ing columns
where b is the slope from a regression V1 (from datset 1) on V1 (dataset 2) and
Var(V1) the variance from V1(from dataset2).
So what I'm looking for is something like a loop function...
- Original Message -
From: arun
To: R help
Cc:
Sent: Thursday, August 29, 2
amp;idp=46
#7 7 task add 1373503996 NA F http://post/add?id=31&idp=45
A.K.
____
From: srecko joksimovic
To: arun
Sent: Thursday, August 29, 2013 5:38 PM
Subject: Re: [R] Add new calculated column to data frame
Hi Arun,
I really
100 D http://
#7 7 task add 1373503996 NA A http://post/add?id=35&idp=99
A.K.
From: srecko joksimovic
To: arun
Sent: Thursday, August 29, 2013 6:04 PM
Subject: Re: [R] Add new calculated column to data frame
"Did
Hi,
You could try:
res<-colMeans(aperm(moo,c(2,1,3)))
resOld<-apply(moo,c(1,3),mean)
identical(res,resOld)
#[1] TRUE
#Speed:
set.seed(285)
moo1<- array(runif(1400*9*15),dim=c(1400,9,15))
system.time({res1<- colMeans(aperm(moo1,c(2,1,3)))})
#user system elapsed
# 0.004 0.000 0.002
syste
Hi,
I am getting the same error with R 3.0.1
SpectrumSearch(y, sigma=3.0, threshold=1.0, background=TRUE, iterations=13,
markov=FALSE, window=3)
#Error in .Call("R_SpectrumSearchHighRes", as.vector(y), as.numeric(sigma), :
# "R_SpectrumSearchHighRes" not available for .Call() for package "Pea
Hi,
You can use 'fill=0' in ?cast()
merge(dat1,cast(melt(dat2, c("ID", "Type")), ID~Type,fill=0))
# ID Name Management Training
#1 1 Jack 1 3
#2 2 John 1 0
#3 3 Jill 0 4
A.K.
- Original Message -
From: PIKAL Petr
To: Mat ; "r-help@r-p
HI,
Also,
dd1<-matrix(cbind(D[,1],(D[-c(1:2)]/D[,2]>4)*1),dimnames=NULL,ncol=7)
identical(dd,dd1)
#[1] TRUE
A.K.
- Original Message -
From: Jose Iparraguirre
To: Mª Teresa Martinez Soriano ;
"r-help@r-project.org"
Cc:
Sent: Friday, August 30, 2013 5:39 AM
Subject: Re: [R] Outlie
Hi,
Better would be to show a reproducible example using ?dput() and the codes you
used. Assuming that you tried something like this:
lst1<- list(1:10,c(5,4,3),4:15)
mean(lst1)
#[1] NA
#Warning message:
#In mean.default(lst1) : argument is not numeric or logical: returning NA
sapply(lst1,mean)
Hi Ramón,
May be this helps:
tags_totals<-matrix(c(15,11,23,7,5),ncol=1,dimnames=list(c("Wikis","Glosarios","Grupos","Bases
de datos","Taller"),NULL))
tags_totals[order(tags_totals[,1],decreasing=TRUE),,drop=FALSE]
# [,1]
#Grupos 23
#Wikis 15
#Glosarios 1
HI,
You could also parse the data by:
input1<- input
library(stringr)
input2<-str_trim(gsub("[=+]","",input1))
dat1<-read.table(text=word(input2[!grepl("---",input2)& input2!="" &
!grepl("RAM|MiB",input2)],8,15),sep="",header=FALSE,stringsAsFactors=FALSE)
lst1<-split(dat1,cumsum(dat1$V3=="uuidd")
Hi,
Using the same datasets:
dat1<- read.table(text="
V1 V2 V3
2 6 8
4 3 4
1 9 8
",sep="",header=TRUE)
dat2<- read.table(text="
V1 V2 V3
6 8 4
2 0 7
8 1 3
",sep="",header=TRUE)
sapply(seq_len(ncol(dat1)),function(i) cor(dat1[,i],dat2[,i],method="spearman"))
#[1] -1.000 0.5
Hi,
You may try:
x1<- c("4/25/71","4/20/64")
fun1<- function(x, year){
m1<-as.numeric(format(as.Date(x1,"%m/%d/%y"),"%y"))
m1<- ifelse(m1>year%%100,1900+m1,2000+m1)
m2<- paste0(gsub("(.*\\/).*$","\\1",x),m1)
as.Date(m2,"%m/%d/%Y")
}
fun1(x1,1950)
#[1] "1971-04-25" "1964-04-20"
str(fun1(x1,1950)
Hi,
In case, some of the list elements are vectors, this procedure would not work.
a1<- list(c(3,5),4,5:6)
as.numeric(a1)
#Error: (list) object cannot be coerced to type 'double'
The OP didn't provide any info as to how the data looks like. So, these are
just assumptions.
mean(unlist(a1)) #if
Hi,
May be this helps:
rle(try)$values
#[1] 1 2 3 1 2 4
#or
aggregate(try,list(cumsum(c(1,abs(diff(try),unique)[,2]
#[1] 1 2 3 1 2 4
#or
res<-tapply(try,cumsum(c(1,abs(diff(try,head,1)
attr(res,"dimnames")<-NULL
res
#[1] 1 2 3 1 2 4
A.K.
