Hi,
Another way might be:
df1 <- merge(crosspries[[1]], crosspries[[2]], all=TRUE)
template1<-read.table(text=outer(unique(df1[,1]),unique(df1[,2]),FUN=paste),sep="",stringsAsFactors=F)
res<-merge(df1,template1,by.x=c("Product","Year_Month"),by.y=c("V1","V2"),all=TRUE)
res1<-do.call(rbind,split(
HI,
If you use ?join(), it will preserve the order of the first dataframe.
library(plyr)
df3<-rbind.fill(crosspries[[1]],crosspries[[2]])
template2<-read.table(text=outer(unique(df3[,1]),unique(df3[,2]),FUN=paste),sep="",stringsAsFactors=F)
names(template2)<-names(df3[1:2])
res1<-join(template2,df
Hi,
set.seed(25)
Z<-rnorm(50,0.5,1)
res<- as.data.frame(table(cut(Z,seq(0,1,by=0.05),labels=seq(0.05,1,by=0.05))) )
head(res)
# Var1 Freq
#1 0.05 1
#2 0.1 1
#3 0.15 0
#4 0.2 0
#5 0.25 0
#6 0.3 1
A.K.
- Original Message -
From: Wim Kreinen
To: r-help@r-project.
Hi,
In the ?cut(), you can specify `labels`.
#or, you can try:
test.df<-read.table(text="
Var1 Freq
1 (0,0.05] 68
2 (0.05,0.1] 87
3 (0.1,0.15] 85
4 (0.15,0.2] 72
5 (0.2,0.25] 65
6 (0.25,0.3] 73
7 (0.3,0.35] 74
",sep="",header=T,stringsAsFactors=F)
test.df[,1]<-as.numeric(gsub("
Hi,
You could also use:
mat <- matrix(1:15, 5)
set.seed(5)
match_df <- data.frame(Seq = 1:5, criteria = sample(letters[1:5], 5, replace =
T),stringsAsFactors=F)
library(plyr)
res<-daply(as.data.frame(mat),.(match_df$criteria),colSums)
res
#match_df$criteria V1 V2 V3
#
HI,
If I understand your question, "dat3" is not you wanted.
Is it something like this you wanted?
library(reshape2)
dcast(dat,rrt~Mnd,value.var="Result")
# rrt 0 3 6 9
#1 0.35 NA NA 0.05 NA
#2 0.36 NA NA NA 0.06
#3 0.44 NA NA 0.40 NA
#4 0.45 0.1 0.2 NA 0.60
#5 0.46 N
HI,
You can get the plot in the same window, but the width seems to be a problem.
stest<- read.table(text="
site year conc
south 2001 5.3
south 2001 4.67
south 2001 4.98
south 2002 5.76
south 2002 5.93
north 2001 4.64
north 2001 6.32
north 2003 11.5
north 200
Hi,
You could do this:
test<-read.table(text="
V1
1 1
2 1
3 1
4 2
5 2
6 2
7 1
8 1
9 1
10 2
11 2
12 2
",sep="",header=T)
test$Group<-cumsum(abs(c(0,diff(test[,1]+1
test
# V1 Group
#1 1 1
#2 1 1
#3 1 1
#4 2 2
#5 2 2
#6 2 2
#7 1 3
# 1
HI,
This might be better.
test$group<-cumsum(abs(c(1,diff(test[,1]
A.K.
- Original Message -
From: Jeffrey Fuerte
To: r-help@r-project.org
Cc:
Sent: Thursday, January 24, 2013 4:20 PM
Subject: [R] ifelse to speed up loop?
Hello,
I'm not sure how to explain what I'm looking for
Hi,
I guess you meant:
c(TRUE, cumsum( v[-length(v)] != v[-1] ))
[1] 1 0 0 1 1 1 2 2 2 3 3 3
this:
cumsum(c(TRUE, v[-length(v)] != v[-1] ))
# [1] 1 1 1 2 2 2 3 3 3 4 4 4
A.K.
- Original Message -
From: William Dunlap
To: arun ; Jeffrey Fuerte
Cc: R help
Sent: Thursday, January 24
Hi,
x<-1:80
y<- x[-(1:77)]
y
#[1] 78 79 80
#or
?tail() #already suggested
If you want only the last element,,
library(pastecs)
last(x)
#[1] 80
A.K.
