R Michael Weylandt michael.weyla...@gmail.com
on Mon, 22 Oct 2012 09:06:35 +0100 writes:
On Monday, October 22, 2012, Christian Hoffmann wrote:
Hi,
Each help.start() generates a new tree of the R help
system, somewhere in 127.0.0.1:xxx, each xxx being
Dear r-users,
Any such a way that we can compute SPI using kernel function (nonparametric
approach) without assuming any parametric distributions?
Any existing package for above method using R?
Thank you for your help.
Fayyad
[[alternative HTML version deleted]]
Well, i'm no expert on these topics, but if its 2.7 gig and R can maximally use
2gig, then the easiest solution would be giving R more memory. Did you read
through help(memory.size) as the error suggested?
try calling memory.size(T) or memory.limit(3000) and see if it works.
I don't have any
Hi,
I got a small problem on how to define the vector index without manually
inspect the vector.
example:
y=c(2,3,5,2,4,6,8,3,6,2,5) #I have ten set of this kind of vectors (with
different values but same length) that I would also like to run the routine
below
#say;
v=the first index in y
Hi,
Is it what you're looking for?
which(y4) ##all indexes for y4
[1] 3 6 7 9 11
which(y4)[1] ##the first index
[1] 3
HTH,
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
I have a large dataset (~1 million rows) of three variables: ID (patient's
name), DATE (of appointment) and DIAGNOSIS (given on that date).
Patients may have been assigned more than one diagnosis at any one appointment
- leading to two rows, same ID and DATE but different DIAGNOSIS.
The
Hello,
Try the following.
y=c(2,3,5,2,4,6,8,3,6,2,5)
first - function(x) min(which(x))
prefix - function(x, v) x[seq_len(v)]
suffix - function(x, v) x[-seq_len(v)]
first(y 4)
prefix(y, first(y 4))
suffix(y, first(y 4))
Hope this helps,
Rui Barradas
Em 23-10-2012 10:21, Al Ehan escreveu:
Hello,
I'm not sure I understand it well, in the solution below the only
returned value is ID == 814 but it's not the first nor the last DATE.
how.many - ave(id.d[,1], id.d[,1], id.d[,2], FUN = length)
id.d[how.many 1, ]
See the help page for ?ave if the repetition of id.d[,1] is confusing.
Hello,
Thinking again, if you just want the first/last in each ID that repeats
the DATE, the following function does the job. Since there were no such
cases in your data example, I've added 3 rows to the dataset.
ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
On 12-10-23 5:39 AM, Rui Barradas wrote:
Hello,
Try the following.
y=c(2,3,5,2,4,6,8,3,6,2,5)
first - function(x) min(which(x))
prefix - function(x, v) x[seq_len(v)]
suffix - function(x, v) x[-seq_len(v)]
first(y 4)
prefix(y, first(y 4))
suffix(y, first(y 4))
Be careful with this: it
Hi
I did not check your code and rather followed your explanation. BTW, thanks for
test data.
small change in data frame to make DATE as Date class
datum-as.Date(as.character(DATE), format=%Y%m%d)
id.d - data.frame(ID,datum )
ordering by date
id.d-id.d[order(id.d$datum),]
two functions to
Hi
Rui's answer brought me to more elaborated solution which still needs data
frame to be ordered by date
fff-function(data, first=TRUE, remove=FALSE) {
testfirst - function(x) x[1,2]==x[2,2]
testlast - function(x) x[length(x),2]==x[length(x)-1,2]
if(first) sel -
I'd greatly appreciate your help in making a bar graph with multiple
variables plotted on it. All the help sites I've seen so far only plot 1
variable on the y-axis
Data set:
I have 6 sites, each measured 5 times over the past year. During each
sampling time, I counted the occurrences of
Dear all,
I've created a new mirror of the HTML versions of the main R manuals
at http://r-manuals.flakery.org. This mirror does not modify the
content of the manuals at all, but simply injects a new visual style
with the aim of making the text easier to read and navigate. It will
be updated
Hi everyone,
I am Evelina and i have just joined the forum and hope to be able to support
soembody as well as finding some help for my stats problem.
