Hi,
May be this helps:
chnew<-CH
chnew[duplicated(toupper(CH))]<-CH[duplicated(toupper(CH),fromLast=TRUE)]
chnew
#[1] "aBd" "sTb" "aBd" "dFDasd" "asd" "dFDasd"
A.K.
- Original Message -
From: Christofer Bogaso
To: r-help
Cc:
Sent: Sunday, August 11, 2013 8:39 AM
Subjec
MRTZIt" "ppbzJG" "ppbzJG"
A.K.
____
From: Christofer Bogaso
To: Bert Gunter
Cc: arun ; R help
Sent: Sunday, August 11, 2013 1:54 PM
Subject: Re: [R] Working with string
Thanks Bert and Arun for your help.
As Bert already pointed out, Arun's co
Hi,
You may try this:
vec1<-unique(data1[,1])
res<-do.call(rbind,lapply(seq_along(vec1),function(i)
{x1<-data1[data1[,1]%in%vec1[-i],]; x2<-mean(x1$value[x1$value>0]); x3<-
data1[data1[,1]%in% vec1[i],]; x3$value[x3$value<0]<-x3$value[x3$value<0]+x2;
x3}))
res1<-res[match(row.names(dat1),row.nam
PPBZJG" "MRTZIt" "MRTZIT"
chnew<-CH1
chnew[duplicated(toupper(CH1))]<- CH1[duplicated(toupper(CH1),fromLast=TRUE)]
chnew
#[1] "ppbzJG" "ppbzJG" "MRTZIt" "MRTZIt"
sort(chnew)
#[1] "MRTZIt" "MRTZIt" &quo
Hi,
You could change ?options()
options("scipen")
#$scipen
#[1] 0
print(1e5)
#[1] 1e+05
op<- options()
options(scipen=20)
print(1e10)
#[1] 100
Using the example from ?lm()
ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,
Hi,
May be this help:
dat1<- structure(list(V1 = c("h", "f", "s", "n", "r", "x", "h", "t",
"u", "g"), V2 = c("p", "j", "r", "r", "i", "x", "f", "b", "n",
"d"), V3 = c("c", "o", "s", "d", "f", "r", "b", "p", "q", "b"
), V4 = c("i", "g", "j", "d", "y", "f", "s", "q", "s", "z"),
V5 = c("m",
Hi,
The question is not clear so not sure this is what you wanted.
dat1<- read.table(text="
SameName Individual Age Gender
1 4 80 M
2 15 56 F
3 1 75 F
4 15 56 F
5 2 58 F
6 4 80 M
",sep="",header=TRUE,stringsAsFactors=FALSE)
reps<-c(4,15)
dat1$Newcol<-as.numeric(dat1$Individual%in
A.K.
From: Kripa R
To: arun
Cc: R help
Sent: Tuesday, August 13, 2013 10:56 AM
Subject: RE: [R] pulling out pairs from data frame
Oops! Ok So I have this file:
SampleName Individual Age Gender
1 4 80 M
2 15 56 F
3 1 75 F
4 15 56 F
5 2 58 F
6 4 80 M
And I want to pull out paired samples,
HI,
Your desired output is not clear. May be this helps:
#dat1 is the dataset
dat1$ID<- 1:nrow(dat1)
library(reshape2)
res1<-melt(dat1,id.vars=c("ID","DSYSRTKY"))
res1$value<-res1$value!=""
res1[,2]<- as.integer(as.character(res1[,2]))
res1[,3]<-as.character(res1[,3])
colnames(res1)[3:4]<-c("CO
2 58 F
The OP asked for:
And I want to pull out paired samples, so the resulting file would look
something like this:
>>> SampleName Individual Age Gender
#>>> 1 4 80 M
>>> 2 15 56 F
>>> 4 15 56 F
>>> 6 4 80 M
Anyway, the question was not cle
Hi,
You could try:
lapply(split(Clean,list(Clean$TERM,Clean$INST_NUM)),function(x)
shapiro.test(x$GRADE))
A.K.
- Original Message -
From: Robert Lynch
To: r-help@r-project.org
Cc:
Sent: Tuesday, August 13, 2013 1:46 PM
Subject: [R] ave function
I've written the following function
C
According to your first post,
"
NewDataFrame <- data.frame(ID=integer(), DSYSRTKY=integer(), CODE=character(),
PRIMAIRY=logical())"
The new output dataset: "Out1"
str(Out1)
'data.frame': 48 obs. of 4 variables:
$ ID : chr "1" "2" "3" "4" ...
$ DSYSRTKY: chr "10005" "10005"
haracter)
res5<-res4[,c(5,1,3,4)]
colnames(res5)[3]<-"CODE"
identical(res5,Out1)
#[1] TRUE
A.K.