- Original Message -
From: hp wan
To: r-help@r-project.org
Cc:
Sent: Thursday, January 24, 2013 7:23 PM
Subject: [R] How to extrac
1$group2<-cumsum(c(TRUE, v[-length(v)] != v[-1] ))# works
test1
# V1 group group2
#1 a 1 1
#2 a 1 1
#3 a 1 1
#4 b 2 2
#5 b 2 2
#6 b 2 2
#7 a 3 3
#8 a 3 3
#9 a 3 3
#10 d 5 4
#11 d
Or,
suu[unlist(lapply(suu,length)!=0)]
#[[1]]
#[1] 1 2
#[[2]]
# [,1] [,2]
#[1,] 1 3
#[2,] 2 4
A.K.
- Original Message -
From: Rui Barradas
To: Tammy Ma
Cc: "r-help@r-project.org"
Sent: Friday, January 25, 2013 8:15 AM
Subject: Re: [R] how to delete the null elements
Hi,
Try this:
Dates1<-c("2005-04-01 BST","2005-04-04 BST","2005-04-05 BST")
Dates2<-as.Date(gsub("\\s+\\w+$","",Dates1))
mat1<-matrix(1:9,nrow=3)
library(zoo)
z1<-zoo(mat1,Dates2)
z1
#
#2005-04-01 1 4 7
#2005-04-04 2 5 8
#2005-04-05 3 6 9
#if you use:
mat2<-matrix(1:9,nrow=3)
which(x==1,arr.ind=TRUE)
row col
#[1,] 1 7
y<-t(matrix(x,nrow=546))
which(y==1,arr.ind=TRUE)
# row col
#[1,] 341 328
A.K.
- Original Message -
From: emorway
To: r-help@r-project.org
Cc:
Sent: Friday, January 25, 2013 5:29 PM
Subject: Re: [R] resizing data
I played a
Hi,
Your question is bit confusing to me.
When you say that "which rrts are the same, and which are the new ones", to
me it looks like "0.35, 0.36" are new addition to Mnd at time points 6 and 9.
Extending Dennis' solution:
Just for understanding the problem:
vec1<-c(0.45,0.48,1.24,1.22,0.44,
HI,
It's not clear why you wanted to take the transpose and resize it afterwards.
It could be done in one step as David suggested.
Suppose, you wanted to get the result after you transposed the matrix:
x<-matrix(1:64,8)
x1<-t(x)
matrix(unlist(split(x1,row(x1))),ncol=4,byrow=T)
# [,1] [,2] [
Hi,
Inline:
- Original Message -
From: emorway
To: r-help@r-project.org
Cc:
Sent: Friday, January 25, 2013 5:29 PM
Subject: Re: [R] resizing data
I played around with your example on the smaller dataset, and it seemed like
it was doing what I wanted. However, applying it to the larg
Hi,
May be this helps:
Matrix[,colSums(diff(Matrix))!=0]
# [,1] [,2] [,3]
#[1,] 1 5 5
#[2,] 2 4 1
#[3,] 3 3 4
#[4,] 4 2 3
#[5,] 5 1 2
A.K.
- Original Message -
From: Benjamin Ward (ENV)
To: "r-help@r-project.org"
Cc:
Sent: Friday, Jan
HI,
You could use ?Reduce() also in the second case:
lapply(vs,function(v){Reduce(f,as.list(v))})
#[[1]]
#[1] 10
#[[2]]
#[1] 6
#[[3]]
#[1] 1
A.K.
- Original Message -
From: Carlos Pita
To: r-help@r-project.org
Cc:
Sent: Friday, January 25, 2013 7:46 PM
Subject: Re: [R] Pass vector
Hi,
I guess this should also work:
Matrix[,apply(Matrix,2,function(x) all(c(TRUE,x[-length(x)]!=x[-1])))]
# [,1] [,2] [,3]
#[1,] 1 5 5
#[2,] 2 4 1
#[3,] 3 3 4
#[4,] 4 2 3
#[5,] 5 1 2
A.K.
- Original Message -
From: Benjamin Ward (ENV)
,] "d" "e" "a" "e"
#[2,] "a" "e" "d" "c"
#[3,] "e" "b" "c" "e"
#[4,] "d" "d" "d" "a"
#[5,] "b" "e" &
Hi,
myData[]<-lapply(myData,as.character)
as.matrix(myData)
# V1 V2 V3 V4 V5
#[1,] "a" "b" "c" "d" "f"
#[2,] "2" "4" "" "" "7"
#[3,] "" "" "" "" ""
#[4,] "" "34" "" "4" ""
#[5,] "2" "" "4" "" "4"
str(as.matrix(myData))
# chr [1:5, 1:5] "a" "2" "" "" "2" "b" "4" "" "34"
Hi,
May be this helps:
wells<-read.table("wells.txt",header=FALSE,stringsAsFactors=F)
wells2<-wells[-grep(".*\\_.*\\_",wells[,1]),]
head(wells2)
# V1 V2
#1 w7_1 0
#2 w11_1 0
#3 w12_1 0
#4 w13_1 0
#5 w14_1 0
#6 w15_1 0
wellsNew<-wells[grep(".*\\_.*\\_",wells[,1]),]
head(wellsNew
HI,
?grep() found to be a bit faster.
