Hello All,
This is an easy fix but I am not able to find the root cause of the error. I
am trying to upload a csv file but it is throwing an error.
Have done a lot of research on google and some tutorial but cant find a
solution hence please advice:-
Syntax is :- aaa<-read.csv(file ="VehicleData.
Hi Shivi,
R is case sensitive and the error message that the argument "Header" is
unused (because unrecognized). Try with "header" (lower case "h") and it
should work.
HTH,
Ivan
--
Ivan Calandra, ATER
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51
Shivi82 writes:
> Hello All,
> This is an easy fix but I am not able to find the root cause of the error. I
> am trying to upload a csv file but it is throwing an error.
> Have done a lot of research on google and some tutorial but cant find a
> solution hence please advice:-
> Syntax is :- aaa
This ate my head like for 2 hours. God thanks for the help.
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Dear Nezahat
In future it would be helpful if you
1 - gave us the data so we can reproduce what you are doing
2 - told us what the error was in case we cannot replicate ti
3 - did not post in HTML as it messes up everything in your post
What did you think x1 <- numeric was going to do?
Try
x1 <-
Hi Nezahat,
First, you are storing the code of the function "numeric" in x1 and
x2. You probably want to use:
x1<-numeric()
x2<-numeric()
Second, you are then storing the output of your aov summary (a list)
in x1, which requires a bit of analysis to get the information you
want (i.e. p value). Th
Hi,
I’m followed an example to fit a GAM with a spline forced through a point, i.e.
(0,0). This works fine from one of Simon’s examples however when it comes to
making a prediction from a new set of x values I’m a bit stumped.
In the example below a smooth term is constructed and the basis and
Hello Experts,
I have couple of questions on the analysis I am creating.
1) How does R adopt to changes. The case I have here is that the excel I
have started initially had to be modified because the data I had was on
hourly basis ranging from 0 to 23 hours. After Changes 0 was modified to 24
in h
Hello Experts,
I have couple of questions on the analysis I am creating.
1) How does R adopt to changes. The case I have here is that the excel I
have started initially had to be modified because the data I had was on
hourly basis ranging from 0 to 23 hours. After Changes 0 was modified to 24
in h
On Fri, May 29, 2015 at 7:53 AM, Shivi82 wrote:
> Hello Experts,
> I have couple of questions on the analysis I am creating.
> 1) How does R adopt to changes. The case I have here is that the excel I
> have started initially had to be modified because the data I had was on
> hourly basis ranging f
Thanks Sarah. This is magical.
Thanks for explaining in such a length.
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Hello R-Users
I apologize in advance if my post is inappropriate. I read the entire
posting guide and found nothing to say so, but you never know. I am seeking
a knowledgable R-user that might be interested (for whatever reason) in
helping out on what I hope would be considered a worthy project.
Can some help me with a question on this bass model, please
As I read some articles on this topic, I understand that
1. the bass formula is
N(t) = pm + (q-p) N(t-1) - (q/m) (N(t-1))^2
2. which is a difference equation with the solution
N(t) = m (1 − exp(−(p+q)t)) / (1 + (q/p)exp(−(p+q)t))
3. So, u
If you are primarily interested in making your R analyses in to a website
you should look in to the 'Shiny' package. It makes generating web pages
very easy. Here is a link to the Shiny Gallery providing some examples (
http://shiny.rstudio.com/gallery/).
Regards,
Charles
On Fri, May 29, 2015 a
AFAICS this has essentially nothing to do with R. Please post elsewhere,
e.g. on a statistics list like stats.stackexchange.com.
Cheers,
Bert
On Fri, May 29, 2015 at 6:44 AM, Abolfazl Saghafi <
abolfazl.sagh...@gmail.com> wrote:
> Can some help me with a question on this bass model, please
>
>
I would simply separate the database connect and disconnect functions from the
query functions.
Mark
R. Mark Sharp, Ph.D.
msh...@txbiomed.org
> On May 28, 2015, at 12:18 PM, Luca Cerone wrote:
>
> Dear all,
> I am writing a package that is a collection of queries to be run
> against a pos
Hi Henrik,
I don't quite get what I should do here. I am not familiar with
R.methodS3. Can you tell me what command exactly do I need to do?
Thanks,
Mike
On Thu, May 28, 2015 at 3:30 PM, Henrik Bengtsson wrote:
> For some unknown reason, you've managed to install R.matlab without
> the depe
Good morning, All
I have a stat question not specifically related to the the programming language.