A.K.
- Original Message -
From: arun
To: R help
Cc:
Sent: Tuesday, August 13, 2013 2:45 PM
Subject: Re: [R] Create rows for columns in dataframe
According to your first
Hi,
You could try:
set.seed(24)
dat1<- as.data.frame(matrix(sample(1:50,29*6,replace=TRUE),ncol=6))
((seq_len(nrow(dat1))-1)%/%6)+1
# [1] 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5
#For a particular column:
aggregate(dat1[,5],list(((seq_len(nrow(dat1))-1)%/%6)+1),FUN=mean)
# Gr
Hi,
You could try:
tmp[,1]<- as.character(tmp[,1])
tmp[,1][-grep(",",tmp[,1])]<-paste0(",,",tmp[,1][-grep(",",tmp[,1])])
tmp2<-data.frame(read.table(text=tmp[,1],sep=",",header=FALSE,stringsAsFactors=FALSE),rID=tmp[,2],stringsAsFactors=FALSE)
colnames(tmp2)[1:3]<-paste("sID",letters[1:3],sep="
Hi,
You could try:
dat2<- read.table(text='
Respondent, Question1
1, "1,4,7"
2, "2,7,5"
3, "6,3,5"
4, ""
',sep=",",header=TRUE,stringsAsFactors=FALSE)
library(stringr)
dat2New<-cbind(dat2,do.call(rbind,lapply(
str_split(str_trim(dat2[,2]),","),as.numeric)))
colnames(dat2New)[3:5]<- paste("Q1",coln
Hi,
Try:
a1<- "...verified email at neu.eduCited by
99853
To: r-help@r-project.org
Cc:
Sent: Wednesday, August 14, 2013 8:14 AM
Subject: [R] Grabbing numbers inside a character string
I have a string that contains something like:
...verified email at neu.eduCited by 99853https://stat.ethz.ch/ma
Hi,
Try:
set.seed(42)
dat1<- as.data.frame(matrix(sample(LETTERS,2*1e6,replace=TRUE),ncol=2))
dat2<- dat1[1:1e5,]
dat3<- dat1
library(data.table)
dt1<- data.table(dat1)
system.time(dat1$sat<- 1*(dat1[,1]==dat1[,2]))
# user system elapsed
# 0.148 0.004 0.152
library(car)
system.time({dat3$
Hi,
I tried the second method on a bigger dataset. This is what I get,
indx<-rep(1:nrow(dat1),6e4)
dat2<- dat1[indx,]
system.time({
vec1<- paste(dat2[,1],dat2[,2],colnames(dat2)[2],sep=".")
res2<-reshape(dat2,idvar="newCol",varying=list(2:26),direction="long")
res3<-res2[order(res2[,4]),]
res4<
at4[dat4[,2]!="",]
row.names(dat5)<- 1:nrow(dat5)
dat5New<-within(dat5,{ID<-row.names(dat5)})[,c(4,1:3)]
})
# user system elapsed
# 12.620 0.684 13.333
dim(dat5New)
#[1] 288 4
A.K.
Hi Arun,
The second method is indeed working much faster. It worked fast for my
12 2 35
#2 17 6 55
A.K.
From: farnoosh sheikhi
To: "smartpink...@yahoo.com"
Sent: Wednesday, August 14, 2013 2:45 PM
Subject: changing colnames
Hi Arun,
I have a data set as follow:
X1 X2 X3
age race stat
12235
17655
The column
Hi,
Try:
dat1<- read.table(text="
Date Time O H L C U D
06/01/2010 1358 136.40 136.40 136.35 136.35 2 12
06/01/2010 1359 136.40 136.50 136.35 136.50 9 6
06/01/2010 1400 136.45 136.55 136.35 136.40 8 7
06/01/2010 1700 136.55 136.55 136.55 136.55 1 0
06/02/2010
Hi,
May be this helps:
library(gsubfn)
as.numeric(strsplit(gsub("[c() ]","",gsubfn("([0-9]+)-([0-9]+)",
~as.numeric(seq(x,y)),x)),",")[[1]])
#[1] 2 5 6 7 10 12 13 14 15
A.K.
- Original Message -
From: Chris Stubben
To: r-help@r-project.org
Cc:
Sent: Wednesday, August 14, 2013 1:41
I had earlier came up with a similar kind of function, though didn't posted.
unlist(lapply(strsplit(x,",")[[1]],function(x)
sapply(strsplit(x,"-"),function(x) {x1<-as.numeric(x);if(length(x1)==2)
seq(x1[1],x1[2]) else x1})))
A.K.
