Hi,
Some speed comparisons:
df <- data.frame(TreeID=rep(1:6000,each=20), Age=rep(seq(1,20,1),6000))
df$Height <- exp(-0.1 + 0.2*df$Age)
df1<- df
df3<-df
library(data.table)
dt1<- data.table(df)
df$HeightGrowth <- NA
system.time({ #Rui's 2nd function
df2 <- data.matrix(df)
for(i in seq_len(nro
ff
# DataComment2 DataComment3
#1 This is comment 2 This is comment 3
#2 This is comment 2 This is comment 3
A.K.
- Original Message -
From: arun
To: Shreya Rawal
Cc: R help
Sent: Monday, June 10, 2013 6:41 PM
Subject: Re: [R] Combining CSV data
Hi,
T
Hi,
It would be better to provide a reproducible example.
set.seed(25)
all_dfs<-
list(df1=data.frame(col1=sample(1:40,20,replace=TRUE),col2=sample(20:40,20,replace=TRUE),col3=
sample(1:3,20,replace=TRUE)),df2=data.frame(col1=sample(30:60,20,replace=TRUE),col2=sample(35:65,20,replace=TRUE),col3=s
HI,
Not sure if this is what you wanted.
mat1<- matrix(c(1, 1, -1, -1, 1, -1, -1, -2, 1, 1, 1, 1), byrow=TRUE, nc=4)
fun1<- function(mat){
matP<- mat
matN<- mat
matP[matP<0]<- NA
matN[matN>0]<- NA
resP<-rowSums(matP,na.rm=TRUE)/ncol(matP)
resN<- rowSums(matN,na.rm=TRU
dd ee ff This is comment 1
# DataComment2 DataComment3
#1 This is comment 2 This is comment 3
#2 This is comment 2
Otherwise, you need to provide an example that matches the real dataset.
A.K.
From: Shreya Rawal
T
Hi,
dataf <- as.data.frame(cbind(Sample,Vals))
str(dataf)
#'data.frame': 20 obs. of 2 variables:
# $ Sample: Factor w/ 2 levels "A","B": 1 1 1 1 1 1 1 1 1 1 ...
# $ Vals : Factor w/ 20 levels "121","154","159",..: 20 12 13 1 2 14 18 5 17
10 ...
ggplot(dataf,aes(x=Vals,colour=Sample))+geom_
Hi,
Try this:
lines1<- readLines(textConnection("gene1 or1|1234 or3|56 or4|793
gene4 or2|347
gene5 or3|23 or7|123456789"))
lines2<-readLines(textConnection(">or1|1234
ATCGGATTCAGG
>or2|347
GAACCTATCAATTTATATAA
>or3|56
ATCGGAGATATAACCAATC
>or3|23
TTAACAAGAGAATAGACAAA
>or4|793
Hi,
Try this:
dat1<- read.table(text="
DEPTH SALINITY DEPTH SALINITY
18 87 39.06 94 39.06
19 173 39.05 141 NA
20 260 39.00 188 39.07
21 312 38.97 207 39.03
22 1 39.36 1 39.35
23 10 39.36 10 39.33
24 20 39.36 20 39.33
25 30
HI,
You could use:
result3<-
data.frame(result2[,-5],read.table(text=as.character(result2$comment),sep="|",fill=TRUE,na.strings=""),stringsAsFactors=FALSE)
colnames(result3)[5:7]<- paste0("DataComment",1:3)
A.K.
____
From: Sh
HI,
May be this helps:
dat1<- read.table(text="
x1 y1 x2 y2 x3 y3 output
2 100 190 99 1430 79 89
2 100 192 63 1431 75 69
2 100 192 63 1444 51 57
3 0 195 99 1499 50 74.5
3 0 198 98 1500 80 89
<- strsplit(x,"\t")[[1]]; x1New<-x1[-1];x2<-
gsub(">(.*)\\n.*","\\1",lines2New);lines3<-lines2New[match(x1New,x2)];write.table(lines3,paste0(x1[1],".txt"),row.names=FALSE,quote=FALSE)})
I didn't had any problems in the output.
It looks like bel
is is comment 2 This is comment 3
#2
A.K.
____
From: Shreya Rawal
To: arun
Cc: R help ; jim holtman
Sent: Wednesday, June 12, 2013 8:58 AM
Subject: Re: [R] Combining CSV data
Great, thanks Arun, but I seem to be running into this error.
0
#8 33 82 870 100 1490 97 93.0
#9 33 0 871 82 1494 85 55.7
A.K.
