1-n_acPro/ 2 2 2 2 2 2 2
#8 aATASIR 1-n_acPro/ 2 0 0 0 0 1 0
#9 aEQQQFYLLLGNLLSPDNVVR 1-<_Carbamoylation/ 2 0 0 0 18 0 0
#10 aEQQQFYLLLGNLLSPDNVVR 1-n_acPro/ 2 18 0 0 0 0 0
#11 aEQQQFYLLLGNLLSPDNVVR 1-n_acPro/ 3 0 18
Hi,
May be this helps.
date1<-seq.Date(as.Date("01/01/2000",format="%d/%m/%Y"),as.Date("31/12/2010",format="%d/%m/%Y"),by="day")
set.seed(24)
value<- rnorm(4018,25)
dat1<- data.frame(date1,value)
dat2<-
do.call(rbind,split(dat1,as.numeric(gsub(".*\\-(.*)\\-.*","\\1",dat1$date1
library(zoo
ntical(datNew,datNew2)
#[1] TRUE
A.K.
From: Zilefac Elvis
To: arun
Cc: R help
Sent: Thursday, May 23, 2013 12:32 AM
Subject: Re: [R] Sort data by month
Hi Ak,
I seems to work correctly with your data. Please try it on my data (attached).
My data is month
Regards,
Arun
From: "zilefacel...@yahoo.com"
To: arun
Cc: R help
Sent: Thursday, May 23, 2013 1:42 AM
Subject: Re: Re: [R] Sort data by month
Hi AK,
Thanks so much. Problem well handled.
Atem.
-- Original Message -
with(dftest,aggregate(cbind(x,y),list(z),FUN=mean))
# Group.1 x y
#1 0 7 1
#2 1 6 2
#or
library(plyr)
ddply(dftest,.(z),numcolwise(mean))
# z x y
#1 0 7 1
#2 1 6 2
A.K.
- Original Message -
From: jpm miao
To: r-help
Cc:
Sent: Thursday, May 23, 2013 3:05 AM
Subject: [R
y
#1 0 7 1
#2 1 6 2
A.K.
- Original Message -
From: Blaser Nello
To: arun ; jpm miao
Cc: R help
Sent: Thursday, May 23, 2013 3:29 AM
Subject: RE: [R] convert a character string to a name
If you want to use the character string:
attach(dftest)
aggregate(cbind(sapply(x_test, get))~z,
You could also do:
library(plyr)
res1<-join(dat1,dat2,type="full")
res1
# Serialno name year outcome disch_type
#1 1 ken 1989 d
#2 2 mary 1989 a
#3 4 john 1989 a
#4 5 tom 1989 a
#5 6 jol
1
#211 41 4.025000 2
#212 41 2.975000 3
#213 41 1.725000 4
#214 42 6.95 1 #t=0 only row
#215 44 4.300000 1
A.K.
From: GUANGUAN LUO
To: arun
Sent: Thursday, May 23, 2013 9:50 AM
Subject: Re:
Hi,
Try:
datNew<- read.table(text="
activity max_dt
A 2013-03-05
B 2013-03-28
A 2013-03-28
C 2013-03-28
B 2013-03-01
",sep="",header=TRUE,stringsAsFactors=FALSE)
datNew$max_dt<- as.Date(datNew$max_dt)
aggregate(max_dt~activity,data=datNew,m
ge -
From: "Mossadegh, Ramine N."
To: arun
Cc:
Sent: Thursday, May 23, 2013 10:44 AM
Subject: RE: [R] Removing rows w/ smaller value from data frame
Thank but I get : Error in is.list(by) : 'by' is missing
When I tried ddply(datNew,.(activity),summarize, max_dt=max(max_dt))
Hi,
May be this helps:
dat1<- read.table(text="
group var1 var2 myvar
group1 1 a 100
group2 2 b 200
group2 34 c 300
group3 5 d 400
group3 6 e 500
group4 7 f 600
",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
ddply(dat1,.(group),summarize, f_myvar=mifunc(myvar))
# group f_myvar
#1 g
group4 7 f 600 NA
A.K.
From: Estefanía Gómez Galimberti
To: arun
Cc: R help
Sent: Thursday, May 23, 2013 12:08 PM
Subject: Re: [R] apply function within different groups
Thanks a lot!!! It works perkectly!
