vec<- scan("stop.txt")
#Read 635 items
vec1<-vec[vec!=21]
length(vec1)
#[1] 584
any(vec1==21)
#[1] FALSE
A.K.
- Original Message -
From: Naser Jamil
To: R help
Cc:
Sent: Wednesday, April 3, 2013 5:33 AM
Subject: [R] scanning data in R
Dear R-user,
May I seek your suggestion. I have
Hi,
suppose, you have 3 files with 2 columns:
named.list<- list(structure(list(col1 = 1:6, col2 = c(0.5, 0.2, 0.3, 0.3,
0.1, 0.2)), .Names = c("col1", "col2"), class = "data.frame", row.names = c(NA,
-6L)), structure(list(col1 = 1:6, col2 = c(0.9, 0.7, 0.5, 0.2,
0.5, 0.2)), .Names = c("col1", "
Hi,
DATA_names<-c("A_mgkg","B_mgkg","C_mgkg","D_mgkg","E_mgkg","F_mgkg","G_mgkg","H
mgkg")
lapply(DATA_names,function(x) if(grepl("_",x)) f(x) else x)[1:2]
#[[1]]
#A ~ (mg ~ kg^{
# -1
#})
#[[2]]
#B ~ (mg ~ kg^{
# -1
#})
f("Na_mgkg")
#Na ~ (mg ~ kg^{
# -1
#})
A.K.
- Original Me
HI,
I assume that you have a data.frame structure.
dat1<- structure(list(Year = c(2000L, 2000L, 2000L, 2000L, 2001L, 2001L,
2001L, 2001L, 2001L), Area = c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L,
2L), Q = c(1L, 1L, 1L, 1L, 25L, 1L, 1L, 1L, 1L), Bin = c(5L,
10L, 15L, 20L, 1L, 5L, 10L, 15L, 20L), FD = c(0
You could also use:
library(data.table)
dat2<- data.table(dat1)
dat2<-dat2[,setdiff(colnames(dat2),"Q"),with=FALSE]
dat2[,lapply(.SD,sum),by=c("Year","Area")]
# Year Area Bin FD
#1: 2000 1 50 36
#2: 2001 1 1 4
#3: 2001 2 50 23
A.K.
----- O
value.var="FD",sum)
names(res)[grep("^\\d+",names(res))]<-paste0("FD_",names(res)[grep("^\\d+",names(res))])
res
# Bin FD_1_2000 FD_2_2000 FD_2_2001
#1 5 4 16 0
#2 10 2 8 8
#3 15 26 12
lt;- names(lst1)
res
# File1 File2
#A 2.4 1.3
#B 2.1 0.4
A.K.
From: Adrian Johnson
To: arun
Sent: Thursday, April 4, 2013 2:49 PM
Subject: Re: [R] Creating data frame from individual files
HI:
sorry for confusion.
I have 72 files and all 72 files
Hi,
You could also use:
library(seqinr)
x1<- rep(c('A','G','C','T'),4)
splitseq(x1,frame=0,word=2)
#[1] "AG" "CT" "AG" "CT" "AG" "CT" "AG" "CT"
splitseq(x1,frame=1,word=2)
#[1] "GC" "TA" "GC" "TA" "GC" "TA" "GC"
splitseq(x1,frame=0,word=3)
#[1] "AGC" "TAG" "CTA" "GCT" "AGC"
A.K.
- Orig
Hi,
You can preserve the order using join(). Though, in this case, merge would be
easier.
library(plyr)
both<- join(currency_df,rate_df,type="left")
names(both)[14:15]<- paste0("currency_",names(both)[14:15])
names(rate_df)[2]<- "other_currency"
both2<- join(both,rate_df,type="left",by=c("other
HI,
"
btw, could this program be modified such that it take direct input from my
input.txt file???"
Lines1<-readLines("utpalmtbl.txt")
Lines1
#[1] "OG1: or10|1345 or10|387 or10|474 or11|1203 or11|182 or10|2158 or12|637"
#[2] "OG2: or10|1562 or10|1584 or10|1977 or11|2263 or11|43"
Hi Utpal,
You can use the same script from my previous email.
Lines1<- readLines("groups.txt")
library(stringr)
res<-paste(gsub("(.*\\:).*","\\1",Lines1),unlist(lapply(str_match_all(Lines1,"or10\\|\\d+"),paste,collapse="
")),sep=" ")
write.table(res,"res1.txt",row.names=FALSE,col.names=FALSE,
Hi,
Try this:
dat1<- read.csv("sample.csv",sep="\t",stringsAsFactors=FALSE)
with(dat1,tapply(COUNTS,list(ACTIVITY),function(x) if (length(unique(x))==1)
NA else shapiro.test(x)))
#$activity1
#[1] NA
#$activity2
# Shapiro-Wilk normality test
#data: x
#W = 0.4588, p-value = 8.025e-11
#
#$ac
88439
#3 activity3 3.760396e-07 0.7282838
A.K.
