-
From: Yao He
To: arun
Cc: R help
Sent: Monday, January 7, 2013 10:57 AM
Subject: Re: [R] how to aggregate T-test result in an elegant way?
Hi,arun
Yes , I just want to do the t.test
I think maybe it is not necessary to generate a 3D array from the raw
data.frame by acast() at first
Than
contr.funs[1 + isOF[nn]]) :
contrasts can be applied only to factors with 2 or more levels
Even though age has 3 levels; time has 14 years & 21 years; HIBP is a
binary response outcome.
2) When you mentioned summary(m1)$mean["p"] what did the p mean? i
used this in one of
r more levels
Even though age has 3 levels; time has 14 years & 21 years; HIBP is a
binary response outcome.
2) When you mentioned summary(m1)$mean["p"] what did the p mean? i
used this in one of the gee command, it produced NA as answer?
Many thanks
On Mon, Jan 7, 2013 at 5:26
HI,
str1<-"x$Expensive"
regexpr("\\$",str1)[1]
#[1] 2
str2<-"x$Exp$Expression"
unlist(gregexpr("\\$",str2))
#[1] 2 6
A.K.
- Original Message -
From: Data Analytics Corp.
To: "r-help@R-project.org"
Cc:
Sent: Monday, January 7, 2013 4:22 PM
Subject: [R] pattern matching
Hi,
I have
-selection-in-gee-in-r
It's not clear to me "reference to write about missing values".
A.K.
- Original Message -
From: Usha Gurunathan
To: arun
Cc:
Sent: Monday, January 7, 2013 6:12 PM
Subject: Re: [R] random effects model
Hi AK
2)I shall try putting exch. and c
HI,
A possible guess ( with no data):
for (i in 1:7) {
subs <- data$skin_color==levels(data$skin_color)[i]
line<-lm(body_weight~body_length, data=subset(data, subset=subs)) #closing
parentheses
abline(line,col=c("yellow","chocolate1","darkorange2",
"red3","saddlebrown","coral4","grey
HI Elaine,
In the data you sent to me, it had 5 levels for skin_color.
data1<-read.csv("skin_color.csv",sep="\t")
data1$skin_color<-factor(data1$skin_color)
levels(data1$skin_color)
#[1] "1" "2" "3" "4" "5"
mypath<-
HI,
This should also work:
set.seed(5)
list1<-lapply(1:3,function(i)
data.frame(col1=sample(c(1:5,""),10,replace=TRUE),
value=rnorm(10),stringsAsFactors=FALSE))
lapply(list1,function(x) {x[x==""]<-NA;x})
A.K.
- Original Message -
From: Dominic Roye
To: R help
Cc:
Sent: Tuesday, J
Hi,
Try this:
set.seed(5)
list1<-lapply(1:3,function(i)
data.frame(col1=sample(c(1:5,""),10,replace=TRUE),
value=rnorm(10),stringsAsFactors=FALSE))
res<-lapply(list1,function(x) {x[apply(x,2,function(y) y=="")]<-NA;x})
res[[1]]
# col1 value
#1 2 -0.6029080
#2 5 -0.4721664
#3
1 value
#1 2 -0.6029080
#2 5 -0.4721664
#3 -0.6353713
#4 2 -0.2857736
#5 1 0.1381082
#6 5 1.2276303
#7 4 -0.8017795
#8 5 -1.0803926
#9 -0.1575344
#10 1 -1.0717600
--
A.K.
From: Dominic Roye
To: arun
Sent:
u can replace this with your
dataset
dat1$d<-factor(dat1$skin_color)
colnames(dat1)[2:3]<-c("y","x")
models<-dlply(dat1,"d",function(df) mod <- lm(y~x,data=df))
models[[1]]
#Call:
#lm(formula = y ~ x, data = df)
#Coefficients:
#(Intercept) x
#
rk. Didn't strike to me
for so long. Does the AIC value come out with the gee output?
By reference, I meant reference to a easy-read paper or web address
that can give me knowledge about implications of missing data.
Ta.
