Hello Rui,
it's very helpful.
Thank you,
Kai
On Wednesday, July 14, 2021, 10:07:57 AM PDT, Rui Barradas
wrote:
Hello,
Just before for(j in 1:nrow(ora)) include the following code line (I
have removed the underscore):
sdif <- vector("list", length = nrow(ora))
In the loop:
sdif[[j
Hello,
Just before for(j in 1:nrow(ora)) include the following code line (I
have removed the underscore):
sdif <- vector("list", length = nrow(ora))
In the loop:
sdif[[j]] <- sqldf(etc)
Also, once again, why noquote? It's better to form file names with
file.path:
rdcsv <- file.path(
Hello List,
I wrote a script below to compare the difference of data frames structure (and
will do something else). First of all I save the file list in a data frame ora,
then I use for loop to 1. load the data from two resource, 2. generate data
structure into two data frames, 3.do the compares
thanks
>
> m
>
>
> - Messaggio originale -
> Da: "Bert Gunter"
> A: "Massimo Bressan"
> Cc: "r-help"
> Inviato: Mercoledì, 22 novembre 2017 17:32:33
> Oggetto: Re: [R] assign NA to rows by test on multiple col
same basic
structure as the simplified example I posted
thanks
m
- Messaggio originale -
Da: "Bert Gunter"
A: "Massimo Bressan"
Cc: "r-help"
Inviato: Mercoledì, 22 novembre 2017 17:32:33
Oggetto: Re: [R] assign NA to rows by test on multiple columns of
Do you mean like this:
mydf <- within(mydf, {
is.na(A)<- !A_flag
is.na(B)<- !B_flag
}
)
> mydf
A A_flag B B_flag
1 8 10 5 12
2 NA 0 6 9
3 10 1 NA 0
4 NA 0 1 5
5 5 2 NA 0
Cheers,
Bert
Bert Gunter
"The trouble with h
Hi--
I too misread the question twice and i may have mistakenly posted non-text
answer earlier. Below is a step by step solution that works provided that
your real data frame has the same structure (alternative columns as in your
example). You could combine all the steps in 2 statements.
Best of
OPS,
Sorry i did not read the post carfully. Mine will not work if you have
zeros on columns A and B.. But you could modify it to work for specific
columns i believe.
EK
On Wed, Nov 22, 2017 at 8:37 AM, Ek Esawi wrote:
> Hi *Massimo,*
>
> *Try this.*
>
> *a <- mydf==0mydf[a] <- NAHTHEK*
>
> On
Hi *Massimo,*
*Try this.*
*a <- mydf==0mydf[a] <- NAHTHEK*
On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan <
massimo.bres...@arpa.veneto.it> wrote:
>
>
> Given this data frame (a simplified, essential reproducible example)
>
>
>
>
> A<-c(8,7,10,1,5)
>
> A_flag<-c(10,0,1,0,2)
>
> B<-c(5,6,2,1,0
;Rui Barradas"
A: "Massimo Bressan" , "r-help"
Inviato: Mercoledì, 22 novembre 2017 11:49:08
Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame
Hello,
Try the following.
icol <- which(grepl("flag", names(mydf)))
mydf[icol] <- lapply
p"
Inviato: Mercoledì, 22 novembre 2017 11:49:08
Oggetto: Re: [R] assign NA to rows by test on multiple columns of a data frame
Hello,
Try the following.
icol <- which(grepl("flag", names(mydf)))
mydf[icol] <- lapply(mydf[icol], function(x){
is.na(x) <- x == 0
Hello,
Try the following.
icol <- which(grepl("flag", names(mydf)))
mydf[icol] <- lapply(mydf[icol], function(x){
is.na(x) <- x == 0
x
})
mydf
# A A_flag B B_flag
#1 8 10 5 12
#2 7 NA 6 9
#3 10 1 2 NA
#4 1 NA 1 5
#5 5 2 0 NA
Given this data frame (a simplified, essential reproducible example)
A<-c(8,7,10,1,5)
A_flag<-c(10,0,1,0,2)
B<-c(5,6,2,1,0)
B_flag<-c(12,9,0,5,0)
mydf<-data.frame(A, A_flag, B, B_flag)
# this is my initial df
mydf
I want to get to this final situation
i<-which(myd
Dear Bert,
This is awesome, thanks a lot!