- Original Message -
Fro
HI,
You could try:
#If all the files are in the working directory:
vec1<-list.files() #Created 3 dummy files in my WD
vec1
#[1] "mergedStatstA.csv" "mergedStatstB.csv" "mergedStatstC.csv"
library(gdata)
lapply(seq_along(vec1),function(i)
{x1<-read.csv(vec1[i],header=TRUE,sep="\t");x2<-rename.va
Hi,One way would be:
df<-
data.frame(case,obsdate=as.Date(obsdate,format="%d/%m/%Y"),score,stringsAsFactors=FALSE)
#using as.data.frame(cbind(... should be avoided
df$date_end<-as.Date(unlist(lapply(with(df,tapply(obsdate,case,FUN=function(x)
x-1)),function(x)
c(x[-1],as.Date("31/03/2011",fo
nd score
#1: a 2001-04-01 2007-05-19 60
#2: a 2007-05-20 2010-10-07 72
#3: a 2010-10-08 2011-03-31 85
#4: b 2001-04-01 2005-11-09 72
#5: b 2005-11-10 2011-03-31 79
#6: c 2001-04-01 2011-03-31 65
identical(df1,as.data.frame(dt1))
#[1] TRUE
A.K.
----- Original
Hi,
You could try:
res<-data.frame(lapply(wkend,function(x) seq(x,as.Date("2013-12-26"),by=7)))
colnames(res)<- paste0("date",1:3)
head(res)
# date1 date2 date3
#1 2013-01-04 2013-01-05 2013-01-06
#2 2013-01-11 2013-01-12 2013-01-13
#3 2013-01-18 2013-01-19 2013-01-20
#4 2013-01-
HI,
xyplot(pct~sampdate|func_feed_grp,data='burns.date.ffg')
#Error in eval(substitute(groups), data, environment(x)) :
# invalid 'envir' argument of type 'character'
xyplot(pct~sampdate|func_feed_grp,data=burns.date.ffg) #no error
#Not sure about your exact specification, so this would get yo
0.764 0.000 0.767
identical(res1,res9)
#[1] TRUE
system.time({
dt1 <- data.table(dat2, key=c('Date', 'Time'))
ans <- dt1[, .SD[.N], by='Date']})
# user system elapsed
# 37.384 0.000 37.454
#separate the data.table creation step:
dt1 <- data.table(
attached packages:
[1] data.table_1.8.8 stringr_0.6.2 reshape2_1.2.2
loaded via a namespace (and not attached):
[1] plyr_1.8
- Original Message -
From: Steve Lianoglou
To: arun
Cc: William Dunlap ; Noah Silverman
; Michael Hannon ; David
Winsemius ; R help
Sent: Thursday, August 1
'2011-07-13','Predator',0.0247
'2011-07-15','Filterer',0.0385
'2011-07-15','Gatherer',0.8469
'2011-07-15','Grazer',0.0147
'2011-07-15','Omnivore',0.0005
'2011-07-15','Parasite',0.
]
names(res1)<-names(res3New)
identical(res1,as.data.frame(res3New))
#[1] TRUE
A.K.
- Original Message -
From: Steve Lianoglou
To: arun
Cc: R help
Sent: Thursday, August 15, 2013 4:48 PM
Subject: Re: [R] How to extract last value in each group
Hi,
On Thu, Aug 15, 2013 at 1:38 PM, a
.K.
- Original Message -
From: Pooya Lalehzari
To: arun
Cc: "r-help@r-project.org"
Sent: Thursday, August 15, 2013 5:09 PM
Subject: RE: [R] How can I create a data.table with 1000 variables
(Var1:Var1000)
Here is the sample code.
Large_Table = data.frame(
ion(x)
{x1<-data.frame(numeric());colnames(x1)<-x; x1}))
identical(Large_Table,Large_Table2)
#[1] TRUE
A.K.
- Original Message -
From: Ista Zahn
To: Pooya Lalehzari
Cc: arun ; "r-help@r-project.org"
Sent: Thursday, August 15, 2013 5:19 PM
Subject: Re: [R] How can I cre
lt;- as.data.frame(res)
system.time(res1<-dat2[c(diff(as.numeric(as.factor(dat2$Date))),1)>0,])
row.names(resNew)<- row.names(res1)
attr(resNew,"row.names")<- attr(res1,"row.names")
identical(resNew,res1)
#[1] TRUE
A.K.
- Original Message -
From: Steve
Slightly modified, also seems to work.
gsubfn( "([[:alpha:]][[:alnum:]]*)((?=\\s*[-+~*)])|$)",function(x,...)
paste0(toupper(x),'z'), test, perl=TRUE )
#[1] "Y1z + Y2z ~ Az*(Bz + Cz) + Dz + Fz * (h == 3) + (sex == 'male')*Iz"
A.K.
