Hi All,
I am completely new to R. I have the below data and want to create a
conditional variable say Prof.H as such that it equals 1 if Prof is > 50 and
0 otherwise and create a scatter plot of Value and Dim conditional on the
new variable.
Area Value Dim Prof
1 1 145.1
I have timestamps from mysql database:
> dput(tstamp)
c(1225221868L, 1225221906L, 1225221906L, 1225230997L, 1225231000L,
1225231003L, 1225231152L, 1225231348L, 1225231351L, 1225231400L
)
How to convert these into normal dates?
Thanks,
jrara
__
R-help@
i got it
use function rank can get the result
2010/1/28 song song
> the question like this :
>
> x=c(2,6,8,11,11,11,6,8,2,8,6,11,8,2,6,11)
>
> x contains values (2, 6, 8, 11) with frequency (3, 4, 4, 5).
>
> now I want to get the vector by replacing (2, 6, 8, 11) --->>> (1 2 3 4)
>
> that is
the question like this :
x=c(2,6,8,11,11,11,6,8,2,8,6,11,8,2,6,11)
x contains values (2, 6, 8, 11) with frequency (3, 4, 4, 5).
now I want to get the vector by replacing (2, 6, 8, 11) --->>> (1 2 3 4)
that is, want to get new x as (1 2 3 4 4 4 2 3 1 3 2 4 3 1 2 4)
is there any function in R
Hi -
I also posted this on r-sig-ecology to little fanfare, so I'm trying
here. I've recently hit an apparent R issue that I cannot resolve (or
understand, actually).
I am using the quantreg package (quantile regression) to fit a vector
of quantiles to a dataset, approx 200-400 observation
I am not able to get spm function in SemiPar to work with plyr. Here's an
example:
library(plyr)
library(SemiPar)
data <- data.frame(id=c(rep("111",100),rep("222",200)),
value=c(rnorm(100,2,1),rnorm(200,10,5)))
#this works
data111 <- data[data$id=="111"
Hi
r-help-boun...@r-project.org napsal dne 28.01.2010 17:40:01:
> Thank you, Dennis and Petr.
>
> One more question: when aggregating to one es per id, how would I go
about
> keeping the other variables in the data.frame (e.g., keeping the value
for
> the first row of the other variables, suc
Dear Rxperts..
what settings in barplot and histogram do I use, to show bars in an
monotonously increasing or decreasing order of the frequency of a
categorical variable?
an example is provided below..
histogram(~factor(sample(letters,200,rep=T)))
I was able to get it in 3 - 4 steps.. l
a1 <- s
I heard form my friend there is a way to run R in system hard disk space not
in the RAM .By that we may not run out of memory and have problem attached
with the same.Someone could help me in this.Thanks.
--
View this message in context:
http://n4.nabble.com/R-on-Hard-drive-memory-tp1401528p14015
The message you are getting is unrelated to reading files. That
probably means you
have loaded a previous work file that is masking something. Type
conflicts(detail=TRUE)
If that doesn't give you the answer, then
try closing R entirely and start a new session from
START > Run > c:\Progra~1\
IGNACIO PEREZ VELEZ escuelaing.edu.co> writes:
>
> Hi,
>
> I migrated to Windows 7 and now rattle does not read files whenever I try to
read a file I get the following error:
>
> "Error en sqrt(ncol(crs$dataset)) :
> Argumento no numérico para función matemática"
>
> In english:
>
> "Err
I have a data set of many rows and many columns in which both the rows and
the columns have associated grouping factors.
Is there a way to do what 'aggregate' does but in the other dimension?
The way I have been doing this is to use 'aggregate' on the data in the
usual way and then rotate the res
Yes I do have zeros in my data. But I m not able to understand y inclusion of
zeros results in error messages, because range for x in weibull distribution
is x>=0. Can you please clarify this doubt?
--
View this message in context:
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Tena koe Gabriele
read.table() creates a data frame not a matrix. The distinction is
important. A data frame is a special kind of list: it is a list of
column vectors all of the same length. Because it is a list, each
column can be of a different type (i.e., integer, numeric, character,
factor
Hi,
I'm trying to create a frequency histogram for all the values in a table, but
when I try to do so, an error is returned, saying that I cannot create a
histogram with an obkect of that class. Here's what I do:
>library(lattice)
>table<-read.table("C:/.../table",header=TRUE,sep="\t")
>histog
you might want to look at the rollmean function in the zoo package if you want
a fixed window. If you want a cumulative mean then you can do something like
> cumsum(x)/seq_along(x)
If neither of those work, then give us some more detail.