I am trying to plot Frequencies against Densities of an ecological community
using a Olmstead-Tukey diagram (O-T diagram). it is based on the
You may also have a look here:
http://flowingdata.com/2010/11/23/how-to-make-bubble-charts/
Note the Circles incorrectly sized by radius instead of area. Large values
appear much bigger.
Similar point made here:
https://stat.ethz.ch/pipermail/r-help/2011-January/265475.html
Message: 44
Dear R-users,
May I seek some suggestions from you. I have a long programme written in R
with several 'for' loops inside. I just want to get them out by any elegant
way (if there is!) to reduce the computational time of the main programme.
For instance, is there any smart way for the following
Ah, no, my method does fail.
Consider an ID that has a duplicate DATE that isn't the first date, but it's
first date is the same as another ID's first date that IS a duplicate.
Test data is all - see below it failing.
So, I remain very grateful for your function!
Stuart
ID -
Yes, I had stripped out the Fortran from the geta subroutine downwards
because is wasn't very important in returning a test output matrix.
I think that I should make explicit what problem I believe I am seeing in
what I have coded, and what form of the output matrix I would prefer.
When I run
Hi there,
I tried it many times but didn't get it worked.
I just want to export the summary of a OLS regression (lm() function) into a
csv-file including the call-formula, coefficients, r-squared,
adjusted r-squared and f statistic.
I know I can export:
i connected Rserver successfully. can do some functions like
rnorm(),print(),etc...
i am connecting R with Hive-hadoop
i have installed library RHive which will take care of connection
but when i am connecting it throws warning..like
Warning: type 7 is currently not implemented in the PHP client.
Thanks Rui - your initial, very elegant suggestion, has spurred me on!
1. As you noticed, my example data had no examples of duplicate first dates
(DOH!)
I have corrected this, and added a test - an ID that has a duplicate which is
not the earliest DATE, but is the same DATE an
I believe that previously could not be understood. To facilitate'll give you
an example. Assuming my table is presented below with the amount received
from each candidate for president in a particular country state.
AL AR CA NY
Doug 250 250 250 NA
Jennifer 20 340 300
On Tue, Oct 23, 2012 at 11:06 AM, Naser Jamil jamilnase...@gmail.com wrote:
Dear R-users,
May I seek some suggestions from you. I have a long programme written in R
with several 'for' loops inside. I just want to get them out by any elegant
way (if there is!) to reduce the computational time
Hi,
I have a data frame with 100 variables (numeric and non numeric types), and
I want to join them in only one column, like a vector, but i want to keep
the non numeric variables like they are.
I know that i can do something like this:
Suppose that my data is in df variable
Hi there.
Not sure I follow what you are doing.
I want a list of all the IDs that have duplicate DATE entries, only when the
DATE is the earliest (or last) date for that ID.
I have refined my test dataset, to include some tests (e.g. 910 has the same
dup as 1019, but for 910 it's not the
Hi
-Original Message-
From: Stuart Leask [mailto:stuart.le...@nottingham.ac.uk]
Sent: Tuesday, October 23, 2012 2:29 PM
To: PIKAL Petr; r-help@r-project.org
Subject: RE: [r] How to pick colums from a ragged array?
Hi there.
Not sure I follow what you are doing.
I want a list
Hi
You indeed cannot use conventional optimisation algorithms as the indicator
function is discontinuous, with no derivative. The search hence has to be
done with a grid search over each potential threshold value, which is
termed conditional/concentrated least square, or also profile
likelihood.
Hi!
Either I don't understand what you want to do, or it doesn't make any sense.
First, a vector cannot have different modes. If you want a single vector
it will most likely be coerced in to characters; probably not what you want.
