On 20/10/19 3:00 PM, Richard M. Heiberger wrote:
Look at
methods(as.data.frame)
Define your specialized columns to have a newly defined class, say "myclass".
Then write as.data.frame.myclass
It will be similar to the function you already have in the lapply statement.
Now your statement
X <-
Look at
methods(as.data.frame)
Define your specialized columns to have a newly defined class, say "myclass".
Then write as.data.frame.myclass
It will be similar to the function you already have in the lapply statement.
Now your statement
X <- X[ok,]
should work.
Rich
On Sat, Oct 19, 2019 at 8:20
I am writing a function that involves a data frame "X" some columns of
which have attributes. I replace X by a data frame with a subset of the
rows of X:
X <- X[ok,]
where "ok" is a logical vector. When I do this the attributes of the
columns (which I need to retain) are lost
Yes, options(knitr.kable.NA = '-') is The answer for kable.
Do you happen to know what are the arguments used for gridExtra grid.draw to
acomplish the same thing?
Sent from Yahoo Mail on Android
On Sat, Oct 19, 2019 at 1:01 PM, Jeff Newmiller
wrote: Then the polite next step is for you
Searching on "lasso penalty with deviance" on rseek.org brought up many
packages.
-- Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along and
sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Oct 19, 2019
Then the polite next step is for you to indicate what that solution was so
people searching the archives can learn from your question. Was it to set the
kable option?
options(knitr.kable.NA = '-')
On October 19, 2019 12:50:20 PM PDT, Felipe Carrillo
wrote:
>You are correct. I didnt explain
You are correct. I didnt explain well and failed to mention that this is for
knitr::kable. I already figured it out.
Sent from Yahoo Mail on Android
On Sat, Oct 19, 2019 at 1:04 AM, Jeff Newmiller
wrote: Data frames are NOT spreadsheets. Don't treat them like spreadsheets.
All elements
Sigma squared or sigma square? Hotelling's T-squared or T-square?
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Sat, Oct 19, 2019 at 7:38 AM Therneau, Terry M., Ph.D. via R-help <
r-help@r-project.org> wrote:
> Martin,
>A fun question.
>
> Looking back at my oldest books, Feller (1950)
Daphne Koller (2009) describes L1 regularization (Chapter 20) as an
efficient way for Markov network (i.e. undirected graphical model)
structure learning and feature parameter estimation.
Her focus, and mine, are log-linear models for high-dimensional
contingency tables (i.e. categorical data).
Martin,
A fun question.
Looking back at my oldest books, Feller (1950) used chi-square.
Then I walked down the hall to our little statistics library and looked at
Johnson and
Kotz, "Continous Univariate Distributions", since each chapter therein has
comments about
the history of the
Hello,
See inline.
On Sat, Oct 19, 2019 at 01:48:44PM +0530, Sri Priya wrote:
> Dear R Users,
> Question:
>
> A chit fund has 25 members. Each month they contribute Rs 2000 each. End of
> the month, the person who bids the lowest for the corpus, gets his bid. The
> group organizer gets paid a
Hello,
Why not use read.xlsx argument 'na.strings', an argument that exists in
many file reading functions? (read.table, and derivatives.)
test <- lapply(sheets,function(i) {
read.xlsx("rainfall.xlsx", sheet = i,
startRow = 8, cols = 1:2,
na.strings = "")
})
Dear R Users,
I am practicing problems in the field of finance and economics. While
searching, I got problem like this. And I want to solve the following
problem using R. I dont have any clue how to incorporate this in R. Please
help me!!
Question:
A chit fund has 25 members. Each month they
Data frames are NOT spreadsheets. Don't treat them like spreadsheets. All
elements in a column are parts of a vector which means they all have the same
data type.
On the other hand, if you want to generate formatted output in HTML, LaTeX, or
Word, there are many tools for generating formatted
Consider the following dataset: I need to replace NAs with "-" but I lose my
numeric formatting fall.estimate <- structure(list(`Salmon` = c("salmon
River", "Ant Creek", "big Creek", "oso River", "linda Creek"), `baseline` =
c(80874.384012, 361.1997, 5012.8311, 638.6912, 402.1044),
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