On 1/8/13, arun kirshna [via R]
wrote:
>
>
> HI,
> BP.sta
Hi,
I forgot to mention:
levels(dat1$d)
#[1] "1" "2" "3" "4" "5"
Suppose, if I use different levels
library(car)
dat1$d1<-recode(dat1$d,"1='A';2='B';3='C';4='D';5='E'")
levels(dat1$d
Hi Pradip,
I didn't check the mode at that time. It generated a "matrix"
"test1$newcols<- sapply()"
You can do this:
test2<-data.frame(test1[,-7],test1$newcols)
str(test2)
#'data.frame': 51 obs. of 9 variables:
# $ ObtMj_P : num 49.6 55 52.5 50.5 51.1 55.1 56.3 53.6 53.5 52.7 ..
Hi,
May be this is creating the problem:
set.seed(15)
dat1<-data.frame(A_00060_3=sample(1:10,5,replace=TRUE),B_00060_3_cd=sample(20:30,5,replace=TRUE),C_00060_3=sample(1:15,5,replace=TRUE),D_00060=sample(1:8,5,replace=TRUE),datetime=as.POSIXct(paste(rep("6/3/2011",5),c("0:00","0:30"
HI,
If you have more than one observation per month, you could do this:
dat1<-read.table(text="
Year Month Sales Customer
2011 Jan 150 35
2011 Jan 125 40
2011 Feb 130 45
2011 Feb 135 25
2012 Jan 100 25
2012 Jan 150 35
2012 Feb
Hi Pradip,
Another way to get the results would be:
res<-cbind(test1,do.call(data.frame,lapply(test1[,seq(1,6,2)],CutQuintiles)))
colnames(res)[7:9]<-paste("newcols_",colnames(res)[7:9],"")
sapply(res,is.factor)
# ObtMj_P ObtMj_SE ExpPrevMed_P
#
4 29 13
A.K.
From: Irucka Embry
To: smartpink...@yahoo.com
Cc: r-help@r-project.org
Sent: Wednesday, January 9, 2013 11:36 AM
Subject: Re: [R] select partial name and full name columns
Hi Arun, thank-you for your sugges
Hi Yao,
You could also use:
library(reshape2)
dd<-dat1[,-(1:4)]
res<-dcast(melt(within(dd,{id=row.names(dd)}),id.var="id"),id~value,length)
head(res)
# id AA AG CC CT GA GG GT TC TG TT
#1 27412 29 10 0 0 13 1 0 0 0 0
#2 27413 0 0 4 9 0 0 0 12 0 28
#3 27414 0 0 0 0 0 0 0 0
0 29 0 77
#3 27414 0 0 0 0 0 0 0 0 0 53 0 0 0 106
#4 27415 0 0 53 0 0 0 0 0 0 0 0 106 0 0
#5 27416 0 0 3 9 0 0 0 12 0 29 0 27 0 79
#6 27417 0 0 53 0 0 0 0 0 0 0 0 106 0 0
A.K.
- Original Message -
From: Yao He
To: arun
Cc: William
(strsplit(allVars,"")[[1]])
parts<-sapply(names(row),function(x){u%in%strsplit(x,"")[[1]]})
mat<-parts%*%row
rownames(mat)<-u
mat
})})
#user system elapsed
#21.553 0.000 21.591
A.K.
- Original Message -----
From: Yao He
To: arun
Cc: Will
HI,
Try this:
mat[match(ind,mat[,2]),]
# [,1] [,2]
#[1,] "y" "c"
#[2,] "x" "b"
#[3,] "z" "d"
#[4,] "w" "a"
A.K.
- Original Message -
From: array chip
To: "r-help@r-project.org"
Cc:
Sent: Thursday, January 10, 2013 1:21 PM
Subject: [R] sort matrix based on a specific o
#8 526 521 431 443 523 472 608
#9 329 534 358 374 382 393 467 429
#10 364 377 393 365 419 420 346 472 489
A.K.
From: eliza botto
To: "smartpink...@yahoo.com"
Sent: Thursday, January 10, 2013 9:13 AM
Subject: merging comma
HI,
You could try this:
library(reshape2)
dcast(melt(tutu,"nam"),nam~value,length)
# nam art deb joy mar seb lio nem tat
#1 da 2 3 1 4 1 1 0 0
#2 fr 2 2 2 3 0 1 1 1
#3 ya 1 2 1 0 0 1 1 0
A.K.
- Original Message -
From: Biau David
To
Hi,
Hope this is what you meant.
#data1
aggregate(.~group+gender,data=data1,mean)
# group gender x
#1 2 f 1.750686
#2 1 m -1.074343
A.K.
- Original Message -
From: Martin Batholdy
To: "r-help@r-project.org"
Cc:
Sent: Friday, January 11, 2013 10:07 AM
Subje
HI David,
I get different results with dcast()
library(reshape2)
dcast(melt(tutu,"nam"),nam~value,length)
# nam art deb joy mar seb lio nem tat
#1 da 2 3 1 4 1 1 0 0
#2 fr 2 2 2 3 0 1 1 1
#3 ya 1 2 1 0 0 1 1 0
tutus <- data.frame(nam=tutu$na
Hi,
May be I misunderstood ur question.
