Hi
r-help-boun...@r-project.org napsal dne 08.07.2010 10:45:04:
Hello all R users,
I have a problems transforming (or maybe better regrouping) a
data.frame.
I have a big data.frame, which I would like to sum up according to a
specific column.
This is an example of my matrix:
ID
Hi
see
?merge
Regards
Petr
r-help-boun...@r-project.org napsal dne 07.07.2010 17:07:34:
Say I have two files file and file2:
file1 contains the following:
Date Price
02/07/2010 53.96597903
03/07/2010 56.92825807
04/07/2010 39.27408645
05/07/2010
(number of true positive,number positive Tests)
This works for all other proportions as well, eg sensitivity=P(T+|D+) :
p-15/24 #sensitivity
p+qnorm(.975)*c(-1,1)*sqrt(p*(1-p)/24)
binom.test(15,24)
hth.
Am 08.07.2010 10:59, schrieb Petr PIKAL:
Dear all
Seems that puzzles always come
Hi
r-help-boun...@r-project.org napsal dne 08.07.2010 12:02:45:
I don't understand it. When I'm doing this example it wirks fine, but
when
I'm adding the GO: to the beginning of the first column (as to see in
my
wanted result table:
GO0042787
GO0016070
GO0016070
I'm getting a list of
Hi
r-help-boun...@r-project.org napsal dne 01.07.2010 17:40:22:
Hi, I run into problem when writing a syntax, I don't know syntax that
will
return true or false if an integer is odd or even.
fff - function(x) as.logical(x%%2)
Regards
Petr
Thanks
OYEYEMI, Gafar Matanmi
Department
Dear all
My question is more on statistics than on R, however it can be
demonstrated by R. It is about pros and cons trying to find a relationship
by aggregated data. I can have two variables which can be related and I
measure them regularly during some time (let say a year) but I can not
Hi
did you try aggregate?
aggregate(data[, 5:8],list(data$Date), sum, na.rm=T)
Group.1 verbal self.harm violence_objects violence
1 0 000
2 01/04/07 251539
3 02/04/07 24 68 13
Hi
r-help-boun...@r-project.org napsal dne 26.06.2010 00:11:33:
Albert -
The message refers to a file specifically called .RData.
Files with subscripts of .rdata are not related.
You can see your current working directory by typing
getwd()
at the R prompt.
I'm not sure
Hi
see
?options and scipen parameter
e.g. options(scipen=9)
Regards
Petr
r-help-boun...@r-project.org napsal dne 24.06.2010 08:39:29:
Hi,
The format problem is really annoying. How to get rid of it?
x
[1] 1e+06
And also when I do barplot,
x=rnorm(100,mean=1000)
Hi
r-help-boun...@r-project.org napsal dne 24.06.2010 09:44:49:
Hello Nabble users,
A question about having two y axes on a chart. I'm trying to show on the
left (first) y axis a measure used in an experiment; the x axis shows
the
experiment number, that has been sorted in a specific
Hi
r-help-boun...@r-project.org napsal dne 24.06.2010 15:18:39:
it works with the 'TRUE' as well but this only shows me the boxplots
width;
it doesn't show me the number of data points used though.. This is what
I
can't figure out
Save your boxplot call to same object
bbb -
Hi
r-help-boun...@r-project.org napsal dne 22.06.2010 08:28:04:
The following dataframe will illustrate the problem
DF-data.frame(name=rep(1:5,each=2),x1=rep(A,10),x2=seq(10,19,by=1),x3=rep
(NA,10),x4=seq(20,29,by=1))
DF$x3[5]-50
# we have a data frame. we are interested in the
on a data frame with all numeric variables
On Tue, Jun 22, 2010 at 12:23 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
r-help-boun...@r-project.org napsal dne 22.06.2010 08:28:04:
The following dataframe will illustrate the problem
DF-data.frame(name=rep(1:5,each=2),x1=rep(A,10),x2
this
[obrázek odebrán]
2010/6/9 Petr PIKAL petr.pi...@precheza.cz
Hi
where did you find parameter add=T.
