I am trying to display a table with traitr. I am using the function tableitem
apart of traitr. I am not sure how to enter it. I enter my data as a table
in R (as list2) but im not sure how to use that function tableitem(list2)
but that doesnt seem to work. I would rather not edit it just display t
On Mon, 24 May 2010, Samuel Dennis wrote:
I am attempting to import dates in the following format to R:
5/20/2010 6:45:32 PM
Unfortunately I am unable to get the AM/PM function (%p) to work correctly
under either 2.11.0 or 2.8.1.
strptime("5/20/2010 6:45:32 PM", "%m/%d/%Y %I:%M:%S %p")
[1] NA
If the days are consecutive with no missing rows then the dates don't
need to be calculated and it could be represented as a ts series with
a frequency of 7. Just aggregate it down to a frequency of 1:
rain <- ts(dat$rain, freq = 7)
aggregate(rain, 1)
If there are missing rows (or even the
HI, Dear R community,
I want to know how to select the optimal decision threshold from the ROC
curve? At what threshold will give the highest accuracy?
Thanks!
--
Sincerely,
Changbin
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On May 23, 2010, at 8:41 PM, Claudia Rodriguez wrote:
Dear Friends.
I am just starting to use R. And in this occasion I want to
construct a
high-dimensional contingency table, because I want to crate a mosaic
plot
with the vcd package.
My table is in this format:
año ac.repcat.gru
Thanks for the suggestions! This will keep me busy for a while.
Tom
2010/5/15 Muenchen, Robert A (Bob) :
>> Thomas Levine wrote:
>>Bob Muenchen says that 'Ralph O’Brien says that
>>in a few years there will be so many students
>>graduating knowing mainly R that [he]’ll need to
>>write, “SAS for R
I know it is not very useful to you, but on Vista with 2.11.patched it
works:
> strptime("5/20/2010 6:45:32 PM", "%m/%d/%Y %I:%M:%S %p")
[1] "2010-05-20 18:45:32"
> strptime("5/20/2010 6:45:32", "%m/%d/%Y %I:%M:%S")
[1] "2010-05-20 06:45:32"
>
> sessionInfo()
R version 2.11.0 Patched (2010-04-26
I am attempting to import dates in the following format to R:
5/20/2010 6:45:32 PM
Unfortunately I am unable to get the AM/PM function (%p) to work correctly
under either 2.11.0 or 2.8.1.
> strptime("5/20/2010 6:45:32 PM", "%m/%d/%Y %I:%M:%S %p")
[1] NA
but
> strptime("5/20/2010 6:45:32", "%m/%d/
Dear Friends.
I am just starting to use R. And in this occasion I want to construct a
high-dimensional contingency table, because I want to crate a mosaic plot
with the vcd package.
My table is in this format:
año ac.repcat.gru conteos
1 2005 Rparejas 253
2 2005 Npa
Thanks for your time with this. Erik's solution works best to deal with the
input... I'll try to reshape the output back into the appropriate columns.
David, fold(sq$s1) only outputs the result for the first sequence in the
list I'm afraid. The 'fold' function doesn't deal well with spaces...
Th
Hello,
I am running R on a server that several people share. Previously we
all had separate libraries for R.
I have set up R so everyone on the server shares the same library and
I downloaded the latest version of R and installed it on the
main drive of our server in the "Program Files" folder (o
This is one way to do it. Suppose your data is in the file "rainfall.txt", as
set out below. Then
> dat <- read.table("rainfall.txt", header = TRUE)
> dat <- within(dat, {
+ date <- as.Date(paste(year, month, day, sep="-"))
+ week <- factor(as.numeric(date - date[1]) %/% 7)
+ })
> wRain <-
Dear list,
I'd like to use path.analysis in the package agricolae in batch format on
many files, retrieving the path coefficients for each run and appending them
to a table. I don't see any posts in the help files about this package or
the path.analysis package. I've tried creating an object out of
Hi r-users,
I have this data below. I would like to obtain the weekly rainfall sum. That
is I would like to find sum for day 1 to day 7, day 8 - day15, and so on.
year month day rain
1 1922 1 1 0.0
2 1922 1 2 0.0
3 1922 1 3 0.0
4 1922 1 4 0.0
5 1922 1
after read.delim:
'data.frame': 60 obs. of 4 variables:
$ Cell : Factor w/ 60 levels "BR:BT_549","BR:HS578T",..: 23 51 20 25 34
16 44 3 60 55 ...
