Hi
'gridSVG' might be one way to get this. For example ...
library(lattice)
# Draw boxplot (with a package that sits on top of 'grid')
bwplot(voice.part ~ height, data=singer, xlab="Height (inches)",
horizontal=FALSE)
library(grid)
grid.ls()
# Looks like boxes are called bwplot.box.poly
On Tue, Dec 6, 2016 at 5:10 PM, Chris Evans wrote:
> {{SIGH}}
>
> You are absolutely right.
>
> I wonder if I am losing some cognitive capacities that are needed to be part
> of the evolving R community. It seems to me that if a tibble is designed to
> be an enhanced replacement for a dataframe
At startup or and using R CMD .. I stumble about
1: Setting LC_CTYPE failed, using "C"
2: Setting LC_TIME failed, using "C"
3: Setting LC_MESSAGES failed, using "C"
4: Setting LC_MONETARY failed, using "C"
Iam am working with
R version 3.3.2 (2016-10-31)
Platform: x86_64-apple-darwin13.4.0 (64-
You really need sleep. Then you need to read
?`[[`
and in particular read about the second argument to the `[[` function, since
you don't seem to understand what it is for. Maybe reread the Introduction to R
document that comes with R.
The simplest solution is to treat `[[` as supporting one i
{{SIGH}}
You are absolutely right.
I wonder if I am losing some cognitive capacities that are needed to be part of
the evolving R community. It seems to me that if a tibble is designed to be an
enhanced replacement for a dataframe then it shouldn't quite so radically
change things.
I notic
Simpler I think: ?all.vars
> all.vars(~A+B)
[1] "A" "B"
Note also:
> all.vars(~log(A))
[1] "A"
Cheers,
Bert
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue, Dec
Not at a computer to check right now, but I believe single bracket indexing
a tibble always returns a tibble. To extract a vector use [[
On Dec 6, 2016 4:28 PM, "Chris Evans" wrote:
>
> I hope I am obeying the list rules here. I am using a raw R IDE for this
and running 3.3.2 (2016-10-31) on x86_
Hi Dagmar,
Having recovered somewhat, I can refine my suggestion a bit:
datframe <- data.frame(Name=c("Kati","Kati","Kati","Leon","Leon","Leon"),
changepoint=as.POSIXct(strptime(c("03.01.2011","05.01.2011", "27.01.2011",
"26.01.2011","28.01.2011", "28.02.2011"), "%d.%m.%Y")),
knownstate =c("br
This might help.
# your example data
trboot3 <- structure(c(0L, -1L, -1L, -1L, -1L, -1L, -1L, -1L, 0L, 0L, -1L,
0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 1L, 1L, 0L, 0L, 0L, 0L,
1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, -1L, -1L, 0L, 1L, 0L, 0L, -1L,
-1L, -1L, -1L, -1L, -1L, -1L, -1L, -1L, 0L, 0L, 0L, 0
I hope I am obeying the list rules here. I am using a raw R IDE for this and
running 3.3.2 (2016-10-31) on x86_64-w64-mingw32/x64 (64-bit)
Here is a reproducible example. Code only first
require(tibble)
tmpTibble <- tibble(ID=letters,num=1:26)
min(tmpTibble[,2]) # fine
max(tmpTibble[,2]) # fine
Hi Greg,
What is happening is easy to see:
ph<-matrix(sample(1:100,40),ncol=4)
colnames(ph)<-c("M1","X1","X2","X3")
ph[sample(1:10,3),1]<-NA
ph
M1 X1 X2 X3
[1,] 34 98 3 35
[2,] 13 66 74 68
[3,] NA 22 99 79
[4,] 94 6 80 36
[5,] 18 9 16 65
[6,] NA 29 56 90
[7,] 41 23 7 5
Hello, there,
I will like to fill the boxplot with gradient color, as exampled below:
Can anyone help me figure out what package I should go with?
Thank you very much for any inputs!
