Hello,
In the particular case you have, to change to NA based on condition, use
`is.na<-`.
Here is some test data, 3 times the same df.
set.seed(2021)
df3 <- df2 <- df1 <- data.frame(
x = c(0, 0, 1, 2, 3),
y = c(1, 2, 3, 0, 0),
z = rbinom(5, 1, prob = c(0.25, 0.75)),
a = letters[1:5]
Hi
you could operate with whole data frame (sometimes)
head(iris)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 5.1 3.5 1.4 0.2 setosa
2 4.9 3.0 1.4 0.2 setosa
3 4.7 3.2 1.3 0.2
t;https://www.precheza.cz/en/01-disclaimer/>
https://www.precheza.cz/en/01-disclaimer/
From: Hesham A. AL-bukhaiti
Sent: Tuesday, September 15, 2020 1:48 PM
To: PIKAL Petr
Subject: Re: [R] Loop for two columns and 154 rows
thanks petr v much
i attached my problem in word and my data
## I start with sim_data_wide
sim_data_wide <- tidyr::spread(sim_data, quarter, pd)
## and calculate wide
wide1 <- with(sim_data_wide, cbind(PC_1 = P_1,
PC_2 = 1-(1-P_1)*(1-P_2),
PC_3 = 1-(1-P_1)*(1-P_2)*(1-P_3),
PC_4 = 1-(1-P_1
Does this help?
sim_wide2 <- (
sim_data
%>% arrange( borrower_id, quarter )
%>% group_by( borrower_id )
%>% mutate( cumpd = 1 - cumprod( 1 - pd ) )
%>% ungroup()
%>% mutate( qlbl = paste0( "PC_", quarter ) )
%>% select( borrower_id, qlbl, cumpd )
%>% spread( qlbl, cumpd )
)
On May 9, 2020 4:4
Dear Axel,
Assuming that you're not wedded to using mutate():
> D1 <- 1 - as.matrix(sim_data_wide[, 2:11])
> D2 <- matrix(0, 10, 10)
> colnames(D2) <- paste0("PC_", 1:10)
> for (i in 1:10) D2[, i] <- 1 - apply(D1[, 1:i, drop=FALSE], 1, prod)
> all.equal(D2, as.matrix(sim_data_wide[, 22:31]))
[1]
Just to add one more option (which is best probably depends on if all
the same dates are together in adjacent rows, if an earlier date can
come later in the data frame, and other things):
df$count <- cumsum(!duplicated(df$Date))
Skill a cumsum of logicals, just a different way of getting the logi
Is this what you're after?
> df <- data.frame(
+ Date = as.Date(c("2018-03-29", "2018-03-29", "2018-03-29",
+ "2018-03-30", "2018-03-30", "2018- ..." ...
[TRUNCATED]
> df$count <- cumsum(c(TRUE, diff(df$Date) > 0))
> df
Date count
1 2018-03-29 1
2 2018
Hi Phillip,
While I really like Ana's solution, this might also help:
phdf<-read.table(text="Date count
2018-03-29 1
2018-03-29 1
2018-03-29 1
2018-03-30 1
2018-03-30 1
2018-03-30 1
2018-03-31 1
2018-03-31 1
2018-03-31 1",
header=TRUE,strings
Hello,
Maybe I am not understanding but isn't this what you have asked in your
previous question and my 2nd post (adapted) does?
If not, where does it fail?
Hope this helps,
Rui Barradas
Às 18:46 de 20/09/19, Phillip Heinrich escreveu:
With the data snippet below I’m trying to increment the
Hi Phillip,
This can be done in several ways as most things in programming. Here is one
posible solution:
dates <- c("2018-03-29", "2018-03-29", "2018-03-29",
"2018-03-30", "2018-03-30", "2018-03-30",
"2018-03-31", "2018-03-31", "2018-03-31")
dates <- as.data.frame(as.Dat
Wow...Great one BOB...Gracias, Merci.