Best,
Marlin
On Sun, 2016-12-11 at 06:52 -0800, Bert Gunter wrote:
> Use list indexing, "[[" not "[" .
>
> > df <- data.frame(a=1:3,b=letters[1:3])
> > x <- "new"
> > df[[x]]<- I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))
> > df
>
> a b
If you see my previous example, I have tried something like
> df[,n3] <- I(mylist)
However, in my case, the name of the new column is in a variable (by
user input) which can not be directly used by the dollar assign. On the
other hand, "[<-" does not work correctly even if I wrap the list into
"
Glad to help.
However, I need to publicly correct my misstatement. Both "[[" and "["
can be used and are useful for list indexing. As ?"[" clearly states,
the former selects only a single column, while the latter can select
several.
Also:
"Both [[ and $ select a single element of the list. The m
Use list indexing, "[[" not "[" .
> df <- data.frame(a=1:3,b=letters[1:3])
> x <- "new"
> df[[x]]<- I(list(1:5,g = "foo", abb = matrix(runif(6),nr=3)))
> df
a b new
1 1 a 1, 2, 3,
2 2 b foo
3 3 c 0.248115
> df$new
[[1]]
[1] 1 2 3 4 5
$g
[1] "foo"
$abb
[,1]
?data.frame says:
"If a list or data frame or matrix is passed to data.frame it is as if
each component or column had been passed as a separate argument
(except for matrices of class "model.matrix" and those protected by
I). "
So doing what Help says to do seems to do what you asked:
> df <- da
Dear all,
I want to assign a list to one column of data.frame where the name of the column
is a variable. I tried the following:
Using R version 3.3.2
> df <- iris[1:3, ]
> df
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
# 1 5.1 3.5 1.4 0.2 setos
Mark:
I do not understand what you mean by plot labels, which generally mean
the x and y axis labels that can be given with xlab and ylab
parameters in the xyplot call. Clearly, however, this is not what you
mean. If you mean the labels in the key for the groups, you should
create the grouping fac
Dear all,
In order to create plots comparing two measurements I am using the "xyplot" and
"make.groups" functions (see script example below).
Now I would like to assign a value to the variable names in the plot. Meaning:
In the beginning of my script I assign a name(value) to a variable (for
ex
Check the size of df_both. It would be that there are no Command fields
that contain both a 't1' and 't2'.
You can so do:
sum(grepl("t1", df$Command) & grepl("t2", df$Command))
to see how many Command fields contain both.
Jim Holtman
Data Munger Guru
What is the problem that you are trying t
now after this:
df_both <- subset(df, grepl("t1", Command) & grepl("t2", Command))
I use factor to apply the subset to df but then the Command level becomes 0
df_both$Command=factor(df_both$Command)
str(df_both)
$ Protocol : Factor w/ 0 levels:
Do you know what is the r
'grepl' returns a logical vector; you have to use this to get your subset.
You can use:
df_tq <- subset(df, grepl("t1", Command))
df_t2 <- subset(df, grepl("t2", Command))
# if you want to also get a subset that has both, use
df_both <- subset(df, grepl("t1", Command) & grepl("t2", Command))
my problem is that in Command I have 2229 levels and I want to do subsets based
on the names I have in Command. for example if the name has t1 or t2 in it or
if it has both of them.and then I need to plot in a way that colors are names
with t1,names with t2 and names with both. But now even t
You never did provide a reproducible example or say how you wanted to
plot. Here is a way to get a subset of t1 or t2, and you can then use it
as input to ggplot:
library(dplyr)
your_subset <- df %>%
mutate(key = grep(".*(t1|t2).*", "\\1", Command, value = TRUE))
%>%
filter(!(
Hi
I have the following df and I created two subsets but I don't know how to use
these subsets as the colors of my plot.
data.frame': 36919 obs. of 162 variables
$TE :int 38,41,11,52,48,75,.