You could do this:
res<-aggregate(.~group,data=data1,mean)
res$gender<-data1$gender[match(res$gender,as.numeric(data1$gender))]
res
# group x gender
#1 1 -1.074343 m
#2 2 1.750686 f
A.K.
- Original Message -
From: Mar
HI,
May be this also works for you:
do.call(cbind,by(df1[,-1],df1[,1],function(x) x))
# A.col2 A.col3 B.col2 B.col3
#1 1 1 3 3
#2 2 2 4 4
A.K.
- Original Message -
From: Johannes Radinger
To: r-help@r-project.org
Cc:
Sent: Friday, January 11,
ce a plot with it( i would
also like to produce a plot). I searched the responses in the relevant
sections in r but could n't find an answer.
Thanks,
On Wed, Jan 9, 2013 at 12:31 PM, arun kirshna [via R] <
ml-node+s789695n465499...@n4.nabble.com> wrote:
> HI,
>
> In your datas
but what I wanted is to find out the percentages
with each level of the variable with my dataset, as in if there is more
missing data in females or males etc?.
I installed "mi" package, but unable to produce a plot with it( i would
also like to produce a plot). I searched t
(is.na(x[,-1]))/nrow(x))*100,"%",sep="")))
From ur reply, it seemed like you were trying different codes:
as data(df,package, package="vmv")
A.K.
____
From: Usha Gurunathan
To: arun
Cc: R help
Sent: Saturday, January 12, 2013 1:42 A
Obese14
#Female "15.2076677316294%" "15.2076677316294%" "24.0894568690096%"
#Male "16.5015974440895%" "16.5015974440895%" "25.814696485623%"
# Overweight14 Overweight21 Obese21
#Female "24.08945
HI Ved,
I was able to get the dotted circle using the same command. I am using R 2.15
on Ubuntu 12.04.
A.K.
- Original Message -
From: Ved P. Sharma
To: r-help@r-project.org
Cc:
Sent: Saturday, January 12, 2013 2:43 AM
Subject: [R] Drawing a dotted circle.
Hi,
I am trying to dr
mur...@tiscali.it"
To: smartpink...@yahoo.com
Cc:
Sent: Saturday, January 12, 2013 1:23 PM
Subject: bind tables
Hi Arun,
Thank you very much for your reply.
I know that I was not clear enough, since my basic knowledge of r language.
To be more clear
I have this df:
year h len fre
1994 5 10.5
- Original Message -
From: "mmur...@tiscali.it"
To: smartpink...@yahoo.com
Cc:
Sent: Saturday, January 12, 2013 1:23 PM
Subject: bind tables
Hi Arun,
Thank you very much for your reply.
I know that I was not clear enough, since my basic knowledge of r language.
To be more clear
nathan
To: arun
Cc: R help
Sent: Saturday, January 12, 2013 5:59 PM
Subject: Re: [R] random effects model
Hi AK
That works. I was trying to get similar results from any other package. Being
a beginner, I was not sure how to modify the syntax to get my output.
lapply(split(BP_2bSexNoMV,BP_
828 2200
Xsq<-chisq.test(M1)
Xsq
# Pearson's Chi-squared test with Yates' continuity correction
#data: M1
#X-squared = 2.5684, df = 1, p-value = 0.109
I will take a look at your second question later.
A.K.
____
From: Usha Gurunathan
HI,
Not sure this helps:
netw<-read.table(text="
lastname_initial, year
Aaron H, 1900
Beecher HW, 1947
Cannon JP, 1985
Stone WC, 1982
van der hoops bf, 1948
NA, 1976
",sep=",",header=TRUE,stringsAsFactors=FALSE)
res1<-sub("^[[:space:]]*(.*?)[[:space:]]*$","\\1",gsub("\\w+$","",netw[,1]))
res1[!
riya
#10 riad biau
res[complete.cases(res),]#removes the NA rows.
A.K.
From: Biau David
To: arun ; r help list
Sent: Sunday, January 13, 2013 12:02 PM
Subject: Re: [R] extracting character values
OK,
here is a minimal working example
Hi,
This should also work:
do.call(data.frame,lapply(netw,function(x) gsub("^ *(\\D+) \\w+$","\\1",x)))
A.K.
From: Biau David
To: arun ; r help list
Sent: Sunday, January 13, 2013 12:02 PM
Subject: Re: [R] extracting character val
fore
#Next line
BPsub3$Categ[BPsub6$Overweight==1&BPsub3$time==1&BPsub3$Obese==0]<-
"Overweight14" #It should be BP.sub3 and what is BPsub6, it was not defined
previously.