On Tue, 6 Aug 2019, 10:46 Bob O'Hara, wrote:
> For a start, try this:
>
> for(i in 1:5) {
> x <- runif(4,0,1)
> }
>
> Which will do what you want, but will over-write x each time (so isn't
> very good). Better (if you want to use the random numbers outside
For a start, try this:
for(i in 1:5) {
x <- runif(4,0,1)
}
Which will do what you want, but will over-write x each time (so isn't
very good). Better (if you want to use the random numbers outside the
loop) is this:
x <- matrix(NA, nrow=5, ncol=4)
for(i in 1:5) {
x[i,] <- runif(4,0,1)
}
But
Thanks guys, I've tried all you're suggesting, both for (x in 1:5) and
break, but I cant seem to ascertain when the loop has generated a vector of
4 random numbers 5 times.
On Tue, 6 Aug 2019, 10:09 Jim Lemon, wrote:
> Hi Tolulope,
> The "in" operator steps through each element of the vector o
Hi Tolulope,
The "in" operator steps through each element of the vector on the
right. You only have one element. Therefore you probably want:
for(x in 1:5)
...
Jim
Jim
On Tue, Aug 6, 2019 at 6:54 PM Tolulope Adeagbo
wrote:
>
> Hey guys,
>
> I'm trying to write a loop that will repeat an action
Is there anything wrong with just doing this?
x <- runif(5, min = 0, max = 1)
Also note that you use x to be at last 2 things: in
for (x in 5) {
you set it to 5, and then in the loop you
x = runif(1:4, min = 0, max = 1)
you make it a vector of length 4.
You also fail to use break to stop the
Thank you so much, Jim. That’s exactly what I need. Sorry for not providing the
data frame. But you created the correct data structure. Thanks again!
From: jim holtman
Sent: Monday, March 25, 2019 2:07 PM
To: Yuan, Keming (CDC/DDNID/NCIPC/DVP)
Cc: R-help@r-project.org
Subject: Re: [R] loop
"Does anyone know how to use loop (or other methods) to create new columns?
In SAS, I can use array to get it done. But I don't know how to do it in R."
Yup. Practically all users of R know how, as this is entirely elementary.
You will too if you make the effort to go through a basic R tutorial, o
R Notebook
You forgot to provide what your test data looks like. For example, are all
the columns a single letter followed by “_" as the name, or are there
longer names? Are there always matched pairs (‘le’ and ‘me’) or can singles
occur?
Hide
library(tidyverse)# create some data
test <- tibble(a
colname.mat[,
> > column]]),3,TRUE)))
> >
> > }
> >
> > > get(samplenames[1])
> > [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
> > year224 0.556 0.667 0.571 0.526 0.629 0.696 0.323 0.526 0.256 0.667
> > year142 0.324 0.32
0.526 0.256 0.667
> year142 0.324 0.324 0.706 0.638 0.600 0.294 0.612 0.688 0.432 0.387
> year237 0.571 0.696 0.629 0.471 0.462 0.471 0.452 0.595 0.333 0.435
>
>
>
>
> --
> *From:* Jim Lemon
> *Sent:* September 11, 2018 1:44 AM
> *To:* Kristi G
.333 0.435
From: Jim Lemon
Sent: September 11, 2018 1:44 AM
To: Kristi Glover
Cc: r-help mailing list
Subject: Re: [R] loop for comparing two or more groups using bootstrapping
Hi Kristy,
Try this:
colname.mat<-combn(paste0("year",1:4),2)
samplenames
Hi Kristy,
Try this:
colname.mat<-combn(paste0("year",1:4),2)
samplenames<-apply(colname.mat,2,paste,collapse="")
k<-1
for(column in 1:ncol(colname.mat)) {
assign(samplenames[column],replicate(k,sample(unlist(daT[,colname.mat[,column]]),3,TRUE)))
}
Then use get(samplenames[1]) and so on to
Hello,
There are now three solutions to the OP's problem.