From: r-help-bo
oki :D
and do you know if there is an inbuilt R pacakage that calculate moving
averages over time?
i found this ,
"Calculate various moving averages (MA) of a series."
Usage
SMA(x, n=10)
EMA(x, n=10, wilder=FALSE)
but when i tried SMA , R wouldnt recognize it! and there isnt such a packa
Sorry, that is a stupid bug in the blockrand function. I will hunt down the
author and slap him upside the head until he fixes it (actually I will probably
just raid his freezer and eat his ice cream).
In the mean time there is a simple work around, if you run blockrand like:
> blockrand(40, b
I have tried blockrand but it works in a way that it keeps altering the block
size.. like it alternates between 8,6,12. I want a constant block size of 8
:(
--
View this message in context:
http://n4.nabble.com/random-permuted-block-randomization-tp1401407p1401496.html
Sent from the R help maili
the command:
> axis(4)
will put tickmarks and labels on the right, you may want to preceed the plot
with something like:
> par(mar=c(5,4,4,4)+0.1)
see ?axis and ?par for details.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Beha
Take a look at the blockrand package, it may do what you want.
> library(blockrand) #after installing
> myrand <- blockrand(40, block.sizes=4)
> myrand
hope this helps,
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
Ayesha
For example I have a time series
Q(t) ~ Q(t-1) + Q(t-2) + Q(t-3)
meaning that my current value is dependent to the 3 previous values.
Can anybody help me express this in a formula that I can use for my neural
network model (I am planning to use packages "nnet" and "MASS")
--
View this message
manipulated
> (by some algorithm or be saved individually etc. etc.)...
>
> I am trying to modify the code at the above link but somehow I can not
> make it to work with zoo time series objects.
>
> Any help would be greatly appreciated.
>
> Thanks in advance,
> Costas
>
hi everyone!
I have a 40-year monthy streamflow record. And i want to fit a neural
network. I already fitted an autoregressive model and found out that an
AR(3) model fits my time series (time.series) the best.
I am currently having problems on how to express the argument "formula" for
my neura
Hi r-users,
I manage to install the Tinn-R_2.3.4.4 successfully. However, I cannot the
execute my code and cannot open that window and get this message:
"The preferred Rterm was not defined "
and it suggest 'Options/main/R//path/Gui'. What that mean and how to do it.
Thank you so mcuh f
No it isn't strange. Please read:
?options digits
?print.default
and then print the results with more digits.
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Len Vir
Sent: Th
Hi!
That is somewhat strange.
§R>2^100
[1] 1.267651e+30
§R> x <- 2^50
§R> y <- x + 1
§R> y-x
[1] 1
§R>x
[1] 1.1259e+15
§R>x+1
[1] 1.1259e+15
len
From: Duncan Murdoch [murd...@stats.uwo.ca]
Sent: Tuesday, January 26, 2010 4:09 PM
To: Blanford, Glenn
Cc: r-help@R-project.o
Hi,
I am very new to R. Just started yesterday. I have to generate a sequence of
40 random treatments using permuted block randomization with a block size of
8. Then I have to plot moving averages for the resulting sequence. I have
tried the sample function but I dont know if wht i did is right or
Did you try it? I said
library(zoo)
na.locf(P, fromLast=TRUE)
Note the 'fromLast=TRUE'; that tells na.locf() to use the
reverse of your vector.
What I meant by
"You'll have to decide what to do if the last value is NA"
is that you'll need to decide what to do if your vector
ends with a NA, say
Hi,
I migrated to Windows 7 and now rattle does not read files whenever I try to
read a file I get the following error:
"Error en sqrt(ncol(crs$dataset)) :
Argumento no numérico para función matemática"
In english:
"Error on sqrt(ncol(crs$dataset)) :
Non numeric argument for a mathematic
On Thu, 28 Jan 2010, Jason Smith wrote:
It wouldn't be guaranteed to produce any usable permutation, but it seems
like it would be much faster and so could be repeated until an acceptable
vector is found. What do you think?
Thanks--
Andy
I think I am not understanding what your ultimate goa
Hi Jim your suggestion doesn´t work properly. I have find some documents on
graphs so let me think till monday and if not I will detail where it fails.
