Hi,May be this helps:
dat1<-structure(list(X.DATE = c("01052007", "01072007", "01072007",
"02182007", "02182007", "02242007", "03252007"), X.TIME = c("0230",
"0330", "0440", "0440", "0440", "0330", "0230"), VALUE = c(37,
42, 45, 45, 45, 42, 45), VALUE2 = c(29, 24, 28, 27, 35, 32, 32
)), .Names
My code was based on the assumption that your dataset was similar to the one I
provided. Please provide an example dataset (use dput(head(dataset),20))
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
A.K.
>Arun,
>
>I have tried applying your su
red by the treatment A or U for all sites.
Maybe something like
>t.test(x[WrackMass$Site=="ALA1" and Treatment=="A"],x[WrackMass$Site=="ALA"
>and Treatment=="U"])$p.value)
>
>Ideas?
- Original Message -
From: arun
To: R help
Cc:
Sent
s
#[1,] 0.6482613 0.9411953 0.7927984 0.3027634
#
#$CLA1
# Algae.Mass Seagrass.Mass Terrestrial.Mass Other.Mass
#[1,] 0.454294 0.02519427 0.5650988 0.2981702
A.K.
- Original Message -
From: arun
To: R help
Cc:
Sent: Thursday, May 2, 2013 3:30 PM
Subject: Re: Multiple
Hi,
May be this helps:
set.seed(24)
dat1<-
data.frame(date1=sample(seq(as.Date("2012-09-14",format="%Y-%m-%d"),length.out=40,by="day"),20,replace=FALSE),
value=sample(1:60,20,replace=TRUE))
dat1$days1<- as.numeric(difftime(dat1$date1,as.Date("1970-01-01")))
#or
library(lubridate)
dat1$days2<- da
Hi,
I am not sure about what you meant.
lapply(1:5,function(i) data.frame())
[[1]]
data frame with 0 columns and 0 rows
[[2]]
data frame with 0 columns and 0 rows
[[3]]
data frame with 0 columns and 0 rows
[[4]]
data frame with 0 columns and 0 rows
[[5]]
data frame with 0 columns and 0 rows
num1<- c(0,NA,1,3)
num1==NA
#[1] NA NA NA NA
num1!=NA
#[1] NA NA NA NA
is.na(num1)
#[1] FALSE TRUE FALSE FALSE
A.K.
- Original Message -
From: jpm miao
To: r-help
Cc:
Sent: Friday, May 3, 2013 11:24 AM
Subject: [R] Why can't R understand if(num!=NA)?
I have a program, when I wri
Hi ST,
Try this:
set.seed(51)
df1<- as.data.frame(matrix(sample(1:40,60,replace=TRUE),ncol=10))
df2<- df1
check<- c("V3","V7","V9")
df1[,match(check,colnames(df1))]<-lapply(df1[,match(check,colnames(df1))],as.factor)
str(df1)
#'data.frame': 6 obs. of 10 variables:
# $ V1 : int 32 9 12 40 9
(ID),nrow)[,2])
#[1] 2.4
#or
mean(with(DF,tapply(ID,ID,FUN=length)))
#[1] 2.4
A.K.
From: Sarah Jo Sinnott <105405...@umail.ucc.ie>
To: arun
Sent: Friday, May 3, 2013 4:35 PM
Subject: Re: selecting certain rows from data frame
Yes - but if I can count t
HI,
May be this helps:
dat1<- read.table(text="
ID Var1 Var2 Group
A1 1 1 BB
A2 1 2 AA
B1 2 1 CC
B2 1 3 DD
C1 1 2 EE
",sep="",header=TRUE)
lib
HI,
Not sure I understand it correctly.
dat1<- read.table(text="
site Year doy fish Feed swim agr_1 agr_2 agr_3 rest hide
3 2012 203 1 1 0 0 0 0 0 0
3 2012 203 1 0 1 0 0 0 0 0
3 2012 203 1 0 1 0 0 0 0 0
3 2012 203 2 0 0 0 0 0 1 0
3 2012 203 2 1 0 0 0 0 0 0
3 2012 203 2 1 0 0 0 0 0 0
4 2012 197 1 0
Hi,
One more possibility:
dateA<-seq.Date(as.Date("1jan2012",format="%d%b%Y"),as.Date("14Feb2013",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=cumsum(rnorm(411)))
set.seed(2
Hi,
If it is to get the first 18 genes (based on rownumbers)
dat1<- read.table("apcall.txt",header=TRUE,sep="",stringsAsFactors=FALSE)
dat1[1:18,]
#or
dat1[as.numeric(rownames(dat1))<19,]
A.K.
>The row numbers represent 7129 different genes
>The 38 columns represent 38 different patients
>
18 As within one row.
>Or should I say, the count for the character "A" is no more than 18 in one
>row.