$ hsa-miR-204: num -4.37 -4.34 -4.33 -4.29 -4.26 ...
$ hsa-miR-210: num -0.223 1.575 1.66 1.668 0.373 ...
$ Tissue : Factor w/ 9 lev
Good evening gentlemen!
I have a test in split-plot with randomized block design where my answer is
a binomial variable. I wonder if there is any way I can calculate the
probability of my factors considering the design errors in the case are two.
I looked at various threads here and elsewhere, a
On May 23, 2010, at 6:32 PM, Zoppoli, Gabriele (NIH/NCI) [G] wrote:
This is what I get:
str(x)
chr [1:60, 1:4] "ME:SK_MEL_5" "ME:SK_MEL_28" "ME:SK_MEL_2" ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:60] "48" "47" "46" "50" ...
..$ : chr [1:4] "Product" "hsa.miR.204" "hsa.miR.210" "Tissu
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Zoppoli,
> Gabriele (NIH/NCI) [G]
> Sent: Sunday, May 23, 2010 3:44 PM
> To: ted.hard...@manchester.ac.uk
> Cc: R-help@r-project.org
> Subject: Re: [R] "order" issue
>
> crazy s
crazy stuff!!! I tried to reload the txt file, and now it's working...
this is the original (attached)
thanks!
Gabriele Zoppoli, MD
Ph.D. Fellow, Experimental and Clinical Oncology and Hematology, University of
Genova, Genova, Italy
Guest Researcher, LMP, NCI, NIH, Bethesda MD
Work: 301-451-85
This is what I get:
str(x)
chr [1:60, 1:4] "ME:SK_MEL_5" "ME:SK_MEL_28" "ME:SK_MEL_2" ...
- attr(*, "dimnames")=List of 2
..$ : chr [1:60] "48" "47" "46" "50" ...
..$ : chr [1:4] "Product" "hsa.miR.204" "hsa.miR.210" "Tissue"
It doesn't make much sense to me...
I would like to have the se
On 23-May-10 21:39:06, Zoppoli, Gabriele (NIH/NCI) [G] wrote:
> Hi everybody, this is a real dummy thing.
>
> I sorted a matrix based on a given column, and what I get is right,
> until it comes to columns of negative and positive values; than,
> "order" orders everything from max to min in the ne
I tried this, but it doesn't change the "division" between negative and
positive values (see that you have first positive from max to min, and then
negative from min to max, as if "order" considered only the absolute values...)
Product hsa.miR.204 hsa.miR.210 Tissue
48 "ME:SK_MEL_5
do 'str' on your object to see if you have factors where you think you
have numerics.
What is the problem you are trying to solve?
Sent from my iPhone.
On May 23, 2010, at 17:39, "Zoppoli, Gabriele (NIH/NCI) [G]" > wrote:
Hi everybody, this is a real dummy thing.
I sorted a matrix based o
Hi everybody, this is a real dummy thing.
I sorted a matrix based on a given column, and what I get is right, until it
comes to columns of negative and positive values; than, "order" orders
everything from max to min in the negative values, and then AGAIN from max to
min in the positive values!
On May 23, 2010, at 1:43 PM, Erik Iverson wrote:
Hello,
Can anyone think of a non-iterative way to generate a decreasing
geometric sequence in R?
For example, for a hypothetical function dg, I would like:
> dg(20)
[1] 20 10 5 2 1
where I am using integer division by 2 to get each subsequ
Erik Iverson ccbr.umn.edu> writes:
> Can anyone think of a non-iterative way to generate a decreasing geometric
> sequence in R?