Sorry again...I feel very silly, really, it is not a joke but I will
continue trying...
I have a question also, in a general plot (please if you
Hi,
I migrated to Windows 7 and now rattle does not read files whenever I try to
read a file I get the following error:
"Error en sqrt(ncol(crs$dataset)) :
Argumento no numérico para función matemática"
In english:
"Error on sqrt(ncol(crs$dataset)) :
Non numeric argument for a mathematic
Hi Peter, thank you for helping. The thing is don't want to it replace it
with the last value but with the next value
-
Anna Lippel
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.co
Well, your bars are not unevenly spaced; you just have
some zero-count intervals. Time to learn about the
str() function which will tell you what's going on.
zh <- hist(your_code)
str(zh)
zh$breaks
zh$counts
You could set breaks with
hist(..., breaks=0:5 + .5)
But a histogram doesn't seem like
Does this help:
library(zoo)
na.locf(P, fromLast=TRUE)
You'll have to decide what to do if the last value is NA.
-Peter Ehlers
anna wrote:
Hello everyone, I have a vector P and I want to replace each of its missing
values by its next element, for example:
P[i] = NA --> P[i] = P[i+1]
To do th
Well, Albyn Jones gave a great solution to my challenge that found the best
reading schedule.
My original thought was that doing an exhaustive search would take too much
time, but Albyn showed that there are ways to do it efficiently.
My approach (as mentioned before) was to use optim with meth
I am sure this is trivial, but I cannot solve it.
I make a histogram. There are 5 categories "1",...,"5" and 80 values and
the histogram does not evenly space the bars.
Bars "1" and "2" have no space between them and the rest are evenly spaced.
How can I get all bars evenly spaced?
The code:
Hello everyone, I have a vector P and I want to replace each of its missing
values by its next element, for example:
P[i] = NA --> P[i] = P[i+1]
To do this I am using the replace() and lag() functions like this:
P <- replace(as.ts(P),is.na(as.ts(P)),as.ts(lag(P,1)))
but here is the error that I ge
If you understand the differences between R lists and R vectors then this
should be easy:
> vec1 <- 1:10
> vec2 <- 2:4
> myListOfVectors <- list( vec1, vec2 )
Now you can pass the single list of 2 different sized vectors to your function.
For more details on working with lists (and vectors an
On Jan 28, 2010, at 4:03 PM, Jonathan Greenberg wrote:
list1=c(1:10) # neither of which really are lists
list2=c(2:4)
lists = list(list1,list2) $ a list of two vectors.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
__
R-help@r-proj
On Jan 28, 2010, at 2:14 PM, DispersionMap wrote:
sorry, i ommited some important information.
this is a documentation question!
i meant to ask how to find out how R calculates the standard error
and how
it differs between the two models
Luke, Use the Code!.
--
View this message
I'm hoping to get some "best practice" feedback for constructing a
function call which takes an undefined set of DIFFERENT length vectors
-- e.g. say we have two lists:
list1=c(1:10)
list2=c(2:4)
lists = data.frame(list1,list2) coerces those two to be the same length
(recycling list2 to fill
I don't know of any existing palettes that meet your conditions, but here are a
couple of options for interactive exploration of colorsets (this is quick and
dirty, there are probably some better orderings, base colors, etc.):
colpicker <- function( cols=colors() ) {
n <- length(cols)
> It wouldn't be guaranteed to produce any usable permutation, but it seems
> like it would be much faster and so could be repeated until an acceptable
> vector is found. What do you think?
>
> Thanks--
> Andy
>
I think I am not understanding what your ultimate goal is so I'm not
sure I can give
On 28/01/2010 3:22 PM, Changyou Sun wrote:
Duncan,
Thank you for your quick reply. Do we users have any options to change
that? I personally become addicted to the navigation panel and feel it
is kind of table of contents.
You could downgrade to 2.9.2, but you'd lose all the other new stuff
Duncan,
Thank you for your quick reply. Do we users have any options to change
that? I personally become addicted to the navigation panel and feel it
is kind of table of contents.
Regards,
Edwin Sun
-Original Message-
From: Duncan Murdoch [mailto:murd...@stats.uwo.ca]
Sent: Thursday,
On 28/01/2010 3:15 PM, Edwin Sun wrote:
All,
I installed the lastest version of R 2.10.1. On the help page for a specific
function, it turns out that the vertical navigation panel on the left does
not appear anymore. For example,
?lm
The help page from this command is a page without navigati
Thank you very much everybody. That worked.