>
>Thanks
- Original Message -
From: arun
To: R help
Cc:
Sent: Saturday, May 4, 2013 11:43 AM
Subject: Re: Help, how to find the genes with A<19?
Hi,
If
A"]<19)
names(head(res2Alt19)) #here 38 is included because the count of "A" is zero.
#[1] "19" "38" "39" "40" "41" "42"
res3Aeq0<-which(res3[,"A"]==0)
names(head(res3Aeq0))
#[1] "38" "42"
Hi,
rownames(a)
# [1] "4" "5" "6" "7" "8" "9" "11" "12" "13" "15" "16" "17"
#[13] "18" "21" "40" "48" "50" "52" "53" "54" "81" "101" "102" "103"
#[25] "108" "118" "147" "803" "805" "806" "807" "808" "809" "810" "812" "814"
#[37] "815" "816" "822" "825"
rownames(a1)
#[1]
Hi,
set.seed(15)
vec1<- sample(1:15,10,replace=TRUE)
mean(vec1)
#[1] 10
sd(vec1)
#[1] 4.346135
stErr<- sd(vec1)/sqrt(length(vec1))
stErr
#[1] 1.374369
library(plotrix)
std.error(vec1)
#[1] 1.374369
A.K.
- Original Message -
From: Xavier Prudent
To: S Ellison
Cc: "r-help@r-proje
set.seed(25)
dat1<- as.data.frame(matrix(sample(c(1:20,NA),10*20,replace=TRUE),ncol=20))
dat2<- dat1
vec1<-colMeans(dat1,na.rm=TRUE)
dat2[]<-lapply(seq_len(ncol(dat1)),function(i) {x<-dat1[,i]; x[is.na(x)]<-
vec1[i];x})
dat2<-signif(dat2,digits=3)
dat2
A.K.
>I have Data frame with 20 variabl
Hi,
library(car)
set.seed(25)
x1<- data.frame(id=sample(1:8,8,replace=FALSE),f1=LETTERS[1:8])
x1$id<-recode(x1$id,'5=1;6=2;7=3;8=4')
x1
# id f1
#1 4 A
#2 1 B
#3 1 C
#4 3 D
#5 2 E
#6 3 F
#7 2 G
#8 4 H
A.K.
- Original Message -
From: Hui Du
To: "r-help@r-project.org"
Hi,
Try this:
dat1<- read.table(text="
ObsNumber ID Weight
1 0001 12
2 0001 13
3 0001 14
4 0002 16
5 0002 17
",sep="",header=TRUE,col
3 0001 14 0001_3
#4 4 0002 16 0002_1
#5 5 0002 17 0002_2
dat2$UniqueID<-unlist(lapply(split(dat2,dat2$ID),function(x)
with(x,as.character(interaction(ID,seq_len(nrow(x)),sep="_",use.names=FALSE)
A.K.
- Original Message -
From: arun
To
(y1$patient_id),scores=mean(y1$scores,na.rm=TRUE))})
) }))
row.names(res1)<-1:nrow(res1)
res1$period<-with(res1,ave(patient_id,patient_id,FUN=seq))
res1
# patient_id scores period
#1 1 2.05 1
#2 2 2.40 1
#3 2 2.05 2
A.K.
___
quot;_",use.names=FALSE)
#[1] "0001_1" "0001_2" "0001_3" "0002_1" "0002_2"
A.K.
From: Ye Lin
To: arun
Cc: R help
Sent: Tuesday, May 7, 2013 2:54 PM
Subject: Re: [R] create unique ID for each group
Thanks A
2.2
#5 5 0002 17 0002_2 0002 2.6
#6 6 18 NA
#7 7 0003 19 0003_1 NA
#8 8 20 NA
#9 9 0003 21 0003_2 NA
A.K.
From: Ye Lin
To: arun
Sent: Tuesday, May 7, 2013 4:05 P
X29 X30
#1 14 27 3 21 6 44 33 42 10 29
#2 48 13 8 47 18 9 23 9 44 3
#3 25 14 31 19 14 6 26 13 6 49
#4 43 28 15 6 9 19 43 21 41 21
#5 1 27 18 3 42 5 16 39 46 47
A.K.
- Original Message -
From: Vivek Das
To: arun
Cc:
Sent:
tringsAsFactors=FALSE)
out_dat2<-merge(dat1[,1:4],dat2,by="ID")
identical(out_dat,out_dat2)
#[1] TRUE
A.K.