Kind regards,
Ace
__
R-help@r-project.org mailing list -- To UNSUBSCRIB
Hi everyone,
I have a very large shapefile with many polygons (17769 polygons, 104.4 Mb),
which I want to convert to a raster file. A grid cell can be covered by either
none, 1 or several polygons. I want to assign the percentage cover of polygons
to the respective grid cell, just like the funct
Will do.
From: Ista Zahn
Sent: Tuesday, December 6, 2016 2:49:30 PM
To: Patrick Casimir
Cc: r-help@r-project.org
Subject: Re: [R] Why is DocumentTermMatrix showing 0 term?
On Tue, Dec 6, 2016 at 2:28 PM, Patrick Casimir wrote:
> Actually, the DTM works now. Th
On Tue, Dec 6, 2016 at 2:28 PM, Patrick Casimir wrote:
> Actually, the DTM works now. This is amazing. Million thanks. Why wasn't it
> working before?
Do as I suggested and
start adding back your tm_map's until you find the thing that
breaks it.
--Ista
>
> See below:
>
>
>> cname <- file.path
Actually, the DTM works now. This is amazing. Million thanks. Why wasn't it
working before?
See below:
> cname <- file.path("C:\\Users\\Desktop\\Text Mining\\Cases\\MyCorpus")
> docs <- Corpus(DirSource(cname))
> dtm <- DocumentTermMatrix(docs)
> dtm
<>
Non-/sparse entries: 920/2144
Sparsity
R is interactive so you can print the intermediate results:
> ph <- data.frame(M1=c(1,NA,3,4,5), X1=1:5, X2=c(1,2,NA,4,5), X3=1:5,
Y=c(11,12,13,14,NA), row.names=paste0("R",1:5))
> ph
M1 X1 X2 X3 Y
R1 1 1 1 1 11
R2 NA 2 2 2 12
R3 3 3 NA 3 13
R4 4 4 4 4 14
R5 5 5 5 5 NA
> miss
Hello,
The first command line produces a logical vector with TRUE if at least
one row element of ph is NA and FALSE otherwise.
The second creates a vector of zeros with length equal to nrow(ph).
Now the third command line. ! negates miss, so TRUE becomes FALSE and
vice-versa. sum(!miss) counts
> On Dec 6, 2016, at 7:33 AM, Rui Barradas wrote:
>
> Perhaps the best way is the one used by library(), where both
> library(package) and library("package") work. It uses
> as.charecter/substitute, not deparse/substitute, as follows.
>
> mydf <-
> data.frame(id=c(1,2,3,4,5),sex=c("M","M","M
Dear All;
I am very new in R and try to understand the logic for a program has been
run sucessfully. Here select[!miss] <- 1:sum(!miss) par is confussing me. I
need to understandand the logic behind this commend line.
Thanks in advance for your help,
Greg
miss <- apply(is.na(ph[,c("M1","X1","X
Perhaps the best way is the one used by library(), where both
library(package) and library("package") work. It uses
as.charecter/substitute, not deparse/substitute, as follows.
mydf <-
data.frame(id=c(1,2,3,4,5),sex=c("M","M","M","F","F"),age=c(20,34,43,32,21))
mydf
class(mydf)
str(mydf)
myf
Ok, that's a way of seeing it.
Rui Barradas
Em 06-12-2016 14:28, John Sorkin escreveu:
Over my almost 50 years programming, I have come to believe that if one
wants a program to be useful, one should write the program to do as much
work as possible and demand as little as possible from the user
Does
cname <- file.path("C:\\Users\\Desktop\\Text Mining\\Cases\\MyCorpus")
docs <- Corpus(DirSource(cname))
dtm <- DocumentTermMatrix(docs)
dtm
work?
If so, add start adding back your tm_map until you find the thing that
breaks it.
Best,
Ista
On Tue, Dec 6, 2016 at 10:25 AM, Patrick Casimir
Fortune Nomination!