I am trying to delete repeated values in an
HI,
May be this helps:
library(Ckmeans.1d.dp)
set.seed(24)
RanNum1<-runif(20,9,12)
Ck1<-Ckmeans.1d.dp(RanNum1,4)
plot(RanNum1,col=Ck1$cluster)
abline(h=Ck1$centers,col=1:4,pch=8,cex=2)
Ck1$cluster
# [1] 2 2 3 3 3 4 2 4 4 2 3 2 3 3 2 4 2 1 3 1
which(Ck1$cluster==2,arr.ind=T)
#[1] 1 2 7
Hi,
You forgot a closing parentheses ")".
x<-matrix(cells,nrow=7,ncol=2,byrow=TRUE,dimnames=list(rnames,cnames) #
#run K-Means
km <-kmeans(x,4,10)
#Error: unexpected symbol in:
#"#run K-Means
#km"
x<-matrix(cells,nrow=7,ncol=2,byrow=TRUE,dimnames=list(rnames,cnames) )
km<-kmeans(x,4,10)
Hi,
May be this helps:
x<- 1:10
set.seed(28)
y1<- rnorm(10)
set.seed(485)
y2<- rnorm(10)
plot(x,y1,col="red",type="b",ylab="y1:y2")
lines(y2,col="blue",type="b")
legend("topleft", legend = c("y1", "y2"),text.col=c("red","blue"))
#or
library(ggplot2)
dat1<- data.frame(x,y1,y2)
ggplot(dat1,ae
HI,
You could try this:
dat1<- read.table(text="
A B C D
A 1 2 3 4
E 5 6 7 8
F 9 10 11 12
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat1$ID<-row.names(dat1)
library(reshape2)
dat1New<-melt(dat1,id.var="ID")
dat2<- data.frame(expand.grid(ID=LETTERS[1:6],variable=LETTERS[1:6]),value=0)
0 0 0 0 0 0
#D 0 0 0 0 0 0
#E 5 6 7 8 0 0
#F 9 10 11 12 0 0
A.K.
- Original Message -
From: arun
To: R help
Cc:
Sent: Saturday, August 31, 2013 9:25 PM
Subject: Re: Insert null columns and rows into a matrix to make it square
HI,
You could try this:
dat1<- read.table(tex
Hi Arnaud,
No problem.
Try,
x<- 1:10
set.seed(28)
y1<- rnorm(10)
set.seed(485)
y2<- rnorm(10,25)
library(plotrix)
twoord.plot(x,y1,y2,lylim=c(-2,2),rylim=c(20,30),ylab="y1",rylab="y2",lcol=2,rcol=4,main="y1,
y2 vs. x")
A.K.
- Original Message
HI,
coma<- read.table(text="
gr T C M B
arcor 6 4 6 5
corfo 24 21 23 24
corma 25 15 26 17
crust 3 2 6 5
fil 15 12 15 15
fol 11 9 6 8
leat 10 11 13 13
seag 2 2 2 2
",sep="",header=TRUE)
coma.x<- as.matrix(coma)
str(coma.x)
# chr [1:8, 1:5] "arcor" "corfo" "corma" "crust" "fil" "fol" ...
# - attr(
HI,
It is better to use ?dput() to show the reproducible example.
dput(korma)
structure(list(T = c(0L, 0L, 1L, 1L, 0L, 1L), C = c(0L, 1L, 1L,
0L, 0L, 0L), M = c(0L, 0L, 0L, 1L, 0L, 1L), B = c(1L, 0L, 1L,
1L, 1L, 1L)), .Names = c("T", "C", "M", "B"), class = "data.frame", row.names =
c("1",
HI,
It is better to provide a reproducible example. From the nature of the error,
it looks like the problem is similar to the one in the link below:
http://r.789695.n4.nabble.com/mlogit-error-td4663601.html
A.K.