From: Vivek Das
To: arun
Cc: R help
Sent: Tuesday, May 7, 2013 6:07 PM
Subject: Re: R help for creating expression data of Different
Hi,
May be this helps:
funFluxSumSeason <- function(DF, FUN = sum){
month <- as.integer(format(DF$Date, format="%m"))
year <- format(DF$Date, format="%Y")
DF$season<- NA
DF$season[month %in% 10:12] <- paste(year[month %in% 10:12], "Fall")
DF$season[month %in% 1:3] <- paste(year[month %in% 1:3
Hi,
dat1<- read.table(text="
Restaurant owner purchase_date
23 Chuck 3/4/2011
23 Chuck 3/4/2011
23 Chuck 3/4/2011
23 Chuck 3/4/2011
23 Bob 1/1/2013
23 Bob 1/1/2013
23 Bob 1/1/2013
1
HI,
Try this:
dat1<- read.table(text="
V1 V2
A 1
B 2
A 1
B 3
",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
ddply(dat1,.(V1),summarize, V2=list(V2))
# V1 V2
#1 A 1, 1
#2 B 2, 3
#or
aggregate(V2~V1,data=dat1,FUN= function(x) x,simplify=FALSE)
# V1 V2
#1 A
Hi,
May be this helps:
fibv =function(n)
{
f1 = f2 = 1
f3<- c(f1,f2)
for(i in seq(2, n-1)) {
if(n == 0 || n == 1) return(n)
if(n == 2) return(1)
f = f1 + f2
f2 = f1
f1 = f
f3<- c(f3,f)
}
f3
}
fibv(0)
#[1] 0
fibv(1)
#[1] 1
fibv(3)
#[1] 1 1 2
fibv(10)
# [1] 1 1 2 3 5 8 13 21 34 55
A.K.
>I
Hi,
vec1<-factor(rep(c("A","B","C"),c(19,8,11)),levels=c("A","B","C"))
sapply(split(vec1,vec1),length)
# A B C
#19 8 11
#or you could change gl() function
gl.new<-function (n, k, labels = 1:n, ordered = FALSE) {
out<- numeric()
for(i in 1:n){
out<- append(out,rep(i,length.out=k[i]))
Hi,
dat1<- read.table(text="
x1 x2 x3
a1 b2 2
a1 b4 4
a2 NA 3
NA b2 6
a3 b1 NA
a1 b2 9
a1 b2 NA
a1 b4 2
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- dat1
dat2$x4<-with(dat2,ave(x3,x1,x2,FUN=f
Hi,
May be this helps:
set.seed(24)
mydata4<- as.data.frame(matrix(sample(1:100,10*38,replace=TRUE),ncol=38))
dim(mydata4)
#[1] 10 38
library(matrixStats)
res<-do.call(cbind,lapply(1:400, function(i) {permutation<-sample(mydata4);
(rowMeans(permutation[,1:27])-rowMeans(permutation[,28:38]))/
Hi,
Try this:
dat1<- read.table(text="
V1 V2 V3
1 Dosis Gewicht Geschlecht
2 0 6.62 m
3 0 6.65 m
4 0 5.78 m
5 0 5.63 m
6 1 6.78 m
7 1 6.45 m
8 1 5.68 m
9 1
[,1:27])+rowSds(permutation[,28:38]))}
)) })
# user system elapsed
# 3.624 0.000 3.632
identical(fourPGCs,res)
#[1] TRUE
A.K.
- Original Message -
From: David Winsemius
To: arun
Cc: R help
Sent: Saturday, May 11, 2013 2:31 PM
Subject: Re: [R] How to repeat 2 functions in suc
Hi,
If the means are based on the combination.
means<-tapply( Daten$weight, list(Daten$Dosis,Daten$sex), mean)
library(reshape2)
points(melt(means)$value,pch=5,col="red",lwd=5)
A.K.
- Original Message -
From: maggy yan
To: R-help@r-project.org
Cc:
Sent: Saturday, May 11, 2013 8:57 P
mean)
points(means, pch=5, col="red", lwd=5)
but only the boxplots for male got that point on them, its really weird
because I don't think that I separated the sex in the codes above
- Original Message -
From: arun
To: maggy yan
Cc: R help
Sent: Saturday, May 11, 2
Hi,
May be this helps:
plot(dataset1~Date,data=mydata[!is.na(mydata$dataset1),],ylim=range(5.7,8),pch=10,cex=0.8,col="black",xlab="Date",ylab="pH")
with(mydata[!is.na(mydata$dataset2),],points(Date,dataset2,col="blue",pch=2,cex=0.8))
with(mydata[!is.na(mydata$dataset3),],points(Date,dataset3,co
Hi,
According to the error, the variables should have the same length.