$TR :int 100,210,548,546,.
$Command :factor W/2229
consistent structure to the strings (a second dash, which may be
implied by the original question).
Best,
John
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Georges
> Monette
> Sent: April 9, 2016 3:04 PM
> To: r-help@r-project.org
> Sub
-
I hope this helps,
John
-
John Fox, Professor
McMaster University
Hamilton, Ontario
Canada L8S 4M4
web: socserv.mcmaster.ca/jfox
From: R-help [r-help-boun...@r-project.org] on behalf of Val
[valkr...@gmail.com]
You are not using that function as it was designed to be used. You should read
the help for gsub...
?gsub
And if you don't know what the term "regular expression pattern" that is
mentioned there means then you will probably need to study one of the many fine
tutorials that are available on th
Professor
McMaster University
Hamilton, Ontario
Canada L8S 4M4
web: socserv.mcmaster.ca/jfox
From: R-help [r-help-boun...@r-project.org] on behalf of Val
[valkr...@gmail.com]
Sent: April 8, 2016 10:21 PM
To: r-help@R-project.org (r-help@r-project.org)
Subject: [
Hi all
I am trying t extract a variable from a column
ASk/20005-01-45/90
Alldatk/25-17-4567/990
I want to assign a variable to the numbers coming the first"-"
x=01 for the first and
x=17 for teh second
I tried using gsub but did not work
x=gsub("-")
any help?
[[altern
> > What I need is this:
> > [[1]]
> > [1] 1 2 3
> > [[1]]
> > [2] 1 2 3
> > [[1]]
> > [2] 1 2 3
Try
rep(list(1:3), 3)
S Ellison
***
This email and any attachments are confidential. Any use...{{dropped:8}}
Hi,
On Wed, Mar 9, 2016 at 10:22 AM, Jan Kacaba wrote:
> Hello I would like to assign a vector to list sequence. I'm trying my code
> bellow, but the output is not what inteded.
>
> # my code
> mls=vector(mode="list") # my list
> cseq=c(1:3) # my vector
> mls[cseq]=cseq
>
> I get following:
> [
Hello I would like to assign a vector to list sequence. I'm trying my code
bellow, but the output is not what inteded.
# my code
mls=vector(mode="list") # my list
cseq=c(1:3) # my vector
mls[cseq]=cseq
I get following:
[[1]]
[1] 1
[[1]]
[2] 2
[[1]]
[2] 3
What I need is this:
[[1]]
[1] 1 2 3
[[1]
Dear all,
I am calculating the optimal cut-off for a test against a gold
standard measure. I can determine the cut-off (in the following
example 5.905) but I would like to assign it to an atomic variable so
that the cut-off is updated every time the data is changed. I am using
the ROC and roc funct
I don't disagree with the other answers, but I'd like to talk about what I
see as a more underlying issue.
Generally speaking, functions return the value of the last expression
within them.
myfun <- function(x) {
z <- sqrt(x)
2
}
Then
y <- myfun(7)
assigns 2 to y because 2 was the last expr
That's it ! Sorry for writing in a hurry, Merm!
Il 16/apr/2015 14:14, "Jim Lemon" ha scritto:
> Hi merm,
> In case Sergio's message is a little cryptic:
>
> return_a_list<-function() {
> a<-"First item of list"
> b<-c(2,4,6,8)
> c<-matrix(1:9,nrow=3)
> return(list(a,b,c))
> }
>
> x<-return_a_
Hi merm,
In case Sergio's message is a little cryptic:
return_a_list<-function() {
a<-"First item of list"
b<-c(2,4,6,8)
c<-matrix(1:9,nrow=3)
return(list(a,b,c))
}
x<-return_a_list()
x
Jim
On Thu, Apr 16, 2015 at 7:21 PM, Sergio Fonda wrote:
> Collect in a vector or dataframe or list the
Collect in a vector or dataframe or list the variables of interest and
output it.