You can use
par(new=T)
before using new plot command
or use
points, lines
Regards
Petr
r-help-boun...@r-project.org napsal dne 09.06.2010 16:42:25:
I forgot to show my
Hi
r-help-boun...@r-project.org napsal dne 16.06.2010 22:14:33:
Thanks for your reply. Possibly I donot have perl. I am not sure
although.
How I can find whether I have it? If I dont have it then where can I
download it from?
Do you have Excel? If yes you can
Open Excel
Select data you
character and you need to do some fiddling to turn it
back to numerics.
Regards
Petr
Petr PIKAL petr.pi...@precheza.cz 2010/06/18 01:42 PM
Hi
r-help-boun...@r-project.org napsal dne 16.06.2010 22:14:33:
Thanks for your reply. Possibly I donot have perl. I am not sure
although
Hi
r-help-boun...@r-project.org napsal dne 15.06.2010 19:51:39:
Hi All,
I am trying to turn a Matrix into a vector for analysis purposes. I
need to
select only certain columns from the entire matrix for the vector
(intraday
time intervals). Also I need to transpose the Matrix (so times
Hi
I use an option with PDF, which can have multi page output. You can
pdf(some_graphs.pdf)
# do all your plotting let say in cycle something like
for( i in 1:n) plot(something[,1], something[,i+1], ...)
dev.off()
You will get one multipage pdf document in working directory which you can
Hi
Would you like to know which colour is for which curve? If yes use
parameter col=1:5 in your plotting call.
If not you can try col = sample(colours(), 5) instead
Regards
Petr
r-help-boun...@r-project.org napsal dne 14.06.2010 07:05:43:
Sir,
I want to plot 5 curve on a single graph. I
Hi
r-help-boun...@r-project.org napsal dne 14.06.2010 11:05:39:
Hi,
I have a data frame that looks like this:
ID
Var
Var2
Var3
xxx
100
909
920
yyy
110
720
710
zzz
140
680
690
I can load the file and produce a plot
Hi
I am not sure if you can do what you want. Segments are not points so your
pch option is (I believe) ignored. You could play with lmitre and lend
parameters, but it probably would not help much.
You cold try to look at
?symbols
but you probably need to change source code to suit your
Hi
Look at the source code.
graphics:::plot.histogram
You can find that boxes are actually drawn by rect
So if you want to use standard graphics, you probably need to modify
source code and set up your version of plot.histogram.
Maybe with ggplot2 package you can find some way how to do what
Hi
Peter Ehlers ehl...@ucalgary.ca napsal dne 09.06.2010 19:05:24:
Soapbox:
Well, if you're just starting out with R it would be
a VERY good idea to learn right away that T is not TRUE
and F is not FALSE, at least not always. Sooner or
later you WILL have problems. So do yourself a favour
Hi
split/sapply can be used besides other options
sapply(split(iris[,1:4], iris$Species), mean)
Regards
Petr
r-help-boun...@r-project.org napsal dne 10.06.2010 00:43:29:
Hi there:
I have a question about generating mean value of a data.frame. Take
iris data for example, if I have a
Hi
you shall use na.action = na.exclude option in your lm call?
Regards
Petr
r-help-boun...@r-project.org napsal dne 10.06.2010 01:12:55:
Hello R help
I have a dataframe, with 71 samples (rows) and 30 variables. I got
linear
models for some of the variables, and I want to join fitted
Hi
Uwe Ligges lig...@statistik.tu-dortmund.de napsal dne 10.06.2010
10:37:05:
On 10.06.2010 10:19, Petr PIKAL wrote:
Hi
Peter Ehlersehl...@ucalgary.ca napsal dne 09.06.2010 19:05:24:
Soapbox:
Well, if you're just starting out with R it would be
a VERY good idea to learn
Hi
r-help-boun...@r-project.org napsal dne 10.06.2010 15:56:06:
Hello everyone,
I have a matrix of over 4 line and about 30 columns.