For example:
set.seed(24)
dat1<-
cbind(RACE=sample(1:10,10,replace=TRUE),as.data.frame(matrix(sample(1:100,20*10,replace=TRUE),ncol=20)))
lapply(dat1[,-1],function(x)
CrossTable(x,dat1$RACE,format="SPSS",prop.chisq=FALSE,digi
Hi,
You can do:
vec1<-scan(text="[1] 18 18 17 14 17 13 13 18 13 16 16 14
[1] 12 19 15 21 13 17 10 13 18 14 17 13
[2] 13 15 20 14 18 15 10 12 20 17 17 17",what="character")
as.numeric(vec1[-grep("\\[",vec1)])
# [1] 18 18 17 14 17 13 13 18 13 16 16 14 12 19 15 21 13 17 10 13 18 14 17 13 13
#[26] 15
HI,
Try:
library(plyr)
res1<-ddply(df2aggregate,.(id),summarize,x=sum(x),y=mean(y),a=head(a,1))
res1
# id x y a
#1 a 3 NA
#2 b 7 2.5 A
#3 c 11 4.5 C
#4 d NA NA E
res1$x<- as.numeric(res1$x)
identical(ag1.2,res1)
#[1] TRUE
A.K.
- Original Message -
From: Spenc
You may also try:
grepl("[.]",c("ad.1","ads","ad.2"))
#[1] TRUE FALSE TRUE
A.K.
- Original Message -
From: Francesco Isotta
To: r-help@r-project.org
Cc:
Sent: Monday, May 13, 2013 4:19 AM
Subject: [R] grepl
Hello,
it is not clear to me, how to search if in a string there is a "." (f
HI,
May be:
dat1<- read.table(text="
male female transsexuals
0 1 0
1 0 0
0 0 1
0 1 0
1 0 0
1 0 0
0 1 0
",sep="",header=TRUE)
dat1$sex<-colnames(dat1)[apply(dat1,1,function(x) which(x==1))]
dat1
# male female transsexuals sex
#1 0 1 0 female
#2 1 0
t1)[RGB]]})
# user system elapsed
# 0.072 0.000 0.071
head(dt1)
# Red Green Blue RGB
#1: 1 0 0 Red
#2: 1 0 0 Red
#3: 0 0 1 Blue
#4: 0 1 0 Green
#5: 0 1 0 Green
#6: 0 0 1 Blue
identical(dt1,data.table(d))
#[1] TRUE
A.K.
In the first case:
mdat$value<-.POSIXct(mdat$value,tz="EST")
mdat$value[1:3]
#[1] "2012-01-01 01:00:00 EST" "2012-01-01 02:00:00 EST"
#[3] "2012-01-01 03:00:00 EST"
mydata$dd1[1:3]
#[1] "2012-01-01 01:00:00 EST" "2012-01-01 02:00:00 EST"
#[3] "2012-01-01 03:00:00 EST"
A.K.
- Original Mess
You can also do:
mdat$value<-structure(mdat$value,class=c("POSIXt","POSIXct"))
#or
as.POSIXct(mdat$value,origin="1970-01-01")
class(mdat$value)
#[1] "POSIXt" "POSIXct"
A.K.
- Original Message -
From: John Kane
To: arun
Cc: R help
S
Hi,
mat1<- matrix(log(1:20),ncol=5)
exp(mat1)
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 5 9 13 17
#[2,] 2 6 10 14 18
#[3,] 3 7 11 15 19
#[4,] 4 8 12 16 20
mat2<- matrix(log2(1:20),ncol=5)
2^mat2
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 5 9 1
teraction(Tdate,symbol,TA)))%in%
as.character(with(datB,interaction(Tdate,symbol,TA))),]
# Tdate symbol TA
#3 12/12/12 WQ B8R
A.K.
- Original Message -
From: "Mossadegh, Ramine N."
To: arun
Cc:
Sent: Monday, May 13, 2013 4:17 PM
Subject: RE: [R] how to merge 2 data
Hi,
Try:
integrate(Pareto,lower=0,upper=(MM-1)+0.5,x0=x0,alpha2=alpha2)
#170.5065 with absolute error < 0.016
A.K.
For a homework assignment I'm constructing a simple code. R keeps returning
that a x0 is missing in evaluating p2.
I can't figure out why.
lab1 <- 50; mu <- 1; sigma <- 1
alpha1
May be:
matrix(c(test[c(TRUE,diff(test)>1)],test[c(which(diff(test)>1),length(test))]),ncol=2)
# [,1] [,2]
#[1,] 1 5
#[2,] 22 29
#[3,] 33 40
A.K.
- Original Message -
From: Lizzy Wilbanks
To: r-help@r-project.org
Cc:
Sent: Monday, May 13, 2013 9:18 PM
Subject: [R] B
1 5
#2 22 29
#3 33 40
A.K.