#Error in BPsub3$Categ[BPsub6$Overweight == 1 & BPsub3$time == 1 & BPsub3$Obese
== :
#object
HI,
YOu could do either:
Lines<-readLines(textConnection("Name: John Smith Age: 35 Address: 32, street,
sub, something
Name Adam Grey Age: 25 Address: 26, street, sub, something"))
Lines[-grep("Name\\:",Lines)]<-gsub("Name","Name:",Lines[-grep("Name\\:",Lines)])
Name<-gsub("Name\\: (.*) Age
Hi,
Do you want to read it from a saved file?
res<-split(dat1,dat1$day)
res[[1]]
names(res)<-paste("data",1:5,sep="")
write.table(res[[1]],file="data1.txt")
read.table("data1.txt",sep="",header=TRUE)
# number day month hours
#1 1 14 10 2
#2 2 14 10 12
#3 3 14
t==0] <-
"Normal21"
BPsub6$Categ[BPsub6$Obese==1&BPsub6$time==2&BPsub6$Overweight==0|BPsub6$Obese==1&BPsub6$time==2&BPsub6$Overweight==1]
<- "Obese21"
BPsub6$Categ <- factor(BPsub6$Categ)
BPsub6$time <- factor(BPsub6$time)
summary(BPsub6$Categ)
BPsub7
could try lmer() from lme4.
library(lme4)
fm1<-lmer(HiBP~time+(1|CODEA), family=binomial,data=BP.stack3) #check codes,
not sure
print(dotplot(ranef(fm1,post=TRUE),
scales = list(x = list(relation = "free")))[[1]])
qmt1<- qqmath(ranef(fm1, postVar=TRUE))
print(qmt1[
HI,
In this case, all the elements in Dat$att are found in Replace.
Dat1<-within(Dat,{att<-as.character(att)})
vec1<-unlist(lapply(strsplit(Dat1$att,""),unique))
vec1
#[1] "b" "b" "b" "d" "b" "b" "a" "c" "a" "d" "a" "b" "b" "b" "b" "d" "b" "b"
"d"
#[20] "b"
Dat1[5:7,2]<-c("uu","tt","vv")
vec1<
Hi,
Check these links:
http://comments.gmane.org/gmane.comp.lang.r.ggplot2/6527
https://groups.google.com/forum/#!msg/ggplot2/nfVjxL0DXnY/5zf50zCeZuMJ
A.K.
From: Usha Gurunathan
To: arun
Cc: R help
Sent: Tuesday, January 15, 2013 6:31 AM
Subject: Re: [R
Hi,
May be this helps:
res<-do.call(rbind,lapply(xaulist,function(x)
as.integer(apply(tata,1,function(i) any(i==x)
res[]<-sapply(res,as.numeric)
identical(res,tutu)
#[1] TRUE
A.K.
- Original Message -
From: Biau David
To: r help list
Cc:
Sent: Tuesday, January 15, 2013 2:41 PM
Hi,
You could also do this:
res1<-do.call(rbind,lapply(xaulist,function(x)
as.numeric(apply(t(mapply(`==`,tata,x)),2,any
identical(res1,tutu)
#[1] TRUE
A.K.
- Original Message -
From: Biau David
To: r help list
Cc:
Sent: Tuesday, January 15, 2013 2:41 PM
Subject: [R] removing
HI,
vec1<-"'asd'f"
vec2<-'"asd"f'
gsub("[\"]","",vec2)
#[1] "asdf"
gsub("[']","",vec1)
#[1] "asdf"
A.K.
- Original Message -
From: Christofer Bogaso
To: r-help
Cc:
Sent: Tuesday, January 15, 2013 4:38 PM
Subject: [R] Regular expression
Hello again,
I am having a problem on Regu
Hi,
Try this:
The last part was not clear.
A.df<-read.table(text="
a c 0.9
b x 0.8
b z 0.5
c y 0.9
c x 0.7
c z 0.6
",sep="",header=FALSE,stringsAsFactors=FALSE)
lst1<-split(A.df[,-1],A.df$V1)
lst1
#$a
# V2 V3
#1 c 0.9
#
#$b
#
HI,
Not sure if this is what you meant.
tapply(iris$Sepal.Length,iris$Species,FUN=function(x)
sum(x)/sum(iris$Sepal.Length)*100)
# setosa versicolor virginica
# 28.55676 33.86195 37.58129
A.K.