To compare distributional consistency / discrepancy between two
samples, we usually use kolmogorov-smirnov test, which is implemented
in R with ks.test() or in SAS with "pro npar1way edf".
I am wonder
Hi everyone.
I tried the (modeest) package on my initial test data and it worked.
However, it doesn't work on the entire data set. I saved one of the
protions that gives error. (Not for all of the values but for some of
them). For example: lines 36 and 37 and 39 correctly show the mode value
but 3
I have a pedigree file as so:
X0001 BYX859 0 0 2 1 BYX859
X0001 BYX894 0 0 1 1 BYX894
X0001 BYX862 BYX894 BYX859 2 2 BYX862
X0001 BYX863 BYX894 BYX859 2 2 BYX863
X0001 BYX864 BYX894 BYX859 2 2 BYX864
X0001 BYX865 BYX894 BYX859 2 2 BYX865
And I was hoping to change
Wensui: There are the multi-response permutation procedures (MRPP) that
readily test the omnibus hypothesis of no distributional differences among
multiple samples for univariate or multivariate responses. There also are
empirical coverage tests that test a similar hypothesis among multiple
sampl
Here is an example to get you started:
mycol <- c('b','a','d','d','b','c')
as.numeric(factor(mycol))
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 5/29/15, 9:58 AM, "Kate Ignatius" wrote:
>I have a pedigree file as
Of course, but I would not recommend it. A factor is a vector of integers with
an attribute containing the labels that those integers correspond to. You seem
to be asking for a factor that has lost the definitions part. But hey,
newvector <- as.integer(factor(oldvector)) should get you what you
match() will do what you want. E.g., run your data through
the following function.
f <- function (data)
{
uniqStrings <- unique(c(data[, 2], data[, 3], data[, 4]))
uniqStrings <- setdiff(uniqStrings, "0")
for (j in 2:4) {
data[[j]] <- match(data[[j]], uniqStrings, nomatch = 0L
C W gmail.com> writes:
>
> Hi Henrik,
>
> I don't quite get what I should do here. I am not familiar with
> R.methodS3. Can you tell me what command exactly do I need to do?
>
> Thanks,
>
> Mike
install.packages("R.methodsS3")
install.packages("R.matlab")
library("R.matlab")
[snip sni
Very nice, Brian
Sincerely appreciate your assistance!
On Friday, May 29, 2015, Cade, Brian wrote:
> Wensui: There are the multi-response permutation procedures (MRPP) that
> readily test the omnibus hypothesis of no distributional differences among
> multiple samples for univariate or multiva
Hi Kate,
I found that matching the character vector to itself is a very
effective way to do this:
x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
"is", "of", "little", "interest")
ids <- match(x, x)
ids
# [1] 1 2 3 4 5 6 7 8 3 10 11
By using this tri
On May 29, 2015, at 9:31 AM, Wensui Liu wrote:
> Good morning, All
> I have a stat question not specifically related to the the programming
> language.
> To compare distributional consistency / discrepancy between two
> samples, we usually use kolmogorov-smirnov test, which is implemented
> in R
I found this helpful. However - the second to forth columns come out
all zero - was this the intention?
That is:
X0001 0 0 0 2 1 BYX859
X0001 0 0 0 1 1 BYX894
X0001 0 0 0 2 2 BYX862
X0001 0 0 0 2 2 BYX863
X0001 0 0 0 2 2 BYX864
X0001 0 0 0 2 2 BYX865
On Fri, May 29, 2015 at 1:31 PM,
Is anyone aware of point.in.polygon giving different results for 32-bit vs.
64-bit R? Our OS is 64-bit Windows 7 Enterprise. I'm working with someone
else's extensive R program and the final results are close but not exactly
matching. We're thinking it might be something with the point.in.polygo
On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès wrote:
> Hi Kate,
>
> I found that matching the character vector to itself is a very
> effective way to do this:
>
> x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
> "is", "of", "little", "interest")
> ids <- match(x, x)
Hi all,
I have a question about using R in a way that may not be correct but I
thought I would ask anyway.
I have an instrument that outputs a text file with comma separated data. A
new line is added to the file each time the instrument takes a new reading.
