psych does not currently have bootstrapped confidence intervals for loadings.
That is a reasonable request and I will try to add it, perhaps in the “real
soon now” version of 1.5.4 (almost finished), perhaps in the next release,
Bill
> On Apr 13, 2015, at 2:38 PM, stephen sefick wrote:
>
> H
Dear all,
i am trying to calculate the Cy0 from a series of PCR runs in the
384-well. I can create the regression model using the modlist
function:
ml<-modlist(
obj,
cyc = 1,
model = l4,
norm = "FALSE",
remove = "none"
)
where obj is the dataframe with the fluo
Hi Alejo,
The color.id function in plotrix will do this, one color at a time:
sapply(rainbow(6),color.id)
#FFFF #00FF #00FF00FF #00FF # #FF00
[1,] "red" "yellow" "green" "cyan""blue""magenta"
[2,] "red1""yellow1" "green1" "cyan1" "blue1" "magent
That is probably the number one reason for requesting a reproducible example
when writing to R-help. In the proce3ss of working that out, you often solve
your own problem.
Best of luck with your bootstrapping,
Dan
Daniel J. Nordlund, PhD
Research and Data Analysis Division
Services & Enterpris
Hi, this is another quesite related to the use of 'metafor' for calculation
of standardized mean change in pre-post design studies.
Essentially, my aim is to compare different method to arrive at the same
conclusion: Does the treatment work?
The Cochrane manual advise not to calculate change score
Hi,
Please search the mailing list archives for this, or type bootstrapped PCA
R into google. Please provide a minimal self-contained example of what you
are trying to solve. Please read the posting guide that is referenced at
the end of every email.
kind regards,
Stephen
On Mon, Apr 13, 2015 at
Hi, I have a quesite on meta-analysis with 'metafor'.
I would like to calculate the standardized mean difference (SMD), as
Hedges' g, in pre-post design studies.
I have data on baseline (sample size, mean and SD in both the experimental
and the control group) and at end of treatment (same as before
That all sounds so straightforward I wonder why you don't just code it up and
try it out.
You might profit from the advice of chapter 2 of the R-Inferno for your "main
problem".
If "some people" who you ask for advice think that arrays in R are
intrinsically slow, you might also want to look fo
Hi all,
Recently I sent an email and I was asked to provide reproducible code of a
simple example of my situation. Instead of providing the code, I decided to
describe what I need in my code.
I've written a function V, which is a function of (r,s); so I have a
function V(r,s) in fact. The output of
Hi Daniel,
Sorry for that, once more ;=(
Here is the reproducible code and this time IT WORKS FINALLY !!!
GDP.LOG <-c(14,12,13,15.5,16,17,16.5,13.5,12.5,12)
Quality.score <-c(12,11,13,14,15,16,12,10,9,9)
Competitivness.score=c(8,6,7,5,6.5,7,8,4.5,6,7)
df=data.frame(GDP.LOG,Quality.score,
Actually all 6 colors in rainbow(6) do have names. I missed the fact that
rainbow() adds an alpha value that we need to strip off before comparing to the
values in clrs$RGB:
> rain <- substr(rain, 1, 7)
> sum(clrs$RGB %in% rain)
[1] 12
So there are two color names for each color in rainbow(6):
On 4/13/2015 9:06 AM, varin sacha wrote:
Hi Michael,
Thanks for your response. About the data frame not necessary. I correct the
code according to your comments. I still get the following warnings :
[1] "All values of t are equal to 5.75620151906917 \n Cannot calculate confidence
intervals"
N
Hi Peter,
I thought it was possible to have multiple subjects in each group, while
still specifying another factor, as is the case for repeated measures
ANOVA. If this is not the case, I guess I should look into Regression
models for count data as suggested by Micheal.
On Mon, Apr 13, 2015 at 1:
> To add slightly to that:
> What you want to do is write a function that returns the named color that has
> the smallest difference to your input hex-triplet. But note that color
> difference is a large topic. Assuming you want to minimize *perceptual*
> differences, you want to calculate your
By coincidence, there actually _is_ enough info to pinpoint the issue:
*Subj Group Emotion Response*94HRHappy 2
119 HC Happy 0
3 HR Sad 4
61 HC Sad 2
64 HC Sad 0
etc
An unreplicated complete block design has exactly 1 observation for each
combination of the
And expanding at a more elementary level. The reason you need to find the
smallest difference is
that all of the possible colors do not have names. There are 256^3 = 16,777,216
possible rgb color designations, but only 657 named colors. You can create a
data frame of the named colors and their
Dear Lindsay
If the problem is that you have an excess of zeroes you might look at
the vignette for the package pscl which is called something like
Regression models for count data.