- Original Message -
From: "Haghpanahan, Houra"
To: "'r-help@r-projec
Hi,
levels(nCourse)<-gsub("^0+","",levels(nCourse))
nCourse
# [1] 2A 2B 2C 7A 7B 7C 101 118A 118B 118C
#Levels: 2A 2B 2C 7A 7B 7C 101 118A 118B 118C
#The second part is not very clear.
res<-setNames(data.frame(lapply(c("B","P","C"),function(x)
paste0(levels(nCourse),x)),stringsAsFa
HI,
If I understand it correctly:
fruit<- read.csv("example.csv",header=TRUE,stringsAsFactors=FALSE,sep="\t")
res<-merge(fruit["reference"],fruit[,-1],by.x="reference",by.y="list")
res
# reference information
#1 grapefruit pink
#2 lemon yellow
#3 pear green
If th
atrix(tags_totals[1,1])
#[1] FALSE
is.matrix(tags_totals[1,1,drop=FALSE])
#[1] TRUE
A.K.
Many thanks Arun,
This is important for me because I will need to do this operation
many times. However, there is one thing that intrigues me. Why it is
necessary to put two commas in a row?
> tags_tota
Hi,
You could try:
A<- matrix(unlist(read.table(text="
1 2 3
4 5 6
7 8 9
9 8 7
6 5 4
3 2 1
",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL)
library(matrixStats)
res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds))
res1
# [,1] [,2] [,3]
#1 4 10 18
#2 6
2] [,3]
1 4 10 18
2 63 64 63
3 18 10 4
4 1 3 5
A.K.
From: Bert Gunter
To: arun
Cc: R help
Sent: Monday, September 2, 2013 10:55 AM
Subject: Re: [R] Product of certain rows in a matrix
These elaborate manipulations are unnec
4 10 18
#[2,] 63 64 63
#[3,] 18 10 4
fun1(Anew)
# [,1] [,2] [,3]
#[1,] 4 10 18
#[2,] 63 64 63
#[3,] 18 10 4
#[4,] 1 3 5
A.K.
- Original Message -
From: arun
To: Bert Gunter
Cc: R help
Sent: Monday, September 2, 2013 11:26
# [,1] [,2] [,3]
#1 252 640 1134
#2 18 30 20
A.K.
From: Edouard Hardy
To: arun
Cc: Bert Gunter ; R help
Sent: Monday, September 2, 2013 11:46 AM
Subject: Re: [R] Product of certain rows in a matrix
Thank you all for your responses.
The real problem i
rdy
To: arun
Cc: R help
Sent: Monday, September 2, 2013 11:58 AM
Subject: Re: [R] Product of certain rows in a matrix
Thank you A.K.
And do you have a solution without installing any package ?
Thank you in advance.
E.H.
Edouard Hardy
On Mon, Sep 2, 2013 at 5:56 PM, arun wrote:
>
&g
Hi,
If you check ?str()
str(zseq)
#List of 6
# $ : int [1:19] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:20] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:21] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:22] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:23] 1 2 3 4 5 6 7 8 9 10 ...
# $ : int [1:24] 1 2 3 4 5 6 7 8 9 10 ...
#
ng.
Do you have a better solution on big data ?
- Original Message -
From: arun
To: Edouard Hardy
Cc: R help ; Bert Gunter
Sent: Monday, September 2, 2013 12:07 PM
Subject: Re: [R] Product of certain rows in a matrix
Hi,
No problem.
n<- 4
t(sapply(split(as.data.frame(Anew),as.nume
er overflow
4: In mat1[j22 - 0, ] * mat1[j22 - 1, ] * mat1[j22 - 2, ] * mat1[j22 - :
head(res,3)
# [,1] [,2] [,3]
#[1,] NA NA NA
#[2,] NA NA NA
#[3,] NA NA NA
A.K.
From: Edouard Hardy
To: arun
Sent: Monday, September 2, 2013 2:32 PM
Subject: Re: [R] Product of certa
nrow(mat2)/n)
vec1<- rep("j22",n)
res<- eval(parse(text=
paste(paste0("mat2","[",paste0(vec1,"-",seq(n)-1),",]"),collapse="*")
))
resNew<-rbind(res,apply(tail(mat1New,10),2,prod)
)})
# user system elapsed
# 0.008 0.00
HI,
You may try this:
dat1<- read.table(text="
CustID TripDate Store Bread Butter Milk Eggs
1 2-Jan-12 a 2 0 2 1
1 6-Jan-12 c 0 3 3 0
1 9-Jan-12 a 3 3 0 0
1 31-Mar-13 a 3 0 0 0
2 31-Aug-12 a 0 3 3 0
2 24-Sep-12 a 3 3 0 0
2 25-Sep-12 b 3 0 0 0
",sep="",header=TRUE,stringsAsFactors=FALSE)
dat2<- d
Hi,
You may try this:
set.seed(285)
dat1<-
as.data.frame(matrix(paste0(sample(1:10,100,replace=TRUE),sample(LETTERS[1:10],100,replace=TRUE)),10,10),stringsAsFactors=FALSE)
set.seed(3490)
dat2<-
data.frame(old=unique(unlist(dat1)),new=sample(1:100,63,replace=FALSE),stringsAsFactors=FALSE)
dat1
t(dat1)),new=sample(1:100,100,replace=FALSE),stringsAsFactors=FALSE)
system.time({dat1New<-as.data.frame(array(dat2[,2][match(as.matrix(dat1),dat2[,1])],dim=
dim(dat1),dimnames=dimnames(dat1)))})
#user system elapsed
# 1.480 0.236 1.719
A.K.