Just one thing, is there a way to prese
Using the previous solution:
dat3<-mutate(dat1,f_myvar=ddply(dat1,.(group),summarize,f_myvar=mifunc(myvar))[,2])
identical(dat2,dat3)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: Estefanía Gómez Galimberti
Cc: R help
Sent: Thursday, May 23, 2013 1:01 PM
Subject: Re: [R] ap
Hi,
ab<- cbind(a,b)
indx<-duplicated(names(ab))|duplicated(names(ab),fromLast=TRUE)
res1<-cbind(ab[!indx],v2=rowSums(ab[indx]))
res1[,order(as.numeric(gsub("[A-Za-z]","",names(res1,]
#v1 v2 v3
#1 3 4 5
#Another example:
a2<- data.frame(v1=c(3,6,7),v2=c(2,4,8))
b2<- data.frame(v2=c(2,6,7
Hi,
Try:
dat1<- structure(list(V2 = c("ALKBH1", "ALKBH2", "ALKBH3", "ANKRD17",
"APEX1", "APEX2", "APTX", "ASF1A", "ASTE1", "ATM", "ATR", "ATRIP",
"ATRX", "ATXN3", "BCCIP", "BLM", "BRCA1", "BRCA2")), .Names = "V2", class =
"data.frame", row.names = c(NA,
18L))
dat2<- structure(list(V2 = c("AL
#or
dat1$V2[is.na(match(dat1$V2,dat2$V2))]
#[1] "ALKBH1" "ALKBH2" "ANKRD17" "ASF1A" "ASTE1" "ATRX" "ATXN3"
#[8] "BCCIP"
a[is.na(match(a,b))]
#[1] 2 2 4
A.K.
- Original Message -
From: Will
You could convert those columns to "Date" class by:
Data[,c(4,6)]<-lapply(Data[,c(4,6)],as.Date,origin="1970-01-01")
#or
Data[,c(4,6)]<-lapply(Data[,c(4,6)],function(x) structure(x,class="Date"))
# dat1 dat2 Dat1a Dat1b Dat2a Dat2b
#1 41327 41327 2013-02-22 2013-02-22 201
.
- Original Message -
From: Denis Chabot
To: arun
Cc: R help
Sent: Thursday, May 23, 2013 10:06 PM
Subject: Re: [R] subsetting and Dates
Thank you for the 2 methods to make the columns class Date, but I would really
like to know why these variables were not in Date class with my code
Hi,
You may also try:
mat: data
lst1<-lapply(split(mat,col(mat)),function(x) {x1<- which(x>=0.8*max(x));x2<-
which.max(x);x1[abs(x1-x2)==1|length(x)-abs(x1-x2)==1]<-NA;rbind(index=x1[!is.na(x1)],values=x[x1[!is.na(x1)]])})
names(lst1)<-NULL
lst2<-apply(mat,2,function(x) do.call(rbind,fun2(x))) #
Lines1<- readLines(textConnection("[1] 1.000 0.000 0.3425584 0.000
[1] 1.000 0.3425584 1.6693396 0.000
[1] 1.00 1.669340 9.918513 0.00
[1] 1.00 9.918513 24.00 0.00
-- more lines-
-
Hi,
Try this:
mat1: 1st matrix
mat2: 2nd matrix
fun1<- function(x){
big<- x>0.8*max(x)
n<- length(big)
startRunOfBigs<- which(c(big[1],!big[-n] & big[-1]))
endRunOfBigs<- which(c(big[-n] & !big[-1], big[n]))
index<- vapply(seq_along(startRunOfBigs),function(i)
which.max(x[st
#1_2 2.5966051
#1_3 1.0267435
#1_4 3.7387830
#1_5 1.8489204
#1_6 6.5233654
#2_3 4.2951411
#2_4 1.9040790
#2_5 2.2874235
#2_6 5.1526016
#3_4 0.9726777
#3_5 2.1359229
#3_6 5.0221450
#4_5 0.9124638
#4_6 8.2604187
#5_6 14.0550864
A.K.
____
From: eliza bot
__
From: eliza botto
To: "smartpink...@yahoo.com"
Sent: Saturday, May 25, 2013 9:17 PM
Subject: RE: QA
thanks arun,
i dont think thANKyou is enough for wat u did. anyway, there is slight
modification that i want to ask to understand the codes more efficient
the code returns Error:
do.call("[", c(list(z),1,1,ix))
#Error in 1:24[1, 1, 1:3, 1:4] : incorrect number of dimensions
May be something is missing.
A.K.
- Original Message -
From: Bert Gunter
To: Martin Ivanov
Cc: r-help@r-project.org
Sent: Sunday, May 26, 2013 10:43 AM
Subject:
Hi,
You could use:
library(abind)
#using Berend's and Bert's example
x1 <- array(runif(9),dim=c(3,3))
x2 <- array(runif(8),dim=c(2,2,2))
z <- array(1:24,dim=2:4)
#applying your code:
eval(parse(text=paste0("x1[1",paste(rep(",",length(dim(x1))-1),collapse=""),"]")))
#[1] 0.6439062 0.7139397 0.6017
]
#[1,] 0.026671344 0.2116831
#[2,] 0.003903368 0.1551140
array(x1[slice.index(x1,1)==1],dim= dim(x1)[1])
#[1] 0.6439062 0.7139397 0.6017418
A.K.