Matrix or data.frame?
For my analysis I would like to add another column with ascending
numbers
(column header should be order, and than 1,2,3,4
Hi
You have got several suggestions. When I try your code I get only errors,
not list or any other objects.
x-unique(Returns.names$date_)
Error in unique(Returns.names$date_) : object 'Returns.names' not found
n-1
while(n=19) {
+ Returns.period-lapply(((n-1)*125+1):((n+1)*125), function
Hi
r-help-boun...@r-project.org napsal dne 09.06.2010 12:44:33:
See ?textConnection, eg
test -1 2.5 3.4
1 2.3 3.1
1 2.6 3.9
2 2.9 2.8
2 2.6 2.9
2 2.7 2.9
3 2.3 3.3
3 2.4 3.0
3 2.7 3.2
x - textConnection(test)
read.table(x)
close(x)
On Wed, Jun 9, 2010 at 10:38 AM, Paul Chatfield
Hi
r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:
Dear R community,
I am puzzled by the following lines:
v - seq(-0.5,0.5,by=0.1)
v
[1] -0.5 -0.4 -0.3 -0.2 -0.1 0.0 0.1 0.2 0.3 0.4 0.5
which(v == -0.4)
[1] 2
which(v == 0)
[1] 6
which(v == 0.1)
Hi
if you do not insist on switch you can use factor to get desired result
set.seed(666)
x-sample(1:5,20, replace=T)
x
[1] 4 1 5 2 2 4 5 3 1 2 4 1 1 1 2 5 1 5 3 3
factor(x, labels=letters[1:5])
[1] d a e b b d e c a b d a a a b e a e c c
Levels: a b c d e
r-help-boun...@r-project.org napsal
$double.eps)
seems to me too cumbersome to write. Any other easier way? all.equal
does not quite work
Nikhil
On Jun 9, 2010, at 7:54 AM, Petr PIKAL wrote:
Hi
r-help-boun...@r-project.org napsal dne 09.06.2010 13:16:40:
Dear R community,
I am puzzled by the following lines
Hi
one option is to create an empty list with
my.list - vector(list, n)
where n is number of elements in list. Then you can populate each element
by
my.list[[x]] - something
where x can be number or
my.list[x] -something
where x is numeric vector
Regards
Petr
Hi
where did you find parameter add=T.
You can use
par(new=T)
before using new plot command
or use
points, lines
Regards
Petr
r-help-boun...@r-project.org napsal dne 09.06.2010 16:42:25:
I forgot to show my code... it is like this
jpeg()
plot(x,y,main=str, xlab=str,ylab=str,
Hi
r-help-boun...@r-project.org napsal dne 08.06.2010 08:44:39:
Hi everybody
I have found something (for me at least) strange with duplicated(). I
will
first provide a replicable example of a certain kind of behaviour that I
find odd and then give a sample of unexpected results from my
Hi
r-help-boun...@r-project.org napsal dne 08.06.2010 11:46:17:
Hi,
You are using if/then/else which is a logical control statement and so
doesn't
return a value, see
?if
for details.
You are probably looking for the ifelse function.
Or use possibility of easy conversion logical
Hi
Hm, maybe you can first make a sequence of all required dates and ids,
construct empty data frame with all possible dates, merge your existing
data frame with empty one just to fill in all dates, get rid of duplicated
dates and ids if necessary and finally use na.locf from zoo library to
Hi
r-help-boun...@r-project.org napsal dne 08.06.2010 14:21:10:
Thank you to all for your help!
I received
two equal alternative solutions that bypassed elegantly the problem
from peter.l.e.koni...@gmail.com
and
rafael.bj...@gmail.com
rr.dia2.corr - rr.dia.2m
Hi
some example would be helpful.
r-help-boun...@r-project.org napsal dne 07.06.2010 09:01:27:
Hey Everyone,
I have been stumped by this all day.