- Original Message -----
From:
Hi,
You could try:
df2<- do.call(cbind,split(df[,-1],df[,1]))
res<-sapply(seq_len(ncol(df2)),function(i) {x<-df2[,i];x[duplicated(x)]<-NA;x})
dimnames(res)<- dimnames(df2)
res
# 0 1 2
#[1,] 1585.274 6462.3288 816.7808
#[2,] 2481.678 344.1781 6031.3356
#[3,] 8871.575
;-cbind(dat1[,c(1:3)],dat2[,i]);colnames(x1)[4]<-
colnames(dat2)[i];x2<-x1[x1[,4]!=0,];within(x2,
{daysbetweentrips<-unlist(tapply(as.Date(x2$TripDate,"%d-%b-%y"),list(x2$CustID),function(x)
c(NA,as.numeric(diff(x);previoustripstore<-ave(x2$Store,x2$CustID,FUN=funct
PANID,FUN=function(x) c(x[-1],""))
x2
}
)
In my previous reply, I used ?within().
A.K.
Hi Arun.. I made the factor to character and eventually date conversion.
I am able to see the res dataframe, but it only has the store
names in it.. I cant see all columns like your out
Hi,
Please dput() the example dataset. When I read from the one shown below, it
looks a bit altered.
library(zoo)
dat1<- read.zoo(text="2009-07-15,#N/A N/A,#N/A N/A,18.96858
2009-07-16,20.30685,20.40664,#N/A N/A
2009-07-17,20.78813,20.03991,20.40664
2009-07-20,21.41278,21.41278,20.03991
2009-07-
HI,
?list.files()
list.files() #created 4 files in my working directory
#[1] "A_hubs.txt" "A_nonhubs.txt" "B_hubs.txt" "B_nonhubs.txt"
#If you want to do wilcox.test in a pairwise manner:
combn(list.files(),2)
# [,1] [,2] [,3] [,4]
#[1,] "A_
Hi,
Please use dput() to show the example dataset:
Not sure this matches with your original example though..
lst1<- structure(list(Contrasts = structure(list(linear = c(-0.437,
-0.378, -0.201, 0.271, 0.743), emax1 = c(-0.799, -0.17, 0.207,
0.362, 0.399), emax2 = c(-0.643, -0.361, 0.061, 0.413
# V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
#2 3 13 13 29 16 29 16 18 12 12
#3 13 22 14 25 9 19 13 30 6 6
#11 23 20 5 27 1 24 18 18 26 14
A.K.
From: Vivek Das
To: arun
Sent: Tuesday, September 3, 2013 10:34 AM
Subject:
Hi Arun,
I would like to know if you
d to sum them up so I used "sum(Z, na.rm=TRUE)" to remove the NaN values
but not for the Inf/-Inf.
Hope it is clear to you.
Cheers,
R.L
- Original Message -
From: arun
To: R help
Cc:
Sent: Tuesday, September 3, 2013 2:47 AM
Subject: Re: remove rows with infinite/nan value
Hi,
May be this helps you in getting started.
set.seed(29)
df1<- as.data.frame(matrix(sample(1:20,5*10,replace=TRUE),5,10))
cond<- c("V1eq2","V8eq2","V6eq4orV8eq7")
fun1<- function(df,prefix,cond){
lst1<- list(df[df$V1==2,],df[df$V8==2,],df[df$V6==4|df$V8==7,])
for(i in seq_along(cond)){
as
Hi,
gsub("#%", "[percentagevalue]", text1)
#[1] "this is a number 23%"
gsub("\\d+%$", "[percentagevalue]", text1)
#[1] "this is a number [percentagevalue]"
gsub("bla", "23", text2)
#[1] "this is not a number 23%"
A.K.
- Original Message -
From: Simon Pickert
To: r-help@r-proje
4 elements"
How could I ignore the rows with NA value when read a csv file?
Thank you.
R.L
- Original Message -
From: arun
To: R help
Cc:
Sent: Tuesday, September 3, 2013 11:49 AM
Subject: Re: remove rows with infinite/nan values from a zoo dataset
Hi,
No problem.
