Hello,
I am trying to graph a regression line using the followings:
Age <- c(39, 47, 45, 47, 65, 46, 67, 42, 67, 56, 64, 56, 59, 34, 42, 48, 45,
17, 20, 19, 36, 50, 39, 21, 44, 53, 63, 29, 25, 69)
BloodPressure <- c(144, 220, 138, 145, 162, 142, 170, 124, 158, 154, 162,
150, 140, 110, 128, 130,
Dear All,
Ages ago I posted to this mailing list asking for advice about to
evangelize the use of R in an international public
administration where the fact that R is free is not a decisive factor
(actually its being "freeware" may even be seen negatively). After a
long time, I think it is
I am having difficulty fitting a mgcv::gamm model that includes both a random
smooth term (i.e. 'fs' smooth) and autoregressive errors. Standard smooth
terms with a factor interaction using the 'by=' option work fine. Both on my
actual data and a toy example (below) I am getting the same
Hi Anne,
I would suggest to change the linear model to lm(BloodPressure~Age), as
this model makes more sense in biological means (you would assume that
age influences pressure, not vice versa) and also obeys the statistical
assumption of weak exogeneity, that age can be measured without error,
at
Hi, Anne,
assign Age and Bloodpressure in the correct order
to the axes in your call to plot as in:
plot(y = Age, x = BloodPressure)
abline(SimpleLinearReg1)
Hth -- Gerrit
-
Dr. Gerrit Eichner
Courses are good for people who are motivated to learn. It sounds like you need
to generate the motivation first. Why not develop 4 - 5 case study examples?
Kinds of analyses currently performed on Excel that would be easier to
replicate and repeat using simple R scripts so that you are showing
To expand on Eik Vettorazzi's answer:
Don't use html emails. When the mailing list converts it to plain text, it
often re-wraps lines so that your code becomes jumbled.
First, you did not define coeff. You need to add
coeff <- coef(SimpleLinearReg1)
Second, you should use the formula you used
Dear Jeff and Eric,
Okay and many thanks.
Best Regards,
Ashim
On Wed, Apr 18, 2018 at 4:56 PM, Jeff Newmiller
wrote:
> Look at
>
> which(x>100)
>
> This is a zero-length vector. The negative of nothing is nothing, not a
> list of all possible index values.
>
> Do you
Look at
which(x>100)
This is a zero-length vector. The negative of nothing is nothing, not a list of
all possible index values.
Do you want
x[ !( x > 100 ) ]
?
On April 18, 2018 6:13:30 AM CDT, Ashim Kapoor wrote:
>Dear All,
>
>Here is a reprex:
>
>> x<- 1:100
>>
I'm not sure how you are incorporating time period into your data structure.
Typically we are looking at plots or assemblages as the rows and taxa as the
columns. Time period adds a third dimension that could be added as blocks of
rows. For example, depending on the resolution of your data, one
Not necessarily. The R-help archives are publicly accessible,
with a "sort by date" option. So if someone sets up a web=page
monitor which reports back when new messages appear there
(at the bpottom end), then their email addresses are readily
copied (subject to " at " --> "@".
Once they have the
Here's a hint:
> y <- which(x>100)
> identical(y,y)
# TRUE
> identical(y,-y)
# TRUE
The '-' is misleading - it is absorbed into the empty y, leaving the
request x[y] to be x for an empty set of indices.
HTH,
Eric
On Wed, Apr 18, 2018 at 2:13 PM, Ashim Kapoor wrote:
>
> On Apr 18, 2018, at 1:04 AM, Hannah Meredith wrote:
>
> Hello,
>
> I am solving a set of ODEs using deSolve and have run into a problem I
> would appreciate some advice on. One of the parameters (m) in the ODEs
> changes between two states when one of the variables
Dear R users,
I need to merge two data frames based on both equal and unequal comparisons.
The "sqldf" package used to work well , but today, I cannot resolve the
following error by reinstallation of the sqldf package. Can anyone suggest a
different way to perform this kind of merge
Hi Gerrit!
Thank you so much! I mistakenly reversed the order of the variables of the
regression. Actually what I meant to write was: lm(BloodPressure ~ Age).
Best,
Anne
-Message d'origine-
De : Gerrit Eichner [mailto:gerrit.eich...@math.uni-giessen.de]
Envoyé : mercredi 18 avril 2018
Seems it must be the R-list. A horde of ‘solicitation’ emails began arriving
about 27 minutes after I posted about not seeing any! Had left work by that
time, so did not encounter them until now.