Basically, I have a data.frame of multiple columns. Of concern are id
date
For some reason, oftentimes there are duplicates of data
Hi
r-help-boun...@r-project.org napsal dne 07.06.2010 14:28:58:
Dear list,
I am getting weired with fitting data with a 1/x-polynomial. Suggest I
have
the following data:
x - c(1,2,3,4,5,6,7)
y - c(100,20,4,2,1,.3,.1)
I may fit this with a linear model
fit1 = lm(y ~ I(x))
fit1 =
Hi
r-help-boun...@r-project.org napsal dne 03.06.2010 18:18:33:
One option:
t - data.frame(x1=c(1,1,0,0,0,1), x2=c(0,0,0,1,0,1),
Count=c(523,23,2,45,3,433))
t.sum - function(df, x1, x2) sum(df[df$x1==x1 df$x2==x2,]$Count)
t.sum(t, 1, 0)
# [1] 546
t.sum(t, 0, 0)
# [1] 5
If this is
Dear all
I encountered strange problem with regexpr replacement
I made this character object
str - 02.06.10 12:40
str(str)
chr 02.06.10 12:40
I read in an object which seems to be quite similar
str(as.character(becva$V1)[1])
chr 02.06.10 12:40
However I can not remove
Meys jorism...@gmail.com napsal dne 02.06.2010 14:35:19:
Could you provide us with dput(becva$V1[1])?
Cheers
Joris
On Wed, Jun 2, 2010 at 2:07 PM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Dear all
I encountered strange problem with regexpr replacement
I made this character object
str
Hi
you have several options
apply(a, 2, as.numeric)
matrix(as.numeric(a),3,2)
b-as.numeric(a)
dim(b)-c(3,2)
Regards
Petr
r-help-boun...@r-project.org napsal dne 02.06.2010 14:15:25:
Hello R experts,
can you tell me a simple way to convert a charactor matrix to numeric
matrix?
if I use
at 9:22 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
dput(bbb)
c(02.06.10 12:40 , 02.06.10 12:00 , 02.06.10 11:00 ,
02.06.10 10:00 , 02.06.10 09:00 , 02.06.10 08:00 ,
02.06.10 07:00 , 02.06.10 06:00 , 02.06.10 05:00 ,
02.06.10 04:00 , 02.06.10 03:00 , 02.06.10 02
or even newline characters.
You
can check that easily with
grep(\t, as.character(becva$V1[1]))
grep(\n, as.character(becva$V1[1]))
Cheers
Joris
On Wed, Jun 2, 2010 at 3:54 PM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
thanks. I am puzzled what was wrong. Now even
sub
Hi
r-help-boun...@r-project.org napsal dne 01.06.2010 10:20:35:
On 31/05/2010, Gabor Grothendieck ggrothendi...@gmail.com wrote:
Use read.csv or read.table(..., sep = ,). Also note that if you
delete the first comma of the header (as in the second example below)
you won't have to specify
Hi
r-help-boun...@r-project.org napsal dne 01.06.2010 13:01:38:
Dear All,
I am newbie to R, and I wanted to plot a barplots with R and in such a
way
that It will also show me position which I can plot on the bar line.
Here is my code that I am using to plot,
chromosome - c(40.2,
Hi
Probably not the best way
read.table(clipboard, header=T)
head.as.character.XY..