- Original Message -
From: arun
To: Martin Ivanov
Cc: R help ; Berend Hasselman ; Bert
Gunter
Sent: Sunday, May 26, 2013 11:48 AM
colnames(dd)
#[1] "col1" "colb"
null_vector<- colnames(dd)
sapply(null_vector,makeNull,dd)
# col1 colb
#[1,] NA 4
#[2,] 2 NA
#[3,] 3 2
#[4,] 4 NA
#[5,] 1 4
#[6,] NA 5
#[7,] 1 6
A.K.
>I am trying to make a column value in a dataframe = NA if there is a 0
May be this helps
levels(x$Species)
#[1] "setosa" "versicolor" "virginica"
x$Species<- factor(x$Species,levels=unique(x$Species))
xa <- ddply(x, .(Species), function(x)
{data.frame(Sepal.Length=x$Sepal.Length, mean.adj=(x$Sepal.Length -
mean(x$Sepal.Length)))})
head(xa)
# Species Sepal
rginica 1.012
A.K.
- Original Message -
From: arun
To: Liviu Andronic
Cc: R help
Sent: Monday, May 27, 2013 10:06 AM
Subject: Re: [R] configure ddply() to avoid reordering of '.variables'
May be this helps
levels(x$Species)
#[1] "setosa" "versicolor" &q
I have a doubt about your New table especially the 3rd row:
Since after "test1" , the test fails, i guess 4,5 should be NA
dat1<-read.table(text="
Device,first_failing_test,test1,test2,test3,test4,test5
1,test2,1,2,3,4,5
2,test4,2,3,4,5,6
3,test1,3,4,5,6,7
",sep=",",header=TRUE,stringsAsFactors=FA
Hi,
Not sure if this is what you expected:
set.seed(24)
mat1<- matrix(sample(1:20,3*4,replace=TRUE),ncol=3)
set.seed(28)
mat2<- matrix(sample(1:25,3*6,replace=TRUE),ncol=3)
set.seed(30)
mat3<- matrix(sample(1:35,3*8,replace=TRUE),ncol=3)
set.seed(35)
mat4<- matrix(sample(1:40,3*10,replace=TRUE),nc
Hi,
Try this:
dat2<-dat[order(as.numeric(gsub("preV(\\d+).*","\\1",colnames(dat]
dat2
# preV15A1b preV59A1b preV1001A1b preV2032A1b preV2035A1b
#1 0.57 0.05 0.59 0.40 0.95
#2 0.62 0.57 0.30 0.80 0.67
#3 0.51 0.03 0
1
#9 53 5 9 2011-02-13 8 0.000 0
#10 54 5 10 2011-03-19 9 -1.200 1
A.K.
___
From: GUANGUAN LUO
To: arun
Sent: Monday, May 27, 2013 8:48 AM
Subject: choose the lines
Hello, Arun,
in this data, i want to c
res1<- xtabs(X3~X1+X2,data=Dat)
res1
# X2
#X1 1 2 3 4
# A 11 12 13 14
# B 15 16 17 18
# C 19 20 21 0
library(reshape2)
dcast(Dat,X1~X2,value.var="X3")
# X1 1 2 3 4
#1 A 11 12 13 14
#2 B 15 16 17 18
#3 C 19 20 21 NA
A.K.
Hello again, let say I have following data-frame:
> Dat
Hi,
Try either:
set.seed(28)
stats1<- as.data.frame(matrix(rnorm(5*1),ncol=5))
pdf(paste("test",1,".pdf",sep=""))
par(mfrow=c(2,1))
lst1<- lapply(names(stats1),function(i)
{hist(stats1[,i],100,col="lightblue",main=paste0("Histogram of ",i),xlab=i
);qqnorm(stats1[,i])})
dev.off()
#or
pdf(p
0 7 1 0.825
#or
library(plyr)
res2<-ddply(dat2,.(patient_id,evnmt_brutal),summarize,basdai_d=mean(basdai_d))
identical(res11,res2)
#[1] TRUE
A.K.
From: GUANGUAN LUO
To: arun
Sent: Tuesday, May 28, 2013 4:53 AM
Subject: Re: choose the li
Hi,
You could use:
which(a%in%b)
#[1] 1 2 3 4 5
a1<-c(1,2,5)
b1<-c(1,3,4,5,7)
which(a1%in%b1)
#[1] 1 3
A.K.