system.time({
indx<-grep(".*\\_.*\\_",wells[,1])
wells2<-wells[-indx,]
wellsNew<-wells[indx,]})
# user system elapsed
# 0.024 0.000 0.023
system.time({
w.sub <- gsub("[^_]+","",wells[,1])
u.2 <- which(w.sub=="__")
u.1 <- which(w.sub=="_")
w.u
Hi,
Anther possibility may be to use:
library(MatrixModels)
model.Matrix(~d-1,sparse=FALSE)
#7 x 4 Matrix of class "ddenseModelMatrix"
# dblue dgreen dred dyellow
#1 0 0 1 0
#2 0 1 0 0
#3 0 0 1 0
#4 1 0 0 0
#5 0
Hi,
You could use library(plyr) as well
library(plyr)
pnew<-colSums(aaply(laply(split(as.data.frame(p),((1:nrow(as.data.frame(p))-1)%/%
25)+1),as.matrix),c(2,3),function(x) x))
res<-rbind(t(pnew),colSums(p))
row.names(res)<-1:nrow(res)
res<- 100-100*abs(res/rowSums(res)-(1/3))
A.K.
- Orig
5.5 million rows, and grep doesn't seem to work with over a million rows.
Any idea on a workaround?
On Sat, Jan 26, 2013 at 9:37 PM, Yasha Podeswa wrote:
> Awesome, thanks Arun, that's exactly what I was looking for!
>
>
> On Sat, Jan 26, 2013 at 9:21 PM, arun kirshn
elapsed
# 1.732 0.108 1.844
system.time(dfNew1<-df[grep("\\w+",df$names),])
# user system elapsed
# 0.896 0.024 0.923
system.time(dfNew2<- df[df$names!="",])
# user system elapsed
# 0.460 0.028 0.490
A.K.
From:
228 32.4000
#7 HM001 1229 NA
#8 HM001 1230 27.4000
#9 HM001 1231 22.4236
#Based on the variance,
You could set up some limit, for example 50 and use:
tempnew$WEIGHT<- ifelse(tempnew$WEIGHT>50,NA,tempnew$WEIGHT)
A.K.
From: 남윤주
To: arun
Sent:
New,imputNew$CTIME),function(x)
{x$WEIGHT<-mean(x[,3]);head(x,1)}))
row.names(res3)<-1:nrow(res3)
identical(res1,res2)
#[1] TRUE
identical(res1,res3)
#[1] TRUE
A.K.
From: 남윤주
To: arun
Sent: Monday, January 28, 2013 9:47 PM
Subject: Re: Thank you your hel
Names = c("ID", "CTIME", "WEIGHT"
), class = "data.frame", row.names = c("1", "2", "3", "4", "5",
"6", "7", "8", "9", "10", "11", "12", &q
uot;6", "7", "8", "9", "10", "11", "12", "13", "14", "15")))
#It could be combined by:
do.call(rbind, imput1_2_3)# But if you do this the total number or rows will be
the sum of the number of rows of each
HI,
You could try:
dat1<-read.table(text="
X Z
x1 102
x2 102
x2 102
x2 77
x3 23
",sep="",header=T,stringsAsFactors=F)
res1<-tapply(dat1[,1],list(dat1[,1],dat1[,2]),length)
res1[apply(res1,2,function(x) is.na(x))]<-0
colnames(res1)<-paste("Z.",colnames(r
]<-NA
Hope it helps.
A.K.
From: 남윤주
To: arun
Sent: Tuesday, January 29, 2013 3:36 AM
Subject: Re: I succeed to get result dataset.
Arun ~ I have a dfficuliting in using R again.
A Dataset 'temp' contatins NA and strange value(like 8 row
Hi,
May be this helps:
x<- list(1:5,NA,20:25,5)
names(x)<-1:4
fun1<-function(lst){
lst[lapply(lapply(lst,Filter,f=Negate(is.na)),length)!=0]}
fun1(x)
#$`1`
#[1] 1 2 3 4 5
#
#$`3`
#[1] 20 21 22 23 24 25
#
#$`4`
#[1] 5
#or
x[lapply(lapply(x,na.omit),length)!=0]
A.K.
- Original Message --
$REACTIVE_KWH)])<
0)]<-NA
A.K.
____
From: 남윤주
To: arun
Sent: Tuesday, January 29, 2013 7:42 PM
Subject: I think you misunderstood my explantation.