1 -448623_854854
2 -448563_854850
3 -448442_854842
4 -448301_854833
5 -448060_854818
6 -446828_854736
test=read.table(clipboard, header=T)
now do the split
Hi
r-help-boun...@r-project.org napsal dne 05.05.2010 07:49:25:
Hi,
I have a data frame where 1 variable is a factor with only 1 level. I
need
the data frame structure to reflect that there are 2 levels for this
factor,
even though this is not the case. I am currently adding extra
Hi
r-help-boun...@r-project.org napsal dne 05.05.2010 11:46:33:
Dear list,
I'm trying to concatenate the values of two columns but im not able to
do it:
i have a dataframe with the following two columns:
X VAR1 VAR2
1 2
2
Hi
r-help-boun...@r-project.org napsal dne 03.05.2010 16:24:11:
I am conducting a very simple t test for two genes using lapply (i try
to
avoid loop since i will have thousands of genes later on). however, I
got
strange error msg like the followings. It looks that R is complaining my
)
x.f
[1] -1.15502538070463 -0.457842907389024 -0.608900689299125
-1.73785992416606
[5] 0.129785271221269
5 Levels: -1.73785992416606 -1.15502538070463 ... 0.129785271221269
as.numeric(x.f)
[1] 2 4 3 1 5
Regards
Petr
Petr PIKAL petr.pi...@precheza.cz 2010/05/04 09:38 AM
Hi
r-help
-1.2319878 -1.8689400 -1.2931785
Petr PIKAL petr.pi...@precheza.cz 2010/05/04 10:12 AM
x-rnorm(5)
x
[1] -1.1550254 -0.4578429 -0.6089007 -1.7378599 0.1297853
x.f-factor(x)
x.f
[1] -1.15502538070463 -0.457842907389024 -0.608900689299125
-1.73785992416606
[5
Hi
r-help-boun...@r-project.org napsal dne 04.05.2010 09:50:28:
Dear all,
In my dataset I have 12 columns and 5824 rows. The second column
contains
information about the height of a claim: it might be zero or positive. I
would like to do an analysis on the positive part of this matrix,
Hi
r-help-boun...@r-project.org napsal dne 04.05.2010 14:54:14:
I need to replace NA occurrences in multiple columns in a data.frame
with 000/000
Be careful if you replace NA in numeric columns.
str(test)
'data.frame': 10 obs. of 3 variables:
$ mp: num 20.9 19.9 20.1 20.2 18.9 ...
Hi
r-help-boun...@r-project.org napsal dne 01.05.2010 00:13:06:
Dear R list,
Our statisticians usually give us results back in a PDF format. I would
like
to be able to copy and past tables from R output directly into a
Microsoft
Word table since this will save us tons of time, be more
Hi
r-help-boun...@r-project.org napsal dne 30.04.2010 23:11:54:
Hello David,
On Apr 30, 2010, at 11:00 PM, David Winsemius wrote:
Note: Loops may be just as fast or faster than apply calls.
How come!? is this true also for other similar functions: lapply, tapply
and sapply?
Then
Hi
r-help-boun...@r-project.org napsal dne 29.04.2010 05:56:23:
Hello,
I have a data.frame:
namecol1col2col3col4
AA23540.9990.78
BB123510.99
AA203980.790.99
I want to get mean value data.frame in terms of name:
name
Hi
I put a search question about nonlinear programming in R site search and
got many answers maybe you could find something which suits your needs.
Maybe you could also look at CRAN task view - Optimisation and
Mathematical programming
Regards
Petr
r-help-boun...@r-project.org napsal dne
Hi
r-help-boun...@r-project.org napsal dne 29.04.2010 08:11:41:
Hi
you could try
do.call('rbind',aa)
No, No, No. rbind and cbind binds vectors as rows or columns of
***matrix***, result is not a data frame
do.call(rbind,aa)
X069rutil X102anatas
105 26.97.9
200
',aa)
df=data.frame(dd)
df
c1 c2 c3c4
AA 113 76 0.8945 0.885
BB 123 5 1. 0.990
Regards
Tengfei
On Thu, Apr 29, 2010 at 1:30 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
r-help-boun...@r-project.org napsal dne 29.04.2010 08:11:41:
Hi
you could try
. this is
name c1 c2 c3c4 c5 c6
AA 23 54 0.999 0.78 87 6
BB123 5 1.000 0.99 98 7
could give advice?