- Original Message -
From: "Mossadegh, Ramine N."
To: arun
Cc:
Sent: Friday, April 5, 2013 4:17 PM
Subject: RE: [R] How to perform a grouped shapiro wilk test on dataframe
The statistic & the p.value . When I
Hi,
Try this:
tbl1<- structure(c(21L, 23L, 127L, 112L, 120L, 0L), .Dim = 2:3, .Dimnames =
structure(list(
labels = c(1, 2), gts = c("A1", "B2", "G3")), .Names = c("labels",
"gts")), class = "table")
dat1<-as.data.frame(tbl1,stringsAsFactors=FALSE)
dat2<-dat1[dat1$gts!="B2" & dat1$Freq!=0,]
Hi,
Try this:
#dat1 is dataset
indx<-apply(dat1,2,function(x) head(which(!is.na(x)),2))
res<-as.data.frame(sapply(seq_len(ncol(indx)),function(i) dat2[indx[,i],i]))
colnames(res)<- colnames(dat1)
res
# 1B 2B 4B 1A 2A 4A 5B 5A C31A C31B C34A C34B C35A
#1 2.518 2.357
ted to have the same number of rows as dat1:
nrow(dat1)
#[1] 145
new1<-as.data.frame(matrix(NA,ncol=23,nrow=5))
colnames(new1)<- colnames(res)
res1<- rbind(res,new1)
nrow(res1)
#[1] 145
A.K.
From: catalin roibu
To: arun
Sent: Saturday, April 6
6 487
A.K.
- Original Message -
From: "Leask, Graham"
To: Rui Barradas
Cc: arun ; "r-help@r-project.org"
Sent: Saturday, April 6, 2013 1:44 PM
Subject: RE: [R] Replace missing value within group with non-missing value
Hi Rui,
I have just pasted this direct and reru
ot; "58" "59" "60"
[61] "61" "62" "63" "64" "65" "66" "67" "68" "69" "70" "71" "72"
[73] "73" "74" "75"
g a specific
column.
tbl1[,!grepl("B2",colnames(tbl1)) & colSums(tbl1==0)==0]
# 1 2
#21 23
tbl1[,colSums(tbl1==0)==0]
# gts
#labels A1 B2
# 1 21 127
# 2 23 112
A.K.
- Original Message -
From: arun
To: Abhishek Pratap
Cc: R help
Sent: Saturday, Apri
Hi,
Try this:
set.seed(25)
el<-as.data.frame(matrix(sample(1:50,75*124,replace=TRUE),ncol=75))
lst1<-lapply(el,function(x) dist(x))
set.seed(28)
q<-as.data.frame(matrix(sample(1:50,75*124,replace=TRUE),ncol=75))
q<-dist(q)
library(ade4)
res<- lapply(lst1,function(x) mantel.rtest(q,x,nrepet=999
Hi,
May be this helps:
input1<-
data.frame(answer=rep(1:4,times=18),p.number=rep(1:18,each=4),session=rep(1:2,each=36),count=rep(1:8,each=9),type=rep(1:3,each=24))
input2<-
data.frame(answer=rep(1:4,times=18),p.number=rep(1:18,each=4),session=rep(1:2,each=36),count=rep(1:8,each=9),type=rep(1:3,e
local guidelines/advice from PCT London 3 - 5 PARTNERS 32
# ALL_P sex disp form set3 set set2 tag1 mode.ids indivs CHOICE
#1.1 0.5295166 1 0 0 1 1 1 4 1 1 FALSE
#1.2 0.5295166 1 0 0 1 1 1 4 2 1 FALSE
A.K.
- Original Message
Hi,
Try:
year1<- 1598:1997
indx<-findInterval(year1,seq(1591,2000,by=10))
group<-seq(1590,2000,by=10)
ind<-seq(1,length(group),by=1)
labl1<-paste(group[ind],group[ind+1],sep="-")[-42]
dat1<- data.frame(year=year1,decade=labl1[indx],stringsAsFactors=FALSE)
head(dat1,5)
# year decade
#1 1598 1
Hi,
DATA[DATA$A!="Blue1",]
# A B C D
#1 Red1 1 1 1
#3 Red2 1 1 1
#4 Red3 1 1 1
#or
DATA[!grepl("Blue1",DATA$A),]
# A B C D
#1 Red1 1 1 1
#3 Red2 1 1 1
#4 Red3 1 1 1
A.K.
- Original Message -
From: Beatriz González Domínguez
To: r-help@r-project.org; R Help
Cc:
Sent: Sunday, A
Hi,
Try this:
Mult3$mode.ids
#[1] 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1 2 3 4 5 6 1
#Levels: 1 2 3 4 5 6
I guess the key is to have equal number of alternatives.
