This is not really a question about R... it is a question about databases
because similar divisions occur in every imperative programming language that
deals with databases. ODBC is a software layer that aims to create a uniform
API for the programmer among different variations of SQL databases
On 21/01/2015 05:51, Lalitha Kristipati wrote:
Hi,
I am trying to connect R to databases.
I found two ways to connect.
1.By using specific package (eg RMySQL to connect to R)
2.By using connectors and RODBC package.
can any one tell me the difference between those two methods.
See the 'R D
Dear all;
I am the new user of R. I want to simulation or prediction the Eutrophication
of a lake. I have weekly data(almost for two years) for Total phosphorus, Total
N, pH, Chlorophyll a, Alkalinity, Silica.
Can I predict the Eutrophication by Neural Network in R?
How can I simulation the Eutro
Hi,
I am trying to connect R to databases.
I found two ways to connect.
1.By using specific package (eg RMySQL to connect to R)
2.By using connectors and RODBC package.
can any one tell me the difference between those two methods.
Regards,
Lalitha K.
Associate Software Engineer
917411291
On Mon, Jan 19, 2015 at 5:38 AM, Jorge I Velez wrote:
> Dear all,
>
> Given vectors "x" and "y", I would like to compute the proportion of
> entries that are equal, that is, mean(x == y).
>
> Now, suppose I have the following matrix:
>
> n <- 1e2
> m <- 1e4
> X <- matrix(sample(0:2, m*n, replace =
Hi Zuzana,
>From your description it seems that you want to repeat the last
"state_new" value when there are repeated zeros in both "state" and
"state_bef". If that is the case, you may need to step through the
data frame like this:
for(drow in 1:(dim(test)[1])) {
if(test$state[drow] != 0) test$s
Hi,
I have probably basic question, but I would be happy if somebody could help
me with this:
I have a dataframe where I want to assign a value to other column copying
the value in previous row under some condition.
Here is an example (dput on the end of message):
> test[278:290,]
Boris,
Thanks very much, this looks just like what I need!
All the best,
Scott
From: Boris Steipe [boris.ste...@utoronto.ca]
Sent: 20 January 2015 20:21
To: Scott Robinson
Cc: r-help@r-project.org
Subject: Re: [R] Rewriting the Biplot Function
Scott -
Hi Joshua,
Chel Hee Lee is correct, you can use rbind. Here is a slight
modification of the script I sent before.
reshape_Rqtl<-function(x) {
newx<-list()
for(dfi in 1:length(x)) {
lenxdfi<-length(x[[dfi]])
x[[dfi]]$trait<-names(x)[dfi]
x[[dfi]]$locus<-rownames(x[[dfi]])
rownames(x[[dfi]
... sorry: to be more clear - you would create the biplot with option type="n",
then use points() and text() to plot whatever you need while using your colour
vector.
B.
On Jan 20, 2015, at 3:21 PM, Boris Steipe wrote:
> Scott -
>
> A few months ago I posted on this list a modified version
Scott -
A few months ago I posted on this list a modified version of biplot that takes
a colour argument (and preserves the axes information, so you can use points()
). I don't have time right now to experiment and look at your code, but perhaps
this does "out of the box" what you need.
Cheers
Hi, Beverly:
In pamr, only numerical features can be accepted. You may check your
dataset to make sure all the data is numerical before training.
On Mon, Jan 19, 2015 at 5:14 PM, Beverly Nguyen
wrote:
> I have recently started learning how to use R, and I am trying to create an
> False Discove
Dear R-Help,
I have been trying to rewrite the base biplot.prcomp function but am coming
across some errors I don't understand. It seems like R is still 'expecting' the
same values despite me rewriting and renaming the methods. My aim is simply to
have an additional biplot function which could
Dear all,
I have a time series (years with n column's) and I want to decompose this
time series in high, low and bandpass components using a n-years filter (in
my case 20 years). Until now I tried many libraries such as mFilter,
signal, spectral.methods, but unfortunately without success.