--- On Thu, 29/4/10, Tengfei Yin yinteng...@gmail.com wrote:
From: Tengfei Yin yinteng...@gmail.com
Subject: Re: [R] by funtion
To: Petr PIKAL
Hi
You have to get rid of thousands separator firsr
as.numeric(gsub(,, , S))
Regards
Petr
r-help-boun...@r-project.org napsal dne 29.04.2010 13:12:44:
Dear group,
I know this issue has been already covered, and before you reply I must
say
I have read the R-FAQ and search the mailing
, Apr 27, 2010 at 2:40 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
r-help-boun...@r-project.org napsal dne 26.04.2010 17:05:54:
I guess my problem was seeing a bunch of examples where they pulled a
variable from a dataframe..
tapply(df$data, index=list(..
df$data results in vector
Hi
r-help-boun...@r-project.org napsal dne 28.04.2010 12:06:04:
Hello all,
I want to plot some probability points (in scale 0-1), but the large
majority have extremely low probabilities.
Imagine I have the following vector
a-c(0.0001,0.004,0.01,0.4)
plot(a)
When you plot it,
problematic.
see
str(df$data)
str(df[, 1])
str(df[,1, drop=FALSE])
str(df[,1:15])
Regards
Petr
Thanks.
On Mon, Apr 26, 2010 at 2:43 AM, Petr PIKAL petr.pi...@precheza.cz
wrote:
Hi
steven mosher mosherste...@gmail.com napsal dne 26.04.2010 10:21:37:
That fails:
The manual
Hi
what about
fit - lm(value~seqMonth+ids+varable, data=theTestLineal)
or similar approach using
?lme
See also
?interaction
Regards
Petr
r-help-boun...@r-project.org napsal dne 27.04.2010 13:48:32:
Hi there,
I am stuck trying to solve what should be a fairly easy problem.
I have a data
of which were mentioned
earlier
by
Petr Pikal.
HTH,
Dennis
On Mon, Apr 26, 2010 at 8:01 AM, steven mosher
mosherste...@gmail.comwrote:
Thanks,
I was trying to stick with the base package and figure out how the
base
routines worked. I looked at plyer and it was very appealing
Hi
r-help-boun...@r-project.org napsal dne 26.04.2010 06:52:55:
Having some difficulties with understanding how tapply works and getting
return values I expect
Data: dataframe. DF DF$Id $D $Year...
Id D Year Jan Feb Mar Apr May Jun Jul Aug Sep
Oct
Nov
for aggregate is quite similar to tapply, only first argument can
be data frame.
Regards
Petr
The length of DF is 2.
Does that mean the list of factors, each of same length as X. would
have to be
2? that doesnt seem to make sense.
On Mon, Apr 26, 2010 at 12:26 AM, Petr PIKAL petr.pi
Hi
r-help-boun...@r-project.org napsal dne 23.04.2010 04:05:00:
Hi all,
I have a dataset similar to the following
Name Date Value
A 1/01/2000 4
A 2/01/2000 4
A 3/01/2000 5
A 4/01/2000 4
A 5/01/2000 1
B 6/01/2000 2
B 7/01/2000 1
B 8/01/2000 1
Hi
r-help-boun...@r-project.org napsal dne 20.04.2010 21:21:26:
Hi all,
I'm a newbie with R and with a very basic question.
Can I define the minor unit for ylim? For example, I have a y scale
ranging
from 1 to 14, jumping automatically every 2 units, but I want it to jump
1
unit at a
Hi
you can also use abline function
x-rnorm(10)
plot(1:10, x)
abline(v=seq(2,10,2), lwd=50, col=lightblue)
lines(1:10, x, type=b)
Regards
Petr
r-help-boun...@r-project.org napsal dne 19.04.2010 20:59:15:
On 04/15/2010 12:36 AM, senne wrote:
hi R gurus
I saw some graphs with
Hi
r-help-boun...@r-project.org napsal dne 16.04.2010 16:15:40:
Hi, I have a need to have 2 variables point to the same dataframe (d1),
I
What does it mean to point to data frame? Seems to me that it is something
from C+.