I have timed them and the results depend on the matrix size.
The solution I thought would be better, Enrico's diag(), is in fact the
slowest. As for the other two, Eric's for loop is 50% fastest than the
matrix index for small matrices
> > Eric Bergeron Wed, 8 Aug 2018 12:53:32 +0300 writes:
>
> > You only need one "for loop"
> > for(i in 2:nrow(myMatrix)) {
> >myMatrix[i-1,i-1] = -1
> >myMatrix[i-1,i] = 1
> > }
Or none, with matrix-based array indexing and explicit control of the indices
to prevent overrun i
> Eric Bergeron Wed, 8 Aug 2018 12:53:32 +0300 writes:
> You only need one "for loop"
> for(i in 2:nrow(myMatrix)) {
>myMatrix[i-1,i-1] = -1
>myMatrix[i-1,i] = 1
> }
>
> HTH,
> Eric
and why are you not using Enrico Schumann's even nicer solution
(from August 6) that I had mention
Thanks a lot ! That's it!
Maija
ke 8. elok. 2018 klo 12.53 Eric Berger (ericjber...@gmail.com) kirjoitti:
> You only need one "for loop"
>
> for(i in 2:nrow(myMatrix)) {
>myMatrix[i-1,i-1] = -1
>myMatrix[i-1,i] = 1
> }
>
> HTH,
> Eric
>
>
> On Wed, Aug 8, 2018 at 12:40 PM, Maija Sirkjärv
You only need one "for loop"
for(i in 2:nrow(myMatrix)) {
myMatrix[i-1,i-1] = -1
myMatrix[i-1,i] = 1
}
HTH,
Eric
On Wed, Aug 8, 2018 at 12:40 PM, Maija Sirkjärvi
wrote:
> Thanks!
>
> If I do it like this:
>
> myMatrix <- matrix(0,5,5*2-3)
> print(myMatrix)
> for(i in 2:nrow(myMatrix))
>
Thanks!
If I do it like this:
myMatrix <- matrix(0,5,5*2-3)
print(myMatrix)
for(i in 2:nrow(myMatrix))
for(j in 2:ncol(myMatrix))
myMatrix[i-1,j-1] = -1
myMatrix[i-1,j] = 1
print(myMatrix)
I get the following result:
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] -1 -1 -1 -1 -1
Hello,
If it is not running as you want it, you should say what went wrong.
Post the code that you have tried and the expected output, please.
(In fact, the lack of expected output was the reason why my suggestion
was completely off target.)
Rui Barradas
On 07/08/2018 09:20, Maija Sirkjärvi w
Thanks, but I didn't quite get it. And I don't get it running as it should.
ti 7. elok. 2018 klo 10.47 Martin Maechler (maech...@stat.math.ethz.ch)
kirjoitti:
>
> > Thanks for help!
> > However, changing the index from i to j for the column vector changes the
> > output. I would like the matrix t
> Thanks for help!
> However, changing the index from i to j for the column vector changes the
> output. I would like the matrix to be the following:
> -1 1 0 0 0 0 0
> 0 -1 1 0 0 0 0
> 0 0 -1 1 0 0 0
> .
> etc.
> How to code it?
as Enrico Schumann showed you: Without any loop, a very nic
Thanks for help!
However, changing the index from i to j for the column vector changes the
output. I would like the matrix to be the following:
-1 1 0 0 0 0 0
0 -1 1 0 0 0 0
0 0 -1 1 0 0 0
.
etc.
How to code it?
Best,
Maija
>> myMatrix <- matrix(0,5,12)
>> for(i in 1:nrow(myMatrix)) {
>>
Hello,
Eric is right but...
You have two assignments. The second sets a value that will be
overwritten is the next iteration by myMatrix[i,i] = -1 when 'i' becomes
the next value.
