read.delim2 did the trick -- many thanks!!!
On Wed, Oct 27, 2010 at 10:01 AM, Jorge Ivan Velez wrote:
> ?read.delim2
>
> HTH,
> Jorge
>
>
> On Wed, Oct 27, 2010 at 9:51 AM, Donald Braman wrote:
>
>> Thanks for your advice! I still get the same er
more columns than column names
Any other thoughts?
--
Donald Braman
http://ssrn.com/author=286206
http://www.culturalcognition.net/braman/
http://www.law.gwu.edu/Faculty/profile.aspx?id=10123
Henrique Dallazuanna
Tue, 26 Oct 2010 09:11:33 -0700
Try this:
read.table('don.5.clusters.txt'
x27;t get that error.
On Tue, Oct 26, 2010 at 10:49 AM, Duncan Murdoch
wrote:
> On 26/10/2010 10:33 AM, Donald Braman wrote:
>
>> I'm importing a lot of text tables of data (from Latent Gold) that
>> includes
>> hashes in some of the column names ("Cluster#1&
I'm importing a lot of text tables of data (from Latent Gold) that includes
hashes in some of the column names ("Cluster#1", "Cluster#2", etc.). Is
there an easy way to strip the offending hashes out before pushing the text
into a table or data frame? I thought I'd use gsub, e.g., but can't figur
Does anyone know of a package (or workaround) for fitting a dirichlet
distribution by maximum likelihood?
(I'm looking for something like this: http://repec.org/bocode/d/dirifit.html,
that allows for both dependent variables summing to 1 & predictive variables
of any sort.)
Don
-
As an alternative to Latent GOLD, I'm wondering if anyone knows of and R
package that can manage Latent Class Analysis with mixed variable types
(continuous, ordinal, and nominal/binary).
[[alternative HTML version deleted]]
__
R-help@r-project.
Thanks! I've figured out how to fix it, but how I got here is still a
puzzle. :-) Cheers, Don
On Sat, Oct 17, 2009 at 5:36 PM, Peter Ehlers wrote:
>
> Donald Braman wrote:
>
>> Can someone help me understand this results?
>>
>> levels(as.factor(miset1$facts
Can someone help me understand this results?
> levels(as.factor(miset1$facts_convict))
[1] "1" "1" "2" "3" "4" "5" "6"
converting to numeric and back doesn't seem to help:
> levels(as.factor(as.numeric(miset1$facts_convict)))
[1] "1" "1" "2" "3" "4" "5" "6"
It's messing up my ologits. Any way
I want the standard error associated with a correlation. I can calculate
using cor & var, but am wondering if there are libraries that already
provide this function.
[[alternative HTML version deleted]]
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R-help@r-project.org mailing list
http
;anotherdv", "yetanotherdv")
for (iv in ivs) {
for (dv in dvs) {
graphname <- paste(iv, dv, ".png", sep = "")
png(file=graphname, width=300, height=300)
plot(dv ~ iv, pch=".")
lines(loess.smooth(iv, dv), lty=1)
dev.off()
}
}
Clearly that doesn't wor
Resolved. It works if I set iter=0.
On Thu, Aug 6, 2009 at 9:03 PM, Donald Braman wrote:
> I was trying to fit a curve to the number of people who identify as liberal
> by age. I got some puzzling results which suggested to me that I don't
> really understand how local poly
I am trying to fit a curve to the number of people who identify as liberal
by age. I got some puzzling results which suggested to me that I don't
really understand how local polynomial fitting works. Why, I am wondering,
is lowess producing a local fit of zero for every age?
> liberal
[1] 0 0 0
I was trying to fit a curve to the number of people who identify as liberal
by age. I got some puzzling results which suggested to me that I don't
really understand how local polynomial fitting works. Why, I am wondering,
is lowess producing a local fit of zero for every age?
> liberal.bin
[1]
Curious to know if recode can work with strings containing colons. I
haven't gotten it to work yet, but perhaps there is a way?
Donald Braman
http://www.culturalcognition.com/braman/
__
R-help@r-project.org mailing list
https://stat.ethz.ch/ma
#x27;, 'florida', 'georgia', 'idaho'
, 'illinois', 'indiana', 'iowa', 'kansas', 'kentucky', 'louisiana', 'maine'
, 'michigan', 'mississippi', 'montana', 'nevada
Thanks for the help everyone! I'm new to vectors, and don't quite get it.