- Original Message -
From: arun
To: Tom Oates
Cc: R help
Sent: Tuesday, June 11, 2013 5:23 PM
Subject: Re: [R] Add a column to a dataframe based on multiple other column
values
HI,
May be this helps:
Hi,
Try this:
set.seed(25)
dat1<-
data.frame(A=sample(1:30,100,replace=TRUE),B=sample(1:35,100,replace=TRUE),C=sample(1:25,100,replace=TRUE))
dat1$pattern<-with(dat1,ifelse(A>20 & B<=2.5 & C<=20,"Normal",ifelse(A <20 & B
>2.5 & C >20,"Increased",ifelse(A >=20 & B >2.5 & C <=20,"Low","Other"
fun3 452 226 226 452
#res20F3 final3New fun3 284 142 142 284
#res5F3_1 final3New1 fun3 288 144 144 288
#res10F3_1 final3New1 fun3 488 244 244 488
#res20F3_1 final3New1 fun3 310 155 155 310
h
uot;,ifelse(A >=20 & B >2.5 & C <=20,"Low","Other"
identical(dat1,dat1New)
#[1] TRUE
dat2<-transform(dat1,pattern=ifelse(A>20 & B<=2.5 & C<=20,"Normal",ifelse(A <20
& B >2.5 & C >20,"Increased",ifel
Hi,
Not clear about the 'Time' column.
dat1<- read.table(text="
Time Var1 Var2
00:00 1 0
00:01 0 0
00:02 1 0
00:03 1 0
00:04 0 0
00:05 1 0
00:06 1 0
00:07 1 0
00:0
HI,
Try this:
ds1<- data.frame(id,time)
which(with(ds1,ave(time,id,FUN=seq))==1)
#[1] 1 4 9
A.K.
- Original Message -
From: Gallon Li
To: R-help@r-project.org
Cc:
Sent: Thursday, June 13, 2013 3:56 AM
Subject: [R] find the position of first observation for each subject
suppose I have
Hi,
Not sure if the OP is concerned about this:
Using David's data
str(DATA)
#'data.frame': 15 obs. of 2 variables:
# $ UnitName_1: Factor w/ 3 levels "lake","pond",..: 3 3 1 3 2 2 3 1 2 3 ...
#$ Var : int 95 97 12 47 54 86 14 92 88 8 ...
toBeRemoved1 <- which(DATA$UnitName_1=="lake"
Also, if the ids are ordered and numeric:
which(c(1,diff(ds[,1]))>0)
#[1] 1 4 9
A.K.
- Original Message -
From: arun
To: Gallon Li
Cc: R help
Sent: Thursday, June 13, 2013 9:40 AM
Subject: Re: [R] find the position of first observation for each subject
HI,
Try this:
Hi,
You could use:
?grep
grep("ARUN",MyDat[,1])
#[1] 2
#or
library(stringr)
which(!is.na(str_match(MyDat[,1],"ARUN")))
#[1] 2
vec1<-c(MyDat[,1],"ARUN003","Arun")
which(!is.na(str_match(toupper(vec1),"ARUN")))
#[1] 2 10 11
A.K.
within(mtcars,{ x<-rep(0,nrow(mtcars));x[gear==4]<-1;x[gear==3]<-2}) #should
fix this;
A.K.
- Original Message -
From: Ista Zahn
To: Bert Gunter
Cc: r-help@r-project.org
Sent: Thursday, June 13, 2013 12:17 PM
Subject: Re: [R] puzzling behavior of within
On Thu, Jun 13, 2013 at 12:11
HI,
Please check this link
(http://r.789695.n4.nabble.com/Remove-levels-td4669441.html). It was posted
today.
df1<-subset(df, Trt!="blank")
df2<- df1
df1$Trt<- factor(df1$Trt)
levels(df1$Trt)
#[1] "Trt1" "Trt2"
#or
df2$Trt<-droplevels(df2$Trt)
levels(df2$Trt)
#[1] "Trt1" "Trt2"
A.K.
Hey
t;-which.max(c(1,diff(index(x)[indx])));index(x)[indx1]
})
lst2[1:3]
#$`2012-12-01`
#[1] "2012-12-01 06:12:00 EST"
#
#$`2012-12-02`
#[1] "2012-12-02 00:40:00 EST"
#
#$`2012-12-03`
#[1] "2012-12-03 04:21:00 EST"
A.K.
From: Ye
0
A.K.
