Hello,
There are many questions about making the limit of the colour key smaller than
the data range, but I have the opposite problem.
Assume one heatmap has data in the range 6 to 12 and another has data in the
range 6 to 9. By providing the same breaks argument to both plots, the heatmaps
ar
Thanks for the reply. Another great option would be "missing" (like in SAS),
especially for factors. I'm struggling to figure out how to do this with
"tables".
Daniel Cher, MD
djc...@gmail.com
+1-650-269-5763
This message and its attachments are confidential.
-Original Message-
From: Du
Thank you dear friends. You have cleared my first doubt. Â
My second doubt:
I have the same data sets "Elder" and "Younger". Elder <- data.frame(
 ID=c("ID1","ID2","ID3"),
 age=c(38,35,31))
Younger <- data.frame(
 ID=c("ID4","ID5","ID3"),
 age=c(29,21,"NA"))
 Row ID3 comes in both
Hi Goran,
thanks for your suggestion, but I believe it's not helpful for me...
phreg statement "Proportional hazards model with parametric baseline hazard(s).
Allows for stratification with dif-ferent scale and shape in each stratum, and
left truncated and right censored data"
I've data whose
Hi, I'm pretty new to R and am trying to develop a reusable set of scripts
that I can use to profile various data types and common fields in our
database. I know that what I'm asking is a can of worms, so please bear
with me. :)
For example, we store a person's first name, last name, phone number,
Jim Lemon píše v Pá 17. 01. 2014 v 13:21 +1100:
> On 01/17/2014 10:59 AM, Marc Schwartz wrote:
> >
> > ...
> > Arggh.
> >
> > No, this is my error for not actually looking at the plot and presuming
> > that it would work.
> >
> > Turns out that it does work for a non-stacked barplot:
> >
> >ba
Thanks for looking at this, I've been tearing my hair out for a day or so
now.
I have done a multiple variable logistic regression in R, and obtained my
coefficients. I am able to make predictions for the training data in R
without problem. But now I would like to create a prediction model in Ruby
On 01/17/2014 10:59 AM, Marc Schwartz wrote:
...
Arggh.
No, this is my error for not actually looking at the plot and presuming that it
would work.
Turns out that it does work for a non-stacked barplot:
barplot(VADeaths, angle = 1:20 * 10, density = 10, beside = TRUE)
However, internally
Indeed it works ! Thanks a lot. But why?
-Original Message-
From: "arun" [smartpink...@yahoo.com]
Date: 01/16/2014 08:44 PM
To: "r-help@r-project.org"
Subject: Re: [R] xts error: number of items to replace is not a multiple of
replacement length
Hi,
Try:
sample.xts["2007-01-02::20
Hi,
Try:
sample.xts["2007-01-02::2007-01-04","Close"]
<-sample.xts["2007-01-02::2007-01-04","Close"] +1
sample.xts["2007-01-02::2007-01-04"]
# Open High Low Close
#2007-01-02 50.03978 50.11778 49.95041 52.11778
#2007-01-03 50.23050 50.42188 50.23050 52.39767
#2007-01-0
Dear all ,
I am getting this error while trying to change columns of an xts object with a
date range as index.
> library(xts)
Loading required package: zoo
Attaching package: ‘zoo’
The following object is masked from ‘package:base’:
as.Date, as.Date.numeric
> data(sample_matrix)
On 2014-01-15 11:00, r-help-requ...@r-project.org wrote:
Date: Wed, 15 Jan 2014 16:39:17 +1000
From: Diana Virkki
To:r-help@r-project.org
Subject: [R] Model averaging using QAICc
Message-ID:
Content-Type: text/plain
Hi all,
I am having some trouble running GLMM's and using model averagin
On Jan 16, 2014, at 5:03 PM, Martin Weiser wrote:
> Marc Schwartz píše v Čt 16. 01. 2014 v 16:46 -0600:
>> On Jan 16, 2014, at 12:45 PM, Martin Weiser wrote:
>>
>>> Dear listers,
>>>
>>> I would like to make stacked barplot, and to be able to define shading
>>> (density or angle) segment-wise
Hi,
It's not clear about the pattern in your rownames.
In the for() loop, I guess you need rownames(df) instead of df.
Using an example dataset (Here the rownames may be different)
set.seed(59)
x <- as.data.frame(matrix(rnorm(110),ncol=2))
set.seed(24)
row.names(x) <- paste0(row.names(x),Reduce(
Marc Schwartz píše v Čt 16. 01. 2014 v 16:46 -0600:
> On Jan 16, 2014, at 12:45 PM, Martin Weiser wrote:
>
> > Dear listers,
> >
> > I would like to make stacked barplot, and to be able to define shading
> > (density or angle) segment-wise, i.e. NOT like here:
> > # Bar shading example
> > b
On 01/16/2014 04:59 PM, Vito Ricci wrote:
Hi guys,
is there in some R package a statement to fit parameters in a 3 parameters
lognormal distribution.