Hi,
Assume that first CTIME value is '20120101'. It means ACTIVE_KWH measured
from '201201
Hi,
set.seed(125)
dat1<-data.frame(BIRTHPLACE=sample(c("AG","AI","AR","BE","ZH","USA","GER","ESP"),20,replace=TRUE),RES_STA=sample(LETTERS[c(1:3,24:25)],20,replace=TRUE))
dat1$VARNEW<-ifelse(dat1$RES_STA=="X" &
dat1$BIRTHPLACE%in%c("AG","AI","AR","BE","ZH"),"swiss","unknown")
A.K.
- Origi
#changed to NA
#Similarly for REACTIVE_KWH
Hope this helps.
A.K.
From: 남윤주
To: arun
Sent: Wednesday, January 30, 2013 12:51 AM
Subject: Re: I think you misunderstood my explantation.
Oh, I forgot to ask about those code.
Can u expain what dose that mean?
Hi,
I guess you could also use:
x[match(min(x$a),x$a[x$a
To: r-help
Cc:
Sent: Tuesday, January 29, 2013 4:11 PM
Subject: [R] Fastest way to compare a single value with all values in one
column of a data frame
Hello!
I have a large data frame x:
x<-data.frame(item=letters[1:5],a=1:5,b=11:15)
Hi,
You could use:
dat1<-data.frame(col1=c("Mercedes_02352","Audi_03555"))
dat1[,1]<-as.numeric(gsub(".*_\\d{1}(.*)\\d{1}$","\\1",dat1[,1])) #if the
number of digits are the same
dat1
# col1
#1 235
#2 355
A.K.
- Original Message -
From: Mat
To: r-help@r-project.org
Cc:
Sent: Wedne
HI,
Sorry, my previous solution doesn't work.
This should work for your dataset:
set.seed(1851)
x<-
data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F)
y<- data.frame(item="f",a=3,b=10,stringsAsFactors=F)
x[x$a%i
tem="z",a=3,b=10,stringsAsFactors=F)
x[intersect(which(x$a < y$a),which.min(x$a)),]
# item a b
#17 c 1 48
x[x$a==which.min(x$a[x$a
To: arun
Cc: R help ; Jessica Streicher
Sent: Wednesday, January 30, 2013 1:49 PM
Subject: Re: [R] Fastest way to compare a single value wit
Hi,
May be this helps:
#dd
res<-data.frame(Include=with(subset(dd,OBS_TYPE == "SBP" & Blood_Pressure >=
140|OBS_TYPE=="DBP" &
Blood_Pressure>=90),apply(tapply(Blood_Pressure,list(PT_ID,OBS_TYPE),length)>=2,1,any,na.rm=T)))
res
# Include
#1900 TRUE
#2900 FALSE
#3900 FALSE
A.K.
---
Hi,
expand.grid(X)
# Var1 Var2 Var3
#1111
#2211
#3121
#4221
#5112
#6212
#7122
#8222 expand.grid(1:2,1:2,1:2)
# Var1 Var2 Var3
#1111
#2211
#3121
#4
Hi,
Perhaps, (#Untested)
do.call(rbind,lapply(split(dat1,dat1$No),function(x) tail(x,1)))
#or
library(plyr)
ddply(dat1,.(No), function(x) x[nrow(x),])
A.K.
- Original Message -
From: Mat
To: r-help@r-project.org
Cc:
Sent: Friday, February 1, 2013 3:04 AM
Subject: [R] Filter according
Hi,
library(reshape)
foo$id<-row.names(foo)
melt(foo,id.var="id")
id variable value
#1 n w 1.51550092
#2 q w 0.09977303
#3 r w 1.17083866
#4 n x 1.43375720
#5 q x 0.81737610
#6 r x 1.24693470
#7 n y 1.27916240
#8 q y 1
Hi,
You could use:
creek <- read.csv("creek.csv",sep="\t")
colnames(creek) <- c("date","flow")
creek$date <- as.Date(creek$date, "%m/%d/%Y")
creek1 <- within(creek, year <- format(date, '%Y'))
library(data.table)
creek2<- data.table(creek1)
creek2[,list(MEAN=.Internal(mean(flow)),MEDIAN=medi
To: r-help@r-project.org
Cc:
Sent: Friday, February 1, 2013 12:19 PM
Subject: Re: [R] cumulative sum by group and under some criteria
Thank you very much for your reply. Your code work well with this example.
I modified a little to fit my real data, I got an error massage.
Error in split.def
k right. I didn't
A.K.
- Original Message -
From: Zjoanna
To: r-help@r-project.org
Cc:
Sent: Friday, February 1, 2013 12:19 PM
Subject: Re: [R] cumulative sum by group and under some criteria
Thank you very much for your reply. Your code work well with this example.