Dana
On Thu, Jan 28, 2010 at 12:23 PM, Henrique Dallazuanna wrote:
> Try this:
>
> ong<-reshape(as.data.frame(dataset), idvar="subject",
> v.names="response", varying=list(2:5), direction="long")
> or
> dataset <- cbind.data.frame(y1, y2, y3, y4)
>
>
All,
I installed the lastest version of R 2.10.1. On the help page for a specific
function, it turns out that the vertical navigation panel on the left does
not appear anymore. For example,
?lm
The help page from this command is a page without navigation panel (which I
prefer to use). I notice
Hi Ram,
As others have pointed out, writing the code is the least of your
problems. In case this isn't sinking in, try the following exercise:
set.seed(10)
P <- vector()
DF <- as.data.frame(matrix(rep(NA, 10), nrow=100))
names(DF) <- c(paste("x",1:999, sep=""), "y")
for(i in 1:1000) {
DF[,i
chipmaney wrote:
typically, the apply family wants you to use vectors to run functions on.
However, I have a function, kruskal.test, that requires 2 arguments.
kruskal.test(Herb.df$Score,Herb.df$Year)
This easily computes the KW ANOVA statistic for any difference across
years
However, my
Hi Jason,
Thanks for you suggestions, I think that's pretty close to what I'd need.
The only glitch is that I'd be working with a vector of ~30 elements, so
permutations(...) would take quite a long time. I only need one permutation
per vector (the whole routine will be within a loop that generat
Regarding the explanation of where the time goes it might be parsing
the statement or the development of the query plan. The SQL statement
for the more complex query is obviously much longer and its generated
query plan involves 95 lines of byte code vs 19 lines of generated
code for the simpler q
Hi
Markus Loecher wrote:
> While I am very happy with and awed by the grid package and its basic
> plotting primitives such as grid.points, grid.lines, etc, I was wondering
> whether the equivalent of a grid.image() function exists ?
No. But a simple implementation based on grid.rect() is not
sorry, i ommited some important information.
this is a documentation question!
i meant to ask how to find out how R calculates the standard error and how
it differs between the two models
--
View this message in context:
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I'm talking about ease of use to. The first line of the Details section in
?"[.data.table" says :
"Builds on base R functionality to reduce 2 types of time :
1. programming time (easier to write, read, debug and maintain)
2. compute time"
Once again, I am merely saying that the
The warning message simply indicates that you have more than one data point
with the same "x" value. So, `approx' collapses over the dulicate x values
by averaging the corresponding "y" values. I am not sure if this is your
problem - it doesn't seem like it. It is doing what seems reasonable for
Try this:
ong<-reshape(as.data.frame(dataset), idvar="subject",
v.names="response", varying=list(2:5), direction="long")
or
dataset <- cbind.data.frame(y1, y2, y3, y4)
On Thu, Jan 28, 2010 at 3:07 PM, Dana TUDORASCU wrote:
> Hello everyone,
> I have a bit of a problem with reshape function in
Why not look into the zoo package na.approx? And related functions.
On Thu, Jan 28, 2010 at 11:29 AM, ogbos okike wrote:
> Happy New Year.
> I have a data of four columns - year, month, day and count. The last column,
> count, contains some missing data which I have to replace with NA. I tried
>
Hello everyone,
I have a bit of a problem with reshape function in R.
I have simulated some normal data, which I have saved in 4 vectors.
y.1,y.2,y.3,y.4 which I combined a dataset:
datasethttps://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/pos
Hi,
I'm having trouble editing the qplot layout. I'm using the geom="tile"
option and I want to do a few things:
1. move the vertical and horizontal gridlines so that they appear on the
edge of each tile (right now they're in the middle)
2. bring the gridlines to the foreground and change thei
vikrant wrote:
Hi,
I want to estimate parameters of weibull distribution. For this, I am using
fitdistr() function in MASS package.But when I give fitdistr(c,"weibull") I
get a Error as follows:-
Error in optim(x = c(4L, 41L, 20L, 6L, 12L, 6L, 7L, 13L, 2L, 8L, 22L,
:
non-finite valu
Hi:
On Thu, Jan 28, 2010 at 8:40 AM, AC Del Re wrote:
> Thank you, Dennis and Petr.