- Original Message -
From: arun
To: Ye Lin
Cc:
Sent: Thursday, June 13, 2013 10:04 PM
Subject: Re: [R] identify data points by certain criteria
"""
e.g. 12/31, the time period where sum(Var1+Var2)=0 are:
from 00:00:00 to 00:55:00 (55 min)
from 01:07:00
HI,
May be this helps:
dat1<- read.table(text="
Col1 Col2 Weight
A D 0.1
B C 0.4
C M 0.6
D P 0.8
E W 1
F D 1.2
G C 3.1
H M 4
",sep="",header=TRUE,stringsAsFactors=FALSE)
vec1<- c(unique(dat1[,1]),unique(dat1[,2]))
datNew<- expand.grid(vec1,vec1)
colnames(datNew)<- colnames(dat1)
Hi,
rename_columns<- function(dat){
for(i in 2:(ncol(dat))){
names(dat)[i]<- paste(names(dat)[1],names(dat)[i],sep="_")
}
dat
}
dat1<- read.table(text="
chr pos ref alt
chr1 5 A G
chr1 8 T C
chr2 2 C T
",sep="",header=TRUE,stringsAsFactors=FALSE
HI,
Try:
dat1<-dat[sapply(dat,function(x) length(unique(x)))>1]
cor(dat1)
# ABC DEF JKL MNO
#ABC 1.000 -0.75600764 0.55245223 -0.2735585
#DEF -0.7560076 1. -0.06479082 0.2020781
#JKL 0.5524522 -0.06479082 1. 0.4564568
#MNO -0.2735585 0.2
.06479082 0.2020781
#JKL 0.5524522 -0.06479082 1. 0.4564568
#MNO -0.2735585 0.20207810 0.45645683 1.000
A.K.
- Original Message -----
From: arun
To: Katherine Gobin
Cc: R help
Sent: Friday, June 14, 2013 10:34 AM
Subject: Re: [R] Removing "NA" from matrix
==1))
#[1] 273
nrow(subset(res5F3High,dummy==0))
#[1] 273
nrow(subset(res10F3,dummy==0))
#[1] 231
nrow(subset(res10F3,dummy==1))
#[1] 231
nrow(subset(res10F3_1,dummy==1))
#[1] 237
nrow(subset(res10F3_1,dummy==0))
#[1] 237
dim(unique(res5F3))
#[1] 278 5
dim(unique(res5F3High))
#[1] 546
Hi,
sapply(colnames(m)[-ncol(m)],function(i) {x1<-cbind(m[,i],m[,ncol(m)]);
length(which(x1[,1]!=0 & x1[,2]!=0))})
#A B C
#2 1 2
A.K.
- Original Message -
From: Hermann Norpois
To: r-help
Cc:
Sent: Friday, June 14, 2013 3:51 PM
Subject: [R] combination of columns in a matrix
Hello
Hi,
You could use ?twoord.plot() from library(plotrix):
library(plotrix)
dat2<- data.frame(month,music,actor)
dat2$month<- factor(month,labels=c(month.abb,"Pag"))
dat2New<-dat2[order(dat2$month),]
with(dat2New,twoord.plot(month,music,actor,
lylim=c(0,1),rylim=c(0,13000)
Hi,
You could try this:
#In one of the list element
SampleSummary$ConcLow
# Min. 1st Qu. Median Mean 3rd Qu. Max. NA's
# 0.4200 0.6100 0.7300 0.7373 0.8150 1.6600 1
#additional entry for NA. You may need to check the data and remove the
missing value. Here, I removed t
Hi,
May be this helps:
dat1<- read.table(text="
pbnr dat dep dys sop ago mis age female
1 10023 1994-02-21 0.75 1.00 0.50 0.50 0.75 35 1
2 10023 1994-05-25 0.75 1.00 0.50 0.50 0.75 35 1
3 10028 1994-02-01 2.00 1.75 3.00 0.50 1.50 42 1
4 10028 1999-01-15 1.25 0.75 2.25
Hi,
It is better to ?dput() the dummy dataset.
I made up some data for female==0
dat1<- read.table(text="
pbnr dat dep dys sop ago mis age female messpunkt2 messpunkt1 tage
eintrittsjahr
1 10023 1994-02-21 0.75 1.00 0.50 0.50 0.75 35 1 8817 8817 0
1994
2 10023
HI,
One way would be:
library(stringr)
tmpstr = "The first number is: 32. Another one is: 32.1.
Here's a number in scientific format, 0.3523e10, and
another, 0.3523e-10, and a negative, -313.1"
pattern<-
"(\\d)+|(\\d+\\.\\d+)|(-\\d+\\.\\d+)|(\\d+.\\d+e\\d+)|(\\d+\\.\\d+e-\\d+)"
str_extract_all
et(res5PercentLow,dummy==1))
#[1] 155
res20PercentHigh<- fun1New(final3New1,0.20,5)
dim(res20PercentHigh)
#[1] 846 5
dim(unique(res20PercentHigh))
#[1] 846 5
nrow(subset(res20PercentHigh,dummy==0))
#[1] 423
nrow(subset(res20PercentHigh,dummy==1))
#[1] 423
A.K.
