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Version 0.11.0
- Enhance type detection in sqliteDataType (dbDataType). The storage
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On Thu, Dec 1, 2011 at 11:54 AM, emorway emor...@usgs.gov wrote:
Hello,
A bit of fairly simple code, yet I don't seem to be able to manipulate it
quite as much as I would like:
1) It would be nice if the objects appearing in the legend were aligned,
and by aligned I mean the boxes are
Thanks Michael, I'll keep that in mind if I want to do anything more
complicated.
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Hello,
I have the following problem. After a PACS upgrade in our hospital, the dynamic
3D images acquired after contrast media injection with a 3T Philips MRI are all
saved in a single 4D file instead of a series of 2D images containing multiple
3D images over time. I used to convert this
Hi, I have a problem with mixed effect model (multilevel).
My model that is written in following formulation using the lme4
package and Zelig package:
mylogit- lmer(OVERFLOW ~ ALTEZZA + INTENSITA + ( 1 | CODICE), family=
binomial(link = probit),data = dati)
OVERFLOW can be 0 or 1 and represents
I ran the following code:
And I run into problems with the last line of code (when it says
hn-..). I keep getting an error code: Error in distsamp(~hab ~ 1,
peldist, keyfun = halfnorm, output = density, :
object 'A' not found
I would appreciate any and all help.
Here you go:
attach(as.data.frame(l_yx))
range(x[y==1])
[1] -22500.746.
range(x[y==0])
[1] -10076.5303653.0228
How do I know what is acceptable?
Also, here are the screen shots of my data that I tried to send earlier
(two screen shots, two pages):
Oops! Please ignore my last post. I mistakenly gave you different data I
was testing with. This is the correct data:
Here you go:
attach(as.data.frame(l_yx))
range(x[y==0])
[1] 0.0 14.66518
range(x[y==1])
[1] 0.0 13.49791
How do I know what is acceptable?
Also, here are the
I'm not proposing this as a permanent solution, just investigating the
warning. I zeroed out the three outliers and received no warning. Can
someone tell me why I am getting no warning now?
I did this 3 times to get rid of the 3 outliers:
mx_dims = arrayInd(which.max(l_yx), dim(l_yx))
Not that it really matters, but
Can someone explain how the row numbers get assigned in the following
sequence? It looks like something funky happens when rbind() coerces
'bar' into a dataframe.
In either sequence of rbind below, once you get past the first two rows,
the row numbers count
Someone is bound to know a better way, but...
subset(unlist(Version1_), subset=names(unlist(Version1_))==First)
LCOG1 wrote
Hi everyone,
I looked around the list for a while but couldn't find a solution to my
problem. I am storing some results to a simulation in a list and for
Those are row *names*, not row *numbers*. It's just that if you don't specify,
numbers are assigned by default when creating a data frame.
Your rbind() statements are implicitly creating a data frame, so the likely
information is in ?data.frame:
check.names: logical. If ‘TRUE’ then the names of
On Dec 1, 2011, at 23:43 , Ben quant wrote:
I'm not proposing this as a permanent solution, just investigating the
warning. I zeroed out the three outliers and received no warning. Can someone
tell me why I am getting no warning now?
It's easier to explain why you got the warning before.
# need zoo to use rollapply()
# your data (I called df)
df - structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
e = c(1L, 5L), f = c(4, 7)), .Names = c(a, b, c, d,
e, f), class = data.frame, row.names = c(NA, -2L))
# transpose and make a zoo object
df2 - zoo(t(df))
#rollapply
Thank you so much for your help.
The data I am using is the last file called l_yx.RData at this link (the
second file contains the plots from earlier):
http://scientia.crescat.net/static/ben/
Seems like the warning went away with pmin(x,1) but now the OR is over
15k. If I multiple my x's by
On 2011-12-01 08:54, emorway wrote:
Hello,
A bit of fairly simple code, yet I don't seem to be able to manipulate it
quite as much as I would like:
1) It would be nice if the objects appearing in the legend were aligned,
and by aligned I mean the boxes are centered over the lines. Do I need
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On Thu, Dec 1, 2011 at 7:13 PM, B77S bps0...@auburn.edu wrote:
# need zoo to use rollapply()
# your data (I called df)
df - structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
e = c(1L, 5L), f = c(4, 7)), .Names = c(a, b, c, d,
e, f), class = data.frame, row.names = c(NA, -2L))
Sorry for that, and thanks Gabor,
I could have sworn that it wouldn't.
Gabor Grothendieck wrote
On Thu, Dec 1, 2011 at 7:13 PM, B77S lt;bps0002@gt; wrote:
# need zoo to use rollapply()
# your data (I called df)
df - structure(list(a = 1:2, b = 2:3, c = c(5L, 9L), d = c(9L, 6L),
e =
Sorry -- I meant to write 'row names,' but the question specifically is
where those unlikely numbers come from. So I guess it comes down to
why, when 'bar' is the first item, the row name is assigned '2' rather
than '1' .