Il 16/apr/2015 10:57, "merm" ha scritto:
> Hi!
>
> So I'm trying as the header suggests to assign the value(s) output by a
> function to a variable, say 'y'
>
> Problem is from what I gather any variables introduced
Hi!
So I'm trying as the header suggests to assign the value(s) output by a
function to a variable, say 'y'
Problem is from what I gather any variables introduced within the function
are contained and the only output I can get is "return(value)" which is
awkward to work with. Any suggestions?
Ch
b
> 12 290 2011-11-19 08:56:54 a b a
> 13 291 2011-11-19 09:19:39 b a b
> >
>
> which results in desired solution.
>
> Cheers
> Petr
>
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-proj
b
>
which results in desired solution.
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of zuzana
> zajkova
> Sent: Tuesday, January 20, 2015 11:15 PM
> To: r-help@r-project.org
> Subject: [R] Assign value to a row in da
Hi Jim,
your code works just perfectly, it is exactly what I was looking for, thank
you very much!!! I'm glad to learn something new, I haven't seen the use of
"drow in 1:(dim(all)[1])" before.
I made the code a bit shorter, excluding the use of "stat_bef" column,
which I created just for my inte
Hi Zuzana,
>From your description it seems that you want to repeat the last
"state_new" value when there are repeated zeros in both "state" and
"state_bef". If that is the case, you may need to step through the
data frame like this:
for(drow in 1:(dim(test)[1])) {
if(test$state[drow] != 0) test$s
Hi,
I have probably basic question, but I would be happy if somebody could help
me with this:
I have a dataframe where I want to assign a value to other column copying
the value in previous row under some condition.
Here is an example (dput on the end of message):
> test[278:290,]
Here's another one: match(d, unique(d)).
Arun
From:Â Greg Snow 538...@gmail.com
Reply:Â Greg Snow 538...@gmail.com
Date:Â March 12, 2014 at 8:41:31 PM
To:Â T Bal studentt...@gmail.com
Cc:Â r-help r-help@r-project.org
Subject:Â Re: [R] Assign numbers in R
Here are a couple more optio
Here are a couple more options if you want some variety:
> d <- c(8,7,5,5,3,3,2,1,1,1)
> as.numeric( factor(d, levels=unique(d)) )
[1] 1 2 3 3 4 4 5 6 6 6
> cumsum( !duplicated(d) )
[1] 1 2 3 3 4 4 5 6 6 6
What would you want the output to be if your d vector had another 8
after the last 1? T
almazó: T Bal [studentt...@gmail.com]
> Küldve: 2014. március 12. 10:13
> To: r-help@r-project.org
> Tárgy: [R] Assign numbers in R
>
> Hi,
> I have the following numbers:
>
> d <- c(8,7,5,5,3,3,2,1,1,1)
>
> I want to convert these into the following numbers:
>
&g
Hi,
Try:
cumsum(c(TRUE,d[-1]!=d[-length(d)]))
A.K.
On Wednesday, March 12, 2014 5:28 AM, T Bal wrote:
Hi,
I have the following numbers:
d <- c(8,7,5,5,3,3,2,1,1,1)
I want to convert these into the following numbers:
r:
1,2,3,3,4,4,5,6,6,6
So if two numbers are different increment it if t
Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ;
meghatalmazó: T Bal [studentt...@gmail.com]
Küldve: 2014. március 12. 10:13
To: r-help@r-project.org
Tárgy: [R] Assign numbers in R
Hi,
I have the following numbers:
d <- c(8,7,5,5,3,3,2,1,1,1)
I want to convert these
Hello,
Try the following.