Second, what you do is build a data.frame with another
Hi
section Arguments of write.table help page clearly says
x the object to be written, preferably a matrix or data frame. If not, it is
attempted to coerce x to a data frame.
summary object from lm is highly structured list an AFAIK can not be easily
coerced to data frame. So either copy
On Tue, Oct 23, 2012 at 12:37 PM, Jon Clayden jon.clay...@gmail.com wrote:
Dear all,
I've created a new mirror of the HTML versions of the main R manuals
at http://r-manuals.flakery.org. This mirror does not modify the
content of the manuals at all, but simply injects a new visual style
with
HI,
I was not following the thread.
May be this is what you are looking for:
new1-id.d[duplicated(id.d)|duplicated(id.d,fromLast=TRUE),]
tapply(new1$ID,new1$DATE,head,1)
#19870508 20040205 20040429 20050421
# 1019 167 814 841
A.K.
- Original Message -
From: Stuart
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Macy Anonuevo
Sent: Tuesday, October 23, 2012 10:23 AM
To: r-help@r-project.org
Subject: [R] plotting multiple variables in 1 bar graph
I'd greatly appreciate your help
Hi,
tapply(new1[,1],new1[,2],head,1) # works.
I used:
id.d-data.frame(ID,DATE)
#In that case,
tapply(new1$ID,new1$DATE,head,1)
#works
On closer look, I think you don't want 814 id. Not sure about the logic
behind that.
A.K.
- Original Message -
From: Stuart Leask
I want to connect R with HTML/PHP pages to take input from user,do
some
statistical processing on it show results to HTML page again.
I search on net,i got Rserve package,but examples are mainly for java
langaure not for PHP
i am wondering how to connect it to PHP-Apache-MySQL
Is there
Hi,
Also one more thing:
This should get the dates which are duplicated. In my first reply, I was
looking for the duplicated rows. Sorry for that!
id.d-data.frame(ID,DATE)
new1-id.d[duplicated(id.d$DATE)|duplicated(id.d$DATE,fromLast=TRUE),]
new2-new1[order(new1$ID,new1$DATE),]
Hi R gurus,
just in case anybody else has a similar problem ... I have programmed a
function that solves this problem by creating a higher dimensional array
out of the individual variables. I have no idea though whether this is
very efficient. Feel welcome to comment in case you think that
Dear All,
I I have one doubt about use of the command when has the same name (name of
command) in two packages.
For example:
package: Stats
package: Signal
Both have the filter command.
How use the command (filter) to a specific package, when I need work with both
packages ?
Thanks
Hi Julio,
Someone correct me if I'm wrong, but I think you can use '::'
For example:
Stats::filter()
HTH,
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogne.fr
Hello,
Use apply/paste.
apply(df, 1, paste, collapse = ) # outputs a vector
Also, df is the name of an R function, you should choose something else,
it can become confusing.
Hope this helps,
Rui Barradas
Em 23-10-2012 12:45, brunosm escreveu:
Hi,
I have a data frame with 100 variables
I too had a parsimonious solution that was also fooled by IDs that had a
duplicate date that wasn't the first date, but was the same as another ID's
duplicate+first.
The right answer
From this data:
ID - c(58,58,58,58,167,167,323,323,323,323,323,323,323
On 23/10/2012 9:42 AM, Ivan Calandra wrote:
Hi Julio,
Someone correct me if I'm wrong, but I think you can use '::'
For example:
Stats::filter()
Yes, but there is no Stats package: R is case-sensitive, and the
package is called stats, and needs to be used that way:
stats::filter()
Well then,,,
On Tue, Oct 23, 2012 at 2:03 AM, Evelina eve.gio...@gmail.com wrote:
Hi everyone,
I am Evelina and i have just joined the forum and hope to be able to support
soembody as well as finding some help for my stats problem.