Hello! I created the following function for calculating which elements
in vector a are existant in vector b. However, all I get is NA NA NA and I
can´t figure out why. =/
fun <- fu
Hi,
May be this helps:
a1<-c(1:5,1:3,rep(1,2),1:5,1:3)
a1
# [1] 1 2 3 4 5 1 2 3 1 1 1 2 3 4 5 1 2 3
a1[sort(c(which(c(diff(a1)<0,TRUE)), which(a1[-length(a1)] == a1[-1])))]
#[1] 5 3 1 1 5 3
a2<-c(1:2,rep(1,4),1:7,1:3)
a2
# [1] 1 2 1 1 1 1 1 2 3 4 5 6 7 1 2 3
a2[sort(c(which(c(diff(a2)<0,TRUE)),
HI,
Possibly R FAQ: 7.31
data1New<-data1
data1New$A<- round(data1New$A,2)
data2New<- data2
data2New$A<- round(data2New$A,2)
merge(data1New,data2New,by="A")
# A B C
#1 0.0 0.9 10.0
#2 1.1 0.6 11.1
#3 1.4 0.7 11.4
#4 3.1 0.4 13.1
#5 4.4 0.8 14.4
A.K.
Hello,
Lets say we have these
elp@r-project.org after
registering at:
https://stat.ethz.ch/mailman/listinfo/r-help
From: Angela Fel Padecio
To: arun
Sent: Wednesday, May 29, 2013 9:05 AM
Subject: bootstrap
hi. i have these dataset:
set.seed(12345)
S=1000
generate <- function(siz
om: Angela Fel Padecio
To: arun
Sent: Wednesday, May 29, 2013 10:33 AM
Subject: Re: bootstrap
hi. thanks for the reply. generally, the command for bootstrapping in R is
sample(x, size, replace=T)
however, i want my x to be the vectors x1 to x10, t and y. i can't do it
because i this
. I must be doing something wrong.
- Original Message -
From: arun
To: R help
Cc:
Sent: Wednesday, May 29, 2013 2:33 AM
Subject: Re: Problems with merge
HI,
Possibly R FAQ: 7.31
data1New<-data1
data1New$A<- round(data1New$A,2)
data2New<- data2
data2New$A<- round(data2
c,100
#x5 Numeric,100 Numeric,100 Numeric,100 Numeric,100 Numeric,100
A.K.
From: Angela Fel Padecio
To: arun
Sent: Wednesday, May 29, 2013 11:05 AM
Subject: Re: bootstrap
thanks for the reply. i think its nearer on the desired result. the previous
synta
May be this helps:
mymatrix<- matrix (1:5, nrow=5, ncol=columnas, byrow=TRUE,
dimnames=(list(myrownames, c(1,2,3,4,5
#Error in matrix(1:5, nrow = 5, ncol = columnas, byrow = TRUE, dimnames =
(list(myrownames, :
# object 'columnas' not found
myrownames[which.max(nchar(myrownames))]
#[1]
Hi,
May be I misunderstood your question:
dat<- read.table(text="
Date Time Var
1 1 2
1 1 4
1 1 5
1 2 8
1 2 8
1 2 9
2 1 3
2 1 4
2 1 4
"
paste(Date,Time,sep="_"))[,c(4,3)]
res
# DT Var
#1 2012-11-01_1 5
#2 2012-11-01_2 5
#3 2012-11-02_1 5
#4 2012-11-02_2 3
A.K.
____
From: Ye Lin
To: arun
Cc: R help
Sent: Wednesday, May 29, 2013 2:40 PM
Subject: Re: [R] combine two columns in
Assuming that you wanted to label '1' for table1 and '4' for table2 (info
column).
Also, not sure why chr2 row is not in the resulted table.
dat1<- read.table(text="
chr pos ref alt
chr1 5 A G
chr1 8 T C
chr2 2 C T
",sep="",header=TRUE,stringsAsFactors=FAL
Hi Tamara,
No problem.
dat3<- rbind(dat1,dat2) #Sorry, forgot this line.
A.K.
From: Tamara Simakova
To: arun
Sent: Thursday, May 30, 2013 12:26 PM
Subject: Re: [R] [BioC] comparing two tables
Hello Arun,
Thanks very much for help. Indeed there i
sendet: Mittwoch, 29. Mai 2013 um 07:04 Uhr
Von: arun
An: "R help"
Betreff: *** GMX Spamverdacht *** Re: [R] Obtaing the maximum
Hi,
May be this helps:
a1<-c(1:5,1:3,rep(1,2),1:5,1:3)
a1
# [1] 1 2 3 4 5 1 2 3 1 1 1 2 3 4 5 1 2 3
a1[sort(c(which(c(diff(a1)<0,TRUE)), which(a1[-length
#x27;s function
# user system elapsed
# 0.856 0.140 0.998
A.K.