Mult3$mode.ids<- as.numeric(as.character(Mult3$mode.ids))
res<-mlogit.data(Mult3[-25,],shape="long",choice="CHOICE",alt.var="mode.id
Hi,
Try this:
dat1<-read.table(text="
X Y1 Y2
0.1 3 2
0.2 2 1
",sep="",header=TRUE)
res<-do.call(rbind,lapply(split(dat1,seq_len(nrow(dat1))),function(x)
{Y=rep(colnames(x)[-1],x[-1]); X=rep(x[,1],length(Y));
data.frame(X,Y,stringsAsFactors=FALSE)}))
row.names(res)<- 1:n
Hi,
Not sure if you have only one "country" or not.
Try this:
dat<- data.frame(val,state,country,stringsAsFactors=FALSE)
dat$country[dat$country==0]<-dat$country[1]
#or
#dat$country[dat$country==0]<- dat$country[dat$country!=0]
res<-do.call(rbind,lapply(split(dat,cumsum(grepl("[A-Za-z]",dat$
","integer"))
as.table(df1)
#Error in as.table.default(df1) : cannot coerce to a table
I am not sure about the problem.
A.K.
- Original Message -
From: Bert Gunter
To: arun
Cc: "dcarl...@tamu.edu"
Sent: Monday, April 8, 2013 1:33 PM
Subject: Re: [R] Reshaping a
Hi,
1)
Grunfeld1951<-Grunfeld[Grunfeld$year==1951,]
Grunfeld1951
# firm year inv value capital
#17 1 1951 755.90 4833.00 1207.70
#37 2 1951 588.20 2289.50 342.10
#57 3 1951 135.20 1819.40 671.30
#77 4 1951 160.62 809.00 203.50
#97 5 1951 80.30 327.30 683.90
#117
748.1000
#6 6 206.8500
#7 7 478.7500
#8 8 165.8750
#9 9 429.
#10 10 11.2525
A.K.
- Original Message -
From: arun
To: jpm miao
Cc: R help
Sent: Monday, April 8, 2013 2:04 PM
Subject: Re: [R] How can I extract part of the data in a panel dataset?
Hi,
1
Hi,
Try ?unique()
You haven't provided reproducible data or information about the package. So,
this may or may not work.
dat1<- data.frame(idvar=c(rep(1,2),2,4),col2=c(7,7,8,9))
dat1
# idvar col2
#1 1 7
#2 1 7
#3 2 8
#4 4 9
unique(dat1)
# idvar col2
#1 1 7
Hi,
Try this:
set.seed(28)
t1<-data.frame(id=rep(1:3,rep(3,3)),dt=rep(1:3,rep(9,3)),var=c('num1','num2','norm'),value=rnorm(27))
head(t1)
# id dt var value
#1 1 1 num1 -1.90215722
#2 1 1 num2 -0.06429479
#3 1 1 norm -1.33116707
#4 2 1 num1 -1.81999167
#5 2 1 num2 0.16266969
#6
2 1 num2 0.30611744
#5 3 1 num1 -0.94940980
#6 3 1 num2 -0.06622304
identical(res,dfNew)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: neal subscribe
Cc: R help
Sent: Tuesday, April 9, 2013 12:59 AM
Subject: Re: [R] (no subject)
Hi,
Try this:
set.seed(28)
t1<-data.fram
Hi,
I tried to read your data from the image:
OPENCUT<- read.table("OpenCut.dat",header=TRUE,sep="\t")
OPENCUT
FC LC SR DM
1 400030.34 1323.5 0 400
2 12680.13 2.5 0 180
3 472272.75 2004.7 3 300
4 332978.03 1301.3 106 180
5 98654.20 295.0 0 180
6 68142.05 259.9
file<- "\"Data labels\""
directory="\"/home/mylaptop/\""
cat("The file", file,"is located in directory",directory,sep=" ")
The file "Data labels" is located in directory "/home/mylaptop/"
A.K.
- Original Message -
From: Daniel Caro
To: r-help@r-project.org
Cc:
Sent: Tuesday, April 9,
Hi,
You could try this:
v <- c("a", "b")
mat1<-diag(length(v))
diag(mat1)<- v
mat1
# [,1] [,2]
#[1,] "a" "0"
#[2,] "0" "b"
v1<- letters[1:5]
mat2<- diag(length(v1))
diag(mat2)<- v1
mat2
# [,1] [,2] [,3] [,4] [,5]
#[1,] "a" "0" "0" "0" "0"
#[2,] "0" "b" "0" "0" "0"
#[3,]
Possibly R FAQ 7.31
length(rep(TRUE,signif(0.29*100,2)))
#[1] 29
A.K.