Start by ditching Nabble and join and post from the r-help
Sorry, I must be a bit thick.!
getRepeat gives me the data with duplicates - but I don't seem to be able to
manipulate the result. It looks like a list of dataframes:
g.r-getRepeat(id.d)
dim(g.r)
NULL
summary(g.r)
Length Class Mode
[1,] 2 data.frame list
[2,] 2 data.frame
Hello,
You're right, getRepeat returns a list of data.frames, one per each ID.
To put them all in the same df use
do.call(rbind, g.r)
Rui Barradas
Em 23-10-2012 13:36, Stuart Leask escreveu:
Sorry, I must be a bit thick.!
getRepeat gives me the data with duplicates - but I don't seem to be
Hi Ivan, thanks for your help!
For example:
df
num letters
1 1 A
2 2 B
3 3 C
4 4 D
5 5 E
What i want is to join num and letters in a single column. Something
like this
new_df
1 1
2 2
3 3
4 4
5 5
6 A
7 B
So I get my list of IDs to exclude from:
g.rr-do.call(rbind, g.r)[1]
dim(g.rr)
g.rr[1:(dim(g.rr)[1]/2)]
Many thanks.
Stuart
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: 23 October 2012 13:42
To: Stuart Leask
Cc: r-help@r-project.org
Subject: Re: FW: [R]
Dear list,
I have a long list of towns in Africa and would need to get their
geographical coordinates. The Google query [/TownName Country coordinates/]
works for most of the TownNames I have and give a nicely formatted Google
output (try Ingall Niger coordinates for an example). I would like to
Thank you for replying.
On Tue, Oct 23, 2012 at 9:11 PM, PIKAL Petr [via R]
ml-node+s789695n4647131...@n4.nabble.com wrote:
Hi
-Original Message-
From: [hidden
email]http://user/SendEmail.jtp?type=nodenode=4647131i=0[mailto:
r-help-bounces@r-
project.org] On Behalf Of Macy
Hi,
Try this:
set.seed(1)
df1-data.frame(col1=rnorm(10,15),col2=rep(c(a,b),5),col3=sample(1:50,10,replace=TRUE),col4=sample(LETTERS[1:10],10,replace=TRUE))
df2-do.call(rbind,lapply(df1,function(x) data.frame(x)))
str(df2)
'data.frame': 40 obs. of 1 variable:
$ x: chr 14.3735461892577
Hi Thomas,
It does appear me and you are working on very similar sediment mixing model
problems. I ran a robust lda using the median by specifying method = t
within the lda function.
So for my data I used:
model-lda(All[,2:12],grouping=All$Catagory, method = t)
Hope this helps.
Regards
Dear all,
I have a 3x2 plot and in addition to the title of the individual plots I
would like to have an overall title
for each row. I managed to get an overall title for the whole plot matrix
with mtext:
par(mfrow=(c(3,2)), mar=c(6.4,4.5,4.2, 1.8), oma=c(0,0,3,0))
for (i in 1:6)
Dear all,
I am trying to create a list, where each list element is a vector of
different length arrays that contain 2by2 matrices. To be more specific
there are 11 treatments that are compared with placebo (we have 11
comparisons) and each comparison is studied by a different number of trials
and
Hello,
Inline.
Em 23-10-2012 14:53, Stuart Leask escreveu:
I too had a parsimonious solution that was also fooled by IDs that had a
duplicate date that wasn't the first date, but was the same as another ID's
duplicate+first.
The right answer
From this data:
ID -
Hi Barry,
On 23 October 2012 14:00, Barry Rowlingson b.rowling...@lancaster.ac.uk wrote:
On Tue, Oct 23, 2012 at 12:37 PM, Jon Clayden jon.clay...@gmail.com wrote:
Dear all,
I've created a new mirror of the HTML versions of the main R manuals
at http://r-manuals.flakery.org. This mirror does
On 10/22/2012 1:10 PM, CMB123 wrote:
Hi all,
I'm working with a large data set (on the order of 300X300) and trying to
apply a function which compares the elements of all possible 2x2
submatrices. There are rc(r-1)(c-1) such submatrices, so obviously the naive
method of looping through the rows
Hello,
Sorry for my earlier post, I misunderstood what you want.
dat - data.frame(letters, 1:26)
as.vector(sapply(d, as.character))
Hope this helps,
Rui Barradas
Em 23-10-2012 14:13, brunosm escreveu:
Hi Ivan, thanks for your help!