From: Mark Leeds [mailto:marklee...@gmail.com]
Sent: April 18, 2018 12:33 AM
To: Rui Barradas
Cc:
Dear All,
Here is a reprex:
> x<- 1:100
> x[-which(x>100)]
integer(0)
In words, I am finding out which indices correspond to values in x which
are greater than 100 ( there are no such items ) . Then I remove those
indices. I should get back the x that I started with since there are no
items
HI All,
I found that other R users experienced the same problem when using sqldf
package. An expert provided solution, adding "method="raw" at the end as below,
one or more of the columns in my data.frame are of class "AsIs" ( I don't know
what it is) . When sqldf retrieves the result from
Hello,
I have a data frame with 400 columns and wanted to filter character columns
with "$" in it.For example: > x <- c("$5", "$89", "$10", "$34") > y <-
c(1:4)> My.Data <- data.frame (x,y)> My.Data x y1 $5 12 $89 23 $10 34 $34 4
I want to detect the columns with $ and remove the $ from
Hi R user,
Would you mind to help me on how I can change a value in a specific column
and row in a big table? but the column of the table is a factor (not
numeric).
Here is an example. I want to change dat[4:5,3]<-"20" but it generated NA>
do you have any suggestions for me?
The simplest would be to convert precip to character and then back to a factor
if you really want it to be a factor. This will also remove the levels that no
longer exist.
str(dat)
# 'data.frame': 5 obs. of 3 variables:
# $ Sites : Factor w/ 5 levels "Site1","Site2",..: 1 2 3 4 5
# $ temp
Hi David and Jeff,
Both ways worked for my table. it helped me a lot (I was really struggling
on it to change the values and thinking to do manually). You are great.
Thank you
On Wed, Apr 18, 2018 at 11:13 AM, David L Carlson wrote:
> The simplest would be to convert precip
I would recommend that you avoid converting (or letting R convert for you) your
textual data values into factors until you have finished doing this kind of
modification.
You can restore the column to character data type either re-reading it with the
stringsAsFactors=FALSE option to
I don’t think that it’s a just a gmail issue. If some of you aren’t receiving
those emails is probably because they get caught in your organization spam
filter and they don’t reach you.
Every time I write to the list I receive those messages.
> On 18 Apr 2018, at 08:12, K. Elo
Hello,
I am solving a set of ODEs using deSolve and have run into a problem I
would appreciate some advice on. One of the parameters (m) in the ODEs
changes between two states when one of the variables (D) crosses a
threshold (D_T) for the first time in either direction. Additionally, when
the
> Peter Langfelder
> on Tue, 17 Apr 2018 11:50:19 -0700 writes:
> I got some spam emails after my last post to the list, and the emails
> did not seem to go through r-help. The spammers may be subscribed to
> the r-help, or they get the poster
Try this:
result <- lapply(71:75, function(x){
# use 'paste0' to add the number to the file name
input <-
read.csv(paste0("C:/Awork/geneAssociation/removed8samples/neuhausen",
x,
"/seg.pr3.csv")
, head=TRUE
)
Hello,
To the OP:
The dollar sign is a meta character, so it must be escaped, that is what
was wrong with your code. The right regular expression would be
grepl("\\$", x)
When a regular expression doesn't work, try reading the help page ?regex.
Another good source you can try is
Your message came through all messed up because you did not tell your email
program to use plain text format. This at best delays a responds and at worst
prevents us from understanding your question as you intended.
1) The columns became factors when you created the data frame because you did
Hi Farnoosh,
Perhaps this will help:
drop_dollar<-function(x) return(as.numeric(as.character(gsub("\\$","",x
sapply(My.Data,drop_dollar)
Jim
On Thu, Apr 19, 2018 at 7:23 AM, Farnoosh Sheikhi via R-help
wrote:
> Hello,
> I have a data frame with 400 columns and wanted
Hola,
Mira la forma de hacerlo aquí:
https://stackoverflow.com/questions/7398998/re-arrange-multiple-columns-in-a-data-set-into-one-column-using-r
Saludos,
Carlos Ortega
www.qualityexcellence.es
El 18 de abril de 2018, 15:34, Diego Iglesias
escribió:
> Hola erreros!
>
Hola erreros!
Acudo de nuevo a ustedes porque ando enredado con algo que debe ser simple
pero que no consigo. Lo que ando buscando hacer es pasar todas las
variables de un dataframe a solo una variable (columna) con la info
apilada. Lo que tengo es un dataframe así (con 200 variables y 30,000
Gracias Carlos y Javier, justo lo que estaba buscando es lo que generan las
funciones stack y unstack
Saludos,
Diego iglesias
El 18 de abril de 2018, 11:26, Javier Marcuzzi <
javier.ruben.marcu...@gmail.com> escribió:
> Estimado Diego Iglesias
>
> Lo que usted desea me genera dudas, por un lado
Estimado Diego Iglesias
Lo que usted desea me genera dudas, por un lado cuándo yo reagrupo armo
algo parecido a lo aportado por Carlos Ortega, pero dos columnas, la
variable y su valor, pero si usted realmente desea una columna donde agrega
todas, puede realizar un rbind, en otras palabras pega
Buenas tardes,
¿Cómo interpretarías el intercepto que da R en un modelo de ceros
inflados? Por un lado en la parte de conteo tengo un intercepto de -4.2 y
en la parte de ceros de 102, ambos salen significativos (***). ¿Qué me
dirían?
Gracias
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