- Ori
Hi,
May be this helps.
set.seed(29)
GLSS<- data.frame(hhid=sample(1:10,30,replace=TRUE),value=rnorm(30))
IDstoSelect<- c(1,5,7)
GLSS[!is.na(match(GLSS[,1],IDstoSelect)),]
# hhid value
#1 1 -0.34216061
#6 1 0.28240263
#14 7 0.04197496
#17 7 0.15005878
#21 7 -1.18640939
.07374,36.64006,48.15659,61.61155,77.95515,100.152)
dat1<- data.frame(X,Y)
dat2<-dat1[order(dat1$Y,dat1$X),]
dat2$X<-factor(dat2$X,levels=dat2$X)
plot(as.numeric(dat2$X),dat2$Y,xaxt="n",xlab="X",ylab="Y",type="b")
axis(1,at=as.numeric(dat2$X),labels=lev
Comment 2, Comment 3 Comment 1 Comment 2 Comment 3
#2 Comment 1a, Comment 2a Comment 1a Comment 2a
A.K.
From: Shreya Rawal
To: arun
Cc: R mailing list ; jim holtman
Sent: Monday, June 17, 2013 9:55 AM
Subject: Re: [R] Combining CSV data
Hi Arun,
Sorry to bother y
May be this helps:
d[cumsum(sapply(d,function(x) any(x>b)))>=1]
#[1] 12 24 36
A.K.
- Original Message -
From: Andras Farkas
To: r-help@r-project.org
Cc:
Sent: Monday, June 17, 2013 2:29 PM
Subject: [R] vector question
Dear All,
would you please help with the following:
let us say I h
HI Andras,
Sorry, I misunderstood your question:
Try this:
sapply(sapply(b,function(x) d[x
To: arun
Cc:
Sent: Monday, June 17, 2013 2:48 PM
Subject: Re: [R] vector question
Arun,
thank you. Looking at it I am wondering if there is a way to get 36 in the
result instead of the 24. The 2nd
Hi,
May be this helps:
dat1<-
data.frame(Ans=c("Friday","Wednesday","Friday,Saturday,Sunday","Saturday","Sat,Sun","Th,F,Sa"),stringsAsFactors=FALSE)
dat1
Ans
1 Friday
2 Wednesday
3 Friday,Saturday,Sunday
4 Saturday
5
F Sa
# 2 0 0 1 1 3 4
#Or Using Bills' solution:
dayNames[pmatch(vec2New,dayNames,duplicates.ok=TRUE)]
# [1] "Friday" "Wednesday" "Friday" "Saturday" "Sunday" "Saturday"
#[7] "Saturday" "Sunday"
Hi,
You didn't provide any information about the package. I guess it is from
"energy".
library(energy)
x <- iris[1:50, 1:4] #examples given in the package
y <- iris[51:100, 1:4]
z<- iris[101:150,1:4]
vec1<-c("x","y","z")
mat1<- combn(vec1,2)
sapply(split(mat1,col(mat1)),function(.dat) dc
gsub("_.*","",myInput)
#[1] "afg3" "alg12" "dbp3" "elp4" "fob1" "gpr1"
A.K.
- Original Message -
From: CompBiol ATL
To: r-help@r-project.org
Cc: Hong Qin
Sent: Monday, June 17, 2013 6:52 PM
Subject: [R] help with string split in R
#I want to remove '_alpha' in a vector of strings
Hi,
Perhaps this works:
a[sapply(b,function(x) which.min(abs(x-a)))]
#[1] 8 32
A.K.
- Original Message -
From: Andras Farkas
To: r-help@r-project.org
Cc:
Sent: Tuesday, June 18, 2013 8:34 AM
Subject: [R] find closest value in a vector based on another vector values
Dear All,
would y
Hi,
Possibly, R FAQ: 7.31
http://www.hep.by/gnu/r-patched/r-faq/R-FAQ_82.html
(1.1%%1)-0.1
#[1] 8.326673e-17
fun1 <- function(x, y, eps = .Machine$double.eps^0.5) abs(x - y) < eps
fun1(1.1%%1,0.1)
#[1] TRUE
A.K.
0.1%%1==0.1
returns TRUE, which is right
But
1.1%%1==0.1
returns FALSE !!
Hi,
May be this helps:
set.seed(24)
dat1<- data.frame(ID=1:200,value=sample(c(5:200,NA),200,replace=TRUE))
which(is.na(dat1$value))
#[1] 56 146 184
sum(which(is.na(dat1$value))) #Not clear about the 2nd part of the question
#[1] 386
sum(is.na(dat1$value))
#[1] 3
table(is.na(dat1$value))
#FALSE
Hi Benjamin,
Thanks for pointing that out.