Please h
...
(just a comment)
and since this appears to be O(m^2 x n), where m,n are the number of
rows and columns (correction requested if I got this wrong), it would
appear that some basically C level functionality -- perhaps the one
Jean suggested? -- would be required for even moderately "large"
matr
Jorge,
I have not used it myself, but you might find the dist() function in the
proxy package to be useful.
http://cran.r-project.org/web/packages/proxy/index.html
Jean
On Mon, Jan 19, 2015 at 7:38 AM, Jorge I Velez
wrote:
> Dear all,
>
> Given vectors "x" and "y", I would like to compute the
Dear Kim,
This seems to be an arules problem. More specifically in the subset
function. I will investigate and fix this as soon as possible. You can
use the following code in the meanwhile:
rr <- rhs(ruleset) %pin% "Product="
inspect(ruleset[rr])
Regards,
Michael
On 01/20/2015 04:55 AM, Kim
I have recently started learning how to use R, and I am trying to create an
False Discovery Rate plot for my data. I loaded the "pamr" package, and am
using a data set with 11 columns: column 1 is a list of Gene IDs, column 2
is a list of Gene Symbols, and columns 3-11 are the dependency rates of
d
Hi there,
I've generated the following data:
library(quantreg)
library(survival)
set.seed(789)
N <- 2000
u <- runif(N)
x1 <- rbinom(N,1,.5)
x2 <- rbinom(N,1,.5)
x1x2<-x1*x2
lambda <- 1 + 1.5*x1 + 1.5*x2 + .5*x1x2
k <- 2
y <- lambda*((-log(1-u))^(1/k));max(y)
c <- runif(N,max=15)
event = as.numer
I hope this toy example gives an idea to understand the problems that
you are facing now. Please see the below:
# 1. Create an empty list
> A <- list()
> A
list()
# 2. Add the first component (vector with two elements).
> A[[1]] <- c(2,6)
> A
[[1]]
[1] 2 6
# 3. Now, add the third component
Hi Joshua,
You may use 'do.call()'. Please see the output in below:
> result <- lapply(names(op), function(x){
+ col1 <- x
+ col2 <- row.names(op[[x]])
+ mat <- op[[x]]
+ row.names(mat) <- NULL
+ rval <- cbind(col1, col2, mat)
+ names(rval) <- c("trait", "locus", names(mat))
+ rva
Hello,
kindly ,
how to catch this Error
Error in A[[i]] : subscript out of bounds
and check that the list is empty is.null( A[[i]] ) do no twork
thanks in advance
R.I
[[alternative HTML version deleted]]
__
Additionally, maybe it's interesting to note that I remember that I first also
tried the apriori function but with support = 0.001 (the rest remains the
same). It gave me the negative length vector error as well but then I thought
to update the packages and it seemed like it worked because the
If the email is still unreadable, please look at the attachment
email.txt. Dear Martin,Thanks for your fast reply. Sorry about
the formatting of the previous email. I actually tested it before sending and
it came fine when sending to my own account but when I send it to the r-help
email address
On 19.01.2015 23:58, Glenda Palacios wrote:
I have any problems in R:
The downloaded binary packages are in
/var/folders/s9/kr21631n3nj30hb_4td3pv_8gn/T//RtmpDDAFmJ/downloaded_packages
Warning messages:
1: In download.file(url, destfile, method, mode = "wb", ...) :
downloaded length 21
> Hi all, I have a question concerning an error that occurs when using the
> inspect() function on a set of rules made by the apriori function from the
> arules package. I have a dataset with 12 million records. It contains some
> basic sales information (such as product, customer data). I want
Hi all, I have a question concerning an error that occurs when using the
inspect() function on a set of rules made by the apriori function from the
arules package. I have a dataset with 12 million records. It contains some
basic sales information (such as product, customer data). I want to use t
27 matches
Mail list logo