- Original Message -
From: Greg Snow <538...@gmail.com>
Hi Steve,
Thanks. The error problem is solved by using quotes.
Will post at data-table mailing list regarding the issue of:
system.time(ans <- dt1[, .SD[.N], by='Date'])
# user system elapsed
# 39.284 0.000 39.352
A.K.
- Original Message -
From: Steve Lianog
You could remove the white space also by:
library(stringr)
levels(bdf$func_feed_grp)
#[1] "Filterer" " Gatherer " " Grazer " " Omnivore "
#[5] " Parasite " " Predator " " Shredder "
levels(bdf$func_feed_grp)<- str_trim(levels(bdf$func_feed_grp))
levels(bdf$f
Hi,
You can upload the dataset using:
library(XLConnect)
wb<-loadWorkbook("excel_data.xlsx")
dat1<- readWorksheet(wb,sheet="excel data",region="A1:DA101") #region can be
specified to read a subset of the dataset. Here, I read the full #dataset.
dim(dat1)
#[1] 100 105
str(dat1)
#'data.frame':
Hi,
May be this helps:
#data1 (changed `data` to `data1`)
set.seed(6245)
data1 <- data.frame(x1=rnorm(5),x2=rnorm(5),x3=rnorm(5),x4=rnorm(5))
data1<- round(data1,digits=3)
data2<- data1
data1[,3:4]<-lapply(data1[,3:4],function(x){x1<-
match(x,sample(unlist(data1[,3:4]),round(0.8*length(unlist(
NA
#3 -1.09734838 0.2069539 -0.6997853
#4 0.75919499 -0.5683809 0.4752002
#5 -0.03063141 -0.7549605 2.6038635
A.K.
From: Richard Kwock
To: arun
Cc: Christopher Desjardins ; R help
Sent: Friday, August 16, 2013 5:55 PM
Subject: Re: [R] Randomly drop a
Hi,
It is better to ?dput() your example dataset.
temp<-
list(structure(c(16.91569, 190.89517, 169.70239, 162.14503, 190.8952,
2573.2651, 1966.6202, 1829.3142, 169.7024, 1966.6202, 2335.3354,
1676.7749, 162.145, 1829.314, 1676.775, 2106.543), .Dim = c(4L,
4L), .Dimnames = list(NULL, NULL)), st
Hi,
May be this helps.
a1<- data.frame(a)
b1<- data.frame(b)
lapply(seq_len(ncol(a1)),function(i) lm(b1[,i]~a1[,i]))
lapply(seq_len(ncol(a1)),function(i) summary(lm(b1[,i]~a1[,i]))$coef)
A.K.
- Original Message -
From: "Folkes, Michael"
To: r-help@r-project.org
Cc:
Sent: Tuesday,
iginal Message -
From: arun
To: "Folkes, Michael"
Cc: R help
Sent: Tuesday, August 20, 2013 7:42 AM
Subject: Re: [R] function on columns of two arrays
Hi,
May be this helps.
a1<- data.frame(a)
b1<- data.frame(b)
lapply(seq_len(ncol(a1)),function(i) lm(b1[,i]~a1[,i]))
lap
Hi,
May be this helps:
A<- matrix(list(),3,3)
A[[1]]<- matrix(1,4,4)
A[[2]]<- matrix(1:15,3,5)
A[[3]]<- matrix(1:10,5,2)
---
str(A)
#List of 9
# $ : num [1:4, 1:4] 1 1 1 1 1 1 1 1 1 1 ...
# $ : int [1:3, 1:5] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:5, 1:2] 1 2 3 4 5 6 7
hod3())
# user system elapsed
# 50.856 0.000 50.948
system.time(method4())
# user system elapsed
# 46.444 0.000 46.526
A.K.
- Original Message -
From: "Folkes, Michael"
To: "Law, Jason" ; arun
Cc: R help
Sent: Tuesday, August 20, 2013 1:57 PM
Su
HI,
nuru<-
read.table("NS.diet.txt",header=TRUE,stringsAsFactors=FALSE,sep="\t",fill=TRUE,check.names=FALSE)
colnames(nuru)
# [1] "Species" "Fish ID" "Date"
#[4] "Time" "Depth (m)" "Haul.No."
#[7] "Longitude
8 2 1.000 1.000
#5 9 2 1.000 1.000
Hope this helps.
A.K.
From: Robert Lynch
To: arun
Cc: R help
Sent: Tuesday, August 20, 2013 8:00 PM
Subject: Re: [R] ave function
I tried
> lapply(split(Clean,list(Clean$TE
610 1106842 1.420787e-11 0.9192428
#2 200610 1324438 2.345177e-12 0.9048394
#3 200610 1557630 4.618117e-10 0.8968445
#4 200701 994474 3.695317e-14 0.7939902
#5 200701 1106842 2.745429e-08 0.9292158
#6 200701 1107019 6.887642e-10 0.9213602
A.K.