Yes, the function 'phreg' in the package 'eha'.
Göran Broström
Many thanks
Vito Ricci
[[alternative HTML version deleted]]
___
On Jan 16, 2014, at 12:45 PM, Martin Weiser wrote:
> Dear listers,
>
> I would like to make stacked barplot, and to be able to define shading
> (density or angle) segment-wise, i.e. NOT like here:
> # Bar shading example
> barplot(VADeaths, angle = 15+10*1:5, density = 20, col = "black",
>
Hi Valerie,
Assuming GfullUA is a column of your data frame task2analyses, you
need to tell R where to look. It's trying to find an object called
GfullUA, and there isn't one.
Here are two ways:
with(task2analyses, cor(GfullUA, GFullUA))
cor(task2analyses$GfullUA, task2analyses$GFullUA)
You mig
Hi,
I am using "bwplot" to depict the box plots for two group by 6 time points.
I need to add 6 p-values in each time point to compare two group at each
time point. P-values are (0.0020, 0.0204, 0.3361, 0.0185, 0.1981, and
0.6677). I could depict the two box plots per each time point using the
c
Hello:
I am a new user, running the latest version of R on my Mac. I have started by
reading a file with the read.csv command:
task2analyses <- read.csv(file="GroupsWithRTsEqualN.csv",head=TRUE,sep=",")
When I print it out in R, the file appears to be intact, with the proper
headers. Yet, whe
Hi Jingxia,
May be this helps:
dat1 <- read.table(text="fatfreemilk fatmilk halfmilk 2fatmilk
A A A A
A B B A
B A A A
C C C C
D . A A
A E A E
C A B A
A . A A
A B . A
A A B
E",sep="",header=TRUE,stringsAsFactors=FALSE,check.names=FALSE,na.strings=".")
dat2 <- dat1
dat2$id <- 1:nrow(dat2)
librar
Also,
You can do the same with the previous solution:
result1 <- result[,-6]
vec1 <- unique(unlist(dat1))
result2 <- as.data.frame(t(sapply(dat1,function(x) {counts<-
table(factor(x,levels=vec1));
percentage<-sprintf("%.1f",(counts/sum(counts))*100);
c(paste0(counts,paste0("(",percentage,")")),
Dear listers,
I would like to make stacked barplot, and to be able to define shading
(density or angle) segment-wise, i.e. NOT like here:
# Bar shading example
barplot(VADeaths, angle = 15+10*1:5, density = 20, col = "black",
legend = rownames(VADeaths))
The example has 5 diffe
Hi,
May be this helps:
x <- data.frame(V1=-1.162877, V2=0.1848928)
set.seed(245)
df <- as.data.frame(matrix(rnorm(5051*2),ncol=2))
cut1 <- cut(df[,1],breaks=c(x[,1]-0.1,x[,1]+0.1))
cut2 <- cut(df[,2],breaks=c(x[,2]-0.1,x[,2]+0.1))
df1 <- df[!is.na(cut1) & !is.na(cut2),]
A.K.
I have a data
My yearly Regression Modeling Strategies course is expanded to 4 days
this year to be able relax the pace a bit. Details are below.
Questions welcomed.
-
*RMS Short Course 2014*
Frank E. Harrell, Jr., Ph.D., Professor and Chai
I guess, you could also do:
names(dat1)[max.col(dat1)]
#[1] "a" "b"
A.K.
On Thursday, January 16, 2014 3:47 PM, arun wrote:
Try:
dat1 <- read.table(text="a b c d
1 0.5 0.1 0.2 0.2
5 0.3 0.5 0.1 0.1",sep="",header=TRUE)
data.frame(Names=apply(dat1,1,function(x) names(
Try:
dat1 <- read.table(text="a b c d
1 0.5 0.1 0.2 0.2
5 0.3 0.5 0.1 0.1",sep="",header=TRUE)
data.frame(Names=apply(dat1,1,function(x) names(x)[x %in% max(x)]))
# Names
#1 a
#5 b
#or
colnames(dat1)[apply(dat1,1,which.max)]
#[1] "a" "b"
A.K.
Hi,
I need a
On 01/17/2014 05:12 AM, Morway, Eric wrote:
As a follow up to this thread started nearly a month ago, I'm in need of
help sorting out the R code that will create a log-scale legend in the call
to "image.plot" below. As the last line of the code provided below shows,
I attempted to force the labe
As a follow up to this thread started nearly a month ago, I'm in need of
help sorting out the R code that will create a log-scale legend in the call
to "image.plot" below. As the last line of the code provided below shows,
I attempted to force the labeling through the argument "legend.lab", to no
Happy New Year (if a little late!). Revolution Analytics staff write
about R every weekday at the Revolutions blog:
http://blog.revolutionanalytics.com
and every month I post a summary of articles from the previous month
of particular interest to readers of r-help.