This works for me:
binary.vars <- c("q1", "q2", "q3", ...)
apply(mydata[binary.vars], 2, tapply, mydata["male"], mean)
but this doesn't:
other.vars <- c("male", "race", "religion")
apply(mydata[other.binary.vars], 2, ta
I've been playing around with various table tools, trying to construct a
fairly simple cross-tab. It shouldn't be hard, but for some reason it
turning out to be (for me).
If I want to see how many men and how many women agree with a agree/disagree
question (coded 1,0), I can do this:
>attach(myd
Is there an easy way to invert a table? (not to solve for the inverted
matrix, just swap rows for columns & vice versa). I've gone through my data
manipulation bible (Phil Spector's book), but to no avail.
[[alternative HTML version deleted]]
rique Dallazuanna <[EMAIL PROTECTED]>wrote:
> Try this:
>
> my.df$my.newvar <- quantile(my.df$my.var, probs = seq(0.01,1, 0.01))
>
>
> On Sat, Sep 27, 2008 at 3:50 AM, Donald Braman <[EMAIL PROTECTED]>
> wrote:
> > I'm wondering if there is a simple way to assig
I'm wondering if there is a simple way to assign a quantile to a vector in a
data frame, much like one could in Stata using centile. Let's say I want 100
slices in my assignation. I can easily see what the limits of each slice by
using quantile:
quantile(my.df$my.var, probs=seq(0, 1, 0.01))
But ho
I'm coming from the AMOS world and am wondering if there is a simple
way to do multiple hypothesis testing in the manner of BIC analyses in
AMOS using the sem package in R. I've read the documentation, but
don't see anything in there except for basic BIC scores. Perhaps
someone has devised a simp
ons
> invisible(apply(xpnd, 1, function(.row) {
> jpeg(paste(paste(.row, collapse="_"),".jpg", sep=''))
>
> my.fit <- lm( my.df[[.row[1]]] ~ my.df[[.row[2]]] + my.df[[.row[3]]] +
>my.df[[.row[2]]]:my.df[[.row[3]]])
> colors <- ifel
ot;.")
curve (cbind (1, 1, x, 1*x) %*% coef(my.fit), add=TRUE, col="black")
curve (cbind (1, 0, x, 0*x) %*% coef(my.fit), add=TRUE, col="gray")
})
dev.off()
}
}
}
# Clearly that's wrong -- why it's wrong is obsc
I'm stumbling my way through manipulating data in multiply imputed datasets,
and have run into a problem translating code I used to run on my pre-imputed
dataset to multiple datasets. The imputation runs just fine, as does the
reading of the mi data sets into an imputationList. I run into trouble
--- snip ---
>
> I hope this helps,
> John
>
> ------
> John Fox, Professor
> Department of Sociology
> McMaster University
> Hamilton, Ontario, Canada
> web: socserv.mcmaster.ca/jfox
>
>> -Original Message-
&g
))
I've also tried some basic loops. I guess I'm also a bit confused as
to when R references the original object and when it creates a new
one. I suppose I could do this in Python and the use PyR, but I'd
really like to learn a bit more about how R syntax.
Any help on this specif
code x, in a
> similar way to what you tried before. This function of x we define will get
> called three times in the above example, once for each of
> reverse_me_varnames. It will then assign those three new columns to the
> left-hand side of the <- operator, which are three n
7;, as.factor.result=FALSE)
While I don't get an error message, the data don't change. Any advice on
reverse coding non-continguous variables?
On Mon, May 19, 2008 at 4:12 PM, Donald Braman <[EMAIL PROTECTED]>
wrote:
> Many thanks --
>
> You are right; I had
Many thanks --
You are right; I had rnorm() and sample() mixed up in my code. I'll work on
generating a normal ordinal sample next.
Cheers, Don
On Mon, May 19, 2008 at 4:07 PM, Erik Iverson <[EMAIL PROTECTED]>
wrote:
> Hello -
>
> Donald Braman wrote:
>
>> # I&#x
esources listed on CRAN.
>
> -- Bert gunter
> Genentech
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
> On
> Behalf Of Donald Braman
> Sent: Monday, May 19, 2008 12:42 PM
> To: r-help@r-project.org
> Subject: [R] recoding data with
# I'm new to R and am trying to get the hang of how it handles
# dataframes & loops. If anyone can help me with some simple tasks,
# I'd be much obliged.
# First, i'd like to generate some random data in a dataframe
# to efficiently illustrate what I'm up to.
# let's say I have six variables as li
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