From: Robe
Hi,
You may try ?merge() or ?join() from library(plyr)
dat1<- read.table(text="
Col1 Col2
red 1.23
blu 4.56
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- read.table(text="
Col1 Col2
red SQU
blu CIR
",sep="",header=TRUE,stringsAsFactors=FALSE)
library(plyr)
res1<-join(dat2,dat1,by="Col1")[
Hi,
Please check:
?FAQ 7.31
http://cran.r-project.org/doc/FAQ/R-FAQ.html#Why-doesn_0027t-R-think-these-numbers-are-equal_003f
intervalloweredge[3]-3.845
#[1] -4.440892e-16
which(as.character(intervalloweredge)=="3.845")
#[1] 3
which(as.character(intervalloweredge)=="3.86")
#[1] 4
#or
which(rou
HI Samuel,
Based on the output you wanted:
(It would be better to use ?dput() to show the example dataset)
dat1<- structure(list(SiteID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L, 1L, 1L), SiteName = c("Big Platte Lake", "Big Platte Lake",
"Big Platte Lake", "Big Platte Lake", "Big
13.716 1 Big Platte Lake TP Grab 7.94 FALSE
A.K.
- Original Message -
From: arun
To: "schris...@wisc.edu"
Cc: R help
Sent: Thursday, August 22, 2013 9:25 AM
Subject: Re: filtering out replicate (duplicate) observations under special
conditions
HI Samuel,
Ba
Hi,
May be this helps:
dat1<- structure(...
dat1$ID<- row.names(dat1)
library(reshape2)
dat2<-melt(dat1,id.vars="ID")
dat2New<-dat2[order(as.numeric(dat2$ID)),]
head(dat2New)
# ID variable value
#1 1 Ramses 4
#11 1 Sheiks 5
#21 1 Trojans 7
#31 1 Unnamed 2
#2 2 R
HI,
May be this works:
row.names(MyDF)[c(0,diff(MyDF[,3]))==0|c(0,diff(MyDF[,3]))>1]
#[1] "1" "5" "8"
A.K.
Hi,
Here i have a dataframe called MyDF.
a<- c(1,1,1,1,1,1,1,0,1,1)
b<-c(1,1,0,1,1,0,0,0,1,1)
c<-c(1,2,3,1,1,2,0,5,6,1)
d<-c(1,1,1,1,1,1,1,1,0,1)
MyDF<-data.frame(DWATT=a,TNH=b,CSG
Hi,
Try:
dat1<- read.table(text="
1 10
3 5
0 0
",sep="",header=FALSE)
dat2<- read.table(text="
2 10
0 0
3 5
",sep="",header=FALSE)
res<-with(rbind(dat1,dat2),aggregate(V2~V1,FUN=sum))
res1<-res[res[,1]!=0,]
res1
# V1 V2
#2 1 10
#3 2 10
#4 3 10
#or
library(data.table)
dt1<- data.table(rbi
as.data.frame(rbind(as.matrix(dat1),as.matrix(dat2))),aggregate(V2~V1,FUN=sum))
res1<- res[res[,1]!=0,]
res1
# V1 V2
#2 1 10
#3 2 10
#4 3 10
A.K.
From: Anindya Sankar Dey
To: arun
Sent: Friday, August 23, 2013 9:40 AM
Subject: Re: [R] Combining two tables
Hi,
May be:
set.seed(24)
dat1<- as.data.frame(matrix(sample(1:50,145*160,replace=TRUE),ncol=160))
dat2<- dat1
dat1$m<-rowSums(dat1[,c(7,12,45,57)])
#or
dat2<-within(dat2,{m<-rowSums(cbind(V7,V12,V45,V57))}) #using column names
identical(dat1,dat2)
#[1] TRUE
A.K.
- Original Message -
#or
dat1<- do.call(rbind,lapply(list.files(),function(x)
read.table(x,header=FALSE,sep="")))
res<-aggregate(V2~V1,data=dat1,FUN=sum)
res1<- res[res[,1]!=0,]
res1
# V1 V2
#2 1 10
#3 2 10
#4 3 15
#5 4 10
Hope this helps.
A.K.
From: Anindy
Hi,
Possibly R FAQ 7.31
vec1[180]
#[1] 18
vec1[180]-18
#[1] 3.552714e-15
vec1[round(vec1,2)%in%vec2]
# [1] 2 4 6 8 10 12 14 16 18 20
A.K.