In my
HI,
May be this helps.
input<- readLines(textConnection("
Private + Shared = RAM used Program
84.0 KiB + 14.5 KiB = 98.5 KiB sleep
108.0 KiB + 11.5 KiB = 119.5 KiB klogd
124.0 KiB + 15.0 KiB = 139.0 KiB hidd
128.0 KiB + 12.5 KiB = 140.5 KiB gpm
116.0 KiB + 28.5
Hi,
replicate(100,length(hist(10,0:10)$counts))
# [1] 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
10
#[26] 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
10
#[51] 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10
10
#
HI,
It is better to provide a reproducible example using ?dput().
you can also check in this link.
http://r.789695.n4.nabble.com/names-attribute-must-be-the-same-length-as-the-vector-td4503946.html
library(plm)
#Using the example from ?plm()
data("Produc", package = "plm")
zz <- plm(log(gsp) ~
Hi,
You could use ?cat()
For ex:
vec1<-1:100
cat(vec1)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55
56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81
82 83 8
t;- dimnames(B2)
all.equal(B2,B3)
#[1] TRUE
A.K.
____
From: Edouard Hardy
To: arun
Sent: Wednesday, September 4, 2013 1:31 PM
Subject: Re: [R] Random products of rows in a matrix
I am not allowed to.
I found a solution :
l<-tapply(rep(M,N),1:N,function(x){A[sample(1 : nr
Hi,
Try:
dat1<-read.table(text="
sex group
F 1
F 2
F 3
",sep="",header=TRUE,colClasses=c("character","numeric"))
dat1
# sex group
#1 F 1
#2 F 2
#3 F 3
#if you are using read.csv()
dat2<-read.csv("new1.csv",sep="",header=TRUE,colClasses=c("character","numer
olumn headings would be
slightly misplaced.
A.K.
Thanks for replying Arun!
I tried the way you suggested. But all the elements are being
clumped together in a single coulumn and i could see no commas
separating them in csv file.
Is it a good idea to recursively collect the each eleme
1
#l15 1 0 0 0 0 0 0 0 1 0
identical(Out2,fn1(N,Out3))
#[1] TRUE
A.K.
- Original Message -
From: arun
To: R help
Cc:
Sent: Thursday, September 5, 2013 4:09 PM
Subject: Re: binary symmetric matrix combination
Hi,
May be this helps:
m1<- as.matrix(read.table(tex
ot;,1:5),function(x) {x1<- get(x);
x2<-paste0(colnames(x1)[col(x1)],rownames(x1)[row(x1)]);
match(x2,vecOut)})
lst2<- list(m1,m2,m3,m4,m5)
N<- length(lst1)
fn1<- function(N,Out){
i=1
while(i<=N){
Out[lst1[[i]]]<-lst2[[i]]
i<-i+1
}
Out
}
fn1(N,Out3)
-
Hi,
May be this helps:
m1<- as.matrix(read.table(text="
y1 g24
y1 0 1
g24 1 0
",sep="",header=TRUE))
m2<-as.matrix(read.table(text="y1 c1 c2 l17
y1 0 1 1 1
c1 1 0 1 1
c2 1 1 0 1
l17 1 1 1 0",sep="",header=TRUE))
m3<- as.matrix(read.table(text="y1 h4 s2 s30
y1 0 1 1 1
h4 1 0 1 1
s2 1
HI,
May be this helps:
set.seed(28)
dat1<-
setNames(as.data.frame(matrix(sample(1:40,10*5,replace=TRUE),ncol=5)),letters[1:5])
indx<-as.data.frame(combn(names(dat1),2),stringsAsFactors=FALSE)
res<-t(sapply(indx,function(x)
{x1<-cbind(dat1[x[1]],dat1[x[2]]);summary(lm(x1[,1]~x1[,2]))$coef[,4]}))
lt;- dcast(dat2,Var1~Var2,value.var="value",fill=0)
row.names(res2New)<- res2New[,1]
res2New<- as.matrix(res2New[,-1])
identical(res2New,res3)
#[1] TRUE
A.K.
Arun,
That does exactly what I wanted to do, but how would I
manipulate into a matrix where the indepedent variable wa
0 0 0 0 0 1
#l16 1 0 0 0 0 0 0 0 0 0 1 0
#or
res2<- xtabs(value~Var1+Var2,data=dat2)
attr(res2,"class")<- NULL
attr(res2,"call")<-NULL
names(dimnames(res2))<-NULL
all.equal(res1,res2)
#[1] TRUE
A.K.
Very many thanks once again...
HI,
The question is not clear.