- Original Message -
From: Karine Charlebois
To: "r-help@r-project.org"
Cc:
Sen
Hi,
Is it this?
aggregate(iris$Sepal.Length,by=list(iris$Species),FUN=function(x)
sum(x)/sum(iris$Sepal.Length)*100)
Group.1 x
1 setosa 28.55676
2 versicolor 33.86195
3 virginica 37.58129
A.K.
From: Karine Charlebois
To: arun
Sent
A.K.
From: Biau David
To: arun
Cc: R help
Sent: Wednesday, January 16, 2013 7:37 AM
Subject: Re: [R] removing loops from code in making data.frame
thanks, it goes a lot faster. Just one thing though, when I apply the code to
my data, both data.frames end up "differente. Or at least
HI,
You could also do this:
lapply(sample1,function(x) {mat1<-cbind(matrix(0,nrow=2,ncol=3),x);
mat1[cbind(rep(1,3),1:3)]<- mat1[cbind(rep(1,3),4:6)];
mat1[cbind(rep(1,3),4:6)]<-0; mat1})
A.K.
- Original Message -
From: Kathryn Lord
To: r-help@r-project.org
Cc:
Sent: Wednesday, Janu
Lines1 <- readLines(con = textConnection(
"informations (unknown count of lines)
... and at some point the table
--
year month mday value
2013 1 16 0 "))
indx<-seq(match(regmatches(Lines1,regexpr("^year.*",Lines1)),Lines1),length(Lines1))
read.table(text=Lines1[indx],sep="",header=TRUE)
# y
Hi,
May be this helps you.
source("Andreadata.txt")
head(occ.data)
melting<- melt(occ.data,id.var=c("Point", "Site", "Rep",
"Año"),measure.var="Pres")
y<-cast(melting,Site~Rep~Point~Año)
dim(y)
#[1] 10 5 25 6
y[,,25,6]
# Rep
#Site 1 2 3 4 5
# 1021 0 0 0 0 0
# 1022 0 0 0 0 0
# 10
HI,
Saw ur post in Nabble.
occ.data<-read.table(text="
Año Punto Especie Pres Ruta_com Point Site Rep guild
1 2012 30 TYSA 1 108 30 1086 5 OTHER
2 2012 26 VACH 1 108 26 1086 1 OTHER
3 2012 27 VACH 1 108 27 1086 2 OTHER
4 2012 26 ZE
Hi,
May be this helps:
library(Matrix)
res1<-lapply(split(x,1:nrow(x)),function(y)
sparseMatrix(i=rep(1:4,each=5),j=1:(4*5),x=y))
do.call(rbind,lapply(seq_along(res1),function(i) res1[[i]][i,]))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
#[1,] 1 5 9
ot;plyr". Still error
message appears with ggplot2
Btw, did you get the attachments with my earlier mail?
Ta.
On Wed, Jan 16, 2013 at 3:16 AM, arun kirshna [via R] <
ml-node+s789695n4655612...@n4.nabble.com> wrote:
>
>
> Hi,
> Check these links:
> http://comments
HI,
May be this helps:
Example$Wi<-unlist(aggregate(Weight~ID,data=Example,function(x)
round(x/sum(x),2))[,2])
res<-do.call(rbind,lapply(split(Example,Example$Specie),function(x) with(x,
{aggregate(Wi,list(Food.item),function(y) sum(y)/length(unique(x[,1])))})))
names(res)<-names(Solution)[2:3]
,] 0 0 0 0 0 0
#[3,] 19 0 0 0 0 0
#[4,] 0 4 8 12 16 20
Not sure if there are any shortcuts with kronecker()
A.K.
- Original Message -
From: Martin Maechler
To: arun
Cc: Kathryn Lord ; R help
Sent: Thursday, January 17
HI,
May be this helps:
mydata_long1<-within(mydata_long,{colorvar<-factor(colorvar,levels=1:3)})
require("ggplot2")
p <- ggplot(data=mydata_long1,
aes(x=variable, y=value,
group=id, colour = colorvar)) +
geom_line()
p
A.K.