You can reference data frame by $ or by square brackets with as
Hi
r-help-boun...@r-project.org napsal dne 19.04.2010 10:12:59:
Hi all,
I'm trying to load a csv file in which all the variables must be of type
number.The object is a dataframe.When i load the file what i get is a
dataframe
You probably have non numeric data in your original CSV.
Hi
If the columns has the same name but different values in them then you
shall either decide which one to keep yourself or you shall keep both. If
they have same name and same values you could select only those which
names do not match.
names(data1) %in% names(data2)
can select which names
Hi
r-help-boun...@r-project.org napsal dne 16.04.2010 14:00:09:
I have a problem with the merge command.
I have to merge two dataframe that looks like the following example:
CODPROD N1 N3 N4
23 3 55 4
24
Hi
Bert Gunter gunter.ber...@gene.com napsal dne 14.04.2010 18:54:37:
Petr Pikal wrote:
...
I mean that you can use
fit- lm(y~x+I(x^2))
coef(fit)[1] + coef(fit)[2]*x + coef(fit)[3]*x^2
but you can not use
fit- lm(y~poly(x,2))
coef(fit)[1] + coef(fit)[2]*x + coef(fit)[3]*x^2
Hi
r-help-boun...@r-project.org napsal dne 15.04.2010 13:00:57:
Peter Ehlers wrote:
You are mixing 'traditional' graphics (par(...)) and
'lattice' graphics.
That won't work. In lattice, you use the 'layout' argument to
select the number of columns/rows. This is easiest if you set
up a
Hi
r-help-boun...@r-project.org napsal dne 14.04.2010 17:12:51:
Hi List,
I can not get my head around the following problem. I want to fit a
quadratic function to some data and stumbled across poly(). What exactly
does it, i.e. why are there different results for fit1 and fit2?
x =
Hi
Bert Gunter gunter.ber...@gene.com napsal dne 14.04.2010 18:01:52:
Below.
-- Bert
Bert Gunter
Genentech Nonclinical Statistics
Coefficients are different as you fit different values. See
?poly
poly(-10:10,2)
I believe that others give you better explanation. So you
Hi
r-help-boun...@r-project.org napsal dne 12.04.2010 07:14:14:
David Winsemius wrote:
I am guessing that the first time through when i= 5200 that i+1 is
indexing an entry that does not exist. What does str( Price[[1]]
[5200+1] ) return? What about str(Ca)?
So what is
Hi
r-help-boun...@r-project.org napsal dne 12.04.2010 12:51:24:
Petr Pikal wrote:
snip
Nobody except you has your data available, so without providing more
clues
you can not expect mor relevant answers.
Try str(your.objects) and maybe you could use debug to see how
Hi
what about
some.data - read.table()[ ,1:2]
Regards
Petr
r-help-boun...@r-project.org napsal dne 08.04.2010 16:05:39:
Hi everyone,
I've got a matrix data with 20 variables (V1, V2, V3, ...) and 215
rows (observations). I'm interested to read only the first and second
variables
Hi
r-help-boun...@r-project.org napsal dne 08.04.2010 16:23:53:
Hi
So my particular problem is this:
I have a row vector of length 5200 elements - specifically created by
x-rbinom(5200,1,0.5)
y-matrix(x,nrow=1,ncol=5200)
y
now, each element is either a 0 or a 1 - e.g. it
Hi
?cumsum
Petr
r-help-boun...@r-project.org napsal dne 07.04.2010 14:26:12:
Dear colleagues,
I have a data frame that looks so:
*x1 x4
1 4.2
2 3.6
3 2.7
.
.