- Original Message -
From: Finlay Scott
To: r-help@r-project.org
Cc:
Sent: Friday, August 23, 2013 11:00 AM
Subject: [R] Unexpected behaviou
HI,
May be this helps:
set.seed(24)
vec1<- sample(1e5,1e4,replace=FALSE)
set.seed(48)
vec2<- sample(1e7,1e6,replace=FALSE)
system.time(res<- sapply(vec1,function(x) which(vec2%in%x)))
# user system elapsed
#212.520 0.500 213.386
#or
system.time({res2<- match(vec1,vec2) #assuming that t
Hi,
Use ?split()
#dat1 is the dataset:
lst1<- split(dat1,dat1$BaseProd)
lst1
#$`2231`
# BaseProd CF OSA
#1 2231 0.5 0.7
#2 2231 0.8 0.6
#3 2231 0.4 0.8
#
#$`2232`
# BaseProd CF OSA
#4 2232 1 2
#5 2232 3 1
#
#$`2233`
# BaseProd CF OSA
#6 2233 0.9 0.5
#7 2233 0.
Hi,
You could use ?paste()
colnames(myData)<-paste(colnames(myData),paste0("var",1:3),sep="_")
myData
# col1_var1 col2_var2 col3_var3
#1 1 2 3
#2 2 3 4
#3 3 4 5
A.K.
- Original Message -
From: Richard Sherman
To:
CF
and OSA for each base productPlease tell me now how to do this thing
- Original Message -
From: arun
To: R help
Cc:
Sent: Saturday, August 24, 2013 11:44 AM
Subject: Re: Divide the data into sub data on a particular condition
Hi,
Use ?split()
#dat1 is the dataset:
lst1<-
Hi,
May be this helps:
#Creating some dummy data.
set.seed(24)
lst1<-lapply(1:8,function(x) ts(sample(1:25,20,replace=TRUE)))
set.seed(49)
lst2<-lapply(1:8,function(x) ts(sample(1:45,20,replace=TRUE)))
??Find_Max_CCF()
#No vignettes or demos or help files found with alias or concept or
#title
#or you could try:
lst2<- split(dat1,as.numeric(gl(69,69,2000)))
# identical(lst1,lst2)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: R help
Cc:
Sent: Sunday, August 25, 2013 1:17 PM
Subject: Re: sbusetting data by rows (every 69 rows)
Hi,
Try:
set.seed(24)
d
Hi,
It's not clear what you really wanted. May be this helps:
dat1<- read.table(text="
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21
V22 V23 V24 V25 V26
1 1 1 3 1 NA NA 1 NA NA 1 2 7 5 3 2 1.0 15 17 5 1
NA 1 2 19 1 1
2
HI,
Couldn't reproduce the problem. I am using R 3.0.1.
time$convert<-as.POSIXct(paste(time[,1],time[,2]),format="%Y/%m/%d %H:%M:%S")
time
date hour convert
1 2012/10/21 00:02:38 2012-10-21 00:02:38
2 2012/10/21 00:11:05 2012-10-21 00:11:05
3 2012/10/21 00:19:33 201
00:11:05
#3 2012/10/21 00:19:33 2012-10-21 00:19:33
#4 2012/10/21 00:28:00 2012-10-21 00:28:00
#5 2012/10/21 00:36:27 2012-10-21 00:36:27
#6 2012/10/21 00:44:57 2012-10-21 00:44:57
A.K.
- Original Message -
From: arun
To: Carlos André Zucco
Cc: R help
Sent: Monday, August 26, 2013
HI,
It may be better to provide an example dataset using ?dput().
dput(head(dataset),20)
Try:
signal3<- ifelse(t.tr=="Pst 24", signal2-17.29, ifelse(t.tr=="Pst 48", signal2
- 43.93256, etc.))
A.K.
- Original Message -
From: "Fethe, Michael"
To: "r-help@r-project.org"
Cc:
Sent:
HI,
You could also try:
res<-reshape(yrmon,varying=!grepl("year",colnames(yrmon)),v.names="Values",timevar="Month",direction="long")[,-4]
res1<- res[order(res$year,res$Month),]
head(res1)
# year Month Values
#1.1 1901 1 -0.2557446
#1.2 1901 2 -0.2318646
#1.3 1901 3 -0.1961822
#1
.abb)))
colnames(yrmon3)<- colnames(res3)
row.names(yrmon3)<- row.names(res3)
identical(res3,yrmon3)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: catalin roibu
Cc: Jim Lemon ; R help
Sent: Monday, August 26, 2013 9:08 AM
Subject: Re: [R] transform variables
HI,
You could also
z = (list[[i]]$a - list[[i]]$b) / list[[i]]$c
}
which gave me only 49 values, rather than 1566 values.