- Original Message -
From: arun
To: Lizzy Wilbanks
Cc: R help
Sent: Tuesday, May 14, 2013 12:11 AM
Subject: Re: [R] Boundaries of consecutive integers
May be:
matrix(c(test[c(TRUE,diff(test)>1)],test[c(which(diff(test)>1),length(tes
Hi,
Not sure if this is what you wanted:
mat1<- as.matrix(read.table(text="
33 45 50
NA NA 54
",sep="",header=FALSE))
mat2<- as.matrix(read.table(text="
24 0.000 0.000
0.000 14 0.000
0.000 0.000
Hi,
Try:
set.seed(24)
dat1<- as.data.frame(matrix(sample(1:60,15*5,replace=TRUE),ncol=15))
colnames(dat1)<- paste0("a",c(3,1,5,7,2,8,11,14:15,10,9,6,12:13,4))
subDat1<-dat1[colnames(dat1)[as.numeric(gsub("[A-Za-z]","",colnames(dat1)))<=10]]
subDat1
# a3 a1 a5 a7 a2 a8 a10 a9 a6 a4
#1 18 56 37 5
#this should also work
within(X,a<- ifelse(b,c,a))
# a b c
#1 2 TRUE 2
#2 2 TRUE 2
#3 1 FALSE 2
#4 1 FALSE 2
#5 1 FALSE 2
#6 2 TRUE 2
A.K.
- Original Message -
From: Pascal Oettli
To: fgrelier
Cc: r-help@r-project.org
Sent: Tuesday, May 14, 2013 4:47 AM
Subject: Re: [R] Dataf
A.K.
#Add line between the points
lines(mydata1$Year,mydata1$MW01)
From: David Doyle
To: arun
Sent: Tuesday, May 14, 2013 11:10 PM
Subject: Re: [R] Broken line questions
Hi,
I'm trying to do something very similar to the graph before and I got the
dict(y.loess, data.frame(x=mydata2$Year))
lines(mydata2$Year,y.predict, lty=2, lwd=2)
legend ("topleft",
c("Smoothing Curve","Detect","Non-Detect"),
col = c(1,1,1),
cex = 1,
text.col = "black",
lty = c(2,-1,-1),
lwd = c(2,-1,-1),
pch = c(-1,19,
Hi,
#dput()
dat1<-
structure(list(a1 = c(432L, 0L, 1295L, 0L, 0L, 0L, 0L, 0L, 0L,
0L, 432L, 0L, 432L, 0L, 0L, 0L, 0L, 432L, 0L, 432L, 0L, 0L, 0L,
0L, 0L, 864L, 0L, 432L, 432L, 1296L, 0L, 432L, 0L, 0L, 432L,
0L, 863L, 432L, 0L, 432L, 432L, 0L, 432L, 0L, 0L, 432L, 864L,
0L, 432L, 0L, 432L, 0L,
Hi Shanthi,
May be this link helps you
(http://stats.stackexchange.com/questions/10328/using-rs-phyper-to-get-the-probability-of-list-overlap).
A.K.
From: Shanthi Mulinti
To: arun
Sent: Wednesday, May 15, 2013 10:25 AM
Subject: Re: Displaying median value
Hi,
May be this helps:
c1<-combn(seq_len(ncol(m)),2)
mat1<- matrix(0,ncol=3,nrow=3,dimnames=list(colnames(m),colnames(m)))
vec1<-unlist(lapply(seq_len(ncol(c1)),function(i) {m1<-m[,c1[,i]];
length(which(m1[,1]!=0 & m1[,2]!=0)) }))
mat1[lower.tri(mat1)]<-vec1
mat1
# rs1 rs2 rs3
#rs1 0 0
HI,
You could use:
(# with 3 files in my folder)
lfiles<-list.files(pattern=".txt")
lfiles
#[1] "file1.txt" "file2.txt" "file3.txt"
lst1<-lapply(lfiles,function(x)
read.table(x,header=TRUE,sep="",stringsAsFactors=FALSE))
lst1
#[[1]]
# col1 col2
#1 1 0.5
#2 2 0.2
#3 3 0.3
#4 4 0.3
- Original Message -
From: "Zilefac, Elvis"
To: arun
Cc:
Sent: Thursday, May 16, 2013 2:25 AM
Subject: RE: Loop through a simulation
Dear AK,
Here is data from 21 simulations or so.
Thanks so much.
____
From: arun [smartpink...@yahoo.com
May be this helps:
mat1<- as.matrix(read.table(text="
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
5 5 5 5 5 5
2 2 2 2 2 2
3 3 3 3 3 3
6 6 6 6 6 6
2 2 2 2 2 2
3 3 3 3 3 3
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
",sep="",header=FALSE))
colnames(mat1)<- NULL
t(sapply(1:3,function(i) colMeans(mat1
Hi,
Try this:
unemp.wy <-
read.table("ftp://ftp.bls.gov/pub/time.series/la/la.data.59.Wyoming";,
header=TRUE, sep="\t",stringsAsFactors=FALSE,na.strings="")
dim(unemp.wy)
#[1] 46692 5
head(unemp.wy)
# series_id year period value footnote_codes
#1 LASST5603 1976 M01 4.2
V3,res$V4),]
row.names(res2)<- 1:nrow(res2)
res2$V5<- round(res2$V5,2)
identical(res1,res2)
#[1] TRUE
A.K.