- Original Message -
From: Jorge Fernando Saraiva de Menezes
To: r-help@r-project.org
Cc:
Sent: Tuesday, April 9, 2013 12:11 PM
Subject: [R] rep() fails at times=0.29*100
Dear list,
I have found an unusual behavior
A NA NA NA NA NA NA NA NA
# 17-234 18 19 20 23 24 25 26 27 HAR133 HAR105 HAR211 HUMH255 HUMH276 major
#99 NA NA NA NA NA NA NA NA NA NA NA NA NA NA 0
#100 NA NA NA NA NA NA NA NA NA NA NA NA NA NA NA
# minor
#99 7
#100
Hi,
library(plyr)
library(reshape2)
df2<-mutate(dcast(melt(tableData),State~Var2),
State=factor(State,levels=rownames(tableData)))
dfNew<-df2[as.numeric(df2$State),c(1,3:4,2)]
dfNew$State<- as.character(dfNew$State)
dfNew
# State Frost Population Area
#1 Connecticut 139 3100
Hi,
"
>Thank You, Arun Kirshna , for your efforts and time.
>I also found an equivalent code:
>library(zoo)
>dat[] <- lapply(dat, function(x){replace(x, x == 0, NA)})
>dat <-na.locf(dat)
"
No problem.
If you have equal replications, you could also us
Hi,
Assuming that this is a data.frame.
dat1<-read.table(text="
X,Y
ab,su-di
ac,pi-tu
ad,tu-tu
",sep=",",header=TRUE,stringsAsFactors=FALSE)
library(stringr)
dat2<-data.frame(X=rep(dat1$X,each=2),Y=
unlist(str_split(dat1$Y,"-")),stringsAsFactors=FALSE)
dat2
# X Y
#1 ab su
#2 ab di
#3 ac pi
#4
quot;-")),side="both")
X<-unlist(str_split(str_trim(str_dup(str_pad(dat1$X,width=3,side="right"),2)),"
"))
data.frame(X,Y,stringsAsFactors=FALSE)
# X Y
#1 ab su
#2 ab di
#3 ac pi
#4 ac tu
#5 ad tu
#6 ad tu
A.K.
- Original Message -
From: arun
Hi,
You could try ?within()
set.seed(25)
ALL<- data.frame(q3=sample(15:30,10,replace=TRUE),
q1=sample(1:14,10,replace=TRUE))
res<-within(ALL,{lower<-q1- 1.5*(q3-q1);upper<-q3+1.5*(q3-q1)})
res
# q3 q1 upper lower
#1 21 5 45.0 -19.0
#2 26 6 56.0 -24.0
#3 17 14 21.5 9.5
#4 29 9 59.
Hi,
TRy this:
set.seed(52)
dat1<- as.data.frame(matrix(sample(1:40,100*60,replace=TRUE), nrow=600))
lapply(split(dat1,((seq_len(nrow(dat1))-1)%/% 60)+1),nrow)
#$`1`
#[1] 60
#$`2`
#[1] 60
#$`3`
#[1] 60
res<-lapply(split(dat1,((seq_len(nrow(dat1))-1)%/% 60)+1),colMeans)
res[1:2]
#$`1`
# V1
Hi,
This could be done in different ways:
tab1<-read.table(text="
V1 V2 V3 V4 V5
14.23 1.71 2.43 15.6 127
13.20 1.78 2.14 11.2 100
13.16 2.36 2.67 18.6 101
14.37 1.95 2.50 16.8 113
13.24 2.59 2.87 21.0 118
",sep="",header=TRUE)
tab2<-read.table(text="
V1 V2 V3 V4 V5
1.23 1.1 2.3 1.6 17
1.20
Hi,
Try this:
which(MAT==0)
#[1] 10 19 22 27 28 34 40 46 51 52
MAT[MAT==0]<-sapply(seq_along(MAT[MAT==0]),function(x)rbeta(1,0.1,1))
MAT
# X1 X2 X3 X4 X5 X6
#1 0.68 9.731207e-14 0.03005021 0.003249449 1.36e+00 1.180602e-03
#2 9.86 1.15e
: arun
Cc: R help
Sent: Wednesday, April 10, 2013 4:50 PM
Subject: Re: [R] how to calculate average of each column
Thanks so much! it works fine!
But when I convert the result to a data frame, it mess up.
so i have:
res <- data.frame(res)
then i get:
X1 X2 X3
Hi,
The pattern is not clear.
In this particular case,
library(stringr)
word1<-c("Shangh i", "Hello here i am","h llo")
word1[str_count(word1," ")==1]<-gsub(" ","a",word1[str_count(word1," ")==1])
word1
#[1] "Shanghai" "Hello here i am" "hallo"
But, this will not work here:
word2<-c(
.7 22.40678
#6 21.86207 21.3
#7 20.81034 17.25000
#8 21.5 21.45763
#9 22.18966 19.7
#10 22.31579 20.58929
A.K.
From: Ye Lin
To: arun
Sent: Wednesday, April 10, 2013 6:02 PM
Subject: Re: [R] how to calculate average of each column
Hey A.
.72
mean(unlist(lapply(lst1,function(x) x[5,2])))
#[1] 20.97
A.K.