Sorry, didn't read the question carefully.
(0.1%%1)-0.1
#[1] 0
(1.1%%1)-0.1
#[1] 8.326673e-17
(1%%1.1)
#[1] 1
A.K.
From: Benjamin
To: arun
Sent: Tuesday, June 18, 2013 11:33 AM
Subject: Re: [R] Problem
Hi,
You could also use:
dat1<- data.frame(Vec1,Vec2,Vec3)
Reduce(paste0,lapply(seq_len(ncol(dat1)),function(i) {x1<-
as.character(factor(dat1[,i],labels=c("",LETTERS[i])))}))
# [1] "ABC" "ABC" "" "AC" "AB" "ABC" "A" "B" "ABC" "AC" "B" "A"
#[13] "AB" "C" "B"
#or
library(plyr)
lib
Hi,
May be this helps:
set.seed(24)
B101<-
data.frame(SID=sample(400:450,20,replace=FALSE),zGRADE=sample(0:10,20,replace=TRUE),col3=rnorm(20))
set.seed(28)
B2A<-
data.frame(SID=sample(400:600,50,replace=FALSE),zGRADE=sample(0:10,50,replace=TRUE),col3=rnorm(50))
set.seed(35)
B2B<-
data.frame(SID=
rror in cor(A, D) : incompatible dimensions
A.K.
____
From: Amanda Li
To: arun
Sent: Wednesday, June 19, 2013 1:11 AM
Subject: Re: [R] write a function to do pairwise calculation
Hello,
Thanks for your help! I am a novice in R, and may I ask how I am suppose
Hi,
May be this helps:
set.seed(28)
x<- sample(1:40,20,replace=TRUE)
qx<-quantile(x,probs=0.10)
qx
#10%
#3.8
qx+1
#10%
#4.8
attr(qx,"names")<-NULL
qx
#[1] 3.8
qx+1
#[1] 4.8
A.K.
- Original Message -
From: Francesco Miranda
To: "r-help@r-project.org"
Cc:
Sent: Wednesday, J
s,dat1$ID))
#[1] 2
A.K.
Hi, Unfortunately somehow it won't help. The unused values are not NA,
the unused values are simply not there. Since these are student Ids, for
instance there is no 4,8,9 etc... I need to find out which of these are
not there.
- Original Message -
From:
Hi,
Please use ?dput().
If `lst1` is the list.
do.call(rbind,lapply(lst1,t))
# Estimate Std. Error t value Pr(>|t|)
#card1 0.000577912 0.003956905 0.1460515 0.88388493
#card2 0.005164347 0.003311546 1.5594972 0.11892405
#card3 0.002682773 0.003683422 0.7283372 0.46643165
#card
856 0.25225436
#resp1 0.005552192 0.007135446 0.7781142 0.43652806
#resp2 0.010337814 0.004848899 2.1319923 0.03304227
#resp3 0.002850401 0.006422570 0.4438100 0.65719368
#respir 0.006212050 0.004121313 1.5072989 0.13177931
A.K.
- Original Message -
From: arun
To: R help
Cc:
Sent: Wednes
Hi,
May be this helps:
library(plyr)
ddply(df,.(Country,Iso),summarize,NbrProj=length(Project))
# Country Iso NbrProj
#1 Burkina Faso BF 2
#2 Ethiopia ET 1
#3 Ghana GH 1
#4 Kenya KE 2
#5 Madagascar MG 2
#6 Mali ML 1
LouisTalbot
137357 115 NoahStorey
137357 92 GraceChapman
etc...
The exact quesiton is the following:
Some ids in the range 1 to 161 are unused (e.g. 4,7,9). In fact,
there are 30 unused pupil Ids between 1 and 161. What is the sum of
these 30 integers?
- Original Message -
Hi,
May bet this also helps:
set.seed(12345)
x <- c(1:30)
x1 <- sample(x, 15, replace = TRUE)
x2 <- sample(x, 15, replace = TRUE)
x3 <- sample(x, 15, replace = TRUE)
x4 <- sample(x, 15, replace = TRUE)
indx<-1+2*(x1==1)+4*(x2==1)+8*(x3==1)+16*(x4==1)
as.numeric(indx!=1)
#[1] 0 1 0 0 0 0 0 0 0 0 0
,"mid","late","wknd"))
with(dat1,tapply(res1,list(Name),FUN=sum))
#early mid late wknd
# 0 1 4 6
#or
sapply(split(res1,names(vec3)),sum)
#early late mid wknd
# 0 4 1 6
A.K.