If you fix the second index and use 'j', you might as well do
myMatrix[] = -1
myMatrix[, ncol(myMatrix)] = 1
H
Quoting Maija Sirkjärvi :
I have a basic for loop with a simple matrix. The code is doing what it is
supposed to do, but I'm still wondering the error "subscript out of
bounds". What would be a smoother way to code such a basic for loop?
myMatrix <- matrix(0,5,12)
for(i in 1:nrow(myMatrix)) {
Both loops are on 'i', which is a bad idea. :-)
Also myMatrix[i,i+1] will be out-of-bounds if i = ncol(myMatrix)
On Mon, Aug 6, 2018 at 12:02 PM, Maija Sirkjärvi
wrote:
> I have a basic for loop with a simple matrix. The code is doing what it is
> supposed to do, but I'm still wondering the err
Here is a simplified example:
dat <- data.frame(x=1:4, y1=runif(4), y2=runif(4), y3=4:1)
for (icol in 2:4) plot(dat[,1] , dat[,icol] )
(not tested, so hopefully all my parentheses are balanced, no typos, etc.)
This shows the basic principle.
An alternative is to construct each column name as a
As Jim says, "No data". R-help is very fussy about what files it will accept.
You might try changing the extention to txt. However the preferred way here is
to use the dput() command and paste the results into the post. See ?dput for
details.
On Monday, May 21, 2018, 1:40:59 a.m. EDT
Hi Steven,
Sad to say that your CSV file didn't make it to the list and I can't
access the data via your Dropbox account. Therefore we don't know the
structure of "mydata". If you are able to plot the data as in your
example, this might help:
genexp<-matrix(runif(360,1,2),ncol=18)
colnames(genexp)
1. Statistically, you probably don't want to do this at all (but
that's another story).
2. Programatically, you probably want to use one of several packages
that do it already rather than trying to reinvent the wheel. A quick
search on rseek.org for "all subsets regression" brought up this:
http:
It's hard to imagine a situation where this makes sense, but of course
you can do it if you want. Perhaps
rhs <- unlist(sapply(1:(ncol(df)-1), function(x)
apply(combn(names(df)[-1], x), 2, paste, collapse = " + ")))
lapply(rhs, function(x) lm(as.formula(paste("y ~", x)), data = df))
--Ista
On Fr
8 148 1 1
>
>
>
> Cheers
>
> Petr
>
>
>
>
>
> *From:* dusa.adr...@gmail.com [mailto:dusa.adr...@gmail.com] *On Behalf
> Of *Adrian Du?a
> *Sent:* Monday, October 10, 2016 12:26 PM
> *To:* Christoph Puschmann
> *Cc:* r-help@r-project.org; PIKAL Petr
&g
il.com [mailto:dusa.adr...@gmail.com] On Behalf Of Adrian
Du?a
Sent: Monday, October 10, 2016 12:26 PM
To: Christoph Puschmann
Cc: r-help@r-project.org; PIKAL Petr
Subject: Re: [R] Loop to check for large dataset
This is an example of how a reproducible code looks like, assuming you have
three columns i
This is an example of how a reproducible code looks like, assuming you have
three columns in your dataset named S (store), P (product) and W (week),
and also assuming they have integer values from 1 to 19, 1 to 22 and 1 to
157 respectively:
#
mydata <- expand.grid(seq(19), seq(22), seq(15
ull set STORE, WEEK and description and merge it
with original data by merge.
Cheers
Petr
From: Christoph Puschmann [mailto:c.puschm...@student.unsw.edu.au]
Sent: Monday, October 10, 2016 9:34 AM
To: PIKAL Petr ; r-help@r-project.org
Subject: Re: [R] Loop to check for large dataset
Dear Petr,
Dear Petr,
I attached a sample file, which contains the first 4 products.
It is more that I have: 157 weeks, 19 different Stores and 22 products:
157*19*22 = 65,626 rows. And as I sated I have roughly 63,127 rows. (so some
have to be missing).