- Original Message -
From: Silvano Cesar da Costa
To: arun
Cc:
Sent: Wednesday, April 10, 2013 6:02 PM
Subject: Re: [R] means in tables
Hi Arun,
I thought with an example with two tables I could generalize to the 100
tab
1]]
# col1 col2
#1 1 0.5
#2 2 0.2
#3 3 0.3
#4 4 0.3
#5 5 0.1
#6 6 0.2
library(abind)
apply(abind(lst2,along=3),c(1,2),mean)
# col1 col2
#[1,] 3 0.500
#[2,] 4 0.467
#[3,] 5 0.567
#[4,] 6 0.267
#[5,] 7 0.400
#[6,] 8 0.267
A
0.012 0.373
dim(res)
#[1] 8000 3
A.K.
- Original Message -
From: Silvano Cesar da Costa
To: arun
Cc:
Sent: Wednesday, April 10, 2013 9:38 PM
Subject: Re: [R] means in tables
I performed all the procedures you described.
I read each file with the command:
lst2[[1]]; lst2[[2]]
HI Silvano,
No problem.
Just wanted to make sure that it worked.
Regards,
A.K.
- Original Message -
From: Silvano Cesar da Costa
To: arun
Cc:
Sent: Wednesday, April 10, 2013 10:34 PM
Subject: Re: [R] means in tables
Arun,
this code work very well:
setwd('c:/Dados/')
,sep=""),collapse="+")))/length(lst1))
# user system elapsed
# 0.060 0.000 0.063 #faster
system.time(res<-apply(abind(lst1,along=3),c(1,2),mean))
# user system elapsed
# 0.356 0.024 0.380
res<- as.data.frame(res)
identical(res,res1)
#[1] TRUE
A.K.
- Ori
HI,
May be this helps:
library(Unicode)
utf2uxxx<- function(x){
a<- tolower(as.u_char(utf8ToInt(x)))
a<- gsub("[+]","",a)
a<- paste(paste("\\",a,sep=""),collapse="")
a<-cat('"',a,'"',sep="")}
utf2uxxx("õäöü")
#"\u00f5\u00e4\u00f6\u00fc"
sessionInfo()
R version 3.0.0 (2013-04-03)
Platform: x86
Hi,
dat1<- read.table(text="
ID Value
AL1 1
AL2 2
CA1 3
CA4 4
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- dat1
dat1$State<-gsub("\\d+","",dat1$ID)
dat1
# ID Value State
#1 AL1 1 AL
#2 AL2 2 AL
#3 CA1 3 CA
#4 CA4 4 CA
#or
library(stringr)
dat2$State<-s
Hi,
May be this helps:
lines1<- readLines("WAT_DEP.DAT.part")
indx<- which(grepl("[*]",lines1))
indx2<-indx[seq(from=indx[2],length(indx),by=2)]+1
lines2<-str_trim(lines1[indx2],side="left")
dat1<-read.table(text=lines2,sep="",header=FALSE)
library(stringr)
lst1<- lapply(split(indx,((seq_along(in
Hi,
May be this helps:
tb[1,]%*%(((val-rep(meansb79[1,],5))^2)/6)
# [,1]
#[1,] 1.47619
tryvarb<-c(1,2,3,4,4,4,4)
var(tryvarb)
#[1] 1.47619
tb[2,]%*%(((val-rep(meansb79[2,],5))^2)/6)
# [,1]
#[1,] 1.904762
sapply(seq_len(nrow(tb)),function(i) tb[i,]%*%(((val-rep(meansb79[i,],5))^
Hi,
table.x<- as.table(as.matrix(read.table(text="
A B C D
2 4 6 5
",sep="",header=TRUE)))
table.y<- as.table(as.matrix(read.table(text="
C D
5 1
",sep="",header=TRUE)))
library(plyr)
library(reshape2)
vec1<-rowSums(join(melt(table.x)[,-1],
melt(table.y)[,-1],by="Var2")[
hoo.com
Cc:
Sent: Friday, April 12, 2013 1:05 PM
Subject: FW: [R] processing matrix equation derived from rows of two matrices
Thanks so much Arun. If you have time, what reference(s) do you use to
learn the techniques like "seq_len(nrow" etc and other aspects of your
code?