- Original Message -
From: William Dunlap
To: arun
Cc:
Sent: Thursday, May 30, 2013 2:28 PM
Subject: RE: [R] Fwd: Obtaing the maximum
Also, your suggestion could avoid the relatively expensive call to sort() if
Hi,
May be this helps:
speciesTime1New<- speciesTime1[,-1]
speciesTime2New<- speciesTime2[,-1]
imm1<-as.data.frame(matrix(0,4,3))
imm1[]<-lapply(seq_len(ncol(speciesTime2New)),function(i)
ifelse(speciesTime2New[,i]==1& speciesTime2New[,i]!=speciesTime1New[,i],1,0))
Ext1<-as.data.frame(matrix(0,4,
library(plyr)
colnames(Y)[2]<- colnames(X)
join(X,Y,type="left",by="k1")
# k1 k2
#1 A 1
#2 NA
#3 C 3
#4 B 2
A.K.
- Original Message -
From: nevil amos
To: r-help
Cc:
Sent: Friday, May 31, 2013 4:07 AM
Subject: [R] merge without NA last
I am trying to create a merge
HI,
Try this:
data1 <- data.frame("var1"=c(1,2,3),"var2"=c("A","B","C"),"var3"=c(4,5,6))
#your example
data2 <- data.frame("var1"=c(1,2,3),"var2"=c("5","6","9"),"var3"=c(4,5,6))
#your example
data3<- data.frame(var1=c(5,6,5,5),var2=c(4,4,5,3),var3=c(1,4,3,5))
data4<- data.frame(var1=c(1,2,3),va
Hi,
You could reduce those steps by:
as.data.frame(do.call(rbind,a2))
A.K.
- Original Message -
From: Abdul Rahman bin Kassim (Dr.)
To: John Kane ; "r-help@r-project.org"
Cc:
Sent: Friday, May 31, 2013 11:47 AM
Subject: Re: [R] Returning combine output from R loop
Dear Kane,
Thanks
HI,You could also try:
set.seed(24)
dat1<-
data.frame(Reaction_time=sample(24:60,80,replace=TRUE),Categ=rep(1:8,each=10))
with(dat1,tapply(Reaction_time,list(Categ),FUN=mean))
# 1 2 3 4 5 6 7 8
#43.7 39.8 37.5 42.7 33.9 42.3 43.2 40.0
#or
library(plyr)
ddply(dat1,.(Categ
Hi,
May be this is what you wanted:
dat2: dataset
library(plyr)
res1<-ddply(dat2,.(subject,conditionNo,state),summarize,MLat=mean(latency))
res1
#subject conditionNo state MLat
#1 1 1 1 674.8947
#2 1 1 2 649.5000
#3 1 1 3 662.63
Hi,
It is not clear.
dta1<-do.call(data.frame,dta)
dta2<-dta1[complete.cases(dta1),]
dta2[,-3]<-lapply(dta2[,-3],as.character)
lstdta2<-split(dta2,dta2$place)
library(plyr)
join_all(lapply(lstdta2,`[`,-1),by="name",type="inner") #none of them are common
#[1] name value value value value value
#<0
Hi,
May be this helps:
res1<-df1[with(df1,unlist(tapply(var,list(subid),FUN=function(x)
c(FALSE,diff(x)!=0)),use.names=FALSE)),]
res1
# subid year var
#3 36 2003 3
#7 47 2001 3
#9 47 2005 1
#10 47 2007 3
#or
library(plyr)
subset(ddply(df1,.(subid),mutate,delta=c(FALSE,di
al Message -
From: David Winsemius
To: arun
Cc: R help
Sent: Tuesday, June 4, 2013 12:37 AM
Subject: Re: [R] Read 2 rows in 1 dataframe for diff - longitudinal data
On Jun 3, 2013, at 7:10 PM, arun wrote:
> Hi,
> May be this helps:
> res1<-df1[with(df1,unlist(tapply(var,list
_brutal
#52 52 5 8 2011-01-11 7 -2.8 0
#53 53 5 9 2011-02-13 8 0.0 0
#54 54 5 10 2011-03-19 9 -1.2 1
#
#[[2]]
# [,1]
#2 0.9
#5 0.0
A.K.
From: GUANGUAN LUO
To: ar
ns.
-ST
- Original Message -----
From: David Winsemius
To: arun
Cc: R help
Sent: Tuesday, June 4, 2013 11:13 AM
Subject: Re: [R] Read 2 rows in 1 dataframe for diff - longitudinal data
On Jun 3, 2013, at 9:51 PM, arun wrote:
> If it is grouped by "subid" (that wou
6 10 2011-03-09 9 0.15 0
#76 76 6 11 2011-04-11 10 -0.45 1
#or
res11<-do.call(rbind,lapply(res1,function(x) tail(x,-1)))
row.names(res11)<-1:nrow(res11)
A.K.