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Tue, Dec 6, 2016 at 6:09 AM, Ista Zahn wrote:
> Hi Patrick,
>
> How could any
Note that library has another argument, character.only=TRUE/FALSE,
to control whether the main argument should be regarded as a variable
or a literal. I think you need two arguments to handle this.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 6, 2016 at 7:33 AM, Rui Barradas wrote:
Buy more memory? Do something different than you were doing before the error
occurred? Use a search engine to find what other people have done when this
message appeared? Follow the recommendations in the Posting Guide mentioned in
the footer of this and every post on this mailing list?
--
Sen
hi everyone,
I tried to run my code in RStudio,but I received this error message,what should
I do?
Error: cannot allocate vector of size 12.1 Gb
In addition: Warning messages:
1: In cor(coding.rpkm[grep("23.C", coding.rpkm$name), -1],
ncoding.rpkm[grep("23.C", :
Reached total allocation of 602
hi everyone,I tried to run my code in RStudio,but I received this error
message,what should I do?Error: cannot allocate vector of size 12.1 Gb
In addition: Warning messages:
1: In cor(coding.rpkm[grep("23.C", coding.rpkm$name), -1],
ncoding.rpkm[grep("23.C", :
Reached total allocation of 6027M
docs has 4 documents and inspect(docs) shows 4 plaintextdocument
> summary(docs)
Length Class Mode
case1.txt 2 PlainTextDocument list
case2.txt 2 PlainTextDocument list
case3.txt 2 PlainTextDocument list
case4.txt 2 PlainTextDocument list
> inspect(doc
What is in docs?
What does
inspect(docs)
say?
--Ista
On Tue, Dec 6, 2016 at 9:29 AM, Patrick Casimir wrote:
> Thanks Ista. See codes below. I am not sure why the DTM is showing 0 term. I
> have 4 documents in the corpus. And I was able to make transformations
>
> to the documents inside the
I basically agree with Rui - using substitute will cause trouble. E.g., how
would the user iterate over the columns, calling your function for each?
for(column in dataFrame) func(column)
would fail because dataFrame$column does not exist. You need to provide
an extra argument to handle this
Thanks Ista. See codes below. I am not sure why the DTM is showing 0 term. I
have 4 documents in the corpus. And I was able to make transformations
to the documents inside the corpus.
> cname <- file.path("C:\\Users\\Desktop\\Text Mining\\Cases\\MyCorpus")
> dir(cname)
[1] "case1.txt" "case2.tx
Over my almost 50 years programming, I have come to believe that if one wants a
program to be useful, one should write the program to do as much work as
possible and demand as little as possible from the user of the program. In my
opinion, one should not ask the person who uses my function to re
The specific error message you are getting says that there are
non-ASCII characters
in one of your .Rd files. You can locate them with the
tools::showNonASCII function.
library(tools)
?showNonASCII
Replace them with ASCII characters and the packaging should work.
If you really need the non-ASCI
What operating system? It is very unlikely anyone will offer help in the
darkness.
JN
On 16-12-06 01:20 AM, Pijush Das wrote:
> Hello Sir,
>
> I am facing a common problem in R. I have searched a lot but nobody gives a
> proper way. The problem is : when I am tying to check my package in R
> cr
This is not really an R problem but a TeX one. Apparently, your input file
contains Unicode characters that LaTeX doesn't know what to do with.
That raises the suspicion is that the file might not be in UTF-8 encoding at
all, however tan kind of issue does occasionally happen, even in UTF-8, if
Hi Patrick,
How could anyone possibly answer this question with only the information
you've provided? It's like showing me an empty cup and asking why it's
empty. Maybe you didn't put anything in it. Maybe you did and then you dog
drank it or your cat knocked it over or your girlfriend drank it. H
Hi Paul,
Not everything people ask for makes sense. If you insist on trying to get R
integration working in SPSS you should reach out to the SPSS company and/or
community for support.
Best,
Ista
On Dec 6, 2016 8:05 AM, "Paul Bernal" wrote:
> Hi Ista,
>
> Your suggestion is great. There is no b
This worked well! Thank you very much.