Again. Th
Hi,
I am not sure I understand your question correctly.
dat1<- read.table(text="
id responsed_at number_of_connection
scores
1 12-01-2010 1
2
1
Hi,
>From your example data,
dat1<- read.table(text="
id1 id2 value
a b 10
c d 11
b a 10
c e 12
",sep="",header=TRUE,stringsAsFactors=FALSE)
#it is easier to get the output you wanted
dat1[!duplicated(dat1$value),]
# id1 id2 value
#1 a b
Hi,
May be this helps:
Not sure how you wanted to select those two letters.
dat1<- read.table(text="
Seq,Output
A B B C D A C,Yes
B C A C B D A C,Yes
C D A A C D,No
",sep=",",header=TRUE,stringsAsFactors=FALSE)
library(stringr)
lapply(str_split(str_trim(dat1$Seq)," ")[dat1$Output=="Yes"],f
x[order(x)],collapse="")); x3<-table(x2); x3[x3%in% max(x3)]})
dat1$MaxCombn[dat1$Output=="Yes"]<-lapply(res1,names)
dat1
# Seq Output MaxCombn
#1 A B B C D A C Yes AB, AC, BC
#2 B C A C B D A C Yes AC, BC
#3 C D A A C D No NA
A
Hi,
dat1<- read.table(text="
Number,TimeStamp,Value
1,1/1/2013 0:00,1
2,1/1/2013 0:01,2
3,1/1/2013 0:03,3
",sep=",",header=TRUE,stringsAsFactors=FALSE)
dat2<-data.frame(Number=dat1[,1],do.call(rbind,strsplit(dat1[,2]," ")),
Value=dat1[,3])
names(dat2)[2:3]<- c("Date","Time")
dat2
# Number D
- Forwarded Message -
From: arun
To: Ye Lin
Cc:
Sent: Friday, April 12, 2013 6:25 PM
Subject: Re: [R] split date and time
Hi Ye,
Is this okay?
dat2<-cbind(dat1[,-2],do.call(rbind,strsplit(dat1[,2],"
")),stringsAsFactors=FALSE)
dat2
# Number Value 1
Hi,
In the example you provided, it looks like the dates in Date2 happens first.
So, I changed it a bit.
DataA<- read.table(text="
ID,Status,Date1,Date2
1,A,3-Feb-01,15-May-01
1,B,15-May-01,16-May-01
1,A,16-May-01,3-Sep-01
1,B,3-Sep-01,
=0]
library(plyr)
dataNew<-do.call(rbind,lapply(lst2,function(x) {colnames(x)[3]<-
colnames(DataB)[2];x}))
res<-join(dataNew,DataB,by=c("Date.Accident","ID"),type="right")
identical(res,res2)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: farnoos
Hi,
Try this;
library(reshape2)
res<-dcast(Input,people~place,value.var="time")
res[is.na(res)]<-0
res
# people beach home school sport
#1 Joe 5 3 0 1
#2 Marc 0 4 2 0
#3 Mary 0 0 4 0
#or
xtabs(time~.,Input)
# place
#people beach home s
Hi,
Using S=1000
and
simdata <- replicate(S, generate(3000))
#If you want both "m1" and "m0" #here the missing values are 0
res1<-sapply(seq_len(ncol(simdata.psm1)),function(i)
{x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1})
res1[,997:1000]
# [,1] [,
te(DataA1$Date2,format="%d-%b-%y")
DataB1<- DataB
DataB1$Date.Accident<- as.Date(DataB1$Date.Accident,format="%d-%b-%y")
dataNew2<-do.call(rbind,lapply(unique(DataA1$ID),function(i) { x1<-
DataA1[DataA1$ID==i,] ;x1New<- t(x1); x2<-DataB1[DataB1$ID==i,];
do.ca
put)
# place
#people school home sport beach
# Marc 2 4 0 0
#Joe 0 3 1 5
#Mary 4 0 0 0
A.K.
From: sylvain willart
To: arun
Cc: R help
Sent: Saturday, April 13, 2013 5:41 PM
Subject: Re: [R] Reshaping Data for
Hi,
lapply(LETTERS[1:2],function(x) {x1<-get(x); x1$Rate<- ROC(x1$population);x1})
#[[1]]
# population Rate
#1 100 NA
#2 300 1.0986123
#3 5000 2.8134107
#4 2000 -0.9162907
#5 900 -0.7985077
#6 2500 1.0216512
#[[2]]
# population Rate
Hi,
It would be better if you provided the output of dput(dataset). I am not sure
about the structure of your dataset.
Just from reading the data as is shown.
dat1<- read.table(text="
separator,tissID
>,>,2
,2,1
,6,5
,11,13
>,>,4
,4,9
,6,2
,7,3
,21,1
,23,58
,25,9
,26,4
>,>,11
,1,12
>,>,21
,4,1
library(plyr)
arrange(df,names,desc(dates))
# names dates values
#1 A 4/15/2013 31
#2 A 4/14/2013 102
#3 A 4/13/2013 31
#4 B 4/15/2013 34
#5 B 4/14/2013 47
#6 B 4/13/2013 17
#7 C 4/15/2013 10
#8 C 4/14/2013 29
#9 C 4/13/2013
/2013 29
#5 C 4/13/2013 11
A.K.