All the best,
Christoph
Hi
see in line
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Christoph
> Puschmann
> Sent: Sunday, October 9, 2016 1:27 AM
> To: Adrian Dușa
> Cc: r-help@r-project.org; Christoph Puschmann
>
> Subject: Re: [R] Loop to
Dear Adrian,
Yes it is a cyclical data set and theoretically it should repeat this interval
until 61327. The data set itself is divided into 2 Parts:
1. Product category (column 10)
2. Number of Stores Participating (column 01)
Overall there are 22 different products and in each you have 19 diffe
It would help to have a minimal, reproducible example.
Unless revealing the structure of your FD object, it is difficult to
understand how a column having 61327 values would be "consistent over an 1
to 157 interval": is this interval cyclic until it reaches 61327 values?
>From your example using F
Hello,
I'm not at all sure if the following is what you need but instead of
for (i in length(FD$WEEK))
try
for (i in 1:length(FD$WEEK))
or even better
for (i in seq_len(FD$WEEK))
And use Control[i, 1], not Control[, 1]
Hope thi helps,
Rui Barradas
Citando Christoph Puschmann :
Hey
Hi
I want to comment something.
When i added the detach(get(yyz)) the RAM consumption was considerable
reduced.
So, i want to declare this issue as solved and thank you all for your
assistance.
Good luck to all.
On Tue, Aug 30, 2016 at 6:24 PM, Juan Ceccarelli Arias
wrote:
> The attach(get(yyz))
The attach(get(yyz)) option i tried and it worked. The only issue, is that
when i'm trying to export the results, it takes a lot of time.
Considerably more than Stata. Also, the computer almost collapses for a
task which isn't so exhausting for my PC station.
Im dubious...
On Tue, Aug 30, 2016 at
Hello,
Try
attach(get(yyz))
Hope this helps,
Rui Barradas
Citando Juan Ceccarelli Arias :
Hi.
I need to loop over rda files.
I generated the list of them.
That's ok. The problem is that the name of the files are as _mm (eg
2010_01 is january or 2010, 2016_03 is march of 2016).
So,
You can attach rda files directly with the attach function, no need to
load them first (see the what argument in the help for attach). This
may do what you want more directly.
In general it is better to not use loops and attach for this kind of
thing. It is better to store multiple data objects
Here's the problem: when you load the object and name it yyz, its simply
storing the name of the data frame as a string. Naturally, when you you
attach the string, it throws an error. The loop actually loads the data
frame but inside yyz there's not a data frame.
One problem with load() is that yo
Ok. Please, declare this issue as solved.
And thanks again for your help.
On Wed, Aug 24, 2016 at 2:18 PM, wrote:
> Maybe it's better to open a new thread.
>
> Rui Barradas
>
>
> Citando Juan Ceccarelli Arias :
>
> The error wasn't in the loop. It was in the file list.
> It's running now because
The error wasn't in the loop. It was in the file list.
It's running now because i added full.names option to TRUE
fuente=list.files("C:/Users/Jceccarelli/Bases/Stata", pattern="dta$",
full.names=T)
Now R can proccess the data. Now it callapses or stops because other kind
of error.
¿Should i open an
Maybe it's better to open a new thread.
Rui Barradas
Citando Juan Ceccarelli Arias :
> The error wasn't in the loop. It was in the file list.
> It's running now because i added full.names option to TRUE
> fuente=list.files("C:/Users/Jceccarelli/Bases/Stata",
> pattern="dta$", full.names=T)
>
I just doesn't work...
Im loading the read,dta13 package already.
When i try to perform a simple table(sex), i received the "File not found"
message.
However, if i load the data using the file.choose() option inside
read.dta13, i can open the stata file.
I don't know what am i doing wrong...
On Tu
Hello,
That means that probably the files are in a different folder/directory.