r <- cumsum(c(TRUE, diff(d) != 0))
Hope this helps,
Rui Barradas
Em 12-03-2014 09:13, T Bal escreveu:
Hi,
I have the following numbers:
d <- c(8,7,5,5,3,3,2,1,1,1)
I want to convert these into the following numbers:
r:
1,2,3,3,4,4,5,6,6,6
So if two numbers are
Hello,
For your example, the following will work:
R> d <- c(8,7,5,5,3,3,2,1,1,1)
R> idx <- 1:length(unique(d))
R> rep(idx, rle(d)$length)
[1] 1 2 3 3 4 4 5 6 6 6
HTH,
Pascal
On Wed, Mar 12, 2014 at 6:13 PM, T Bal wrote:
> Hi,
> I have the following numbers:
>
> d <- c(8,7,5,5,3,3,2,1,1,1)
>
>
Hi,
I have the following numbers:
d <- c(8,7,5,5,3,3,2,1,1,1)
I want to convert these into the following numbers:
r:
1,2,3,3,4,4,5,6,6,6
So if two numbers are different increment it if they are same then assign
the same number:
r <- NULL
for (i in 1:length(d)) {
if (d[i] != d[i+1]) {
r[i]
Hi Weijia,
This will give you the rownames of the split variables.
lst1 <- split(a,list(a$COUNTRY,a$SITEID))
res <- t(sapply(lst1,function(x) {
x$SCRNDT <- as.Date(x$SCRNDT, "%d-%b-%y")
unlist(lapply(split(b,b$TMPT),function(y){
sum(x$SCRNDT >=
HI Weijia,
Please check whether this is what you wanted.
Weijia <- load("/home/arunksa111/Downloads/arun_help.RData" )
a[sapply(a,is.factor)] <-lapply(a[sapply(a,is.factor)],as.character)
str(a)
b[sapply(b,is.factor)] <- lapply(b[sapply(b,is.factor)],as.character)
str(b)
b$DT_ETP <- as.Date(b$D
Hi,
Please use ?dput() to show the dataset. Also, it is not clear about how you
store the time interval.
dat <- read.table(text="
GroupID Date
1 1 10-Dec-12
2 1 11-Dec-12
3 2 13-Dec-12
4 2 15-Dec-12
5 3 06-Dec-12
6 3 19-Dec-12",s
Hi, I have something very interesting:
Say I have this:
GroupID Date
1 1 10-Dec-12
2 1 11-Dec-12
3 2 13-Dec-12
4 2 15-Dec-12
5 3 06-Dec-12
6 3 19-Dec-12
Now, I have time interval,
week 1: from 9-Dec-12 to 15-Dec-12,
wee
Hi, is there a way to assign observations from a dataset to nodes of a tree? Â
For example, if I have a data frame like the following:
ID Var1 Var2 Var3 ⦠VarX
1 Â Â Â Â Â
. Â Â Â Â Â
. Â Â Â Â Â
. Â Â Â Â Â
n     Â
And then run a classification tree with th
This works too! Thank you.
Such a relief after few days spent on trying to solve it.
Lilia
On 13 March 2013 19:52, jim holtman wrote:
> I forgot the 'seq':
>
> > data=data.frame(row=seq(1:10),beh=c(1,1,1,2,2,2,1,1,2,2))
> > data$tripid <- cumsum(c(TRUE, diff(data$beh) != 0))
> > data$seq <- ave
#10 10 2 2 4
> A.K.