Reduce("%/%",rep(2,4),init=20,accum=TRUE)
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Rick Reiss exponent.com> writes:
>
>
> I'm using the plotCI function and I'd like to overlay additional means
> with CIs onto an existing plotCI-created plot in a different color. Is
> this possible? Thanks.
>
> Rick
>
>
Assuming you mean the one from the plotrix package: use add=TRUE
Erik Iverson writes:
> Hello,
>
> Can anyone think of a non-iterative way to generate a decreasing
> geometric sequence in R?
>
> For example, for a hypothetical function dg, I would like:
>
>> dg(20)
> [1] 20 10 5 2 1
>
> where I am using integer division by 2 to get each subsequent value in
> t
I'm using the plotCI function and I'd like to overlay additional means
with CIs onto an existing plotCI-created plot in a different color. Is
this possible? Thanks.
Rick
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Erik Iverson wrote:
Hello,
Can anyone think of a non-iterative way to generate a decreasing geometric
sequence in R?
For example, for a hypothetical function dg, I would like:
> dg(20)
[1] 20 10 5 2 1
where I am using integer division by 2 to get each subsequent value in the
sequence.
Hello,
Can anyone think of a non-iterative way to generate a decreasing geometric
sequence in R?
For example, for a hypothetical function dg, I would like:
> dg(20)
[1] 20 10 5 2 1
where I am using integer division by 2 to get each subsequent value in the
sequence.
There is of course:
d
On May 23, 2010, at 10:00 AM, Kang Min wrote:
Hi,
I have a dataset that looks like the one below.
data
plot plantno.species
H 31 ABC
D 2 DEF
Y 54 GFE
E 12 ERF
Y 98 FVD
H
Thanks, but what I want is not 100 groups of 7 samples. Let's say in
my samp2 I get
[[1]] "D" "H" "K" "S" "E" "U" "O"
[[2]] "H" "S" "R" "V" "A" "L" "B"
etc...
I want to select all rows from 'data' containing "D" "H" "K" "S" "E"
"U" "O" first, then "H" "S" "R" "V" "A" "L" "B" and so on.
On May
try this:
> x <- read.table(textConnection("plot plantno.species
+ H 31 ABC
+ D 2 DEF
+ Y 54 GFE
+ E 12 ERF
+ Y 98 FVD
+ H 4 JKU
+ J 7 J
Hi,
I have a dataset that looks like the one below.
data
plot plantno.species
H 31 ABC
D 2 DEF
Y 54 GFE
E 12 ERF
Y 98 FVD
H 4 JKU
J 7 J
On May 22, 2010, at 10:48 PM, Mohan L wrote:
Dear All,
I have an array some thing like this:
avglog
January February March April May June July
August September
60102 83397 56774 48785 49010 40572 38175
47037 51402
The class of "avglog" a
On 05/23/2010 06:29 AM, Marc Carpentier wrote:
Thanks for the answer.
Unfortunately, I'm not yet skilled enough to do such a thing. I had a look on
the code and I'll try to understand it, as a good exercise.
I thought about sending fake fit objects to nomogram() derived from the
original one :
On May 23, 2010, at 3:27 AM, Erik Iverson wrote:
Hello,
sedm1000 wrote:
Sorry - I figured that this to be a more common defined error than
anything
specific to the data/function... Thanks for looking at this.
The data and function are below. Creating a single line of the
data.frame at
a
As Frank mentioned in his reply, expecting to estimate tens of
thousands of fixed-effects parameters in a logistic regression is
optimistic. You could start with a generalized linear mixed model
instead
library(lme4)
fm1 <- glmer(resp ~ 1 + (1|f1) + (1|f2) + (1|f1:f2), mydata, binomial))
If you
On Sun, May 23, 2010 at 5:09 AM, Peter Ehlers wrote:
> On 2010-05-23 0:56, john smith wrote:
>>
>> Hi,
>> I am trying to implement Higham's algorithm for correcting a non positive
>> definite covariance matrix.