I modified
Hi,
Not sure this helps:
ml <- data.frame(matrix(sample(1:50,80, replace=TRUE),20,4))
mm <- apply(ml, 2, cumsum)
starts<- data.frame(matrix(0,600,4))
starts1<- data.frame(matrix(0,600,4))
for (i in 1:4){
starts1[,i][mm[,i]] <-1
}
starts2<-as.data.frame(do.call(cbind,lapply(1:4,function(i)
{star
--- Original Message -
From: Zjoanna
To: r-help@r-project.org
Cc:
Sent: Friday, February 1, 2013 12:19 PM
Subject: Re: [R] cumulative sum by group and under some criteria
Thank you very much for your reply. Your code work well with this example.
I modified a little to fit my real data, I got a
Hi,
May be this helps:
df1<-read.table(text="
V1 V2 V3 V4 V5 V6
chr1 18884 C C 2 0
chr1 135419 TATACA T 2 0
chr1 332045 T TTG 0 2
chr1 453838 T TAC 2 0
chr1 567652 T TG 1 0
chr1 602541 TTTA T 2 0
",h
Hi,
Try this:
library(reshape2)
dat1<-read.table(text="ID Dx
A nausea
A diabetes
A kidney_failure
A heart_attack
A fever
B fever
B pneumonia
B heart_attack
B nausea
B cough
C kidney_failure
C nausea
C foot_pain",header=T,stringsAsFactors=F,sep="")
dcast(dat
Hi,
library(reshape2)
dcast(DF,a~b,value.var="c",sum)
# a 1 2 3 4
#1 1 900 100 500 300
#2 2 300 900 200 100
A.K.
- Original Message -
From: Benjamin Gillespie
To: "r-help@r-project.org"
Cc:
Sent: Monday, February 4, 2013 4:29 AM
Subject: [R] Script for conditional sums of vect
ssage -
From: Zjoanna
To: r-help@r-project.org
Cc:
Sent: Friday, February 1, 2013 12:19 PM
Subject: Re: [R] cumulative sum by group and under some criteria
Thank you very much for your reply. Your code work well with this example.
I modified a little to fit my real data, I got an error massage
Hi,
I am not sure about what your end result should be:
>From your code, it looks like you want to replace the rows of dat1 that
>contain at least a `3` with the corresponding row of dat2. Or is it only the
>cell with number `3` replaced by corresponding row in dat2.
dat1<- read.table(text="
1
#2 1 0 0 0 0
#3 0 0 1 0 1
#4 0 1 0 0 0
identical(dat1,dat2)
#[1] TRUE
A.K.
- Original Message -
From: bibek sharma
To: arun
Cc:
Sent: Monday, February 4, 2013 12:54 PM
Subject: Re: [R] longitudinal study
Dear Arun,
I have a data set with dim 1200 by 56 where the var
HI,
If the `Date` column is not ordered:
Date1=as.Date(c("01/05/2012","01/07/2012","01/15/2012","01/09/2012","01/14/2012","01/25/2012",
"01/08/2012","01/24/2012","01/03/2012"),format="%m/%d/%Y")
dat1<-data.frame(ID=rep(1:3,each=3),Date1)
aggregate(Date1~ID,data=dat1,function(x) min(x))
# ID
Hi,
You could try:
df[df>0]<-1
#or
df[]<-as.integer(df!=0)
df
# V1 V2 V3
#yes 1 0 1
#no 0 1 1
A.K.
- Original Message -
From: Wim Kreinen
To: r-help@r-project.org
Cc:
Sent: Monday, February 4, 2013 1:33 PM
Subject: [R] Transform every integer to 1 or 0
Hello,
I have a da
2 0.00032 0.0025
#3 2 3 0.01952 0.00048125
res3[res3[,3]<0.01 & res3[,4]<0.01,]
# Group.1 Group.2 cterm1_P1L cterm1_P0H
#2 3 2 0.00032 0.0025
Hope it helps.
A.K.
____
From: Joanna Zhang
To: arun
Sent: Tuesday, Febr
For converting from first dataset to second,
library(reshape2)
#dat1 is data
dcast(dat1,person~group,value.var="group",length)
# person a b c d
#1 Greg 1 0 0 0
#2 Mary 0 1 0 1
#3 Sam 1 1 1 0
#4 Tom 0 1 1 1
A.K.
- Original Message -
From: Sebastian Haunss
To: r-help@r-project.
m1 n1 cterm1_P1L cterm1_P0H
#1 5 4 3 2 0.00032 0.0025
#2 5 5 3 2 0.00032 0.0025
#3 5 6 3 2 0.00032 0.0025
#4 5 7 3 2 0.00032 0.0025
#5 5 8 3 2 0.00032 0.0025
#6 5 9 3 2 0.00032 0.0025
A.K.