- Original Message -
From: "Zilefac, Elvis"
To: arun
Cc:
Sent: Thursday, May 16, 2013 12:05 PM
Subject: RE: Loop through a simulation
Hi AK,
The output looks a bi
2011-07-17 0 NA 2.8 0.5 0.6
A.K.
From: GUANGUAN LUO
To: arun
Sent: Thursday, May 16, 2013 12:05 PM
Subject: Re: how to calculate the mean in a period of time?
Hello, AK,
now i have a problem really complicated for me,
Now my table is like this:
Hi,
The example dataset only shows 1 level for ID.
chm
id site plot rx rxg rxl t w l spp inid inih d09 d10 d11 d12 h09 h10
1 1 C-H 2002 1 1 Mn N N 14.55 ac 9.6 74.5 9.6 13.0 13.5 14.2 96 109
2 1 C-H 2002 1 1 Mn N N 14.55 ac 7.4 69.5 6.0 9.8 NA 10.7 72 77
3 1 C-H 200
HI,
methods(xyplot)
#[1] xyplot.formula* xyplot.ts*
#
# Non-visible functions are asterisked
lattice:::xyplot.formula
function (x, data = NULL, allow.multiple = is.null(groups) ||
outer, outer = !is.null(groups), auto.key = FALSE, aspect = "fill",
panel = lattice.getOption("panel.x
Hi,
Try either:
tolerance <-
read.csv("http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt";)
aggregate(exposure~male,data=tolerance,mean)
# male exposure
#1 0 1.246667
#2 1 1.12
#or
library(plyr)
ddply(tolerance,.(male),summarize,exposure=mean(exposure))
# male expos
Hi,
rep(seq_along(v),v)
#[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6
A.K.
- Original Message -
From: Stefan Petersson
To: r-help@r-project.org
Cc:
Sent: Friday, May 17, 2013 6:53 AM
Subject: [R] Repeating sequence elements
I want to create a sequence, repeating each element according
12 2 2 2011-09-22 3 65 2.6 0.8 0.5
#13 2 3 2011-10-26 4 34 2.7 0.8 0.5
#14 3 1 2011-07-17 0 NA 2.8 0.5 0.6
A.K.
From: GUANGUAN LUO
To: arun
Sent: Friday, May 17, 2013
Hi,
May be this helps:
dat<- read.table(text="
Date,Parameter,Weight
2012-01-31,90,200
2012-01-31,80,400
2012-01-31,70,500
2012-01-31,60,800
2012-02-29,120,220
2012-02-29,110,410
2012-02-29,75,520
2012-02-29,65,840
2012-03-31,115,210
2012-03-31,100,405
2012-03-31,70,500
2012-03-31,60,800
",sep=",",
Hi,
library(plyr)
res<-dcast(dataset,Date~ScowNo,sum,value.var="EstimatedQuantity")
rownames(res)<- res[,1]
res[,-1]
# 3002 4001 4002 BR 8
#9/7/2010 2772 3535 6763 6685
#9/8/2010 0 3305 0 0
A.K.
- Original Message -
From: Tim
To: r-help@r-project.org
Cc:
Sent: Friday,
Hi,
dat1<- read.table(text="
Var Time
1 51
2 151
3 251
4 234
5 331
6 351
",sep="",header=TRUE)
dat1[!is.na(match(gsub(".*(\\d{2})$","\\1",dat1$Time),51)),]
# Var Time
#1 1 51
#2 2 151
#3 3 251
#6 6 351
#or
dat1[substr(dat1$T
HI,
May be this helps:
set.seed(24)
dat1<-
data.frame(age=sample(30:70,120,replace=TRUE),income=sample(4:8,120,replace=FALSE),country=rep(c("USA","GB","France"),each=40),stringsAsFactors=FALSE)
library(plyr)
ldply(dlply(dat1,.(country),lm,formula=income~0+age),function(x) coef(x))
# cou
Hi,
xl<- vector("list",7)
for(i in 5:7){##loop over numeric vector
xl[[i]] <- rnorm(i)
}
xl
#[[1]]
#NULL
#
#[[2]]
#NULL
#
#[[3]]
#NULL
#
#[[4]]
#NULL
#
#[[5]]
#[1] 0.3266762 0.4316069 1.2290551 -0.6725783 1.6159861
#
#[[6]]
#[1] -2.8560618 -0.5694743 -0.7325862 1.6786160 0.3883842 -0.3
Hi,
May be this helps:
lev1<- unique(as.vector(z))
lapply(lapply(as.data.frame(z),factor,levels=lev1),table,lev1)
#$V1
#
# A B C
#A 1 1 0
#B 0 0 1
#C 0 0 0
#
#$V2
#
# A B C
#A 0 0 0
#B 1 0 0
#C 0 1 1
#
#$V3
#
# A B C
#A 1 0 0
#B 0 1 0
#C 0 0 1
#or
library(plyr
May be this helps:
set.seed(28)
mat1<- matrix(sample(0:50,20,replace=TRUE),nrow=5)
A<- dist(mat1)
A[2:4]<- 0
set.seed(35)
mat2<- matrix(sample(1:50,20,replace=TRUE),nrow=5)
B<- dist(mat2)
B[which(A==0)]<-0
B
# 1 2 3 4
#2 49.07138
#3 0.00
Hi,
Try either:
dat2<- structure(list(Operator = structure(c(1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L,
4L, 5L, 1L, 2L, 3L, 4L, 5L, 1L, 2L, 3L, 4L,
Hi,
Try:
H<-function(x) H1(x)+H2(x)
H1(2)
[#1] 4
H2(2)
#[1] 7
H(2)
#[1] 11
A.K.