Thank you very much!
Kind regards,
R.L
- Original Message -
From: arun
To: R help
Cc:
Sent: Sunday, August 25, 2013 1:28 PM
Subject: Re: sbusetting data by rows (every 69 ro
2 val signal3
#1 Pto 24 658 -29.39 628.61
#2 Pm 48 608 -0.20 607.80
#3 Pst 72 689 -43.47 645.53
#4 Pto 72 624 -68.17 555.83
#5 Pto 72 666 -68.17 597.83
#6 Pto 24 620 -29.39 590.61
#or ?merge()
A.K.
- Original Message -
From: Bert Gunter
To: arun
Hi,
May be this helps:
dat1<- as.data.frame(matrix(1:30,6,5))
lapply(seq_len(ncol(dat1)),function(i) dat1[,i])
[[1]]
[1] 1 2 3 4 5 6
[[2]]
[1] 7 8 9 10 11 12
[[3]]
[1] 13 14 15 16 17 18
[[4]]
[1] 19 20 21 22 23 24
[[5]]
[1] 25 26 27 28 29 30
#from excel dataset (the same data)
library(X
Hi,
Suppose you created a dataframe like this:
set.seed(28)
dat1<-as.data.frame(simplify2array(list(letters[1:5],sample(1:20,5,replace=TRUE),6:10)),stringsAsFactors=FALSE)
str(dat1)
#'data.frame': 5 obs. of 3 variables:
# $ V1: chr "a" "b" "c" "d" ...
# $ V2: chr "1" "2" "10" "18" ...
#
Hi,
Your attachment didn't came through..
Using the same example, that I used before:
set.seed(549)
dat1<- data.frame(t.tr=sample(c(paste("Pst", c(24, 48, 72)), paste("Pto",
c(24, 48, 72)), paste ("Pm", c(24, 48, 72))), 50, replace=TRUE), signal2 =
sample(600:700, 50, replace=TRUE))
dat2<- data
onvert the whole list to data frame rather
than each element? If so, how could I do so? ( I tried
'as.data.frame(list)' which didn't work since the last element had less
rows)
Your contribution is highly appreciated!
Best regards,
R.L.
- Original Message -
From
is for people
in condition 1 who are under 25. Hope this makes sense.
The data set is actually 158 rows with 25 columns for now, and will include
300+ more participants.
- Original Message -
From: arun
To: R help
Cc:
Sent: Sunday, August 25, 2013 9:20 PM
Subject: Re: Sorting
Hi,
May be this helps:
lst1<-dlply(data,.(FileNo,ChannelNo),transform,filtered=applyfilter(ChannelB))
data2<-do.call(rbind,lapply(lst1,function(x){x$filtered<-
as.numeric(x$filtered);x}))
row.names(data2)<- row.names(data)
head(data,3)
# Time ChannelA ChannelB FileNo ChannelNo
#116
#or
data2<-ddply(data,.(FileNo,ChannelNo),transform,filtered=as.numeric(applyfilter(ChannelB)))
A.K.
- Original Message -
From: arun
To: Ingo Reinhold
Cc: R help
Sent: Tuesday, August 27, 2013 10:03 AM
Subject: Re: [R] [plyr] Moving average filter with plyr
Hi,
May be this he
m: Ingo Reinhold
To: arun
Cc: R help
Sent: Tuesday, August 27, 2013 10:32 AM
Subject: RE: [R] [plyr] Moving average filter with plyr
Thank you very much.
But why is this? The function applyfilter should return a numeric result,
shouldn't it?
Best regards,
Ingo
Hi,
May be this helps:
#dat1 is dataset.
with(dat1,tapply(row.names(dat1),list(samp.depth),FUN=head,1))
# 1 2 3
#"1592" "1596" "1600"
row.names(dat1)[tapply(seq_len(nrow(dat1)),list(dat1$samp.depth),FUN=head,1)]
#[1] "1592" "1596" "1600"
#or
row.names(dat1)[!duplicated(dat1$sam
Hi,
May be this helps:
dat1<- read.table(text="
id user action
1 12 login
2 12 view
3 12 view
4 12 view
5 12 login
6 12 view
7 12 view
8 12 login
",sep="",header=TRUE,stringsAsFactors=FALSE)
split(dat1,cumsum(da
Hi,
set.seed(29)
myVector<- rnorm(100)
seq(1,100,by=2)
# [1] 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
#[26] 51 53 55 57 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99
myVector[seq(1,100,by=2)]
rev(myVector)
sum(myVector>0)
#[1] 46
#or
tabl
Hi,
If you wanted the 'fitted' values:
source("cat1.txt")
#dat1 is the dataset
dat2<-dat1$fitted
A.K.