From: GUANGUAN LUO
To: arun
Sent: Tuesday, J
Hi,
Try this:
set.seed(24)
b1<- sample(1:40,20,replace=TRUE)
set.seed(28)
a1<- sample(30:50,20,replace=TRUE)
t_test_apparie <- function(x,y)
{
t.test(x,y,paired=TRUE,alternative = "greater")$p.value
}
t_test_apparie(a1,b1)
#[1] 6.571404e-08
A.K.
>Hi,
>
>I have written this function :
>
>t
Since you mentioned this > half-hour running time, it would be good to check
your data.
?str()
A.K.
- Original Message -
From: arun
To: R help
Cc: David Winsemius
Sent: Tuesday, June 4, 2013 1:18 PM
Subject: Re: [R] Read 2 rows in 1 dataframe for diff - longitudinal data
Hi,
By co
Hi,
May be this helps:
I duplicated your dataset (only one was attached) and changed the dates.
lstf1<- list.files(pattern=".txt")
lstf1
#[1] "dt3031093-1.txt" "dt3031093-2.txt" "dt3031093-3.txt"
#3rd one has less number of observations.
fun1<- function(lstf){
lst1<-lapply(lstf,function(x
0 0.00 NA
dim(res)
#[1] 16740 3
(2005-1960)*12*31
#[1] 16740
A.K.
- Original Message -
From: arun
To: Zilefac Elvis
Cc: R help
Sent: Tuesday, June 4, 2013 8:14 PM
Subject: Re: dates and time series management
Hi,
May be this helps:
I duplicated your data
0 540 540 540 540 540 540 540 540 540 540 540 540
# [55] 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540
# [73] 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540
# [91] 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540 540
#[109] 540 540
Hello,
May be this helps:
dta1<-do.call(data.frame,dta)
dta2<-dta1[complete.cases(dta1),]
dta2[,-3]<-lapply(dta2[,-3],as.character)
lstdta2<-split(dta2,dta2$place)
library(plyr)
#Some names are common in a few, but not in all the places. If you are looking
for names common in 2 places, 3, places,
24
#1559 1559 26
#8069 8069 26
#--
res[ which(res==-.99,arr.ind=TRUE)]<-NA
#or
res[res==-.99]<-NA
which(res==-.99)
#integer(0)
A.K.
From: Zilefac Elvis
To: arun
Sent: Wednesday, June 5, 2013 10:56 AM
Subject: Re:
Hi,
Try:
fun1<- function(dat,Col1,Col2,number){
lst1<- split(dat,list(dat[,Col1],dat[,Col2]))
lst2<- lst1[lapply(lst1,nrow)>0]
res<- lapply(lst2,function(x) sample(x[,1],if(nrow(x)< number) nrow(x) else
number,replace=FALSE))
res}
head(fun1(Gpool,"LngtClas","SpCode",5),4)
#$`40_49.MERLMER`
#[
num NA NA NA NA NA NA NA NA NA NA ...
$ dt3011887.txt: num 0.17 0.28 0 0.3 0 0 1.78 0 0.3 0 ...
$ dt3012205.txt: num 0.34 0.21 0 0.51 0 0 2.82 0 0.3 0 ...
---
res1$dates<-as.Date(res1$dates)
res2<-res1[!is.na(res1$dates),]
res2
-01 14.61 40.79 127.74 25.07
Looks like this is the same result as above.
A.K.
____
From: Zilefac Elvis
To: arun
Sent: Wednesday, June 5, 2013 11:04 PM
Subject: Re: dates and time series ma
Hi,
Try:
hsb2 <- read.csv("http://www.ats.ucla.edu/stat/data/hsb2.csv";)
varlist<-names(hsb2)[8:10]
fun2<- function(varName){
res<- sapply(varName,function(x){
model1<-
lm(substitute(cbind(female,race,ses)~i,list(i=as.name(x))),data=hsb2)
sM<- summary(model1)
HI,
Not sure I understand your question:
a <- 2
b <- 3
f1<- function(x) a+b*x
f1(2)
#[1] 8
f1(3)
#[1] 11
f<- function(x) 2+3*x
f(2)
#[1] 8
f(3)
#[1] 11
A.K.