- Original Message -
From: Honest Chipoyera
To: "r-help@r-project.org"
Cc:
Sent: Monday, May 20, 2013 8:09 AM
Subject: [R] DEfining and plotting the sum of two functions
I have two functions H1(x) and
Hi,
Check this link
https://stat.ethz.ch/pipermail/r-help/2008-May/163274.html
A.K.
>i have a mer object named "model" :
>
>I want to extract t statistics of the coeeficients from "model".
>Please help me out
>
>package used lme4
__
R-help@r-p
Hi,
You may need to add "dec=","" in the read.csv.
dat1<- read.table(text="
Time;Mid
31/01/2013 00:00;1,35679
31/01/2013 00:01;1,35678
31/01/2013 00:02;1,356785
31/01/2013 00:03;1,35689
31/01/2013 00:04;1,3569
31/01/2013 00:05;1,3569
31/01/2013 00:06;1,356885
31/01/2013 00:07;1,35691
31/01/2013 00:
Hi,
May be this helps.
lines<- readLines(textConnection("NM_019397 // Egfl6 // EGF-like-domain,
multiple 6 // X F5|X 71.5 cM // 54156
NM_019397 // Egfl7 // EGF-like-domain, multiple 6 // X F5|X 71.5 cM // 54158"))
library(stringr)
word(lines,2,sep=" // ")
#[1] "Egfl6" "Egfl7"
lines1<- readLines(
You could also use:
sapply(seq_len(ncol(mdat)),function(i) mdat[,i]*100/x[i])
# [,1] [,2] [,3] [,4] [,5]
#[1,] 7.142857 13.3 30 36.36364 27.8
#[2,] 78.571429 80.0 130 90.90909 55.6
#[3,] 14.285714 20.0 40 45.45455 33.3
#[4,] 71.428571 60.0
Hi,
You could try:
set.seed(24)
mat1<- matrix(sample(0:5,10*20,replace=TRUE),ncol=20)
mat2<- mat1
mat2[2,1:19]<-0
sapply(split(mat2,row(mat2)),function(x) sum(x[x!=0][1:3],na.rm=TRUE))
# 1 2 3 4 5 6 7 8 9 10
# 5 2 12 8 5 14 6 10 10 8
mat3<- mat2
mat3[2,]<- 0
sapply(split(mat3,row(ma
Hi,
Try:
set.seed(28)
dat1<-
as.data.frame(matrix(c(rep(c(5,10,15,20,25),each=3),sample(1:20,15,replace=TRUE),sample(15:35,15,replace=TRUE),sample(20:40,15,replace=TRUE)),ncol=4))
names(dat1)[1]<- "concentration"
lapply(seq_len(ncol(dat1[,-1]))+1,function(i) {x1<-
cbind(dat1[,1],dat1[,i]);colnam
Hi,
library(stringr)
b[str_detect(colnames(b),"^y")]
# y y.1 y.2
#1 0.0 0.00 0.00
#2 19.55811 17.023812 15.354880
#3 10.74991 9.024250 8.177128
#4 5.91924 4.789331 4.367188
#or
b[,!is.na(match(gsub("\\..*","",names(b)),"y"))]
# y y.1 y.2
May be this helps:
dat<- read.table(text="
ID Var
1 1
2 4
3 6
4 7
5 7
6 8
7 9
",sep="",header=TRUE)
dat$Date<-cumsum(seq_len(nrow(dat))%%2)
dat
# ID Var Date
#1 1 1 1
#2 2 4 1
#3 3 6 2
#4 4 7 2
#5 5 7
Hi,
May be this helps:
dat1<-read.table(text="
Date Var day
1/1/2013 1 Tue
1/2/2013 2 Wed
1/3/2013 3 Thu
1/4/2013 4 Fri
1/5/2013 5 Sat
1/6/2013 6 Sun
1/7/2013 7 Mon
1/8/2013 8 Tue
1/9/2013 9 Wed
1/10/2013 10 Thu
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat1$days<-as.num
Hi,
Try this:
lst1<-lapply(1:5,function(i) {pdf(paste0(i,".pdf"));
hist(rnorm(100),main=paste0("Histogram_",i));dev.off()}) #you can change the
numbers
A.K.