For example:
df
num letters
1 1 A
2 2
I don't think so. The OP wanted the list by columns, not rows, if I
understand correctly.
unlist(df)
provides this. Of course, the request probably does not makes sense in
the first place, as the different types/classes in the data.frame will
be coerced to one type/class. Moreover, factors will
Hi
and what is wrong?
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of noobmin
Sent: Tuesday, October 23, 2012 2:52 PM
To: r-help@r-project.org
Subject: Re: [R] How to use tapply with more than one variables grouped
Hi again,
It seems that you want the modes to be conserved. This is just not
possible; everything will be coerced to characters, as Rui highlighted.
But the question is why do you want to do this? You would loose helpful
information (i.e. modes). Maybe there is a better way to do what you
Hello,
Inline.
Em 23-10-2012 15:23, Bert Gunter escreveu:
I don't think so. The OP wanted the list by columns, not rows, if I
understand correctly.
unlist(df)
provides this. Of course, the request probably does not makes sense in
the first place, as the different types/classes in the
You are right.Thanks a lot for the help Rui.
best,
/Shahab
On Mon, Oct 22, 2012 at 5:52 PM, Rui Barradas ruipbarra...@sapo.pt wrote:
Hello,
Your error message means that the arrays have different dim attributes.
check with
dim(M1) == dim(M2)
They must be the same.
If they are, the
Rather than requiring manual tweaking,
library(XML)
readHTMLTable(http://www.worldatlas.com/aatlas/populations/usapoptable.htm;)
will do the job for us.
D.
On 10/22/12 8:17 PM, David Arnold wrote:
All,
A friend of mine would like to use this data with his stats class:
On 23-10-2012, at 12:33, paulfjbrowne wrote:
Yes, I had stripped out the Fortran from the geta subroutine downwards
because is wasn't very important in returning a test output matrix.
But then any results are not reproducible. So there is no purpose in trying to
reproduce your results.
See
useRs –
I’m working with the attached data that contains one year’s worth of
sub-daily observations of flow (“Q”) and specific conductance (“SC”, a
surrogate for concentration) at a point in a stream. The R code posted
below shows the extent of data processing thus far. My goal is to create a
I want to test whether the proportional odds assumption for an ordered
regression is met.
The UCLA website points out that there is no mathematical way to test the
proportional odds assumption (http://www.ats.ucla.edu/stat//R/dae/ologit.htm),
and use graphical inspection (We were unable to locate
Don't try to integrate R with PHP, as you'll end up writing spaghetti code.
Use AJAX to send requests to the server with RApache scripts. OpenCPU will
do the job just fine, but make sure you use JSONP/CORS to bypass the same
origin policy. Long story short, it's something like this:
LAMP -AJAX- R
If you have your data organized in a matrix like the one printed at the
bottom of your jpg, barplot will produce the same graph:
set.seed(42)
a - matrix(round(runif(45)*100, 0), nrow=5, ncol=9)
a
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 91 52 46 94 90 51 74 83
R-helpers:
I'm sure this question has been asked and answered through the ages, but
given there are some new textbooks out there, I wanted to re-pose it. For
a course that will cover the application of R for general computing and
spatial modeling, what textbook would be best to introduce
Hi all,
How does R know to regard a variable as a factor and not a character?
For example, consider the following table:
ObservationGenderDosage
Alertness
1 ma
8
2
Hi
I want help for repeat because it takes a long time
repeat{
for (i in 1:n){
probb[i]=sum(Wc[z[,j]=yb[i]])
}
deltab[,b]=rbinom(n,1,probb)
if(length(which(is.na(deltab[,b])==T))==0){break}
}
This code works but takes much computation time.