sessionInfo()
R version 3.0.0 (2013-04-03)
Platform: x86_64-unknown-linux-gnu (64-bit)
locale:
[1] LC_CTYPE=en_CA.UTF-8 L
Perhaps this also helps:
library(plyr)
do.call(rbind,alply(aperm(laply(list(A,B),as.matrix),c(1,3,2)),3)) #Using
Berend's example
1 2 3 4 5
# [1,] -0.591 -0.934 -0.828 0.012 -0.683
#[2,] 1.000 6.000 11.000 16.000 21.000
#[3,] 0.027 1.324 -0.348 -0.223 -0.016
HI,
May be this helps:
dat1<- read.table("sampledata.txt",header=TRUE,sep=",",stringsAsFactors=FALSE)
pdf("Barplots.pdf")
lst1<-lapply(seq_len(ncol(dat1)),function(i) {Ctdat<-
table(dat1[,i]);Ctdat1<-(Ctdat/sum(Ctdat))*100;barplot(Ctdat1,ylim=c(0,100),xlab=colnames(dat1)[i],ylab="Relative
Frequ
ylab="Relative Frequency",
rylab="Frequency",main=paste("Bar
plot:",colnames(dat1)[i],sep=" "),
type=c("bar","l"),lcol=2,rcol=4,xtickpos=Var1,xticklab=Var1))
})
HI,
Try:
?split()
source("http://www.openintro.org/stat/data/cdc.R";)
str(cdc)
#'data.frame': 2 obs. of 9 variables:
# $ genhlth : Factor w/ 5 levels "excellent","very good",..: 3 3 3 3 2 2 2 2 3
3 ...
# $ exerany : num 0 0 1 1 0 1 1 0 0 1 ...
# $ hlthplan: num 1 1 1 1 1 1 1 1 1 1 ...
HI,
Please dput() the example data.
"""
Could anyone give a help ASAP ?"
You have been posting for long time and your many posts show this "ASAP". I
know that you got comments to that and also advised to dput() the data.
Formatting your data took some time. Would it be better to follow th
Hi,
You would get better response if you post at Bioconductor mailing list.
http://www.bioconductor.org/help/mailing-list/
A.K.
- Original Message -
From: Payal Urs
To: r-help@r-project.org
Cc:
Sent: Friday, June 7, 2013 6:12 AM
Subject: [R] Bionconductor help
Hi
I am trying to do so
Hi,
?which()
which(A==2)
#[1] 3 7 10
A.K.
Suppose there is a vector
A <- c(1,3,2,6,7,8,2,1,3,2).
Now, I want to get the subscript of elements of A which equal 2.
How can I do it ?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman
Hi,
May be this helps:
lst1<-split(final3,list(final3$year,final3$industry))
lst2<-lst1[lapply(lst1,nrow)>0]
lst3<-lapply(lst2,function(x) lapply(x$dimension,function(y) x[(y<
(x$dimension+x$dimension*0.1)) & (y> (x$dimension-x$dimension*0.1)),]))
lst4<-lapply(lst3,function(x) x[lapply(x,nrow)==2
industry dummy dimension group
#1 1 2000 20 0 2120 2000.20 #1 group
#2 5 2000 20 1 2189 2000.20 #1
#3 4 2000 20 0 3178 2000.20 #2
#4 7 2000 20 1 3245 2000.20 #2
A.K.
- Original Message -
From: Cecilia Carmo
my==0))
#[1] 22
nrow(subset(res,dummy==1))
#[1] 22
A.K.
- Original Message -
From: Cecilia Carmo
To: arun
Cc: R help
Sent: Friday, June 7, 2013 12:03 PM
Subject: RE: [R] matched samples, dataframe, panel data
Sorry,
Something is not ok, because when I do
> nrow(subset(res,res
.K.
- Original Message -
From: Cecilia Carmo
To: R help
Cc: arun
Sent: Friday, June 7, 2013 5:51 PM
Subject: RE: [R] matched samples, dataframe, panel data
Thank you very much.
I apologize but I want to ask only one more thing:
How to do if I want in addition to impose that the diffenc
t;0]})
res2<-do.call(rbind,lapply(lst7,function(x) do.call(rbind,x)))
identical(res1,res2)
#[1] TRUE
A.K.
----- Original Message -
From: Cecilia Carmo
To: arun
Cc: R help
Sent: Friday, June 7, 2013 6:10 PM
Subject: RE: [R] matched samples, dataframe, panel data
Maybe the final result is
HI,
May be this helps:
res<-data.frame(x=x,read.table(text=gsub("[A-Za-z]","",x[,1]),sep="_",header=FALSE),stringsAsFactors=FALSE)
res
# x V1 V2 V3
#1 aaa1_bbb1_ccc3 1 1 3
#2 aaa2_bbb3_ccc2 2 3 2
#3 aaa3_bbb2_ccc1 3 2 1
A.K.
- Original Message -
From: Dimitri Liakho
user system elapsed
# 2.712 0.016 2.732
head(resNew)
# x V1 V2 V3
#1 ccc12_ggg2_jjj14 12 2 14
#2 ccc7_ddd15_aaa11 7 15 11
#3 hhh12_ddd14_fff12 12 14 12
#4 fff11_bbb15_aaa6 11 15 6
#5 ggg12_ccc9_ggg8 12 9 8
#6 jjj8_eee12_eee4 8 12 4
A.K.
- Original Message -
From:
mmy==0))
#[1] 151
nrow(subset(res5percent1,dummy==1))
#[1] 151
Hope it helps.