- Original Message -
From: Mary
To: r-help@r-proje
HI,
May be this helps:
Lines1<-readLines(textConnection('key1=23, key2=67, key3="hello there"
key1=7, key2=22, key3="how are you"
key1=2, key2=77, key3="nice day, thanks"'))
res<-read.table(text=gsub("key{0,1}\\d","",gsub("[\",]","",Lines1)),sep="=",header=FALSE,stringsAsFactors=F)[-1]
names(re
Hi,
Sorry, there was a mistake. I didn't notice comma in key3
.Lines1<-readLines(textConnection('key1=23, key2=67, key3="hello there"
key1=7, key2=22, key3="how are you"
key1=2, key2=77, key3="nice day, thanks"'))
res1<-read.table(text=gsub("key{0,1}\\d","",gsub("[\"]","",Lines1)),sep="=",header=
Hi,
May be this helps:
frames<-list(data.frame(c1=1:3,day1=17,hour1=c(10,11,6)),data.frame(c1=6:7,day1=19,hour1=8),data.frame(c1=8:10,day1=21,hour1=c(11,15,18)),data.frame(c1=12:13,day1=23,hour1=7))
par(mfrow=c(2,2))
lapply(seq_along(frames),function(i) plot(frames[[i]][,3]))
A.K.
- Origina
HI,
May be this helps:
dat1<-read.table(text="
id status week
1 no 1
1 no 2
1 no 3
1 no 4
1 no 5
1 no 6
1 no 7
2 no 1
2 no 2
2 no 3
2 no 4
2 yes 5
2 yes 6
2 na 7
2 na 8
2 na 9
3 no 1
3 no 2
3 no 3
3 Unknown 4
3 unknown 5
3 na 6
3 na 7
3 na 8
",sep="",header=TRUE,stringsAsFactors=FALSE,na.strings=
apply(f,1,function(x) all(duplicated(x)|duplicated(x,fromLast=TRUE)&!is.na(x)))
#[1] TRUE FALSE FALSE FALSE
A.K.
- Original Message -
From: Sam Steingold
To: r-help@r-project.org
Cc:
Sent: Friday, January 18, 2013 3:53 PM
Subject: [R] select rows with identical columns from a data
Hi,
Not sure what format you wanted the dates:
gsub("^\\w+ ","",gsub("[_]"," ",Text))
#[1] "May 09 2009" "01-01-2001"
#Another way is:
gsub("^\\w+ |\\w+_","",Text)
#[1] "May 09 2009" "01-01-2001"
res<- gsub("^\\w+ |\\w+_","",Text)
res1<-c(as.Date(res[grep(" ",res)],format="%b %d %Y"), as.D
HI,
Assuming a matrix:
set.seed(15)
mat1<-matrix(sample(-10:10,40,replace=TRUE),ncol=5)
apply(mat1,2,function(x) ifelse(x<0,x+24, x))
# [,1] [,2] [,3] [,4] [,5]
#[1,] 2 4 23 1 0
#[2,] 18 7 10 3 16
#[3,] 10 16 16 16 0
#[4,] 3 3 6 17 3
#[5,] 21
Hi,
This could be also done by:
#Using Arun's example:
res<- Reduce('+', split(df, grp))/length(levels(grp))
> res
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
#1 417.3 792.2 504.2 506.1 513.9 480.7 545.4 564.4 473.7 486.2
#2 585.8 416.6 409.5 417.8 480.1 586.4 436.1 615.1
HI,
Not clear what you are trying to do:
set.seed(25)
df1<-as.data.frame(matrix(sample(1:40,20,replace=TRUE),ncol=4))
set.seed(15)
df2<-as.data.frame(matrix(sample(1:40,20,replace=TRUE),ncol=4))
do.call(rbind,lapply(list(df1,df2),`[`,c(2,4)))
# V2 V4
#1 40 6
#2 26 22
#3 14 30
#4 3 20
#
on to R" from the
http://cran.r-project.org/manuals.html.
A.K.
- Original Message -
From: IlyaNovikov
To: r-help@r-project.org
Cc:
Sent: Sunday, January 20, 2013 1:21 AM
Subject: Re: [R] applying a formula from text
Dear Arun,
I am a novice in R bu some my friends that use R
HI,
Your example data had only 4 columns
mydata<-read.table(text="
3.014505 62.96425 3.014505 138.0673
2.817503 56.03400 2.817503 133.3411
2.976227 47.12192 2.976227 139.2438
2.825495 75.05284 2.825495 129.2959
2.793500 52.75190 2.793500 130.9874
3.006333 54.56210 3.006333 136.2982
odd,
critical1<-c(rep(c(1.61,75.89),(ncol(mydata)-1)/2),1.61)
colSums(mapply("<",mydata,critical))
#V1 V2 V3 V4
# 0 19 0 0
A.K.