Lines1<- readLines(textConnection("Year, Day, Hour, Value
2010, 001, 0, 15.9
2010, 001, 1, 7.3
2010, 001, 2, 5.2
2010, 001, 3, 8.0
2010, 001, 4, 0.0
2010, 001, 5, 12.1
2010, 001, 6, 11.6
2010, 001, 7, 13.9
2010
Hi,
example<- data.frame(id1,id2,GENDER,ETH,stringsAsFactors=FALSE)
res<-unique(example[!(grepl("UNK",example$GENDER)|grepl("UNK",example$ETH)),])
res
# id1 id2 GENDER ETH
#1 1 22 G-M E-VT
#3 2 34 G-M E-AF
#5 3 15 G-M E-AF
#7 4 76 G-F E-VT
#8 5 45 G-F E-VT
#
Hi,
summary(pres.gam)
Family: gaussian
Link function: identity
Formula:
prestige ~ s(income) + s(education)
Parametric coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 47.3276 0.6914 68.45 <2e-16 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05
G-F E-AF
#7 7 37 G-F E-AF
#8 8 52 G-F E-AF
#9 9 66 G-F E-AF
#10 10 91 G-F E-VT
A.K.
- Original Message -
From: arun
To: Robert Lynch
Cc: R help
Sent: Saturday, September 7, 2013 11:30 AM
Subject: Re: [R] finding both rows that are duplicated in a data frame
Hi,
sini<- sin(i)
is.vector(sini)
#[1] TRUE
lni<- log(i)
is.vector(lni)
#[1] TRUE
x<-cbind(int=1,sini,lni)
is.matrix(x)
#[1] TRUE
t(x)*x
#Error in t(x) * x : non-conformable arrays
t(x)%*%x
# int sini lni
#int 20.000 0.9982219 42.3356165
#sini 0.9982219 10.29
Hi,
The expected output is not clear.
dat1<- read.table(text="ID diag1 diag2 diag3 proc1 proc2 proc3
1 k23 i269 j123 u123 u456 u123
2 k69 i80 u456 z456 z123 z456
3 l91 i801 g678 u456 u123 u123
4 i80 i90 h983 z123 z456 z456",sep="",header=TRUE,stringsAsFactors=FALSE)
vec1<- c("i80
$diag),function(x) {x1<- x$code; str1<-
paste(colnames(dat3)[grepl("diag",colnames(dat3))],"%in%","x1",collapse="|");
x2<- subset(dat3,eval(parse(text=str1)));nrow(x2) })
# Broken elbow Broken finger Broken foot Broken hip Broken legs
#
HI Matt,
I changed the dates a little bit to show dates that are outside the range in
dataset B.
A<- read.table(text="
ID Date Depth Temp
1 2002-05-12 10 12
1 2003-05-13 10 12
1 2003-05-14 10 12
1 2004-04-15 10 12
Hi,
You could try:
set.seed(48)
dat1<- as.data.frame(matrix(sample(1:40,189*130,replace=TRUE),ncol=189))
res<-do.call(rbind,lapply(split(colnames(dat1),((seq_len(ncol(dat1))-1)%/%21)+1),function(x)
{x1<- dat1[,x]; colnames(x1)<- paste("V",1:21);x1}))
row.names(res)<- 1:nrow(res)
dim(res)
#[1]
"paste" to "paste0"
row.names(res)<- 1:nrow(res)
all.equal(res,res2)
#[1] TRUE
A.K.
- Original Message -
From: arun
To: "w...@dataanalyticscorp.com"
Cc: R help
Sent: Sunday, September 8, 2013 9:29 AM
Subject: Re: [R] melting a data frame
Hi,
You could tr
Forgot:
res2<-subset(res1,select= -c(time,id))
A.K.
- Original Message -
From: arun
To: "w...@dataanalyticscorp.com"
Cc: R help
Sent: Sunday, September 8, 2013 9:49 AM
Subject: Re: [R] melting a data frame
Hi,
You may also try ?reshape()
dat2<- dat1
names(dat2)[-
Hi,
The ?as.numeric() in 'indx' is not needed.
indx1<-(as.Date(AB$Start)<= as.Date(AB$Date)) & (as.Date(AB$Date) <=
as.Date(AB$End))
identical(indx,indx1)
#[1] TRUE
AB[indx1,-c(5:7)]
A.K.
- Original Message -
From: arun
To: R help
Cc: Matthew Guzzo
Se
Hi,
Try:
df1$CatEch<-paste0(df1[,1],".",sprintf("%02d",df1[,2]))
identical(df1,df2)
#[1] TRUE
A.K.