- Original Message -
From: "Crombie
Hi,
May be this helps:
xyz<-function(x,y){
z<-x+y
print(paste("the sum of", deparse(substitute(x)), "and",
deparse(substitute(y)),"is", sep=" "))
z
}
xyz(a,b)
[1] "the sum of a and b is"
[,1] [,2] [,3] [,4] [,5]
[1,] -1.38805146 -1.15706888 -1.0942
HI,
You could save it in a .txt file. (e.g. "vec1.txt")
source("vec1.txt")
length(drg_470)
[1] 4495
head(drg_470)
#[1] 7558 9485 10481 10609 10650 10891
str(drg_470)
# num [1:4495] 7558 9485 10481 10609 10650 ...
A.K.
Hi there,
I am having trouble successfully reading this vector into R:
Hi,
Not sure if you wanted the entries with "0".
library(reshape2)
dfMelt<-melt(df,id.var=c("Country","Iso"))
#subset those with "1"
dfNew<- subset(dfMelt,value==1,select=-4) row.names(dfNew)<- 1:nrow(dfNew)
dfNew
# Country Iso variable
#1 Zimbabwe ZW Abaco
#2 South Afri
Hi,
dat1<- read.table(text="
name number
a 2
a 3
b 5
b 7
c 9
c 1
",sep="",header=TRUE,stringsAsFactors=FALSE)
aggregate(number~name,data=dat1,sum)
# name number
#1 a 5
#2 b 12
#3 c 10
#or
library(plyr)
HI,
You could also use:
beta_results[!is.na(match(beta_results[,1],instru)),]
# instrument beta_values
#3 JKL 0.529
#6 STU -1.080
#8 XYZ 0.420
#If there are no duplicates, this could also work
beta_results[match(instru,beta_results[,1]),]
# instrument bet
Hi,
Try:
dat1<- read.csv("mydata.csv",sep="\t",row.names=1) #sep could be different in
your case.
dat1
# A B C D E
#Jhon 1 3 6 9 8
#Kelly 2 3 4 6 9
#Audrey 3 5 6 9 7
A.K.
Hi,
I am calling my csv data with 37 columns and 16 rows into R. However I really
need to keep coloumns name
HI,
May be this also works:
plot(x,col=as.numeric(factor(x)))
A.K.
- Original Message -
From: Rui Barradas
To: Suparna Mitra
Cc: r-help@r-project.org
Sent: Friday, June 21, 2013 11:32 AM
Subject: Re: [R] How to define desired numbers to a vector based on the
present numbers
Hello,
Hi,
May be this helps:
###Added more lines of fake data
fn_hp<- read.table(text="
id date wl_m wet cuml_day
585 fn 2012-03-03 0.1527048 1 1
586 fn 2012-03-04 0.2121408 1 2
587 fn 2012-03-05 0.1877568 1 3
Hi,
May be this helps:
set.seed(24)
A <- matrix(sample(c(NA,rnorm(1500)),1500,replace=FALSE),nrow=500)
par(mfrow=c(ncol(A),1))
lapply(split(A,col(A)),function(x) acf(na.omit(x)))
A.K.
- Original Message -
From: Zilefac Elvis
To: "r-help@r-project.org"
Cc:
Sent: Sunday, June 23, 2013
HI,
dat1<- read.csv("AMS3.csv",sep="\t",row.names=1)
pdf("AMS3_acf.pdf")
lapply(dat1,function(x)acf(na.omit(x)))
dev.off()
A.K.
____
From: Zilefac Elvis
To: arun
Cc: R help
Sent: Sunday, June 23, 2013 1:51 PM
Subject:
HI,
May be this helps:
dat1<- read.table(text="
beforesmall aftersmall
63.5 512.0
54.5 237.5
52.5 161.5
78.0 153.5
53.5 68.0
50.5 65.5
69.0 52.0
76.0 59.0
68.0 66.5
75.5 66.5
67.0 45.5
81.0 54.5
49.0 44.0
51.0 42.5
53.0
Hi,
Try this:
res1<-do.call(cbind,t(sapply(res,c))[,1]) #get the acf values only. Assuming
that is what you wanted.
write.csv(res1,"AMSacf.csv",row.names=FALSE)
A.K.
From: Zilefac Elvis
To: arun
Sent: Monday, June 24, 2013 1:00 PM
Su
f1<- function(x){env<- parent.frame();env$a<-5; g(x)}
f1(2)
#[1] 7
f1(7)
#[1] 12
f1(5)
#[1] 10
A.K.
From: David Kulp
To: "r-help@r-project.org"
Sent: Monday, June 24, 2013 4:27 PM
Subject: [R] Lexical scoping is not what I expect
According to
http://cr
HI,
Try this:
merge(data,data1["NAME"])
A.K.