____
From: Joanna Zhang
To: arun
Hi,
I got an error message with:
vlist <- apply(mm, list)
Error in match.fun(FUN) : argument "FUN" is missing, with no default
#assuming that
vlist <- apply(mm,2,list)
mapply("%*%",mlist,vlist[1:2],SIMPLIFY=FALSE)
#[[1]]
# [,1]
#[1,] 19
#[2,] 22
#[3,] 25
#[4,] 28
#
#[[2]]
# [,1]
(n1+2):(maxN-m),0:m1,0:n1,
x1:(x1+m-m1),y1:(y1+n-n1)) # I couldn't find "m" or "n"
A.K.
____
From: Joanna Zhang
To: arun
Sent: Wednesday, February 6, 2013 12:48 PM
Subject: Re: cumulative sum by group a
3 2 0.00032 0.0025
nrow(resF)
#[1] 6300
I am not sure what you want to do with this.
A.K.
From: Joanna Zhang
To: arun
Sent: Wednesday, February 6, 2013 10:29 AM
Subject: Re: cumulative sum by group and under some criteria
Hi,
Thanks! I need to do some c
Hi,
I didn't fully understand the logic.
You could get the result by:
list1<-lapply(mapply(cbind,lapply(1:2,function(i)
E[,,i]),lapply(c(1,3),function(i) C[,i,]),SIMPLIFY=FALSE),function(x)
x[,c(TRUE,apply(matrix(!x[,-1]%in% x[,1],nrow=5),2,all))])
Fnew<-array(unlist(list1),dim=c(dim(list1[[1]]
Hi,
You could get the result:
dat1 <- read.table(text = "
v1 v2 v3
1 0 0
0 1 0
0 0 1
",sep="",header=TRUE)
dat1$v4<-apply(mapply(`==`,dat1,1),2,which)
dat1
# v1 v2 v3 v4
#1 1 0 0 1
#2 0 1 0 2
#3 0 0 1 3
A.K.
- Original Message -
From: Winfried Moser
To: r-
Hi,
You could use ?merge() or ?join() from library(plyr)
merge(df1,df2,all=TRUE)
# col2 col3 col6 col8 col9 col1
#1 2 3 6 NA NA 1
#2 NA NA NA NA NA NA
A.K.
- Original Message -
From: Anika Masters
To: r-help@r-project.org
Cc:
Sent: Thursday, February 7, 2
Hi,
library(lubridate)
ymd_hms(testtime)
1 parsed with %Y-%m-%d %I:%M:%S %p
#[1] "2013-01-01 13:00:01 UTC"
A.K.
- Original Message -
From: e-letter
To: r-help@r-project.org
Cc:
Sent: Friday, February 8, 2013 4:44 AM
Subject: [R] convert 12 time stamp to 24 hour
Readers,
For a 12 h
Hi,
Try:
R[,2:4]<-F[,2:4][match(R$x,F$x),]
R$p<- unique(F$p)
R
x y z w k p
1 -1 1 3 8 9 18
2 -2 2 4 9 9 18
3 -3 3 5 10 9 18
4 -1 1 3 8 9 18
5 -2 2 4 9 9 18
6 -3 3 5 10 9 18
7 -1 1 3 8 9 18
8 -2 2 4 9 9 18
9 -3 3 5 10 9 18
10 -1 1 3 8 9 18
11 -2 2 4 9 9 18
12 -3 3 5 10
Hi,
mood<- data.frame(Mood1=1:5,Mood2=6:10,Mood3=11:15)
lapply(seq_along(mood),function(i) i*mood[i])
A.K.
- Original Message -
From: christel lacaze
To: r-help@r-project.org
Cc:
Sent: Thursday, February 7, 2013 11:16 AM
Subject: [R] help with creating new variables using a loop
Hi
Hi,
Just to add:
If the columns are not in the order (Mood1:Mood10) in the dataset.
mood<- data.frame(Mood2=6:10,Mood1=1:5,Mood10=11:15,Mood3=16:20)
do.call(cbind,lapply(names(mood),function(i)
mood[i]*as.numeric(gsub("\\D+","",i
# Mood2 Mood1 Mood10 Mood3
#1 12 1 110 48
#2
You should have used dput().
Assuming the "," are "."
#Using a subset of your example
dat1<-read.table(text="
proc cls
7271 568,03338 0,5
7270 554,68458 0,5
7269 510,20638 0,5
7268 485,59969 0,5
7267 421,92852 0,5
7272 410,12101 0,5
3414 409,71429 0,5
3452 402,78699 0,5
3451 401,28
Hi Elisa,
It seems also that there are spaces between the second '.' and the number
(1901.11. 1).