From: Roslina Zakaria
To: arun
Sent: Monday, January 21, 2013 2:45 AM
Subject: Re: [R] compare and count dat
Hi,
If 'commNo` is factor.
set.seed(5)
data1<-data.frame(commNo=sample(786:789,10,replace=TRUE),Col2=rnorm(10,10))
set.seed(5)
data2<-data.frame(commNo=sample(c(786:789,NA),10,replace=TRUE),Col2=rnorm(10,10))
data4<-within(data1,{commNo<-factor(commNo)})
data4[data4$commNo==786, "commNo"]<-"Name
Hi,
I guess there should be missing values.
set.seed(5)
data1<-data.frame(commNo=sample(786:789,10,replace=TRUE),Col2=rnorm(10,10))
set.seed(5)
data2<-data.frame(commNo=sample(c(786:789,NA),10,replace=TRUE),Col2=rnorm(10,10))
data1[data1$commNo==786, "commNo"]<-"Name of the Community"
data2[data
Hi,
This could be done with ?aggregate()
res<-aggregate(aa$Eaten,by=list(ID=aa$ID),FUN=function(x) x)
res1<-data.frame(ID=res[,1],data.frame(res[[2]]))
names(res1)[2:3]<-unique(aa$Target)
res1
# ID TPP GPA
#1 1 0 9
#2 2 1 11
#3 3 3 8
#4 4 1 8
#5 5 2 10
A.K.
- Orig
HI,
Please check the link:
https://stat.ethz.ch/mailman/listinfo/r-help
At the end, there is an option to unsubscribe:
"To unsubscribe from R-help, get a password reminder,
or change your subscription options enter your subscription
email address:"
Hope it helps:
A.K.
- O
Hi,
I am not sure about the logic behind creation of groups, especially, how do you
want to assign the group number to a particular combination of Feature and OS.
One possible way would be:
dat1$Group<-paste(dat1[,1],dat1[,2],sep="")
> dat1
# Feature OS Group
#1 4 2 42
#2 4 1
Hi,
You could also do this by:
set.seed(15)
tr.df<-data.frame(ID=rep(1:29,each=3),prep=runif(87,1,3),postp=runif(87,0.5,1.5))
tr.df$time<-1:87
res<- reshape(tr.df, varying=2:3, v.name="value",
times=c("prep","postp"),idvar="time",timevar="prepost",direction="long")
res<-res[order(res$ID,res$time)
Hi,
You could also try:
dat1<-read.table(text="
Feature OS
4 2
4 1
4 3
1 2
4 1
",sep="",header=TRUE)
dat1$Group<- as.numeric(factor(Reduce(paste0,dat1)))
A.K.
- Original Message -
From: Tammy Ma
To:
Hi,
dateA<-seq.Date(as.Date("1jan2012",format="%d%b%Y"),as.Date("31Dec2012",format="%d%b%Y"),by="day")
dateB<-seq.Date(as.Date("1Mar2012",format="%d%b%Y"),as.Date("30Nov2012",format="%d%b%Y"),by="day")
set.seed(15)
A<-data.frame(dateA,value=sample(1:300,366,replace=TRUE))
set.seed(25)
B<-data
Hi,
You could also try this:
dateA<-seq.Date(as.Date("1jan2012",format="%d%b%Y"),as.Date("31Dec2012",format="%d%b%Y"),by="day")
dateB<-seq.Date(as.Date("1Mar2012",format="%d%b%Y"),as.Date("30Nov2012",format="%d%b%Y"),by="day")
set.seed(15)
A<-data.frame(dateA,value=sample(1:300,366,replace=TRUE
e(raw_time,format="%d%B%Y")
Could you just dput() a few lines of your dataset if this is not working?
Tx.
A.K.
----- Original Message -
From: "Yuan, Rebecca"
To: 'arun'
Cc:
Sent: Tuesday, January 22, 2013 2:08 PM
Subject: RE: [R] plot two time series with
in Anew, I
guess, it would remove that from plotting)
If you want to remove the NA rows:
use, na.omit() or complete.cases()? #as I did in the previous email.
Could you dput() an example dataset?
A.K.
- Original Message -
From: "Yuan, Rebecca"
To: 'arun'
Cc
UE)]<-as.Date(gsub("(.*-).*(-.*)","\\102\\2",B[,1][duplicated(B[,1],fromLast=TRUE)]))
#this step may not be needed in ur data. In the month of march, there were
two values
library(xts)
Anew<-as.xts(A[,-1],order.by=A[,1])
Bnew<-as.xts(B[,-1],order.by=B[,1])
res<-m
sample(1:72,10,replace=TRUE)))
A[,1]<-as.Date(gsub("\\d+$","28",A[,1]))
library(xts)
library(zoo)
Anew<-as.xts(A[,-1],order.by=A[,1])
Bnew<-as.xts(B[,-1],order.by=B[,1])
res<-merge(Anew,Bnew)
plot.zoo(res)
>From your reply, it seems like dateB day didn'
HI,
You can check this link:
http://r.789695.n4.nabble.com/leap-years-in-temporal-series-command-ts-td3309014.html
Also, this may help you:
library(lubridate), ?leap_year()
leap_year(2008)
#[1] TRUE
ymd("2008-2-29")
1 parsed with %Y-%m-%d
#[1] "2008-02-29 UTC"
A.K.