>I'm trying to generate a pdf called 1.pdf, 2.pdf, 3.pdf etc and it isn't
>working. My code is:
>x <- 0
>for(i in 1:1000){
>x <- x +
HI,
I am not sure about what you expect as output.
dat1<- read.table(text="
Offense Play
Y A
N B
Y A
Y C
N B
N C
",sep="",header=TRUE,stringsAsFactors=FALSE)
with(dat1,tapply(Play,list(Offense),table))
#$N
#
#B C
#2 1
#
#$Y
#
#A C
#2 1
#or
with(dat1
#3 2 2.05 2
#4 3 1.20 1
#5 4 1.55 1
A.K.
From: GUANGUAN LUO
To: arun
Sent: Wednesday, May 22, 2013 5:42 AM
Subject: calcul of the mean in a period of time
Hello, AK, This is the code zhich you have
Hi,
You didn't provide a reproducible example
(http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example)
So, it is difficult to comment.
Using the dataset(sweetpotato)
library(agricolae)
data(sweetpotato)
m1<-aov(yield~virus, data=sweetpotato)
df1<-df.residual(m1)
!as.logical(with(dat1,ave(idx,patient_id,FUN=function(x)
cumsum(is.na(x),]
#count
sum(with(datSub,idx[idx>1 & !is.na(idx)])-1)
#[1] 5
A.K.
From: GUANGUAN LUO
To: arun
Sent: Wednesday, May 22, 2013 11:30 AM
Subject: Re: how to calculate the mean in a
Hi,
May be this helps:
month.dayslong<-rep(31,7)
names(month.dayslong)<-
c("January","March","May","July","August","October","December")
month.dayslong
# January March May July August October December
# 31 31 31 31 31 31 31
unname(month.
Hi,
Couldn't reproduce your error.
It is better to dput() the example data:
v <- read.table("/home/tiago/matrix.txt", header=FALSE)
dput(v)
v<- read.table("matrix1.txt",header=FALSE,sep="")
v<-as.matrix(v)
v
# V1 V2 V3 V4 V5 V6
#[1,]
Hi Joe,
Not sure about your expected result
blah<- paste0("x",1:5)
which(blah=="xyz")
#integer(0)
blah=="xyz"
#[1] FALSE FALSE FALSE FALSE FALSE
any(blah=="xyz")
#[1] FALSE
sum(blah=="xyz")
#[1] 0
sum(blah=="x1")
#[1] 1
A.K.
- Original Message -
From: Joseph Trubisz
To: r-help@r-p
"
if(any(blah=="xyz")) blah else as.numeric(factor(blah))
#[1] 1 2 3 4 5
A.K.
- Original Message -
From: Joseph Trubisz
To: arun
Cc:
Sent: Wednesday, May 22, 2013 2:37 PM
Subject: Re: [R] How to test if something doesn't exist
OK...got it...thanks
Joe
On M
Hi,
May be this helps:
test1<- test[1:5,]
test0<- test1
for(i in 1:5) test0[,i]=0
test0
# Y X1 X2 X3 X4
#1 0 0 0 0 0
#2 0 0 0 0 0
#3 0 0 0 0 0
#4 0 0 0 0 0
#5 0 0 0 0 0
A.K.
- Original Message -
From: Roni Kobrosly
To: r-help@r-project.org
Cc:
Sent: Tuesday, May 2
HI,
Try:
lab2<- function(var1,var2,data1){
a<- anova(lm(var2~var1,data=data1))
cat("df(between) is",a[1,1],"\n")
cat("df(within) is", a[2,1],"\n")
cat("ss(between) is", a[1,2],"\n")
cat("ss(within) is", a[2,2],"\n")
}
#or
lab3<- function(var1,var2,data1){
a<- anova(lm(var2~var1,data=data
Hi,
Try:
dat<- read.table(text="
Var
0
0.2
0.5
1
4
6
",sep="",header=TRUE)
res1<-within(dat,group<-factor(findInterval(Var,c(-Inf,0.1,0.6),rightmost.closed=TRUE),labels=LETTERS[1:3]))
res1
# Var group
#1 0.0 A
#2 0.2 B
#3 0.5 B
#4 1.0 C
#5 4.0 C
#6 6.0 C
#or
res2<-with
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