You could do something like:
Lines1<-readLines(textConnection("1901.11. 1 447.000
1901.11. 2 445.000
1901.11. 3 445.000
1924.11. 4 445.000
1924.11. 5 449.000
1924.11. 6 442.000
Hi,
You could try:
dat1<- read.table(text="
site date year precipitation temp_max temp_min
1 castlepeak Jan-71 1971 26.2903226 38.29032 18.06452
2 castlepeak Feb-71 1971 9.1071429 39.60714 17.5
3 castlepeak Mar-71 1971 36.3548387 38.87097 17.77419
4 castlepeak Apr
Hi,
bg<- read.table(text="
Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125 Otu00131
Otu00137 Otu00155 Otu00158 Otu00172 Otu00181 Otu00185 Otu00190 Otu00209 Otu00218
Gi20Jun11 0.001217 0 0.001217 0 0.00 0 0
0 0.001217 0 0
Hi,
Lines1 <- readLines(textConnection("Date:10.09.19 Time:21:39:05 Lat:N62.37.18
Long:E018.07.32
-0010 | 28| 28
0010-0020| 302| 302
0020-0030| 42| 42
0030-0040| 2| 2
0040-0050| 1| 1
0060-0070| 1| 1
_
Date:10.09.19 Time:21:44:52 Lat:N62.38.00 Long:E018.09.07
Hi,
mapply(`[`,mylist,list(1,2,0),SIMPLIFY=FALSE)
#[[1]]
# a
#1 1
#2 2
#[[2]]
# b
#1 5
#2 6
#[[3]]
#data frame with 0 columns and 2 rows
A.K.
- Original Message -
From: Dimitri Liakhovitski
To: r-help
Cc:
Sent: Tuesday, February 12, 2013 4:33 PM
Subject: [R] grabbing from eleme
Hi,
Sorry, in my previous reply, I omitted one of the columns.
Lines1 <- readLines(textConnection("Date:10.09.19 Time:21:39:05 Lat:N62.37.18
Long:E018.07.32
-0010 | 28| 28
0010-0020| 302| 302
0020-0030| 42| 42
0030-0040| 2| 2
0040-0050| 1| 1
0060-0070| 1| 1
_
Date
Hi Dimitri,
neededcolumns<- c(1,2,0)
mapply(`[`,mylist,lapply(neededcolumns,function(x) x),SIMPLIFY=FALSE)
#[[1]]
# a
#1 1
#2 2
#
#[[2]]
# b
#1 5
#2 6
#[[3]]
#data frame with 0 columns and 2 rows
A.K.
From: Dimitri Liakhovitski
To: arun
Cc: R h
Hi,
For the dataset A,
Assuming that this is what you meant: the maximum value of "lig" in each 2 rows
excluding the first row:
#dataset: A
A[-1,]<-do.call(rbind,lapply(split(A[-1,],((1:nrow(A[-1,])-1)%/%2)+1),function(x){x[4]<-max(x[4]);x}))
A
# ok time secs lig geo
Hi,
#data: dat1
mat1<-matrix(dat1[,2],ncol=3,nrow=7)
colnames(mat1)<-unique(dat1[,1])
library(Hmisc)
rcorr(mat1)$P
# TTK CTJ PKR
#TTK NA 0.0151874 0.6277163
#CTJ 0.0151874 NA 0.5214368
#PKR 0.6277163 0.5214368 NA
A.K.
- Original Message -
From:
r(res1)
#'data.frame': 2192 obs. of 2 variables:
# $ x : Date, format: "1991-01-01" "1991-01-02" ...
# $ V2: num 0.314 0.314 0.273 0.273 0.236 0.236 0.236 0.236 0.273 0.314 ...
A.K.
From: eliza botto
To: "smartpink...@yaho
uot;
#[2190,] "1996.12.29" "0.236"
#[2191,] "1996.12.30" "0.236"
#[2192,] "1996.12.31" "0.236"
A.K.
From: eliza botto
To: "smartpink...@yahoo.com"
Sent: Wednesday, February 13, 2013 9:01 AM
S
0.236 0.264 0.236
0.295 0.315 ...
head(res3[[1]])
# date1 val
#1 1991.01.01 0.314
#2 1991.01.02 0.314
#3 1991.01.03 0.273
#4 1991.01.04 0.273
#5 1991.01.05 0.236
#6 1991.01.06 0.236
I am not sure whether you could do it in one step as memory might be an issue
here. You could split this
ames")=List of 2
.. ..$ : NULL
.. ..$ : chr [1:2] "date1" "val" # both are now of class "chr"
A.K.