>
>
>
>
> - Original Message -
> From: Lilia Dmitrieva
> To: r-help@r-project.org
> Cc:
> Sent: Wednesday, March 13, 2013 3:05 PM
> Subject: [R] Assign the number to each group of multiple rows
>
> Dear R users,
>
>
>
&
I forgot the 'seq':
> data=data.frame(row=seq(1:10),beh=c(1,1,1,2,2,2,1,1,2,2))
> data$tripid <- cumsum(c(TRUE, diff(data$beh) != 0))
> data$seq <- ave(data$beh, data$tripid, FUN = function(x) seq_along(x))
> data
row beh tripid seq
11 1 1 1
22 1 1 2
33 1 1
try this:
> data=data.frame(row=seq(1:10),beh=c(1,1,1,2,2,2,1,1,2,2))
> data$tripid <- cumsum(c(TRUE, diff(data$beh) != 0))
> data
row beh tripid
11 1 1
22 1 1
33 1 1
44 2 2
55 2 2
66 2 2
77 1 3
88 1 3
9
1 2
#5 5 2 2 2
#6 6 2 3 2
#7 7 1 1 3
#8 8 1 2 3
#9 9 2 1 4
#10 10 2 2 4
A.K.
- Original Message -
From: Lilia Dmitrieva
To: r-help@r-project.org
Cc:
Sent: Wednesday, March 13, 2013 3:05 PM
Subject: [R
Dear R users,
My data have repeating "beh" parameter : 1 or 2 - type of animal behavior
in subsequent locations. I need to assign unique number to each sequence of
locations.
My data is:
>data=data.frame(row=seq(1:10),beh=c(1,1,1,2,2,2,1,1,2,2))
>attach(data)
>data
row beh
111
ary 23, 2013 1:35 AM
> To: r-help@r-project.org
> Subject: [R] assign index to colnames(matrix)
>
> Hello, I’m trying to follow the syntax of a script from a journal
> website. In order to create a regression formula used later in the
> script, the regression matrix must have c
, 2013 7:22 PM
To: Prew, Paul
Cc: r-help@r-project.org
Subject: Re: [R] assign index to colnames(matrix)
On 02/23/2013 11:34 AM, Prew, Paul wrote:
> Hello, I’m trying to follow the syntax of a script from a journal
> website. In order to create a regression formula used later in the
> sc
On 02/23/2013 11:34 AM, Prew, Paul wrote:
Hello, I’m trying to follow the syntax of a script from a journal website.
In order to create a regression formula used later in the script, the
regression matrix must have column names “X1�, “X2�, etc. I have tried
to assign these column
Hello, Iâm trying to follow the syntax of a script from a journal website.
In order to create a regression formula used later in the script, the
regression matrix must have column names âX1â, âX2â, etc. I have tried
to assign these column names to my matrix ScoutRSM.mat using a for
Hello,
Em 11-02-2013 01:26, Pete Brecknock escreveu:
malaka wrote
Hi,
I want to assign the ar1 , ma 1 and the intercept estimated by the
following code to three variables a, b and c respectively.
Can anyone help me with this please?
code:
a0 = 0.05; a1 = 0.1; b1 = 0.85
nu = rnorm(2500)
epsi
malaka wrote
> Hi,
> I want to assign the ar1 , ma 1 and the intercept estimated by the
> following code to three variables a, b and c respectively.
>
> Can anyone help me with this please?
>
> code:
>
> a0 = 0.05; a1 = 0.1; b1 = 0.85
> nu = rnorm(2500)
> epsi = rep(0, 2500)
> h = rep(0, 2500)
>
On Aug 1, 2012, at 8:50 AM, John linux-user wrote:
Hi everyone,
I try to add many vectors (L1,L2,L3) to multiple list objects
(a.list,
b.list) in a workspace. Somethings like below, but it is not
working. Any suggestions will be appreciated. Best, John
lf=ls(pattern=".lst")
for
I'm afraid I don't quite understand what you're getting at: can you
say what you're trying to do big-picture wise.
This advice might help making a reproducible example:
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
Best,
Michael
PS -- Just a hunch -- but
Hi everyone,
I try to add many vectors (L1,L2,L3) to multiple list objects (a.list,
b.list) in a workspace. Somethings like below, but it is not
working. Any suggestions will be appreciated. Best, John
lf=ls(pattern=".lst")
for (x in listfiles) {
dat=read.delim(x,header=F)
Dear
Rui Barradas
David Winsemius
ONKELINX, Thierry
David Carlson
I really appreciate your helps.