Use getwd() to see what is your current directory and
setwd("path/to/files") to set the right place where the files can be found.
Rui Barradas
Citando Juan Ceccarelli Arias :
> I just doesn't work...
> Im loading t
Or maybe a print() statement on the table() in the loop.
print(table(...))
Rui Barradas
Citando David Winsemius :
>> On Aug 23, 2016, at 10:01 AM, Juan Ceccarelli Arias
>> wrote:
>>
>> Im running this but the code doesn't seem work.
>> It just hangs out but doesn't show any error.
>>
>> fo
Compare what happens with these two command:
for (i in 1:3) { table(letters[1:4]) }
for (i in 1:3) { print(table(letters[1:4])) }
Then try modifying your loop similarly.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 8/2
Hello,
Where does read_dta come from? You should also post the library() instruction.
Try to run the code without the loop, with just one file and inspect
xxx to see what's happening.
xxx <- read_dta(fuente[1])
str(xxx)
table(xxx$cise, xxx$sexo)
Rui Barradas
Citando Juan Ceccarelli Arias :
> On Aug 23, 2016, at 10:01 AM, Juan Ceccarelli Arias wrote:
>
> Im running this but the code doesn't seem work.
> It just hangs out but doesn't show any error.
>
>
> for (i in 1:length(fuente)){
>
> xxx=read_dta(fuente[i])
>
> table(xxx$cise, xxx$sexo)
>
> rm(xxx)
>
> }
I still find the
Im running this but the code doesn't seem work.
It just hangs out but doesn't show any error.
for (i in 1:length(fuente)){
xxx=read_dta(fuente[i])
table(xxx$cise, xxx$sexo)
rm(xxx)
}
On Tue, Aug 23, 2016 at 6:31 AM, wrote:
> Hello,
>
> The op could also use package sos to find that and oth
Hello,
The op could also use package sos to find that and other packages to
read stata files.
install.packages("sos")
library(sos)
findFn("stata")
found 374 matches; retrieving 19 pages
2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19
Downloaded 258 links in 121 packages
The first package is re
Dear Juan
If this is a Stata 13 file the package readstata13 available from CRAN
may be of assistance.
On 22/08/2016 18:40, Juan Ceccarelli Arias wrote:
I removed the data,frame=True...
I obtain this warnings...
Error in read.dta(fuente[i]) : not a Stata version 5-12 .dta file
In addition: Th
> On Aug 22, 2016, at 10:40 AM, Juan Ceccarelli Arias wrote:
>
> I removed the data,frame=True...
> I obtain this warnings...
> Error in read.dta(fuente[i]) : not a Stata version 5-12 .dta file
Well, that seems fairly self-explanatory. What version of Stata are you using
and does it have capac
I removed the data,frame=True...
I obtain this warnings...
Error in read.dta(fuente[i]) : not a Stata version 5-12 .dta file
In addition: There were 50 or more warnings (use warnings() to see the
first 50)
the warnings() throws this
Warning messages:
1: In `levels<-`(`*tmp*`, value = if (nl == nL)
Hello,
That argument doesn't exist, hence the error.
Read the help page ?read.dta more carefully. You will see that already
read.dta reads into a data.frame.
Hope this helps,
Rui Barradas
Citando Juan Ceccarelli Arias :
> Hi
> I need to apply some code over some stata files that are in fol
You seemed to have re-written over the "b" object in your code.
This might work for you.
library(MASS)
for (i in 58:1){
for(j in 58:i){
str.temp <- paste("y1 ~ x", i, "* x", j, sep = "")
univar <- glm.nb(as.formula(str.temp), data=df)
b <- summary(univar)$coeffients[4, 4]
if(b <
Thank you!
> On Jun 20, 2016, at 12:41 PM, David L Carlson wrote:
>
> It does not test the first column, but a vector must have consecutive
> indices. Since you did not assign a value, R inserts a missing value. If you
> don't want to see it use
>
>> results.pc.all[, -1]
> [,1] [,2]
It does not test the first column, but a vector must have consecutive indices.