For the record, the stat_bin was used in a solution I found on stackexchange
for something similar, but I was having issues adapting.
Thanks again to all who responded. I appreciate it greatly.
Shawn Way, PE
-Original Message-
From: Jeff Newmill
Hi Ista,
Your suggestion is great. There is no better way to work with R than
working with R directly, however, I have been asked to generate forecasts
in SPSS Modeler using the R integration that comes with it, so
unfortunately, I need to find a way around. If I find the way to do it, I
will make
<>
Non-/sparse entries: 0/0
Sparsity : 100%
Maximal term length: 0
Weighting : term frequency (tf)
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
https://stat.eth
Hello Sir,
I am facing a common problem in R. I have searched a lot but nobody gives a
proper way. The problem is : when I am tying to check my package in R
created by me it shows an error given below
* checking PDF version of manual ... WARNING
> LaTeX errors when creating PDF version.
This typi
No attachment. R-help; is very fussy about the type of file it will accept.
Try a text document with .txt extention or just put the code and the sample
data in the actual email. Use dput(), see ?dput for details, to provide the
data.
On Tuesday, December 6, 2016 6:51 AM, michael tsagris
On 06 Dec 2016, at 11:17 , Jim Lemon wrote:
> Hi Paul,
> The easy to understand way is:
>
> n <- c(1:10)
> # Create empty list to store vectors
> list_of_vecs <- list()
>
> # Create n vectors of random numbers - length 10. This works ok.
> for (i in n){
> list_of_vecs[[i]]<-rnorm(10,0,1)
> }
Well personally I would not have a clue on how to approach the problem but have
a look at
http://stackoverflow.com/questions/6644997/showing-data-values-on-stacked-bar-chart-in-ggplot2.
which seems to do what you want only using geom_text()
Best of luck.
On Monday, December 5, 2
Hello, I have encountered a problem when running a glm with the inverse
gaussian as distribution and the log as a link function.
Do you think this could be a bug, or something else?Attached is a document to
run the example. I have written my own functions and they seem to work fine.
___
Hi Paul,
The easy to understand way is:
n <- c(1:10)
# Create empty list to store vectors
list_of_vecs <- list()
# Create n vectors of random numbers - length 10. This works ok.
for (i in n){
list_of_vecs[[i]]<-rnorm(10,0,1)
}
If you really want to use "assign":
for (i in n){
vecname<-paste('v
Hi Dagmar,
I think you want something like a gantt.chart. I know this is wrong in
some ways, but it is late and I must retire:
datframe <- data.frame(Name=c("Kati","Kati","Kati","Leon","Leon","Leon"),
changepoint=as.Date(c("03.01.2011","05.01.2011", "27.01.2011",
"26.01.2011","28.01.2011", "28.
# Dear all,
# I hope someone can help me with this, I am kind of desperated even
though it doesn't sound to complicated at all. Let's see:
# I have a data frame like this:
datframe <- data.frame(Name=c("Kati","Kati","Kati","Leon","Leon","Leon"),
changepoint =as.POSIXct
Thank you Jim.
On 6 December 2016 at 9:17:21 pm, Jim Lemon (drjimle...@gmail.com) wrote:
Hi Paul,
The easy to understand way is:
n <- c(1:10)
# Create empty list to store vectors
list_of_vecs <- list()
# Create n vectors of random numbers - length 10. This works ok.
for (i in n){
Hi,
As an exercise, I am trying to create a list of 10 random number vectors in a
loop. I can create the vectors but am unsure how to assemble them in the list
as part of the loop. Any advice?
# Number of vectors to create
n <- c(1:10)
# Create empty list to store vectors
list_of_vecs <- list
Hello,
Just to say that I wouldn't write the function as John did. I would get
rid of all the deparse/substitute stuff and instinctively use a quoted
argument as a column name. Something like the following.
myfun <- function(frame, var){
[...]
col <- frame[, var] # or frame[[
52 matches
Mail list logo