- Original Message --
Hi,
May be this helps:
barchart(npp,origin=0,box.width=1,
panel=function(x,y,...){
panel.barchart(x,y,...)
panel.abline(v=2,col.line="red",lty=3)})
A.K.
- Original Message -
From: Jonathan Greenberg
To: r-help
Cc:
Sent: Tuesday, January 22, 2013 5:41 PM
Subject: [R] Adding a lin
Hi,
try this:
unlist(lapply(split(test,((seq_along(test)-1)%/% 10)+1),mean))
# 1 2 3 4 5 6 7 8
#0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.146375
# 9 10 11
#0.00 0.194500 0.00
A.K.
-
Hi,
May be this also helps:
s <- c(" ", " ",
" ",
" ",
" ",
" ",
" ",
" ",
" ",
" ",
" ",
" ")
Lines1<-gsub("^\\s+| \\s+$","",gsub("[^0-9A-Z]"," ",s))
dat1<-read.table(text=Lines1[Lines1!=""],sep="",header=F,stringsAsFactors=F)
dat1New<-dat1[,seq(2,ncol(dat1),by=2)]
colnames(dat1New
Hi,
Try this:
x1<-x[rev(order(x$Names,x$Values)),]
do.call(rbind,tapply(x1$Values,list(x1$Names),head,2))
# [,1] [,2]
#CK113234 223.2966 222.6737
#CK113298 192.5964 187.7486
#CK114042 236.3939 232.0223
#CK116292 237.5936 228.0037
#CK116296 223.6372 210.6630
#The average
tapply(x1$
Hi,
It's not clear regarding those blanks especially, the num_daughter. I guess
the father and mother would be the same as the previous row.
Deleting those rows:
df1 <- read.table(text="father mother num_daughter daughter
291 3906 0 NA
275 4219 0 NA
273 4236 1
Hi,
May be this helps:
df1<-read.table(text="
father,mother,num_daughter,daughter
291,3906,0,
275,4219,0,
273, 4236,1,49410
281,4163,1,49408
274, 4226,1,49406
295, 3869,2,49403
295,3869,2,49404
287,4113,0,
295, 3871,1,49401
292, 3895,4,49396
292,3895,4, 49397
292,3895,4,49398
292,3895,4,49399
29
Hi,
You could try this:
dat1<-read.table(text=pub,sep=",",fill=TRUE,stringsAsFactors=F)
dat2<- as.data.frame(do.call(cbind,lapply(dat1,function(x) gsub(" $","",gsub("^
|\\w+$","",x,stringsAsFactors=F)
dat2
# V1 V2 V3 V4
#1 Brown Santos Ro
Hi,
If the `spaces` in "father", "mother", "num_daughter" columns needs to be
replaced by the values in the previous row,
dat1<-read.table(text="
father, mother, num_daughter, daughter
291, 3906, 0,
275, 4219, 0,
273, 4236, 1, 49410
281, 4163, 1, 49408
274, 4226, 1, 49406
295, 3869, 2, 49403
HI,
Not sure this is what you wanted.
for (i in 731:732) {
SEQ <- (i - 5):(i + 5)
print(SEQ)
SEQ <- SEQ[SEQ > 730 & SEQ < 1096]
print(SEQ)
vec1<-731:741
print(vec1[!vec1%in%SEQ])
}
#[1] 726 727 728 729 730 731 732 733 734 735 736
#[1] 731 732 733 734 735 736
#[1] 737 738 739 740 741
#
Arstra Van den Hoops lamarque D # initial present.
I tried this case with Rui's solution:
fun2(pubnew)
#[[1]]
#[1] " Brown" "Santos" "Rome" "Don Juan"
#[[2]]
#[1] "Benigni"
#
#[[3]]
#[1] "Arstra" "Van den Hoo
HI,
You could use:
library(reshape)
res<-melt(dat,id.var=c("region","state"))
names(res)[3:4]<-c("species","presence")
res<-res[rev(order(res$region,res$state)),]
row.names(res)<- 1:nrow(res)
res
# region state species presence
#1 sydney nsw species3 1
#2 sydney nsw species2
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