- Original Message -
From: Nico Met
To: Rui Barradas
Cc: R help
Sent: Tuesday, June 25, 2013 6:00 AM
Subject: Re: [R] Fetch and merge from a merge(dat1["NAME"],dat2)
data set
Many thanks Rui,
However If I want to extract only first colum
May be this also works:
exp(mean(log(abs(x
#[1] 0.4985282
A.K.
- Original Message -
From: PIKAL Petr
To: Shane Carey
Cc: "r-help@r-project.org"
Sent: Tuesday, June 25, 2013 8:50 AM
Subject: Re: [R] Calculate geometric mean with tapply
Hm
> -Original Message-
> From: r
dta[,3]<-as.numeric(as.character(gsub(",",".",dta[,3])))
library(reshape2)
dcast(dta,id~name,value.var="value")
# id AreaCycl AreaEmer AreaMax HeightMax nrfilledseed
#1 TN00016-003A -1.3998944 -3.2771509 -3281.077 8.434493 1.825
#2 TN00016-014A -0.5843794 -0.8428928 -
Hi,
Please check this link:
http://stackoverflow.com/questions/4835202/error-with-custom-aggregate-function-for-a-cast-call-in-r-reshape2
A.K.
- Original Message -
From: Johannes Radinger
To: R help
Cc:
Sent: Wednesday, June 26, 2013 4:36 AM
Subject: [R] Error when using median as ag
Hi,
Not sure whether there is any pattern in adding the small number.
dat1<- read.table(text="
X Y Z
15 20 30
15 20 40
",sep="",header=TRUE)
dat1[,1:2]<
Hi,
Try:
roweqv<- function(m,v)
which(!is.na(match(interaction(as.data.frame(m),drop=TRUE),paste(v,collapse="."
v<- c(2,5,8)
roweqv(m,v)
#[1] 2
set.seed(24)
m1<-matrix(sample(1:15,3e5,replace=TRUE),ncol=3)
v1<- c(10,12,4)
system.time(res<- roweqv(m1,v1))
# user system elapsed
#0.132
0.312 1.076
identical(res,res2)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: Sachinthaka Abeywardana
Cc: R help
Sent: Wednesday, June 26, 2013 3:26 PM
Subject: Re: [R] match rows of R
Hi,
Try:
roweqv<- function(m,v)
which(!is.na(match(interaction(as.data.fr
e"
Data1[,1]
#[1] Residential-future Open space-managed Mixed use
#[4] Residential-existing Residential-existing Residential-future
#[7] Residential-existing Mixed use
#4 Levels: Mixed use Open space-managed ... Residential-future
A.K.
Hi,
May be this helps:
#You can use ?read.csv() to read the two files.
set.seed(24)
dat1<- as.data.frame(matrix(sample(20:40,40,replace=TRUE),ncol=4))
set.seed(285)
dat2<- as.data.frame(matrix(sample(35:60,40,replace=TRUE),ncol=4))
sapply(colnames(dat1),function(i) t.test(dat1[,i],dat2[,i],paired=
sv("file1.csv")
> data2 <-read.csv("file2.csv")
> dat1<- as.data.frame(matrix(data1))
> dat2<- as.data.frame(matrix(data2))
> sapply(colnames(dat1),function(i)
> t.test(dat1[,i],dat2[,i],paired=TRUE)$p.value)
As far as I can see all my values are
Hi,
set.seed(24)
dat1<-data.frame(X=sample(letters,20,replace=TRUE),Y=sample(1:40,20,replace=TRUE),stringsAsFactors=FALSE)
mat1<-as.matrix(dat1)
sapply(dat1,class)
# X Y
#"character" "integer"
sapply(split(mat1,col(mat1)),class)
# 1 2
#"character" "chara
Hi,Without any reproducible example or code, it is difficult to understand the
problem. Did you load the package?
Using the example in ?LSD.test()
library(agricolae)
data(sweetpotato)
model<-aov(yield~virus, data=sweetpotato)
out <- LSD.test(model,"virus", p.adj="bonferroni",
ma
Hi,
Try:
library(zoo)
rollapply(t(mymatrix),width=12,FUN=sum,by.column=T,fill=NA,partial=FALSE,align="left")
# [,1] [,2] [,3] [,4] [,5]
#[1,] 342 354 366 378 390
#[2,] 402 414 426 438 450
#[3,] 462 474 486 498 510
#[4,] 522 534 546 558 570
#[5,] 582 594 606 618 6
Hi,
May be this also helps:
library(data.table)
dt1<- data.table(dat)
dt1[,cor(x,y),by=g]
# g V1
#1: A -0.05643063
#2: B 0.16465040
dt1[,cor(x,y),by=g]$V1
#[1] -0.05643063 0.16465040
A.K.