- Original Message -
From: Arnaud Michel
To: R help
Cc:
Sent: Sunday, September 8, 2013 12:22 PM
Subject: [R] to avoid a do loop
Hello
I have a large dataframe (nrow=55000).
This belo
HI,
df$NewPrices<- unsplit(lapply(split(df,df$Stocks),function(x)
{do.call(rbind,lapply(seq_len(nrow(x)),function(i) {if(x[i,]$Offsets==2)
x[i+2,]$Prices
else if(x[i,]$Offsets==1)
--- Original Message -
From: arun
To: R help
Cc:
Sent: Saturday, September 7, 2013 10:25 PM
Subject: Re: Subsetting isolating a group of values in a group of variables
Hi,
Using the same example:
str1<-paste(colnames(dat1)[grepl("diag",colnames(dat1))],"%in%","vec1&
Hi,
You may try:
unlist(lapply(sara.data[,4:length(sara.data)],function(x)
{x1<-tapply(is.na(x),list(sara.data$Groups),FUN=sum); if(x1[1]!=x1[2]) NULL
else
wilcox.test(x~sara.data$Groups,paired=TRUE,alternative="two.sided")$p.value}))
# Bcl2 Ccl5 Cd27 Cd28
#0.1250 0.1875 0.8125 0.8125
Hi,
Please ?dput() your example dataset. Not sure this helps or not.
u3s<- read.table(text="Current-ID visit AUC Wight ID1
101 3 . . 1
101 4 10 13 2
101 5 . . 3
102 3 . . 4
102 4 4 10 5
102 5 . . 6
103 3 . . 7
103 4 6 9 8
103 5 . . 9",sep="",header=TRUE,na.strings=".",check.names=FALSE)
u3d<-
NA 10.0 1 1 0.5010404 4
#8 4.5817 1.41 0 10.0 1 1 0.5010404 8
#9 NA NA NA 10.0 1 1 0.5010404 16
A.K.
- Original Message -
From: "El-Tahtawy, Ahmed"
To: arun
Cc:
Sent: Monday,
404 16
A.K.
- Original Message -
From: "El-Tahtawy, Ahmed"
To: arun
Cc:
Sent: Monday, September 9, 2013 6:06 PM
Subject: RE: replacing Na's with values on different records
Dear Arun,
Thanks a million for the sophisticated code- it is little above my skill level.
I
ric(row.names(dat4))),]
dim(dat5)
#[1] 639 32
A.K.
____
From: Vivek Das
To: arun
Sent: Monday, September 9, 2013 2:30 PM
Subject: Re: Duplicated genes
actually these are all differentially expressed genes. So the one with the most
differentially expressed
om: David Chertudi
To: arun
Cc: R help
Sent: Monday, September 9, 2013 2:29 PM
Subject: Re: [R] Mann-Whitney by group
Hello Arun,
Thanks so much--while I haven't tried it yet, this seems as though it
will be an excellent way to skip the categories (Actb, etc) that have
missing values (N
0.4600 4.5817 9.1634 18.3268
#[10] 5. 10. 20. 10.1150 20.2300 40.4600 4.5817 9.1634 18.3268
#[19] 2. 4. 8. 10.1150 20.2300 40.4600 4.5817 9.1634 18.3268
#[28] 2. 4. 8. 10.1150 20.2300 40.4600 4.5817 9.1634 18.3268
A.K.
- Original Messag
Hi,
Try:
vec1<- c(2,4,7,10)
df1New<-df1[!df1$Mat %in% vec1,]
dim(df1New)
#[1] 43 3
A.K.
- Original Message -
From: Arnaud Michel
To: R help
Cc:
Sent: Tuesday, September 10, 2013 11:03 AM
Subject: [R] to delete lines by means of a vector
Hi
I would like to eliminate a large number o
ion
Hi,
Thanks for responding to my post, data set attached...
R
> Date: Tue, 10 Sep 2013 08:04:30 -0700
> From: smartpink...@yahoo.com
> To: rhys.mann...@hotmail.co.uk
> Subject: Re: Looping an lapply linear regression function
>
> Hi,
> Any chance you could email me the dat
1634 NA
#[10] NA NA 20. NA 20.2300 NA NA 9.1634 NA
#[19] 2. 4. 8. 10.1150 20.2300 NA NA 9.1634 18.3268
#[28] 2. NA 8. NA 20.2300 NA NA 9.1634 NA
A.K.
- Original Message -
From: "E
:
Sent: Tuesday, September 10, 2013 5:39 PM
Subject: Re: [R] Subtracting elements of a vector from each other stepwise
arun yahoo.com> writes:
>
> Hi,
> Not sure this is what you wanted:
>
> sapply(seq_along(x), function(i) {x1<- x[i]; x2<- x[-i];
x3<-x2[whi
Hi,
Try:
library(reshape2)
sample.dat1<-ddply(sample.dat,.(pop),mutate, valueper=(value/sum(value))*100)
test<-ggplot(sample.dat1, aes(x=response.category,y=valueper, group=pop))
test+geom_bar(stat='identity', position='dodge',aes(fill=pop))
A.K.