Is there any way to configure R such th
Dear all,
I would appreciate a suggestion on the following : I am working with a
data.frame (below) :
EXPCT row_names col_names
1 test -5B4:B5:B6B1:B2:B3
2 test -2B7:B8:B9B1:B2:B3
3 test -2D4:D5:D6H4:H5:H6
4 test -2D10:D11:D12 F10:F11:F12
5 test -2
Hi Sarah,
On 05/29/2015 12:04 PM, Sarah Goslee wrote:
On Fri, May 29, 2015 at 2:16 PM, Hervé Pagès wrote:
Hi Kate,
I found that matching the character vector to itself is a very
effective way to do this:
x <- c("a", "bunch", "of", "strings", "whose", "exact", "content",
"is", "o
Hi,
Please use dput() to provide your data, as it can get somewhat mangled
by copy and pasting, especially if you post in HTML (as you are asked
not to do in the posting guide).
What is a unique element? is "B4:B5:B6" an element, or are "B4" and
"B5" each elements? That is, what is the result you
>I'm not sure why which particular ID gets assigned to each string would
>matter but maybe I'm missing something. What really matters is that each
>string receives a unique ID. match(x, x) does that.
I think each row of the OP's dataset represented an individual (column 2)
followed by its mother a
Hi Sarah,
thank you for your help. I have simplified the example, by reading the
elements in a data frame, eg :
df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
"D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
col_names = c
("B1:B2:B3","B1:B2:B3","H4:H5:
Wow, thanks Ben. That worked very well.
I guess I didn't have R.methodS3? But that doesn't make sense, because I
was using R.matlab few weeks ago. I believe I was on R 3.1.
Maybe it's in R 3.1 folder? I am using a Mac, btw.
Cheers,
-M
On Fri, May 29, 2015 at 1:55 PM, Ben Bolker wrote:
>
Bogdan, the request was for data in dput() format.
Type ?dput for more information.
Do dput(myfile) copy the ouput and paste into the email
You should get something like:
structure(list(c1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L,
2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 5L, 5L
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
I think Henrik's point (which I merely clarified) was that something
funky (we'll probably never know what, and it's not worth figuring out
unless it happens again/to other people) had gone wrong and that the
easiest thing to do was just to reinstall.
A lot will depend on how frequently data is added to the file, how big the
file gets, and how important it is to see updated plots quickly.
I have R doing exactly what you describe, and have found logic like this
(which might be described as crude) to be sufficient
while( {some condition} ) {
{
Hi John,
thanks for clarifications, yes, of course, the dput() output is the
following :
dput(dataframe_matches_ddCT)
structure(list(FIGURE = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label
= "test", class = "factor"), ddCT = c(-5.4595, -2.7467,
-2.7467, -2.7467, -2.7467, -2.7467, -4.5927, -
Hi Jim,
yes, thank you, that is the desired output. one more question please :
after using the dataframe :
df <- data.frame (row_names = c("B4:B5:B6", "B7:B8:B9", "D4:D5:D6",
"D10:D11:D12", "D10:D11:D12", "E10:E11:E12", "A1:A2:A3", "B10:B11:B12"),
col_names = c
("B1:B2:B3","B1:B2:B3","H4:H5:H6","
I'm still not really clear on what you need (format, etc), but this
may help you get started:
> with(df, table(CT, row_names))
row_names
CT A1:A2:A3 B10:B11:B12 B4:B5:B6 B7:B8:B9 D10:D11:D12 D4:D5:D6 E10:E11:E12
20 001 21 1
4
Dear Sarah,
thank you very much, it is very helpful. please may I ask one more question
about a quick and easy tutorial about the loading multiple files (from a
folder) in R, and processing one file at a time ? thanks very much again,
bogdan
On Fri, May 29, 2015 at 2:55 PM, Sarah Goslee
wrote:
Hi Ben,
Thanks for the fun clip. I love it. Have a wonderful day!
-M
On Fri, May 29, 2015 at 5:10 PM, Ben Bolker wrote:
> -BEGIN PGP SIGNED MESSAGE-
> Hash: SHA1
>
> I think Henrik's point (which I merely clarified) was that something
> funky (we'll probably never know what, and it'
Don't use Nabble when posting to the R-Help forum.
Responses inline.
On May 29, 2015, at 7:54 AM, Shivi82 wrote:
> Hello Experts,
> I have couple of questions on the analysis I am creating.
> 1) How does R adopt to changes. The case I have here is that the excel I
> have started initially had
LMGTFY:
http://stackoverflow.com/questions/11433432/importing-multiple-csv-files-into-r
On Fri, May 29, 2015 at 5:58 PM, Bogdan Tanasa wrote:
> Dear Sarah,
>
> thank you very much, it is very helpful. please may I ask one more question
> about a quick and easy tutorial about the loading multiple
On 29/05/2015 2:36 PM, Lensing, Shelly Y wrote:
> Is anyone aware of point.in.polygon giving different results for 32-bit vs.