On 12/1/11 6:26 PM, Sarah Goslee wrote:
Those are row *names*, not
Hi everyone,
I'm having problems with plotting my data. I have a set of positions with
different attributes and I'm wondering if I can plot it, as x,y plot, with
gradient colours according to a 3rd factor. Is it possible to show gradient
segments between postions and not one-coloured segment
Thanks Michael! Yeah, that dealt with one of the problems.
I still get the following error message:
Error in cat(list(...), file, sep, fill, labels, append) :
argument 1 (type 'list') cannot be handled by 'cat'
I know that this has something to do with writing the names of the output
file,
Hi everyone,
I looked around the list for a while but couldn't find a solution to my
problem. I am storing some results to a simulation in a list and for each
element i have two separate vectors(is that what they are called, correct my
vocab if necessary). See below
Version1_-list()
for(i in
How about:
lapply(Version1_, subset, subset=c(TRUE, FALSE))
or sapply() depending on what you want the result to look like.
Thanks for the reproducible example.
Sarah
On Thu, Dec 1, 2011 at 5:17 PM, LCOG1 jr...@lcog.org wrote:
Hi everyone,
I looked around the list for a while but couldn't
I used within and vapply:
x - data.frame(Col1 = c(abc/def, ghi/jkl/mno), stringsAsFactors = FALSE)
count.slashes - function(string)sum(unlist(strsplit(string, NULL)) ==
/)within(x, Col2 - vapply(Col1, count.slashes, 1))
Col1 Col21 abc/def 12 ghi/jkl/mno 2
On Thu, Dec 1, 2011
Resending my code, not sure why the linebreaks got eaten:
x - data.frame(Col1 = c(abc/def, ghi/jkl/mno), stringsAsFactors = FALSE)
count.slashes - function(string)sum(unlist(strsplit(string, NULL)) == /)
within(x, Col2 - vapply(Col1, count.slashes, 1))
Col1 Col2
1 abc/def1
2
I really would like to be able to read about this in a document but I
cannot find my way around the documentation properly
Given the code...
M - matrix(runif(5*20), nrow=20)
colnames(M) - c('a', 'b', 'c', 'd', 'e')
ind - c(1,2,3,4)
dep - 5
I can then do...
l2 - lm(M[,dep]~M[,ind]) ## Clearly
I have this. I have modified your input structure to be a matrix. I
think it is generally recommended to use matrices over data.frames
when your data allows it, i.e., when you only have one type of data,
here character(). Matrices are easier to work with.
x - matrix( c('BMW', '', '',
Similarly, this might work:
unlist(lapply(Version1_, `[`,First))
Michael
On Thu, Dec 1, 2011 at 9:41 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
How about:
lapply(Version1_, subset, subset=c(TRUE, FALSE))
or sapply() depending on what you want the result to look like.
Thanks for the
strsplit is certainly an alternative, but your approach is
unnecessarily complicated and inefficient. Do this, instead:
sapply(strsplit(x,/),length)-1
Cheers,
Bert
On Thu, Dec 1, 2011 at 7:44 PM, Florent D. flo...@gmail.com wrote:
Resending my code, not sure why the linebreaks got eaten:
x -
Tena koe Lara
If I understand your question correctly, I use the colorspace package for that
sort of thing, but you could also use the built-in colour palettes such as
rainbow and topo.colors
HTH
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org
Inefficient, maybe, but what you suggest does not work if a string
starts or ends with a slash.
On Thu, Dec 1, 2011 at 11:11 PM, Bert Gunter gunter.ber...@gene.com wrote:
strsplit is certainly an alternative, but your approach is
unnecessarily complicated and inefficient. Do this, instead:
On Dec 1, 2011, at 11:11 PM, Bert Gunter wrote:
strsplit is certainly an alternative, but your approach is
unnecessarily complicated and inefficient. Do this, instead:
sapply(strsplit(x,/),length)-1
Definitely more compact that the regex alternates I came up with, but
one of these still
On Dec 1, 2011, at 10:50 PM, Worik R wrote:
I really would like to be able to read about this in a document but I
cannot find my way around the documentation properly
Given the code...
M - matrix(runif(5*20), nrow=20)
colnames(M) - c('a', 'b', 'c', 'd', 'e')
ind - c(1,2,3,4)
dep - 5
I can
On Dec 1, 2011, at 1:13 PM, lglew wrote:
Hi R-users,
I'm trying to produce decompositions of a multiple time-series,
grouped by a
factor (called area). I'm modifying the code in the STLperArea
function of
package ndvits, as this function only plots produces stl plots, it
does not
Hi All,
I get the message failed with message Dimension out of range when using
cuhre in package R2Cuba. Does anyone know what this mean? Or would I need
to email the package author?
The funny thing is it does give a result and comparing it to
adaptIntegrate in package cubature, the two
hi
I need some help in dbWriteTable.
I'm not able to insert the rows in the table if the column order are not
same in the database and in the dataframe which i'm inserting. Also facing
issue if the table is already created externally and inserting it thru
dbWrite.
is there some way that we
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