Since you did not assign a value, R inserts a missing value. If you don't want
to see it use
> results.pc.all[, -1]
[,1] [,2]
results.212
results.323
-
Da
> On Feb 29, 2016, at 6:24 AM, Fernando McRayearth wrote:
>
> Need to create ascii maps for 10 species by writing a loop. So i have to have
> the vectors ready in the Global Environment, and the "raster map" so the
> information can be added.
>
> when writing the loop I am using the "paste"
Thank you David and Thierry, your answers helped a lot!
Kind regards,
RK.
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or 1.0536478 0.1712595 6.152348 1.41e-07
> virginica 0.6314052 0.1428938 4.418702 5.647610e-05
>
> David C
>
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L
> Carlson
> Sent: Monday, February 16, 2015 8:52 AM
&
inica 0.6314052 0.1428938 4.418702 5.647610e-05
David C
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L Carlson
Sent: Monday, February 16, 2015 8:52 AM
To: Ronald Kölpin; r-help@r-project.org
Subject: Re: [R] Loop over regression results
In R y
In R you would want to combine the results into a list. This could be done when
you create the regressions or afterwards. To repeat your example using a list:
data(iris)
taxon <- levels(iris$Species)
mod <- lapply(taxon, function (x) lm(Sepal.Width ~ Petal.Width,
data=iris, subset=Specie
Hi Patricia
You are somewhat circling around solution.
Is this what you wanted?
for (i in 5:7) {
plotname = paste("Graph", names(scores)[i], sep="")
png(paste0(plotname,".png"))
p <- ggplot(scores, aes(x=scores[,i], fill=gender ))
print(p+ geom_density(alpha=.3)+xlab(names(scores)[i]))
d
Please don't post in HTML since your code was all messed up. You did
not mention what problems you were having with your code. Now a
couple of things to check is to look at what the structure of "r" that
you are trying to add to "sum" (which should have been "Sum" according
to your assignment ear
Jim, Thanks for the comment about else!
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PLEASE do read the posting guide http://www
Hi
Slightly better but still html scrambled.
see in line
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Frank S.
> Sent: Tuesday, September 30, 2014 2:55 PM
> To: r-help@r-project.org
> Subject: [R] Loop does not work: Erro
Dear Berend and Petr,
I do apologise for the disorderly code I posted. I have tried to solve it in a
new mail.
Frank S.
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Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Frank S.
> Sent: Monday, September 29, 2014 9:17 PM
> To: r-help@r-project.org
> Subject: [R] Loop does not work: Error in else statement
>
> Hi to all members of R list,
>
>
Please, please do not post in HTML as the Posting guide requests. See the tail
of each message to R-help.
Your code is completely messed up and unreadable.
Berend
On 29-09-2014, at 21:17, Frank S. wrote:
> Hi to all members of R list,
>
>
>
> I�m working with data.table package, and with 6
etr
> -Original Message-
> From: Jonas Ulbrich [mailto:jonas.ulbr...@ruhr-uni-bochum.de]
> Sent: Wednesday, June 04, 2014 10:48 AM
> To: PIKAL Petr
> Subject: Re: [R] Loop Autoregression
>
> Thanks for the answer. The class of NSS Parameter is data frame.
>
> On 2014
more info from
your side (at least what object is NSSParameter).
Regards
Petr
> -Original Message-
> From: Jonas Ulbrich [mailto:jonas.ulbr...@ruhr-uni-bochum.de]
> Sent: Tuesday, June 03, 2014 3:02 PM
> To: PIKAL Petr
> Subject: RE: [R] Loop Autoregression
>
>
>
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Jonas Ulbrich
> Sent: Sunday, June 01, 2014 1:51 PM
> To: R-help@r-project.org
> Subject: [R] Loop Autoregression
>
> Hello everybody, I have to confess that I am relatively n
Hi,
You may also try:
fun1 <- function(n, repl, val1, val2) {
mat1 <- suppressWarnings(replicate(repl, log(runif(n, val1, val2
mat1[!is.na(mat1)][seq(n)]
}
#Jim's function
fun2 <- function(init, final, val1, val2) {
i <- init
while (i < final) {
u <- runif(1, val1, val2
On Thu, 22 May 2014 09:11:43 PM Ricardo Rocha wrote:
> Hi everybody.