From: eliza botto
To: "smartpink...@yahoo.com"
Sent: Wednesday, February 13, 2013 11:44 AM
Subject: RE: date
2001 2002 2003 2004
#18055.442 20063.238 21481.157 20138.238 3141.716
In this dataset, some year doesn't have all the Seasons, so the value is less.
I hope it hleps.
A.K.
- Original Message -
From: "iruc...@mail2world.com"
To: smartpink...@yahoo.com
Hi,
May be this helps:
library(plyr)
res<-aaply(laply(my_list, as.matrix),c(2,3),function(x) x)
attr(res,"dimnames")<- NULL
identical(res,foo)
#[1] TRUE
A.K.
- Original Message -
From: Murat Tasan
To: r-help@r-project.org
Cc:
Sent: Thursday, February 14, 2013 1:13 AM
Subject: Re: [R
HI,
#ag data was created
bg<- as.matrix(read.table(text="
Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125 Otu00131
Otu00137 Otu00155 Otu00158 Otu00172 Otu00181 Otu00185 Otu00190 Otu00209
Otu00218
Gi20Jun11 0.001217 0 0.001217 0 0.00 0 0
- colnames(res2)
identical(res1,res2)
#[1] TRUE
A.K.
- Original Message -
From: Ozgul Inceoglu
To: arun
Cc:
Sent: Thursday, February 14, 2013 6:17 AM
Subject: re:Re: [R] spearman correlation and p-value as a matrix
Dear Arun,
I bothered you this week alot, but it seems like that y
Hi,
names(Data)[-grep(".*\\s+.*",names(Data))]<-paste(names(Data)[-grep(".*\\s+.*",names(Data))],"-XXX",sep="")
names(Data)[grep(".*\\s+.*",names(Data))]<-gsub("[
]","-",names(Data)[grep(".*\\s+.*",names(Data))])
names(Data)
#[1] "ada-XXX" "fsdfs-XXX" "asd-fd" "asdsa-XXX" "dsada-XXX"
Dear Ozgul,
I made a comment in the previous email, which I didn't check it.
Thank you for the ?cor2m().
I checked the results of the example dataset with cor2m(). Does it have an
option for spearman correlation?
The values below seem to be pearson
#cor2m
cor2m(bg,ag,trim=TRUE,alpha=0.05) ##
NaN NaN NaN NaN NaN -0.8901029
#4 NaN 0.000 NaN NaN NaN NaN NaN 0.000
#5 NaN 0.000 NaN NaN NaN NaN NaN 0.000
# Otu00209 Otu00218
#1 NaN 0.000
#2 NaN 0.000
#3 NaN -0.8901029
HI,
If the dataset is similar to this:
dat1<- read.table(text="
cu.nr. name value
A 1 Evo 100
B 1 Mer 80
C 2 Ford
HI,
May be this helps:
(assuming that there are only twins)
dat1<- read.table(text="
FamilyID ParticipantID IQ Digit_span
1 1 95 6
1 2 93 7
2 3 102 8
2 4 101 9
3 5 106 7
3 6
Hi,
Try
vec1<- c(10, 1000153846, 1000307692, 1000461538, 1000615385)
round(vec1/1e6,2)
#[1] 1000.00 1000.15 1000.31 1000.46 1000.62
#or
as.numeric(sprintf("%.2f",vec1/1e6))
#[1] 1000.00 1000.15 1000.31 1000.46 1000.62
A.K.
- Original Message -
From: Alaios
To: R help
Cc:
S
HI,
You could either get the results by:
unlist(lapply(groups,function(x) paste(x,collapse=" ")),use.names=FALSE)
#[1] "6" "4 4" "5 3 2"
#or
gsub("[c(),]","",paste(groups))
#[1] "6" "4 4" "5 3 2"
str(gsub("[c(),]","",paste(groups)))
# chr [1:3] "6" "4 4" "5 3 2"
str(unlist(lapply(gro
HI,
You could also use these:
sapply(groups,paste,collapse=" ")
# 1 3 6
# "6" "4 4" "5 3 2"
#or
gsub("^ | $","",gsub("\\D+"," ",paste(groups)))
#[1] "6" "4 4" "5 3 2"
A.K.
- Original Message -
From: Nikola Janevski
To: r-help@r-project.org
Cc:
Sent: Thu
Hi,
Try this:
dat1<- read.table(text="
Task_No Team
A 49397 1
B 49396 1
",sep="",header=TRUE)
dat2<- read.table(text="
ID Team
A 5019 1
B 472 1
C 140 1
D 5013 1
",sep="",header=TRUE)
res1<- as.matrix(merge(dat1,dat2,by="Team"))
ro
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