From: catalin roibu
To: arun
Sent: Wednesday, August 28, 2013 8:26 AM
Subject: Re: [R] SPI package error
I have another quest
Oct Nov Dec
#1 1.7191440 1.1226632 1.3777239 0.1114739 -0.51120062
#2 -0.1091491 -0.1324877 -0.1511746 -0.3828513 -0.08032998
A.K.
----- Original Message -
From: arun
To: catalin roibu
Cc: R help
Sent: Wednesday, August 28, 2013 8:31 AM
Subject: Re: [R] SPI package err
ot;Error: very large number of arguments"
I think it would be better to contact the package maintainer regarding this:
maintainer("spi")
#[1] "Josemir Neves "
A.K.
___
From: catalin roibu
To: arun
Sent: Wednesday, August 28, 2013
HI,
'odd' and 'oddVector' are the same.
identical(odd,oddVector)
#[1] TRUE
identical(odd,myVector[seq(1,100,by=2)]) myVector#
#[1] TRUE
A.K.
From: Zsurzsa Laszlo
To: arun
Cc: R help
Sent: Wednesday, August 28, 2013 8:58 A
.15921 0.3515647 0.04890923 0.03665184
# Aug Sep Oct Nov Dec
#1 1.7191440 1.1226632 1.3777239 0.1114739 -0.51120062
#2 -0.1091491 -0.1324877 -0.1511746 -0.3828513 -0.08032998
A.K.
From: catalin roibu
To: arun
Sent: Wednesday,
HI,
I was talking about another method to get the same result:
myVector[1:100%%2==1]
identical(myVector[1:100%%2==1],myVector[seq(1,100,by=2)])
#[1] TRUE
Hope it helps.
A.K.
From: Zsurzsa Laszlo
To: arun
Cc: R help
Sent: Wednesday, August 28, 2013 5:21
Hi,
The question is not clear.
May be this helps:
set.seed(29)
dat1<- as.data.frame(matrix(sample(c(0:7,NA),10*10,replace=TRUE),ncol=10))
which(is.na(dat1))
#[1] 11 23 24 28 37 43 47 72 77 87
#missing value index for each column
sapply(dat1,function(x) which(is.na(x)))
A.K.
how to Keep a
Hi,
set.seed(249)
dat1<- as.data.frame(matrix(sample(1:20,40*20,replace=TRUE),ncol=20))
#Based on your code:
vecNew<-sort(t(as.data.frame(lapply(dat1,sum))),decreasing=TRUE)[1:10]
#or
vec1<-sort(unlist(lapply(dat1,sum),use.names=FALSE),decreasing=TRUE)[1:10]
identical(vec1,vecNew)
#[1] TRUE
#
Hi,
Try:
iris1<-data.frame(iris,annot=c(""),stringsAsFactors=FALSE)
str(iris1)
#'data.frame': 150 obs. of 6 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.Length: num 1.4 1.4 1.3 1.5 1.4 1.7
HI,
May be this helps:
ts1<- ts(1:20)
ts2<- ts(1:25)
ts1[-(1:3)]<- ts1[-(1:3)]+ts2[1:17]
as.numeric(ts1)
# [1] 1 2 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37
A.K.
Hey everyone,
I`m an absolut beginner in R and need some help for an exercise:
I want to do ordinary calculati
HI,
Your code is not clear:
for( i in 1: nrow(D))
{
for( j in 5:ncol(D))
{
D[((D[i,j]/D[i,2])>1.5)]15999)]<-100
## "1.5)]15999)]"
}
^^^
}
D<- structure(list(X. = c(1108L, 1591L,
Hi,
Try:
dat1<- read.table(text="
id event time
1 add 1373502892
2 add 1373502972
3 delete 1373502995
4 view 1373503896
5 add 1373503996
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat1$time_on_task<- c(NA,diff(dat1$time))
dat1
# id event time
3502892 80 A
#3 3 task add 1373502972 23 A
#4 4 sys login 1373502892 80 B
#5 5 list delete 1373502995 901 C
#6 6 list view 1373503896 100 D
#7 7 task add 1373503996 NA A
A.K.
__
Hi,
Try:
dat1<- read.table(text="
V1 V2 V3
2 6 8
4 3 4
1 9 8
",sep="",header=TRUE)
dat2<- read.table(text="
V1 V2 V3
6 8 4
2 0 7
8 1 3
",sep="",header=TRUE)
res1<- as.matrix(dat1-dat2)
res1
# V1 V2 V3
#[1,] -4 -2 4
#[2,] 2 3 -3
#[3,] -7 8 5
res2<-t(t(dat1)-colMeans(dat2))
res2
#
201 - 300 of 3327 matches
Mail list logo
