sep="")
png(paste0(plotname,".png"))
p <- ggplot(scores, aes(x=scores[,i], fill=gender ))
print(p+ geom_density(alpha=.3)+xlab(names(scores)[i]))
d
Hi
Most probably
years.before.initiated.cat
is a factor. You have basically two options.
change it to character by ?as.character
or
add level named "Total" by
levels(years.before.initiated.cat)<-c(levels(years.before.initiated.cat),
"Total")
Cheers
Petr
> -Original Message-
> From:
Hi
Your example is not reproducible as we do not have Biota_subset and have no
idea what is its structure. You also do not mention any error.
Most probably x and y are not what you think it is.
This works as expected.
testx<-sample(1:10,300,TRUE)
testw<-seq(1,4,by=0.01)
weighted.hist(testx,tes
Hi
instead of this paste/assign/get stuff use list for keeping cycle result
something like
nam <- vector("list", degree)
for (i in 1:degree) {
nam[[i]] <- lm(y~poly(x,i,raw=TRUE))
}
Cheers
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> proje
Hi
You want a function with input of either single value or a data frame? This is
very strange. One option is to make a method for function but it is quite
beyond my capabilities and it is used by "professional" developers.
methods(c) or methods(plot)
So you need to set method for your functio
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Frederic Ntirenganya
> Sent: Tuesday, November 04, 2014 3:45 PM
> To: r-help@r-project.org
> Subject: [R] Function that creates a Date column
>
> Dear All,
>
> I would like to
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of CJ Davies
> Sent: Tuesday, November 04, 2014 2:50 PM
> To: Jim Lemon; r-help@r-project.org
> Subject: Re: [R] Variance of multiple non-contiguous time periods?
>
> On 04/11/14
ou can for single column option use
only the first value, specifying required column.
dt1=yday(as.Date(data[,which[1]]))
Cheers
Petr
PS. No HTML post please.
From: Frederic Ntirenganya [mailto:ntfr...@gmail.com]
Sent: Wednesday, November 05, 2014 9:35 AM
To: PIKAL Petr
Cc: r-help@r-project.o
Hi
slightly more general if first number is quarter and last two are year
> index <- c(406, 107, 207, 307, 407, 108, 208, 308, 408, 109, 209, 309, 409,
+ 110, 210, 310, 410, 111, 211)
year<-substr(as.character(index), 2,3)
qrt<-substr(as.character(index), 1,1)
as.numeric(factor(paste(year, qrt,
Hi
Your original numeric data probably contain something which prevents read.* to
accept them as numeric (decimal point, white space)
what is result of
str(imported.data)
Petr Pikal
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On B
Hi
1. Do not post in html
2. Post example of data, preferably by dput function
3. Based on posted data explain what do you want to achieve
4. As it seems to be biological issue did you look at Bioconductor?
Cheers
Petr
> -Original Message-
> From: r-help-boun...@r-
It seems rather complicated. AFAIK cor gives you correlation matrix, you can
check items in this matrix but I do not understand your rules.
¨
Cheers
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Cox Lwaka
> Sent: Frida
Dear maintainers
I experienced problems with undeliverable messages to R-help
Here is header, which basically says that mail to r-help@r-project.org was
undelivered and that phil2.ethz.ch refused to accept it.
*
Doručení těmto příjemcům nebo skupinám se nezda
thz.ch]
> Sent: Tuesday, November 18, 2014 12:17 PM
> To: PIKAL Petr
> Cc: r-help
> Subject: Re: [R] problems with mail message
>
> >>>>> PIKAL Petr
> >>>>> on Tue, 18 Nov 2014 10:48:28 + writes:
>
> > Dear maintainers
>
>
Hi
Plot calls function pretty which calculates values for annotating axes.
You can get rid of this behaviour by
plot(..., axes=FALSE)
and calling
axis(1, ...)
see
?plot
?pretty
?axis
Cheers
Petr
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> p
Hi
Error message seems to be clear
> Error in if (data$Rain[i_NA] == 60) { :
> missing value where TRUE/FALSE needed
data$Rain[i_NA] produces probably NA
> x<-NA
> if(x==60) print(1+1) else print("Errrorrr")
Error in if (x == 60) 1 + 1 : missing value where TRUE/FALSE needed
> x<-10
> if(x==6
Huh
and what is the result of e.g.
data$Rain[indicNAs[1]]
I bet that you will see
[1] NA
and your if question askes if (NA eqals 60) which results in NA and "if" is
telling you that it expects TRUE or FALSE but not NA.
I do not see how much clearer the error message shall be.
Cheers
Petr
>
Hi
You say
> in other words, experienced R users do not let their workspace be saved
> automatically (to '.RData') and hence do not load any .RData
> automatically at startup.
I save/load .RData for years without any issues (except of not installed
packages when working on different PCs).I usua
Dear all
I encountered strange behaviour of ggplot with combination of facet and
subsetting. I use for creating plots sometimes a for cycle, something like this
for (i in n:m) { p<-ggplot(data, aes(x=x, y=data[,i], colour=f))), ...}
However I found strange result with this combination
This is
nt: Thursday, November 27, 2014 9:29 AM
> To: PIKAL Petr; R help
> Subject: RE: ggplot facet and subsetting
>
> Dear Petr,
>
> You need to use aes_string() instead of aes().
>
> The.cols <- colnames(data)[n:m]
> for (i in The.cols) { p<-ggplot(data, aes_string(x=&quo
Hm
> -Original Message-
> From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us]
> Sent: Wednesday, November 26, 2014 8:09 PM
> To: PIKAL Petr; Martin Maechler
> Cc: R help
> Subject: Re: [R] plot.hclust point to older version
>
> Short answer to your question i
4 6:13 PM
> To: PIKAL Petr; R help
> Subject: RE: [R] ggplot facet and subsetting
>
> Below
>
> John Kane
> Kingston ON Canada
> > This is OK but only in BW
> > p<-ggplot(vec.c, aes(x=fi, y=nad1mi))
> > p+geom_point(size=5)+geom_line()+facet_grid(.~stroj)
my abilities.
If I want several graphs sharing one axis I use facet_grid.
You can also check twoord.plot from plotrix, which is easier.
Cheers
Petr
From: JAVAD LESSAN [mailto:jlessa...@ku.edu.tr]
Sent: Friday, November 28, 2014 4:30 PM
To: PIKAL Petr
Subject: Re: [R] HELP
Hi Sir!
Hope you are
Hm. What do you meen by
"to run a non stationary time series data"
http://cran.r-project.org/web/views/TimeSeries.html
lead you to some packages for testing stationarity.
Did you check them?
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Beha
Hi
Can you be more specific? How do you want parametrize order?
There is no sort.list argument in order.
For sort list, the default method is "radix", AFAIK.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of
> Pierrick Bruneau
> Sent: Tu
Hi
Your data are rather wild so I am not sure if following function can do what
you want. I designed it primarily for intagration of peaks from different
sources.
integ1 <- function (x, y, dm = -Inf, hm = +Inf)
{
ifelse(dm == -Inf, dm <- min(x), dm <- dm)
ifelse(hm == +Inf, hm <- max(x)
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