308 n.a.
x4 is a vector of percentages, sorted in descending value. I would
like to create a new variable that
Hi
Without some insight about foo, list or counts it is impossible to say
what is wrong.
mat-matrix(1:12, 3,4)
colnames(mat)-letters[1:4]
DF-as.data.frame(mat)
fac-factor(names(DF))
fac
[1] a b c d
Levels: a b c d
ff-fac[3]
ff
[1] c
Levels: a b c d
DF[[ff]]
[1] 7 8 9
ff-fac[1]
DF[[ff]]
Hi
r-help-boun...@r-project.org napsal dne 02.04.2010 12:12:02:
Dear useRs,
I'm having a slight problem with plotting on 2 axes. While the following
code works alright on screen, the saved output does not turn out as
desired i.e. the secondary y-axis does not display fully.
Just run
Hi
r-help-boun...@r-project.org napsal dne 30.03.2010 18:46:35:
Hello,
Thomas Jensen wrote:
Dear R-list,
Sorry for spamming the list lately, I am just learning the more
advanced
aspects of R!
I have some data that looks like this:
Out Country1 Country 2 Country 3 ...
Hi
r-help-boun...@r-project.org napsal dne 31.03.2010 06:15:38:
Hi r-users,
I would like to display my legend with fill box and line symbol. The
first
color will be in the box and the second colour will just be a line.
legend(topright, c(observed,gamma sum fit), col= c
Hi
r-help-boun...@r-project.org napsal dne 27.03.2010 11:53:30:
Hi:
Does this do what you want?
# Create some fake data...
df - data.frame(id = factor(rep(c('cell1', 'cell2'), each = 10)),
cond = factor(rep(rep(c('A', 'B'), each = 5), 2)),
time =
Hi
r-help-boun...@r-project.org napsal dne 29.03.2010 08:41:15:
Hi Leo,
see the matrix function e.g.
m - matrix(0, nrow=1, ncol=3)
then you can use functions like rbind or cbind to create bigger ones.
I am a newbie so double check everything :)
HTH,
Best regards,
Giovanni
Hi
One option (probably not the best one) is index an aggregate matrix
mat - rbind(mat1, mat2, mat3)
idx - seq(1,12, 4)+rep(0:3, each=3)
mat - mat[idx, ]
Regards
Petr
r-help-boun...@r-project.org napsal dne 29.03.2010 09:44:12:
Dear all,
Ket say I have 3 matrices :
mat1 -
Hi
It is really a nice example. You managed to break probably all rules
specified in posting guide
No reproducible example
No structure of data
No explain what you really want
No effort on your side even to look to docs.
You maybe want some object of type list.
obj - vector(98, mode=list
you
Hi
r-help-boun...@r-project.org napsal dne 29.03.2010 16:13:31:
Hi,
Your question is really vague.
What about legend(lty=, pch=)?
Well, it is probably not mentioned explicitly in ?legend but he is
probably seeking
legend(..., lty=c(1,NA, 2,3), pch=c(NA, 16, NA,NA))
Regards
Petr
Ivan
Message-
From: Petr PIKAL [mailto:petr.pi...@precheza.cz]
Sent: 26 March 2010 14:00
To: Jessica Cathro
Cc: r-help@r-project.org
Subject: Odp: [R] Help with assigning a value based on existing numbers
Hi
again a work for cut.
See ?cut and notice a labels option.
Regards
Petr
Hi
r-help-boun...@r-project.org napsal dne 26.03.2010 10:41:29:
Hi,
I have a column in a data frame looking something like:
$sex $language $count
male english 0
male english 0
female english 32
male spanish 154
female english 11
female norweigan 7
and so on.
What I
Hi
again a work for cut.
See ?cut and notice a labels option.
Regards
Petr
r-help-boun...@r-project.org napsal dne 26.03.2010 11:37:20:
Hi All
I have a column/variable called time difference. It has a whole list of
numbers from 0 through to the hundreds eg 236. I want to assign a
Hi
There is also mosaicplot in base graphics
?mosaicplot
Petr
r-help-boun...@r-project.org napsal dne 22.03.2010 14:16:58:
Hi,
maybe vcd package is what you are looking for.
Example from vcd library
library(vcd)
?mosaic
data(Titanic)
mosaic(Titanic)
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