I did not realize that there were such many ways to do it.
Among them, I just pick up the simple one and it worked out.
It was like this.
<>
n <- 10
p <- 9
dataset <- data.frame(matrix(rep(seq
-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of ONKELINX, Thierry
> Sent: Friday, June 29, 2012 10:11 AM
> To: David Winsemius; lynx
> Cc: r-help@r-project.org
> Subject: Re: [R] assign object with loop (translation from SAS to R)
>
> You can use a combi
.
~ John Tukey
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Namens
David Winsemius
Verzonden: vrijdag 29 juni 2012 16:18
Aan: lynx
CC: r-help@r-project.org
Onderwerp: Re: [R] assign object with loop (translation from SAS to R)
On Jun 28
On Jun 28, 2012, at 9:18 PM, lynx wrote:
I have a dataset named DM with p1, p2, , p9 (9 columns,
numerical values)
I would like to calculate to multify each pair of columns (p1p2,
p1p3,...
p1p9, p2p3, p2p4 p8p9) and assign them in p1p2, p1p3,... p1p9,
p2p3,
p2p4 p8p9
In SAS,
Hello,
See inline.
Em 29-06-2012 02:18, lynx escreveu:
I have a dataset named DM with p1, p2, , p9 (9 columns, numerical values)
I would like to calculate to multify each pair of columns (p1p2, p1p3,...
p1p9, p2p3, p2p4 p8p9) and assign them in p1p2, p1p3,... p1p9, p2p3,
p2p4 p8p9
I have a dataset named DM with p1, p2, , p9 (9 columns, numerical values)
I would like to calculate to multify each pair of columns (p1p2, p1p3,...
p1p9, p2p3, p2p4 p8p9) and assign them in p1p2, p1p3,... p1p9, p2p3,
p2p4 p8p9
In SAS,
l=0;
p_int_sum=0;
do i=1 to 8;
do j=(i
Hi folks,
I am trying to recurse through several subfolders in a directory. Each
folder has many entries that I want to display summary values for. I have
been expanding data frames using code with the structure name <- rbind(name,
newvals) to produce a data frame with n rows equal to the number o
On Apr 10, 2012, at 8:59 AM, David Winsemius wrote:
On Apr 10, 2012, at 3:16 AM, aajit75 wrote:
I have got solution using within function as below
dd$Seg <- 1
dd <- within(dd, Seg[x2> 0 & x3> 200] <- 1)
In this instance the first of your assignments appears superfluous.
dd <- within(dd,
On Apr 10, 2012, at 3:16 AM, aajit75 wrote:
I have got solution using within function as below
dd$Seg <- 1
dd <- within(dd, Seg[x2> 0 & x3> 200] <- 1)
In this instance the first of your assignments appears superfluous.
dd <- within(dd, Seg[x2> 100 & x3> 300] <- 2)
dd <- within(dd, Seg[x2>
On 10-04-2012, at 08:44, aajit75 wrote:
> Hi Experts,
>
> This may be simple question, I want to create new variable "seg" and assign
> values to it based on some conditions satisfied by each observation.
>
> Here is the example:
> ##Below are the conditions
>
> ##if variable x2 gt 0 and x3 gt
I have got solution using within function as below
dd$Seg <- 1
dd <- within(dd, Seg[x2> 0 & x3> 200] <- 1)
dd <- within(dd, Seg[x2> 100 & x3> 300] <- 2)
dd <- within(dd, Seg[x2> 200 & x3> 400] <- 3)
dd <- within(dd, Seg[x2> 300 & x3> 500] <- 4)
I sthere any better way of doing it!!
--
View th
Hi Experts,
This may be simple question, I want to create new variable "seg" and assign
values to it based on some conditions satisfied by each observation.