- Original Message -
From: David Carlson
To: 'Jannis' ; 'r-help'
Cc:
Sent: Thursday,
Hi,
May be this helps:
res1<-as.matrix(transform(expand.grid(k1=0.3*(1:10),k2=1:10),k3=k1*k2))
A.K.
Dear all,
I'm trying to generate a matrix dataset following 2 loops analysis, such as the
follows:
for (i in 1:10) {
k1 <- 0.3 x i
for (j in 1:10) {
k2 <- j
k3 <- k1*k2
}}
Hi,
set.seed(45)
testframe <- data.frame(a=rnorm(100), b = rnorm(100))
indices <- rep(c(1,2), each = 50)
library(plyr)
ddply(testframe,.(indices),summarize, Cor1=cor(a,b))
# indices Cor1
#1 1 0.002770524
#2 2 -0.10173
A.K.
- Original Message -
From: J
-04
A.K.
thanks very much, you're help is much appreciated.
Just another small question, what's the best way to deal with missing data? If
i want to do a paired t-test?
- Original Message -
From: arun
To: R help
Cc:
Sent: Thursday, June 27, 2013 1:47 PM
Subject: Re: mul
HI,
May be this helps:
dat1<- read.table(text="
date1 time date timeSec topic pupilId correct
02/01/2013 14:58 02/01/2013 140323 fdp.fdp 40 TRUE
02/01/2013 14:59 02/01/2013 140372 fdp.fdp 150 TRUE
03/01/2013 11:23 03/01/2013 213833 fdp.percentage_calc_foundation 15 TRUE
03/0
HI,
Not sure why df2 didn't include Cat3, Debut3, Fin3, instead included columns
with all missing values (Cat5, Debut5, Fin5).
library(plyr)
library(reshape2)
df1New<-ddply(df1,.(Mat),transform,castCat=paste0("Cat",Cat),castDébut=paste0("Début",Cat),castFin=paste0("Fin",Cat))
res<-
join_all(list
Hi,
One problem with your example dataset is that you have 11 list
elements with the last one having different dimensions. I think it was
based on `summarized`.
Then, the second problem I encountered for testing is that the dataset you
provided doesn't fulfill the criteria.
lst1<- list(structur
Hi,
Not sure how your data looks like. May be this helps.
dat1<- read.table(text="
col1
AA-50
AT-34
TT-57
TT-45
TA-42
",sep="",header=TRUE,stringsAsFactors=FALSE)
vec1<-gsub("\\-.*","",dat1[,1])
vec2<- ifelse(vec1=="AA",-1,ifelse(vec1=="AT",0, ifelse(vec1=="TT",1,NA)))
library(stringr)
abs(vec2
Hi,
May be this helps:
Kgeno<- read.table(text="
SNP_ID SNP1 SNP2 SNP3 SNP4
Maj_Allele C G C A
Min_Allele T A T G
ID1 CC GG CT AA
ID2 CC GG CC AA
ID3 CC GG nc AA
ID4 _ _ _ _
ID5 CC GG CC AA
ID6 CC GG CC AA
ID7 CC GG CT AA
ID8 _ _ _ _
ID9 CT
5
A.K.
- Original Message -
From: A M Lavezzi
To: r-help
Cc:
Sent: Tuesday, July 2, 2013 10:22 AM
Subject: Re: [R] matching similar character strings
Dear Arun,
please excuse me for this late reply, we had to stop working on this
temporaririly.
Let me reproduce here tw
sign(a)*abs(a)^(1/3)
# [,1]
#[1,] -1.00
#[2,] -1.259921
#[3,] -1.442250
A.K.
- Original Message -
From: Sachinthaka Abeywardana
To: "r-help@r-project.org"
Cc:
Sent: Tuesday, July 2, 2013 11:11 PM
Subject: [R] nth root of matrix
Hi all,
I want to do the following:
a=
Hi,
I am not sure how you describe combine.
Try this:
df1<-subset(df, !is.na(var1) &!is.na(var2))
df1$new<-paste0(df1$var1,df1$var2)
> head(df1)
# var1 var2 new
#1 b a ba
#2 c b cb
#3 b b bb
#5 a a aa
#6 b b bb
#7 a b ab
A.K.
- Original Message
Hi,
You can also use:
within(small,{order(value)})
# file.name value
#1 /storage/storage0/59Off.Rtable 0.2203863
#3 /storage/storage0/5912314ff.Rtable 0.2886594
#2 /storage/storage0/5912On.Rtable 0.4052370
A.K.
- Original Message -
From: Berend Has
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