>> I found this code in R:
>>
>> http://projects.cs.kent.ac.uk/projects/cxxr/trac/brows
Thanks for the answer.
Unfortunately, I'm not yet skilled enough to do such a thing. I had a look on
the code and I'll try to understand it, as a good exercise.
I thought about sending fake fit objects to nomogram() derived from the
original one :
- orignal : f2<- cph(Surv(d.time,death) ~
sex*(r
On Sun, 2010-05-23 at 00:56 -0700, dusadrian wrote:
> This might help, depending on your exact needs:
> > v1 <- sample(letters[1:2], 10, replace=TRUE)
> > v2 <- sample(letters[3:4], 10, replace=TRUE)
> > v3 <- sample(letters[5:6], 10, replace=TRUE)
> > aa <- data.frame(v1=v1, v2=v2, v3=v3)
And now
Hello Jim,
It sounds like a good time to go read about the packages
bigmemory
and/or
ff
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.
On 2010-05-23 0:56, john smith wrote:
Hi,
I am trying to implement Higham's algorithm for correcting a non positive
definite covariance matrix.
I found this code in R:
http://projects.cs.kent.ac.uk/projects/cxxr/trac/browser/trunk/src/library/Recommended/Matrix/R/nearPD.R?rev=637
I managed to un
You are trying to create an object with 1G elements. Given that these
are integers, this will require about 4GB of space. If you are
running on a 32-bit system, which has a total phyical limit of 2-3GB
depending on what options you are running (at least on Windows), then
you have exceeded the lim
This might help, depending on your exact needs:
> v1 <- sample(letters[1:2], 10, replace=TRUE)
> v2 <- sample(letters[3:4], 10, replace=TRUE)
> v3 <- sample(letters[5:6], 10, replace=TRUE)
> aa <- data.frame(v1=v1, v2=v2, v3=v3)
> aa
v1 v2 v3
1 a d e
2 a d e
3 a c e
4 b d e
5
thmsfuller...@gmail.com wrote:
Hello All,
sample() only sample on one variable x. But I'm interested in sampling
more than one variable without replacement.
Suppose I have 3 vectors x, y, z. I want to draw samples from all
three vectors such that the combination of the three elements in each
dr
Hello,
sedm1000 wrote:
Sorry - I figured that this to be a more common defined error than anything
specific to the data/function... Thanks for looking at this.
The data and function are below. Creating a single line of the data.frame at
a time will work (i.e. fold(s))
For multiple line data.f
Thanks a lot, it works!
On May 23, 3:10 pm, Erik Iverson wrote:
> > "[" is a function, and you want to use it on each element of the list,
> > so...
>
> > lapply(x, "[", c(1:7))
>
> and the call to c() is of course not necessary, since ":" will generate a
> vector.
>
> __
Caitlin Sadowski wrote:
I have a large csv table I am trying to read into R. I would like each
column to be of type factor. However, most columns have only numeral
entries (e.g. likert scales), so are automatically imported as type
numeric. Is there a way to convert ALL columns to be of type fact
Hello All,
sample() only sample on one variable x. But I'm interested in sampling
more than one variable without replacement.
Suppose I have 3 vectors x, y, z. I want to draw samples from all
three vectors such that the combination of the three elements in each
draw is not the same as any previou
"[" is a function, and you want to use it on each element of the list,
so...
lapply(x, "[", c(1:7))
and the call to c() is of course not necessary, since ":" will generate a
vector.
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Kang Min wrote:
Hi,
I have a list of 100, each list has 20 elements, and I would like to
select the first 7 elements in each list.
Let's take the alphabet as an example.
x <- lapply(1:100, function(i) sample(LETTERS))
I tried x[[1:7]], but it doesn't work. Can anyone enlighten me on how
to do
I have a large csv table I am trying to read into R. I would like each
column to be of type factor. However, most columns have only numeral
entries (e.g. likert scales), so are automatically imported as type
numeric. Is there a way to convert ALL columns to be of type factor,
without having to conv
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