Anyone could help me to find a solution to take a
Hi,
I'm Pina and I'm a student in geology. I'm working with spectral profile of
sand and I have to find the similarity between one spectral profile selected
by hyperspectral image anche one that I created to mix different percentage
of 4 mineral component. I have to find the best mix of percentage
Hello,
I have the contour of the shoreline of a lake. I have measured data points
away from the lake shore signifying a particular depth. The data is not
ordered as a polygon and hence trying to draw it creates a series of lines
all over the place. I would like to turn the points into a ordered
Hi,
If the criteria is to pick which among the following states are the top 2
contributors for each candidate,
dat1-read.table(text=
AL AR CA NY
Doug 250 250 250 NA
Jennifer 20 340 300 100
Michele 250 500 250 60
Obama 15 45 520 600
Thank you for your help. In any case, I eventually realized that I had
reversed the Fortran/R column major order in the Fortran, which is why
elements were returned incorrectly. As you point out, I did not run into
memory access issues because of the self-similar array sizes.
--
View this
Hi Stuart,
This also should get you the IDs you wanted.
new1-id.d[duplicated(id.d[,2])|duplicated(id.d[,2],fromLast=TRUE),]
earliest - tapply ( DATE, ID, min)
rownames(earliest[earliest%in% new1])
#[1] 167 841 1019
A.K.
- Original Message -
From: Stuart Leask
So helpful, Arun -- Thank you!
Asaf
--
View this message in context:
http://r.789695.n4.nabble.com/Data-type-in-a-data-frame-tp4647161p4647171.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hi,
I want to make a barplot with the following datasets:
I have a file as following:
name chr position A1 A2 pop1 pop1 pop2 pop2
I have calculated a measure using all the values in columns pops, the
values are saved in a vector.
Now I want to make a barplot using the values in this vector
HI,
I have created a dataframe df and try to sort it by its date using the
order() as below:
df-read.csv(constr,header=T)
sorted.df-df[order(as.Date(df$Date), decreasing = F),]
print(sorted.df)
The dataframe was sorted, but the output from the command console shows
reserved line
To take this example I reduced the number of records absurdly. In the
original database there are 48 000 candidates and dozens of states. There is
no way to analyze data visually. I would not put 400 mb of tables here. But
based on the example how could list the states where obama received more
Hello,
Regarding imbalanced data: When using sampsize correction to balanced
inbalanced data in Random Forests, what are the implications of the
algorithm no longer using a bootstrapped sample? For instance, if I set
sampsize to 25, 25 for a binary response, in a dataset of N=800, how does
The data isn't in a matrix yet. The data printed at the bottom of the jpg
are the means of the variables for each sampling time and I haven't
calculated the means yet. Does this mean that I have to get the means first
then use cbind to make a new matrix? The data looks like this:
[site] [time]
Hi,
res1-data.frame(col=sapply(tapply(DATE,ID,function(x)
duplicated(head(x,2))),function(x) x[2]))
row.names(subset(res1,col==TRUE))
#[1] 167 841 1019
#assuming that dates are sorted
A.K.
- Original Message -
From: Stuart Leask stuart.le...@nottingham.ac.uk
To: Rui Barradas
Hi,
Suppose if you have a threshold (say 500), then:
dat1-read.table(text=
AL AR CA NY
Doug 250 250 250 NA
Jennifer 20 340 300 100
Michele 250 500 250 60
Obama 15 45 520 600
,header=TRUE,stringsAsFactors=FALSE,sep=)
Hi,
Is there a function I can use on my dataframe to give me a concise summary of
variables that are NA,blank,etc? Basically all Null values, Empty strings,
white space, blank values. Ideally it would look something like the below:
# it should only includes the fields with NAs, blanks, etc.
Hi,
The program below work very well.