In case you missed them, here a
Hi guys,
is there in some R package a statement to fit parameters in a 3 parameters
lognormal distribution.
Many thanks
Vito Ricci
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r
Not quite:
> rbind(Elder, Younger)
ID age
1 ID1 38
2 ID2 35
3 ID3 31
4 ID4 29
5 ID5 21
6 ID3 31
Note that ID3 is duplicated.
Should be:
> merge(Elder, Younger, by = c("ID", "age"), all = TRUE)
ID age
1 ID1 38
2 ID2 35
3 ID3 31
4 ID4 29
5 ID5 21
He wants to do a join on both
IN inspect(removeSparseTerms(dtm, 0.4)) does anyone knows how the sparse
term
"A numeric for the maximal allowed sparsity" works? ie what is the
difference between say 0.2, 0.4 & 0.6?
Thanks for your help
--
View this message in context:
http://r.789695.n4.nabble.com/DTM-Package-r
Ups, sorry that should have been
mer <- rbind(Elder, Younger)
/frede
Oprindelig meddelelse
Fra: Frede Aakmann Tøgersen
Dato:16/01/2014 15.54 (GMT+01:00)
Til: "Adams, Jean" ,kingsly
Cc: R help
Emne: Re: [R] Doubt in simple merge
No I think the OP wants
mer <- merge(Elder, Yo
No I think the OP wants
mer <- merge(Elder, Younger)
Br. Frede
Oprindelig meddelelse
Fra: "Adams, Jean"
Dato:16/01/2014 15.45 (GMT+01:00)
Til: kingsly
Cc: R help
Emne: Re: [R] Doubt in simple merge
You are telling it to merge by ID only. But it sounds like you would like
it
On 16/01/2014 8:46 AM, ONKELINX, Thierry wrote:
You want
y <- ifelse(x == 'a', 1, 2)
or use if, rather than ifelse, i.e.
if (x == 'a') {
y <- 1
} else {
y <- 2
}
ifelse() is mainly used when you want to work with whole vectors of
decisions, e.g.
x <- 1:10
ifelse(x > 5, 1, 0)
Duncan M
You are telling it to merge by ID only. But it sounds like you would like
it to merge by both ID and age.
merge(Elder, Younger, all=TRUE)
Jean
On Thu, Jan 16, 2014 at 6:25 AM, kingsly wrote:
> Dear R community
>
> I have a two data set called "Elder" and "Younger".
> This is my code for simp
Dear R community
I have a two data set called "Elder" and "Younger".
This is my code for simple merge.
Elder <- data.frame(
 ID=c("ID1","ID2","ID3"),
 age=c(38,35,31))
Younger <- data.frame(
 ID=c("ID4","ID5","ID3"),
 age=c(29,21,31))
mer <- merge(Elder,Younger,by="ID", all=T)
Output
You want
y <- ifelse(x == 'a', 1, 2)
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
+ 32 2 525 02 51
+ 32 54 43 61 85
thierry.onk
Hi,
Sorry for the newbie question! My code:
x <- 'a'
ifelse(x == 'a',y <- 1, y <- 2)
print(y)
Shouldn't this assign a value of 1? When I execute this I get:
> x <- 'a'
> ifelse(x == 'a',y <- 1, y <- 2)
[1] 1
> print(y)
[1] 2
Am I doing something really daft???
thanks!
> sessionInfo()
R ve
Dear UseRs of R,
My sincere apologizes in advance if my question isn't relevant to the
operations in R. I actually have the following two columns data, with 12 rows
in it.
> dput(el)
structure(c(-1.42607687227285, -1.0200762327862, -0.736315917376129,
-0.502402223373355, -0.293381232121193, -0
Search for 'where' in ?rpart.object (linked from ?rpart).
On 16/01/2014 08:50, Jérémy Lambert wrote:
Hello everyone,
I just completed a recursive pratitionning analysis, using rpart, and have a
beautiful tree with 6 terminal nodes. Each terminal node containing a precise
number of patients (
Hello everyone,
I just completed a recursive pratitionning analysis, using rpart, and have a
beautiful tree with 6 terminal nodes. Each terminal node containing a precise
number of patients (it's a clinical study), I'd like to create a new variable
informing in which terminal node are locating
Hi,
May be this helps:
small <- read.table(text="monthend_n ticker wgtdiff ret interval b1 b2 b3 b4 b5
b6
1 19990228 AA 0.7172 -2.58 0.33896 -0.5868 -0.24784 0.09112 0.43008 0.76904
1.108
2 19990228 AAPL -0.0828 -15.48 0.33896 -0.5868 -0.24784 0.09112 0.43008 0.76904
1.108
3 19990228 ABCW 0.0
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