> 64-bit R? Our OS is 64-bit Windows 7 Enterprise. I'm working with someone
> else's extensive R program and the final results are close but not exactly
> matching. We're
Hi Mohammad,
I have no idea what is happening but for some reason your new data (renamed df1
since df is a reserved word in R) is outputting a list whereas dff1 (your
original test data) is giving a vector as you wanted.
It may be obvious but I don't see why df1 is giving us a list. As far as I
thanks a lot Sarah, very much appreciate it !
On Fri, May 29, 2015 at 3:18 PM, Sarah Goslee
wrote:
> LMGTFY:
> http://stackoverflow.com/questions/11433432/importing-multiple-csv-files-into-r
>
> On Fri, May 29, 2015 at 5:58 PM, Bogdan Tanasa wrote:
> > Dear Sarah,
> >
> > thank you very much, i
Hi Bogdan,
Sarah has already suggested this, but doesn't:
table(df$row_names,df$CT)
table(df$col_names,df$CT)
give you what you want?
Jim
On Sat, May 30, 2015 at 7:11 AM, John Kane wrote:
> Bogdan, the request was for data in dput() format.
>
> Type ?dput for more information.
>
> Do dput(myf
Has anyone found a solution to this? I am having the same issue? thanks!
On Thursday, November 15, 2012 at 10:35:48 PM UTC-8, abcd1234 wrote:
>
> Hi all,
>
> The TWS on my system is unable to connect to my R session. Here is the
> error that I'm getting:
>
> /> tws<-twsConnect()
> Error in
HI
I was working on online example, where virus is spread through a graph. The
example is sufficient for small graph i.e. small number of edges and nodes.
But I tried it on very large graph i.e. 1 nodes and 2 edges, but
the below function is not sufficient for large graph because its slow.
Hi Bill,
On 05/29/2015 01:48 PM, William Dunlap wrote:
I'm not sure why which particular ID gets assigned to each string would
matter but maybe I'm missing something. What really matters is that each
string receives a unique ID. match(x, x) does that.
I think each row of the OP's dataset repre
Hi Bogdan,
If you mean "How can I verify that "B1:B2:B3" is paired with all of
the values 2, 4 and 5"
apply(table(df$col_names,df$CT),1,all)
and if you mean "How can I verify that "B1:B2:B3" is paired with at
least one of the values 2, 4 and 5"
apply(table(df$col_names,df$CT),1,any)
Jim
Hi Ji
Thanks a lot Jim. If I may ask one more little question please,
shall I ask the question ""How can I verify that "B1:B2:B3" is paired with
ALL of the values 2, 4 and 5 ",
regardless of the pairing value (in our case, for the code below, the
"pairing value" for "B1:B2:B3" is 1, but it can be 2,3,4
Hi Jim,
thanks again. now I see : the answer to my previous question seems to be
yes, as "all" functions works on logical vectors ... best wishes,
-- bogdan
On Fri, May 29, 2015 at 4:29 PM, Bogdan Tanasa wrote:
> Thanks a lot Jim. If I may ask one more little question please,
>
> shall I ask t
I've been getting a 403 when I try pulling from the Toronto CRAN mirror
today.
http://cran.utstat.utoronto.ca/
Is there a contact list for mirror managers?
--
Cheers, Mark
*Mark Drummond*
m...@markdrummond.ca
When I get sad, I stop being sad and be Awesome instead. TRUE STORY.
[[alte
Hi Mohammad,
It looks like you are still having problems with this. Given your
latest data set, as below, here is something that might do what you
want. From David's message, I'm not sure whether you are operating on
a single data frame or a list.
# this is the data set as taken from your message
This is why there are mirrors. You don't have to wait for them or tell them to
do their jobs.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
On May 29, 2015, at 7:12 PM, Mark Drummond wrote:
> I've been getting a 403 when I try pulling from the Toronto CRAN mirror
> today.
>
> http://cran.utstat.utoronto.ca/
Right. It's been out for the last 2.7 days:
http://cran.r-project.org/mirmon_report.html#ca
>
> Is there a contact list for
On Fri, May 29, 2015 at 10:12 PM, Mark Drummond wrote:
> I've been getting a 403 when I try pulling from the Toronto CRAN mirror
> today.
>
> http://cran.utstat.utoronto.ca/
>
> Is there a contact list for mirror managers?
>
See the cran_mirrors.csv file in
R.home("doc")
of your R distribution.
Thanks you Sarah. This was very impressive and really helped me out.
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