>
> Consider the following exampling code:
>
> x=numeric()
> for(i in 1:10){
> u=runif(1,-1,2)
> x[i]=log(u)
> }
> This code, in each interation, generates a random value in the (-1,2)
> interval and then calculates
Hi Ricardo
Assuming you have a good reason to use such approach (what are you
trying to do ultimately?), you can just increment your counter when
you get a good value, i.e.:
x <- numeric()
n <- 0
while (n < 10) {
u <- log(runif(1, -1, 2))
if (is.finite(u)) {
n <- n+1
x[n] <
t.
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-Original Message-
From: Beatriz R. Gonzalez Dominguez [mailto:aguitatie...@hotmail.com]
Sent: 21. april 2014 16:27
To: Frede Aakmann Tøgersen; r-help@r-project.org
Subject:
er.
> -Original Message-
> From: Beatriz R. Gonzalez Dominguez [mailto:aguitatie...@hotmail.com]
> Sent: 21. april 2014 16:27
> To: Frede Aakmann Tøgersen; r-help@r-project.org
> Subject: Re: [R] Loop to extract from variables in the workspace
>
> Hi Frede,
>
> Many thanks
Hi Frede,
Many thanks for your reply.
1. The first argument in extract is a Formal class RasterLayer in the
Workspace (e.g RR_1981_1 ).
2. I created an intermediate name to hold the result fromthe extract
function because I'd like to create several dataframes with the output
of the iterative
Hi Beatriz
Did you read the help for extract{raster} carefully?
Several things can be wrong.
1) First argument to extract is not a file name but a raster object.
2) In the loop you name an object extract as an intermediate name to hold the
result from the extract function. Do you think there co
Very helpful, many thanks.
On 12 November 2013 16:09, Rui Barradas wrote:
> Hello,
>
> Once again, use lapply.
>
> mlist <- lapply(seq_along(m2), function(i) m2[[i]])
> names(mlist) <- paste0("mod", seq_along(mlist))
>
> slist <- lapply(mlist, summary)
>
>
> plist <- lapply(slist, `[[`, 'p.table'
Hello,
Once again, use lapply.
mlist <- lapply(seq_along(m2), function(i) m2[[i]])
names(mlist) <- paste0("mod", seq_along(mlist))
slist <- lapply(mlist, summary)
plist <- lapply(slist, `[[`, 'p.table')
Hope this helps,
Rui Barradas
Em 12-11-2013 13:28, Kuma Raj escreveu:
Thanks for the s
Thanks for the script which works perfectly. I am interested to do
model checking and also interested to extract the coefficients for
linear and spline terms. For model checkup I could run this script
which will give different plots to test model fit: gam.check(m2[[1]]).
Thanks to mnel from SO I co
Hello,
Use nested lapply(). Like this:
m1 <- lapply(varlist0,function(v) {
lapply(outcomes, function(o){
f <- sprintf("%s~ s(time,bs='cr',k=200)+s(temp,bs='cr') +
Lag(%s,0:6)", o, v)
gam(as.formula(f),family=quasipoisson,na.action=na.omit,data=df)
})})
m1 <-
On 10/23/2013 09:51 PM, THIRU MANIAM wrote:
Hi,
I need kind help from you. I'm doing my assignment in IR and need to do script
in R programming and using R studio tool.I don't have any knowledge in R but
learning by Youtube. After so long,i successfully came out with below script
for precision
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