Here is the example:
##Below are the conditions
##if variable x2 gt 0 and x3 gt 200 then seg should take value 1,
##if variable x2 gt 100
Hi,
I have troubles to assign names to the assets in the portfolio frontier plot
using frontierPlot() in fPortfolio package. Can anyone please help me? If I
use tailoredFrontierPlot in fPortfolio the assets assign names but I dont
want to plot capital market line e.t.c. Someone maybe know how to d
ahoo.fr
> Sent: Sun, 18 Mar 2012 18:24:34 + (GMT)
> To: r-help@r-project.org
> Subject: [R] assign a value to an element
>
> Assign can be used to set a value to a variable that has name as a value
> of another variable. Example:
>
>> name<-"essai"
>&
On Mar 18, 2012, at 2:24 PM, Marc Girondot wrote:
Assign can be used to set a value to a variable that has name as a
value of another variable. Example:
name<-"essai"
assign(name, "plouf")
essai
[1] "plouf"
OK.
But how to do the same when it is only an element of a vector, data
frame an
h 18, 2012 11:25 AM
> To: r-help@r-project.org
> Subject: [R] assign a value to an element
>
> Assign can be used to set a value to a variable that has name as a value of
> another
> variable. Example:
>
> > name<-"essai"
> > assign(name, "
Assign can be used to set a value to a variable that has name as a value of
another variable. Example:
> name<-"essai"
> assign(name, "plouf")
> essai
[1] "plouf"
OK.
But how to do the same when it is only an element of a vector, data frame and
so on that must be changed.
> vec<-1:10
> vec
[1
Your code isn't reproducible...(note the "legend " block -- that's
not code) and (if I had to guess) it looks like you are using
attach(). Don't do that...
What do you mean it doesn't work? You say something about plotting
lines, but your "working" code (the first block) doesn't do that...
If
Dear all,
Thanks all of u for your help. Now I've another similar problem. I want to
plot within the same plot, different lines, each one in a different color
depending on the factor level. I've been able to do it like this, but if i
try with rainbow colors it doesn't work. Can anybody help me wi
On 02/16/2012 01:35 AM, agent dunham wrote:
Dear community,
I've plotted data and coloured depending on the factor variable v3.
In the legend, I'd like to assign properly the same colors than in the
factor (the factor has 5 levels).
I've been trying this but it doesn't work.
plot(var1, var2,
This is certainly not a reproducible example, but this works fine for me.
lets = factor(sample(letters[1:3], 10, TRUE))
plot(1:10, 1:10, col = lets)
legend("topleft", legend = levels(lets), col = seq.int(length(lets)), lty = 1)
Michael
On Wed, Feb 15, 2012 at 9:35 AM, agent dunham wrote:
> Dear
Dear community,
I've plotted data and coloured depending on the factor variable v3.
In the legend, I'd like to assign properly the same colors than in the
factor (the factor has 5 levels).
I've been trying this but it doesn't work.
plot(var1, var2, xlab = "var1", ylab = "var2", col =var3 , b
On 12-01-06 12:18 PM, Michael Pearmain wrote:
Hi All,
I've just recently discovered the cmpfun function, and was wanting to to
create a function to assign this to all the functions i have created, but
without explicitly naming them.
I've achieved this with:
foo<- function(x) { print(x)}
bar<-
Hi All,
I've just recently discovered the cmpfun function, and was wanting to to
create a function to assign this to all the functions i have created, but
without explicitly naming them.
I've achieved this with:
foo <- function(x) { print(x)}
bar <- function(x) { print(x + 1)}
> foo <- function
On Thu, 01-Dec-2011 at 10:13AM -0800, lglew wrote:
|> Hi R-users,
|>
|> I'm trying to produce decompositions of a multiple time-series, grouped by a
|> factor (called "area"). I'm modifying the code in the STLperArea function of
|> package ndvits, as this function only plots produces stl plots, i
On Dec 1, 2011, at 1:13 PM, lglew wrote:
Hi R-users,
I'm trying to produce decompositions of a multiple time-series,
grouped by a
factor (called "area"). I'm modifying the code in the STLperArea
function of
package ndvits, as this function only plots produces stl plots, it
does not
retu
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