On 13/04/2015 17:17, lindsay hanford wrote:
Hello R Community,
I am using the friedman.test() function to test
To add slightly to that:
What you want to do is write a function that returns the named color that has
the smallest difference to your input hex-triplet. But note that color
difference is a large topic. Assuming you want to minimize *perceptual*
differences, you want to calculate your differenc
Dear All,
I am relatively new in R.
Im working with the 'psych' package and 'principal' function.
I would like to know how to generate the bootstraped conf.intervals
for loadings,
looking for sth similar to setting 'n.iter' argument for the 'fa' function.
If in 'psych' can't work and suggest
We really need " commented, minimal, self-contained, reproducible code' as
asked for in the note at the end of each R-help message.
Have a look at http://adv-r.had.co.nz/Reproducibility.html and/or
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
for some h
Hello R Community,
I am using the friedman.test() function to test differences in a non-normally
distributed dataset, with a dependent variable that either a
continuous variable or a ratio and has 2+ groups.
I am using the friedman.test instead of a repeated measures ANOVA because
my dataset viol
Hi Michael,
Thanks for your response. About the data frame not necessary. I correct the
code according to your comments. I still get the following warnings :
[1] "All values of t are equal to 5.75620151906917 \n Cannot calculate
confidence intervals"
NULL
I have found this on the Net :
"Note
Since 2008, Revolution Analytics staff and guests have written about R
every weekday at the Revolutions blog:
http://blog.revolutionanalytics.com
and every month I post a summary of articles from the previous month
of particular interest to readers of r-help.
(By the way, Revolution Analytics is
A combination of rgb(), col2rgb() and colors() can gives hex values for the
named colors.
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Be
Below is what your code looked like in R-helpl.
I looked at it for about 5 seconds and said, "If the poster cannot understand
something as simple as "DO NOT post in HTML" it's not worth my time to decypher
the mess.
You are lucky Peter is more tolerant.
Please do not post in HTML.
code
This is almost unreadable due to HTML posting (do read the posting guide!).
However, it would seem that you somehow got your data converted to factors of
which one level is NA.
This may be surprising, but the difference is like that of
> factor(c("NA", 18))
[1] NA 18
Levels: 18 NA
> factor(c("
Hi all, I want to convert the output of:
> rainbow(6)
> [1] "#FFFF" "#00FF" "#00FF00FF" "#00FF" "#"
"#FF00"
To a vector of color names. Any tip?
Thanks in advance
C.
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R-help@r-
Jean gave you an excellent pointer for your original question.
To continue -
- break down your task into small steps
- implement your steps piece by piece
- actually read (and follow) the posting guide for this list
(http://www.R-project.org/posting-guide.html)
- see here for some additional hin
See in line
On 13/04/2015 14:38, varin sacha wrote:
Hi Jean,
Many thanks, I got it but there is still a problem. When trying to bootstrap
the confidence intervals, I get these messages.
boot.ci(results,type="bca",index=1)
[1] "All values of t are equal to 5.75620151906917 \n Cannot calculat
My AirQuality[4,1] is equal to 18 and AirQuality[5,1] is equal to NA.
When I type the following
> is.na(AirQuality[5,1])[1] FALSE> AirQuality[5,1][1] NA68 Levels: 1 10 108 11
> 110 115 118 12 122 13 135 ... NA> is.na(AirQuality[4,1])[1] FALSE>
> AirQuality[4,1][1] 1868 Levels: 1 10 108 11 110 115
You should cc r-help in your reply to keep everyone in the loop.
Jean
On Mon, Apr 13, 2015 at 8:02 AM, rsm wrote:
> Hi Jean
> Truly appreciate your guidance.
> honestly, finding it bit challenging to plot.
> before plotting.
> i have some basic q.