(snps = c('rs621782_G', 'rs8087639_G', 'rs8094221_T', 'rs7227515_A',
'rs537202_C'))
Selec = todos[ , colnames(todos) %in% snps]
head(Selec)
But, I have a data set with 1.000 columns and I need extract 70 to use
(like snps in command above).
This 70 snps
Hello,
When read into a data.frame, R defaults to reading character strings as
factors. If you don't want that, use option stringsAsFactors = FALSE.
Using your dataset,
dat1 - read.table(text =
Observation Gender Dosage Alertness
1 m a 8
2 m
Stephanie,
I just realized this is already implemented. For your problem use p -
Predict(f,x2); plot(p, ~ x2, nlines=TRUE)
Frank
Frank Harrell wrote
Stephanie,
I'm working on an option for the plot method for Predict that will allow
you to do this. Note that this approach will not result
Random effects models of the type fitted by ordinal assume something akin to
compound symmetry, which is not realistic when time between measurements is
long or irregular.
Frank
Rune Haubo-2 wrote
lmer is not designed for ordered categorical data as yours are. You could
take a look at the
One possibility:
Estimate the center of the polygon with the mean of the x coords and
the mean of the y coords.
Calculate the angle of each point from that center point using the
atan2 function
sort the data by the angle calculated.
This will not be perfect if you have inlets and peninsulas, but
When read into a data.frame, R defaults to reading character strings as
factors. If you don't want that, use option stringsAsFactors = FALSE.
This is somewhat tangential, but if you plan on using
predict(fit,newdata=nd)
after fitting a model like
fit - lm(y~x, data=d)
be sure you have
Hi All,
I'm trying to do a Scatterplot (package: car), and add a line (just for
reference).
There is my code:
#Code---
library(car)
library(calibrate)
G_T-c(car,bike,boat)
ave-c(80,10,45)
perf-c(100,80,75)
I have a rather large data set (about 30 predictor variables)
I need to preform a logistic regression on this data. My response variable
is binary.
My code looks like this:
mylogit - glm(Enrolled ~ A + B + C + ... + EE, data = data, family =
binomial(link=logit))
with A,B,C, ... as my
The criteria is to list where Obama has a higher number of contributions. The
table shows the number of contribution that each presidential candidate
received in a state of the country.
The table shown is an example, the query should be generic to a database
with hundreds of candidates and dozens
Thank you! This seems to work, just do not understand why you used a
threshold?
I will study your solution, thanks again!
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I meant where obama has higher value compared to other candidates. Looking at
the column NY, Obama has the highest. So to state that he wins. Looking for
AR column, Michelle wins. I JUST want to list where obama wins.
Thank you! This seems to work, just do not understand why you used a
threshold?
Hi,
Your question is not clear.
Suppose if you want to find the highest two contributions for each candidate:
dat1-read.table(text=
AL AR CA NY
Doug 250 250 250 NA
Jennifer 20 340 300 100
Michele 250 500 250 60
Obama 15 45 520 600
You're posting on Nabble, so we don't see earlier messages in the thread here.
-- Bert
On Tue, Oct 23, 2012 at 11:36 AM, noobmin pseudov...@hotmail.com wrote:
The criteria is to list where Obama has a higher number of contributions. The
table shows the number of contribution that each
Hello,
I have a data as follow:
ID Visit
xa1
xa2
yb1
yc23
yb33
I want to look at frequency of visit for ID and create a new column as response
.
For example my response would be 2 for x and 3 for y.
I think I need to write a loop, but I don't know how.
I really appreciate your help.
Thanks
Ah, thanks for the info. I'll go ahead and use rms and see how far I
get.
Joost
On Mon, Oct 22 2012, Frank Harrell f.harr...@vanderbilt.edu wrote:
Incorrect. The code is 90% compatible. Look at
http://biostat.mc.vanderbilt.edu/Rrms for differences.
Frank
Joost Kremers wrote
On Mon,
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