A.K.
From: Cecilia Carmo
To: arun
Sent: Friday, June 7, 2013 7:30 PM
Subject: data
I'm sending the data.
Thank you very much.
Cecília
The code
final3<-read.table(
RowSel <-c(0,1,0,1,2,3,0,5,5)
set.seed(24)
DF<- as.data.frame(matrix(sample(1:40,45,replace=TRUE),ncol=5))
RowSel[!as.logical(RowSel)]<-NA
DF[RowSel,]
# V1 V2 V3 V4 V5
#NA NA NA NA NA NA
#1 12 11 21 24 30
#NA.1 NA NA NA NA NA
#1.1 12 11 21 24 30
#2 9 25 6 26 26
#3 29 15 4 2 28
#
HI,
You could also use:
set.seed(24)
aaa <- matrix(rnorm(60), ncol=3)
bbb <- matrix(runif(15), ncol=3)
ccc1<- mapply(quantile,as.data.frame(aaa),as.data.frame(bbb))
ccc <- sapply(1:dim(aaa)[2], function(i) quantile(aaa[, i], bbb[, i])) #Jean's
solution
colnames(ccc1)<-NULL
identical(ccc,ccc1)
ccc8_ccc1_fff11 8 1 11
#3: hhh15_ggg2_hhh13 15 2 13
#4: fff9_bbb3_ccc9 9 3 9
#5: ggg4_eee2_jjj14 4 2 14
#6: jjj7_ddd9_bbb15 7 9 15
A.K.
From: Dimitri Liakhovitski
To: arun
Cc: R help
Sent: Saturday, June 8, 2013 5:59 PM
Subject: Re: [R] splitting a string colu
Hi,
You could try this:
dat2<- read.table(text='
case pin some_data
"A" "1" "data"
"A" "2" "data"
"A" "1" "data"
"A" "2" "data"
"B" "1" "data"
"B" "2" "data"
',sep="",header=TRUE,stringsAsFactors=FALSE)
dat2[!duplicated(dat2[,1:2]),]
# case pin some_data
#1 A 1 data
Hi,
dat1$colIndex<-apply(dat1,1,function(x) match(max(x),x))
dat1
# v1 v2 v3 v4 max colIndex
#1 0 0 1 2 2 4
#2 0 1 2 0 2 3
#3 1 2 3 4 4 4
A.K.
- Original Message -
From: Camilo Mora
To: r-help@r-project.org
Cc:
Sent: Saturday, June 8, 2013 8:1
lt;- sapply(varName,function(x){
model1<-
gam(get(x)~pm10+pm25+s(trend,k=35)+s(temp,k=6),quasipoisson,na.action=na.omit,data=chicago1)
sM<- summary(model1)$p.coeff[-1]
})
res
}
fun2gam(varlist)
# death cv resp
#pm10 0.001346275 0.0003902584 -0.000844756
Hi,If your dataset is similar to the one below:
set.seed(24)
temp1_df<-
data.frame(names=rep(c('foo','foo1'),each=6),variable=rep(c('w','x'),times=6),value=sample(25:40,12,replace=TRUE),stringsAsFactors=FALSE)
library(reshape2)
res<-dcast(within(temp1_df,{Seq1<-ave(value,names,variable,FUN=seq_
(res10Percent[-indx10,])
#[1] 452
res10PercentNew<-res10Percent[-indx10,]
nrow(subset(res10PercentNew,dummy==1))
#[1] 226
nrow(subset(res10PercentNew,dummy==0))
#[1] 226
nrow(unique(res10PercentNew))
#[1] 452
A.K.
- Original Message -
From: Cecilia Carmo
To: arun
Cc:
Sent: Monday
b4)<-gsub(".*\\.","",row.names(res10PercentSub4))
res10PercentSub5<-res10PercentSub4[order(as.numeric(res10PercentSub4$id)),]
- Original Message -
From: Cecilia Carmo
To: arun
Cc:
Sent: Monday, June 10, 2013 1:41 PM
Subject: RE: please check this
I think it coul
dummy==0))
#[1] 227
nrow(subset(res10PercentFinal,dummy==1))
#[1] 227
nrow(unique(res10PercentFinal))
A.K.
- Original Message -
From: Cecilia Carmo
To: arun
Cc:
Sent: Monday, June 10, 2013 5:48 PM
Subject: RE: please check this
Error message:
Error in row.names(res10PercentSub5) :
Hi,
Try this:
dat1<-read.table(text="
Row_ID_CR, Data1, Data2, Data3
1, aa, bb, cc
2, dd, ee, ff
",sep=",",header=TRUE,stringsAsFactors=FALSE)
dat2<-read.table(text="
Row_ID_N, Src_Row_ID, DataN1
1a, 1,
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