> have multiple data sets spanning over 15 yea
Hi
Truly appreciate your guidance.
honestly, finding it bit challenging to plot.
before plotting.
i have some basic q.
have multiple data sets spanning over 15 years of One Y , the
Independent, X about 30 dependent variables.
am trying to build a forecasting program for Y , given changes in X.
a
Hi Jean,
Many thanks, I got it but there is still a problem. When trying to bootstrap
the confidence intervals, I get these messages.
boot.ci(results,type="bca",index=1)
[1] "All values of t are equal to 5.75620151906917 \n Cannot calculate
confidence intervals"
NULL
> boot.ci(results,type
There are some examples of how to plot correlation matrices at this link.
http://stackoverflow.com/questions/5453336/plot-correlation-matrix-into-a-graph/26637268#26637268
Perhaps that will help get you started.
Jean
On Mon, Apr 13, 2015 at 5:11 AM, ravimantha wrote:
> Hi just started in R, t
If you write a function that takes a data frame as an argument and returns
a data frame, you can use lapply to carry out the tasks that you want. For
example, if your list of data frames is called mydat ...
mon2date <- function(df) {
if ("Month" %in% names(df)) {
df$Month<- as.POSIXct(df$Mo
Why did you submit this post?
Are you asking for help with something?
Reporting an update on an earlier post?
Jean
On Mon, Apr 13, 2015 at 5:43 AM, thanoon younis
wrote:
> Hi
> I have a small problem with my code in R. The problem is the replication of
> simulation didn't work as a sequence fro
Hi just started in R, this is the first time.
I have one Y & Multiple X variables.
was trying to do Correlation, managed it using cov( Y,X)
unable to plot the above.
further since the data sets is very large of about 20 years * 30 variables,
end objective is to build a forecasting model Y.
have
Hello R folks,
I have recently discovered the power of working with multiple data frames in
lists. However, I am having trouble understanding how to perform operations
on individual columns of data frames in the list. For example, I have a
water quality data set (sample data included below) that c
You will likely get more help answering your question if you provide
reproducible code of a simple example of your situation.
Jean
On Fri, Apr 10, 2015 at 3:42 PM, Mahdiyeh Erfaniyan <
mahdiyeh.erfani...@gmail.com> wrote:
> Hi,
>
>
> Consider the line below:
>
>
>
> for(r in a)for (s in a) x=rbi
S,
There is no mention of a type="bca" argument on the ?confint help file.
You can look here for an example of using the boot.ci() function in the
boot package:
http://www.statmethods.net/advstats/bootstrapping.html
Jean
On Fri, Apr 10, 2015 at 11:01 AM, varin sacha wrote:
> Dear R-Expert
Hi
I have a small problem with my code in R. The problem is the replication of
simulation didn't work as a sequence from r=1 to 100 but the results stop
at 1 and when i close the first results with r=1 the second results run and
so on. i want to leave my computer to run until finish 100 replication
On 13/04/2015 10:12, Isabel Natario wrote:
Hello!
I was wanting to find out which random seed could generate the characters
in the word "love", for example, when I sample with replacement from the
vector of the letters. So I've written the code below:
seed=0
set.seed=0
x<-sample(letters,4,repla
you need set.seed(seed) instead of set.seed = seed
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie & Kwaliteitszorg / team Biometrics & Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
To call in the statistician afte
Thanks Jeff,
and OK I'll move next questions on the topic to the devel list :)
I was hoping there were packages that already dealt with this sort of
things, that's why I posted my question here in the first place..
Thanks a lot for helping me with this,
Cheers,
Luca
[[alternative HTML v
Hello!
I was wanting to find out which random seed could generate the characters
in the word "love", for example, when I sample with replacement from the
vector of the letters. So I've written the code below:
seed=0
set.seed=0
x<-sample(letters,4,replace=T)
while (sum(x==c("l","o","v","e"))<4){
s
Hi
I presume your data frames are not big. What about merging them by hour and
comparing appropriate columns?
something like
windMerged<-merge(windHW, spring, by = "hour", all=TRUE)
sel <- which(windMerged[, xx] >= windMerged[,yy])
windMerged[sel, xx] <- NA
Untested because lack of data.
Chee
Hi,
I was trying a plm model with around 400 variables, but after passing that
to the plm function I am getting coefficients for 265 variables.
Can anyone explain me the reason? Is there a size restriction in plm?
--
Anindya Sankar Dey
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