>
> One more question: when aggregating to one es per id, how would I go about
> keeping the other variables in the data.frame (e.g., keeping the value for
> the first row of the other variables, such as mod2)
You'll probably need to consult a suitable text on linear models/applied
regression, as this is a statistics, not an R question -- or look for a
suitable tutorial on the web. You might also try one of the statistics
mailing lists or Google on some suitable phrase.
Bert Gunter
Genentech Nonclinica
On Jan 28, 2010, at 12:08 PM, Marc Schwartz wrote:
On Jan 28, 2010, at 10:04 AM, David Winsemius wrote:
On Jan 28, 2010, at 10:55 AM, Marc Schwartz wrote:
Ivan,
The default behavior for print()ing objects to the console in an R
session is via the use of the print.* methods. For real nu
On Jan 28, 2010, at 10:42 AM, Gopikrishna Deshpande wrote:
Hi,
I have a matrix of size 19x512x20 in R.
No, you don't. Matrices are only 2 dimensional in R. You may have an
"array", however.
I want to export this file into
another format which can be imported into MATLAB.
write.xls or wr
On Jan 28, 2010, at 10:04 AM, David Winsemius wrote:
>
> On Jan 28, 2010, at 10:55 AM, Marc Schwartz wrote:
>
>> Ivan,
>>
>> The default behavior for print()ing objects to the console in an R session
>> is via the use of the print.* methods. For real numerics, print.default() is
>> used and
I need to find out the difference between the way R calculates weighted
regression and standard regression.
I want to plot a 95% confidence interval around an estimte i got from least
squares regression.
I cant find he documentation for this
ive looked in
?stats
?lm
?predict.lm
?weights
?res
Thanks. My mistake was that I used c(dbs.final$Days,dbs.final$Place) instead of
list(... when I tried to follow that part of the documentation.
>>> David Winsemius 1/28/2010 11:49 AM >>>
On Jan 28, 2010, at 10:26 AM, GL wrote:
>
> Can you make tapply break down groups similar to bwplot or suc
I think one would only be concerned about such internals if one were
primarily interested in performance; otherwise, one would be more
interested in ease of specification and part of that ease is having it
independent of implementation and separating implementation from
specification activities. A
Dear Abraham,
If I follow correctly what you want to do, the following should do it:
> f <- factor(c(1, 1, 5, 5, 8, 8, 9, 9, 0, 0))
> f
[1] 1 1 5 5 8 8 9 9 0 0
Levels: 0 1 5 8 9
> recode(f, " '1'=3; '5'=1; '0'=2; else=NA ")
[1] 331122
Levels: 1 2 3
I think that your
On Jan 28, 2010, at 10:26 AM, GL wrote:
Can you make tapply break down groups similar to bwplot or such?
Example:
Data frame has one measure (Days) and two Dimensions (MM and
Place). All
have the same length.
length(dbs.final$Days)
[1] 3306
length()
[1] 3306
length()
[1] 3306
Hi all,
Note:
lm(Yield ~ Block + C(Variety, base = 2), Alfalfa)
equals
i <- 2; lm(Yield ~ Block + C(Variety, base = i), Alfalfa)
However,
lme(Yield ~ C(Variety, base = 2), Alfalfa, random=~1|Block)
which is fine, does not equal
i <- 2; lme(Yield ~ C(Variety, base = i), Alfalfa, random=~1|Block)
Thank you, Dennis and Petr.
One more question: when aggregating to one es per id, how would I go about
keeping the other variables in the data.frame (e.g., keeping the value for
the first row of the other variables, such as mod2) e.g.:
# Dennis provided this example (notice how mod2 is removed f
VAR 980490
Some people have suggested placing new limits on foreign
imports in order to protect American jobs. Others say
that such limits would raise consumer prices and hurt
American exports.
Do you FAVOR or OPPOSE placing new limits on imports, or
haven't you thought much about this?
Ivan,
Now I'm no longer sure of just what you want. Are you concerned about
the *internal* handling of numbers by R or just about the *printing*
of numbers? As Marc has pointed out, internally R will use the full
precision that your input allows.
Perhaps you're using the F-value from the output
Hi everyone,
I am trying to install the RMySQL package under windows xp. I've got the MySQL
installed on the computer (MySQL server 5.1). I went through the steps
presented on the webpage http://biostat.mc.vanderbilt.edu/wiki/Main/RMySQL and
googled around and still can't find the answer. With
I guess the easiest solution for me would therefore be to set
options(digits) to a high number, and then round down if I need to!