- Original Message -
From: arun
To: Katherine Gobin
Cc: R help
Sent: Monday, April 15, 2013 8:57 AM
Subject: Re: [R] Sorting data.frame and again sorting within data.frame
library(plyr)
arrange(df,names,desc(dates))
# names dates values
#1
Hi,
set.seed(25)
myFile1<-as.data.frame(matrix(sample(1:40,50,replace=TRUE),nrow=10))
row.names(myFile1)<- LETTERS[1:10]
groups <- rep (0:1, c(3,2))
kruskal<-apply(myFile1,1,kruskal.test,groups)
p_kruskal <- sapply(kruskal, function(x) x$p.value)
p_kruskal
# A B C
HI,
You could use:
gsub("(^|\\s+)([a-z])","\\1\\U\\2",z,perl=TRUE)
#[1] "R Project" "Hello World" "Something Else"
#or
gsub("\\b([a-z])","\\U\\1",z,perl=TRUE)
#[1] "R Project" "Hello World" "Something Else"
A.K.
- Original Message -
From: Liviu Andronic
To: r-help
Cc:
Hi,
TRy this:
set.seed(30)
mat1<- matrix(sample(1:50,12*12,replace=TRUE),ncol=12)
mat1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12]
[1,] 5 28 13 4 47 3 30 16 3 39 42 9
[2,] 25 44 15 21 9 5 32 5 4 33 30 21
[3,
Hi,
Try:
x<- c("2/26/13 11:59 PM", "2/25/13 10:25 AM")
strptime(x,"%m/%d/%y %I:%M %p")
#[1] "2013-02-26 23:59:00" "2013-02-25 10:25:00"
#or
library(lubridate)
parse_date_time(x,"%m/%d/%y %I:%M %p")
# 2 parsed with %m/%d/%y %I:%M %p
#[1] "2013-02-26 23:59:00 UTC" "2013-02-25 10:25:00 UTC"
A.K.
Hi,
May be this helps:
ZZ<-
data.frame(Index=c("2006-04-07","2006-04-08","2007-01-01","2007-01-01","2008-01-04","2008-01-04"),Open=c(17.5,17.6,16.8,17.2,17.3,17.5),High=c(18.2,17.6,17.2,17.2,17.5,17.6),Low=c(17.3,16.8,16.8,16.8,16.2,15.8),Close=c(17.5,16.8,17.1,16.8,15.9,16.2),Volume=c(23834500,29
Adjusted
#2006-04-07 02:30:00 17.5 18.2 17.3 17.5 23834500 16.8
#2007-01-01 02:30:00 17.2 17.2 16.8 16.8 991400 16.2
#2008-01-04 03:00:00 17.3 17.5 16.2 15.9 443522 16.5
A.K.
- Original Message -
From: arun
To: "Sparks, John James"
Cc: R help
Sent: Monday,
Hi,
You could do this:
dat1<- read.table(text="
ID Value
AL1 1
AL2 2
CA1 3
CA4 4
",sep="",header=TRUE,stringsAsFactors=FALSE)
lst1<-split(dat1,gsub("\\d+","",dat1$ID))
res<-do.call(rbind,lapply(seq_along(lst1),function(i) {x1<-lst1[[i]];
x1$group<- paste0("Key",i);x1}))
res
# ID Value gro
Hi,
vec1<- letters
vec1[!grepl("b|r|x",alp)]
# [1] "a" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "s" "t"
"u"
#[20] "v" "w" "y" "z"
vec1[!vec1%in% c("b","r","x") ]
# [1] "a" "c" "d" "e" "f" "g" "h" "i" "j" "k" "l" "m" "n" "o" "p" "q" "s" "t"
"u"
#[20] "v" "w" "y" "z"
alp<-lapp
You can use na.strings="" in read.table() or read.csv()
library(stringr)
vec1<-unlist(str_split(readLines(textConnection("3,7,11,,12,14,15,,17,18,19")),","))
vec1[vec1==""]<- NA
vec1
# [1] "3" "7" "11" NA "12" "14" "15" NA "17" "18" "19"
#or
scan(text="3,7,11,,12,14,15,,17,18,19",sep=",
Hi,
May be this helps you.
#Using
set.seed(12345)
S=10
simdata <- replicate(S, generate(250))
lstpshat<-lapply(seq_len(ncol(simdata)),function(i)
{glm.t<-glm(t~x1+x2+x3+x4+x5+x6+x7+I(x2^2)+I(x4^2)+I(x7^2)+x1:x3+x2:x4+x3:x5+x4:x6+x5:x7+x1:x6+x2:x3+x3:x4+x4:x5+x5:x6,family=binomial,data=simdata[
HI,
May be this helps:
dat1<- read.table(text="
site1 depth1 year1 site2 depth2 year2
10 30 1860 NA NA NA
NA NA NA 50 30 1860
10 20 1850 11 20 1850
11 25 1950 12 25 1960
10 NA 1870 12 30 1960
11 25 1880 15 22 1890
14 22 1890 14 25 1880
",sep="",header=TRUE,stringsAsFactors=FALSE)
res<-merg
Hi,
Try:
df = data.frame(currency_type = c("EURO_o_n", "EURO_o_n", "EURO_1w", "EURO_1w",
"USD_o_n", "USD_o_n", "USD_1w", "USD_1w"), rates = c(0.47, 0.475, 0.461, 0.464,
1.21, 1.19, 1.41, 1.43),stringsAsFactors=FALSE)
df$currency<-unlist(lapply(str_split(df[,1],"_"),`[`,1))
df$tenor<-unlist(lapp
EURO o_n 0.475
#3 EURO 1w 0.461
#4 EURO 1w 0.464
#5 USD o_n 1.210
#6 USD o_n 1.190
#7 USD 1w 1.410
#8 USD 1w 1.430
A.K.