- Original Message -
From: Simon Kiss
Hi,
Not sure this is what you wanted:
sapply(seq_along(x), function(i) {x1<- x[i]; x2<- x[-i];
x3<-x2[which.min(abs(x1-x2))];c(x1,x3)})
# [,1] [,2] [,3] [,4]
#[1,] 17 19 23 29
#[2,] 19 17 19 23
A.K.
- Original Message -
From: Michael Budnick
To: r-help@r-project.
Sorry, a mistake:
library(plyr) #instead of library(reshape2)
- Original Message -
From: arun
To: Simon Kiss
Cc: R help
Sent: Tuesday, September 10, 2013 3:08 PM
Subject: Re: [R] ggplot2 percentages of subpopulations
Hi,
Try:
library(reshape2)
sample.dat1<-ddply(sample.dat,.(
Hi,
set.seed(24)
mat1<-
matrix(sample(0:1,20*100,replace=TRUE),ncol=100,dimnames=list(paste0("Species",1:20),paste0("Species",1:100)))
which(mat1[1,]==1)
#or
which(!is.na(match(mat1[1,],1)))
A.K.
- Original Message -
From: Elaine Kuo
To: "r-help@r-project.org"
Cc:
Sent: Wednesday,
Hi,
Using ur code:
dat1<- read.table(text="A B
5 257 259
10 257 259
10.1 257 259
4 257 259
9 257 259
2 257 259
8 257 259
1 257 259
8.1 257 259
8.2 257 259
7
Hi,
Try:
set.seed(24)
yall<- as.data.frame(matrix(sample(1:1e5,5000*10,replace=FALSE),ncol=10))
set.seed(49)
lst1<-replicate(100,yall[sample(1:nrow(yall),100,replace=FALSE),],simplify=FALSE)
names(lst1)<- paste0("yall",1:100)
lapply(seq_along(lst1),function(i)
write.csv(lst1[[i]],file=paste0
Hi,
Try:
library(plyr)
ddply(df,.(x1),summarize,x=cumsum(x2))
# x1 x
#1 1 1
#2 1 3
#3 1 6
#4 1 10
#5 1 15
#6 1 21
#7 1 28
#8 1 36
#9 1 45
#10 1 55
#11 2 11
#12 2 23
#13 2 36
#14 2 50
#15 2 65
#16 2 81
#17 2 98
#or using a1
df2<- data.frame(x1=rep(a1$x1,sapply(a1
botto
To: "smartpink...@yahoo.com"
Sent: Wednesday, September 11, 2013 11:37 AM
Subject:
Dear Arun,
Thanks for your previous reply. i have another question. if i have to matrix A
and matrix B. matrix A is of dimension 124 row * 5 columns and B has a
dimension of 12*5.
What i want
Hi,
Try:
set.seed(48)
lst1<-replicate(3,data.frame(y=rnorm(50),z=runif(50),x=sample(10:15,50,replace=TRUE)),simplify=FALSE)
t(sapply(lst1,function(u) coef(lm(y~0+x+z,data=u #change accordingly
# x z
#[1,] -0.01020553 0.3852990
#[2,] -0.01157726 0.3986898
#[3,] 0.0178
Hi,
May be this helps:
mat1<- matrix(c(1,1,0,0,0,1,0,1,0),ncol=3,dimnames=list(paste("Plant",1:3),
paste("Pollinator",1:3)),byrow=TRUE)
mat2<-
matrix(c(1,1,0,1,0,1,0,1,0),ncol=3,dimnames=list(paste("Plant",c(1,4,5)),
paste("Pollinator",c(1,2,4))),byrow=TRUE)
dat1<- as.data.frame(mat1)
dat2<-
Hi,
Try with:
read.csv(file="sales.csv",header=TRUE,check.names=FALSE)
A.K.
- Original Message -
From: Charles Thuo
To: r-help@r-project.org
Cc:
Sent: Thursday, September 12, 2013 3:15 AM
Subject: [R] how to retain dimnames while exporting from excel into r
v<- read.csv(file="sales.
Hi,
res<- ddply(.data=df1,
.variables='Taxa',
.fun=transform,
Class=find.class(Taxa))
#Warning messages:
#1: In grep(x, df2$Taxa) :
# argument 'pattern' has length > 1 and only the first element will be used
#2: In grep(x, df2$Taxa) :
# argument 'pattern' has length > 1 and on
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