- Original Message -
From: arun
To: Katherine Gobin
Cc: R help
Sent: Tuesday, April 16, 2013 9:00 AM
Subject: Re: [R]
Hi,
I am not sure about the problem.
If your non-numeric vector is like:
a,b,,d,e,,f
vec1<-unlist(str_split(readLines(textConnection("a,b,,d,e,,f")),","))
vec1[vec1==""]<- NA
vec1
#[1] "a" "b" NA "d" "e" NA "f"
If this doesn't work, please provide an example vector.
A.K.
>Thanks for the re
8 obs. of 4 variables:
# $ patient_id : num 2 2 2 2 3 3 3 3
# $ responsed_at: Date, format: "2010-05-26" "2010-07-07" ...
# $ number : num 1 2 3 4 1 2 3 4
# $ scores : num 1 1 2 3 1 5 4 5
A.K.
From: GUANGUAN LUO
To: arun
Sent: Tuesday
Hi,
vec1<- c(2465, 2255, 2085, 1545, 1335, 1210, 920, 210, 210, 505, 1045)
vec1[order(vec1)]
#[1] 210 210 505 920 1045 1210 1335 1545 2085 2255 2465
order(vec1)
#[1] 8 9 10 7 11 6 5 4 3 2 1
sort(vec1,index.return=TRUE)
#$x
#[1] 210 210 505 920 1045 1210 1335 1545 2085 2255 2465
HI,
Not sure if this helps:
library(plyr)
res<-mutate(Xa,var1=round(Sepal.Length),var2=round(Sepal.Width))
str(res)
#'data.frame': 150 obs. of 7 variables:
# $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
# $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
# $ Petal.L
# $ var2 : num 4 3 3 3 4 4 3 3 3 3 ...
# - attr(*, "label")= chr "Some df label"
A.K.
- Original Message -
From: arun
To: Liviu Andronic
Cc: R help
Sent: Tuesday, April 16, 2013 2:40 PM
Subject: Re: [R] avoid losing data.frame attributes on cbind()
HI,
N
"label")= chr "Some df label"
A.K.
- Original Message -
From: arun
To: Liviu Andronic
Cc: R help
Sent: Tuesday, April 16, 2013 4:45 PM
Subject: Re: [R] avoid losing data.frame attributes on cbind()
Hi,
Another method would be:
Xc<- Xa
Xc$var1<-NA; Xc$var2<-
566 41.2566
# Removed.SST.AGC.QC16200 Removed.SST.AGC.Kurtosis.Skewness
#1 NA NA
#2 46.1658 NA
#3 41.2566 NA
# Removed.SST.AGC.Kurtosis
#2 2 1 yes
#3 3 22
#4 4 15
#5 5 3 no
#6 6 6
#7 7 8 yes
A.K.
From: farnoosh sheikhi
To: "smartpink...@yahoo.com"
Sent: Wednesday, April 17, 2013 12:52 AM
Subject: Merge
Hi Arun,
I want to merge a data set with another data frame with 2 columns and keep the
sam
HI,
test.date<- rep(c("14/05/2012","01/10/2012","28/09/2012"),each=6)
as.Date(test.date,"%d/%m/%Y")
# [1] "2012-05-14" "2012-05-14" "2012-05-14" "2012-05-14" "2012-05-14"
#[6] "2012-05-14" "2012-10-01" "2012-10-01" "2012-10-01" "2012-10-01"
#[11] "2012-10-01" "2012-10-01" "2012-09-28" "2012-09-28
Hi,
You may also try:
dfab<-data.frame(A,B)
library(plyr)
dfcd<-subset(mutate(dfab,C=A*2,D=B*3),select=-c(A,B))
#or
dfcd1<-subset(within(dfab,{D<-B*3;C<-A*2}),select=-c(A,B))
dfcd$C
#[1] 2 4 6
dfcd$D
#[1] 12 18 21
A.K.
- Original Message -
From: jpm miao
To: r-help
Cc:
Sent: Wedn
Hi,
dat1<- read.table(text="
V1 V2
A 1
B 2
C 1
D 3
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- read.table(text="
V3 V2
AAA 1
BBB 2
CCC 3
",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
join(dat1,dat2,by="V2",type="full")
# V1 V2
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