Thanks you both for your input!
Ivan
Le 1/28/2010 17:02, Peter Ehlers a écrit :
Looks like I didn't read your post carefully enough.
If you want some sort of global
Looks like I didn't read your post carefully enough.
If you want some sort of global option to set the
display of numbers from any operation performed by R
then that's not likely to be possible without
capturing all output and formatting it yourself.
As the saying goes 'good luck with that'.
Note
On Jan 28, 2010, at 10:55 AM, Marc Schwartz wrote:
Ivan,
The default behavior for print()ing objects to the console in an R
session is via the use of the print.* methods. For real numerics,
print.default() is used and the format is based upon the number of
significant digits, not the num
First things first: thanks for your help!
I see where the confusion is. With formatC and sprintf, I have to store
the numbers I want to change into x.
I would like a way without applying a function on specific numbers
because I can shorten the numbers that way, but it won't give me more
deci
On Thu, 28 Jan 2010, Benilton Carvalho wrote:
Hi Duncan,
On Tue, Jan 26, 2010 at 9:09 PM, Duncan Murdoch wrote:
On 26/01/2010 3:25 PM, Blanford, Glenn wrote:
Has there been any update on R's handling large integers greater than 10^9
(between 10^9 and 4x10^9) ?
as.integer() in R 2.9.2 lists
Ivan,
The default behavior for print()ing objects to the console in an R session is
via the use of the print.* methods. For real numerics, print.default() is used
and the format is based upon the number of significant digits, not the number
of decimal places. There is also an interaction with p
On Wed, 27 Jan 2010, David Winsemius wrote:
On Jan 27, 2010, at 9:09 PM, GlenB wrote:
I have an additive model of the following form :
zmdlfit <- lm(z~ns(x,df=6)+ns(y,df=6))
I can get the fitted values and plot them against z easily enough, but I
also want to both obtain and plot the two
Ivan Calandra wrote:
It looks to me that it does more or less the same as format().
Maybe I didn't explain myself correctly then. I would like to set the
number of decimal by default, for the whole R session, like I do with
options(digits=6). Except that digits sets up the number of digits
(i
Hi,
I have a matrix of size 19x512x20 in R. I want to export this file into
another format which can be imported into MATLAB.
write.xls or write.table exports only one dimension.
please send a code if possible. I am very new to R and have been struggling
with this.
Thanks !
Gopi
[[altern
While I am very happy with and awed by the grid package and its basic
plotting primitives such as grid.points, grid.lines, etc, I was wondering
whether the equivalent of a grid.image() function exists ?
Any pointer would be helpful.
Thanks !
Markus
[[alternative HTML version deleted]]
I'm looking for a scheme to generate a default color palette for
plotting points, lines and text (on a white or transparent background)
with from 2 to say 9 colors with the following constraints:
- "red" is reserved for another purpose
- colors should be highly distinct
- avoid light colors (like
On Fri, Jan 22, 2010 at 2:08 AM, Dieter Menne
wrote:
>
>
> Deepayan Sarkar wrote:
>>
>> With a restructuring of the data:
>>
>> df1 = data.frame(x=0:n, y1=((0:n)/n)^2, y2=1-((0:n)/n)^2, age="young")
>> df2 = data.frame(x=0:n, y1=((0:n)/n)^3, y2=1-((0:n)/n)^3, age="old")
>> df = rbind(df1, df2)
>>
That works great. Thanks!
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hi,
i have a longitude vector (x) a latitude vector (y) and a matrix of bathymetry
(z) with the dimensions (x,y). I have already succeeded in plotting it with the
image.plot (package 'field') and the contour functions.
But now, I want to make a grid in order to extract easily the bathymetry
c
Can you make tapply break down groups similar to bwplot or such? Example:
Data frame has one measure (Days) and two Dimensions (MM and Place). All
have the same length.
> length(dbs.final$Days)
[1] 3306
> length(dbs.final$Place)
[1] 3306
> length(dbs.final$MM)
[1] 3306
Doing the followi
On Thu, Jan 28, 2010 at 6:25 AM, GL wrote:
>
> I have beautiful box and whisker charts formatted with lattice, which is
> obviously calculating summary statistics internally in order to draw the
> charts. Is there a way to dump the associated summary tables that are being
> used to generate the ch
It looks to me that it does more or less the same as format().
Maybe I didn't explain myself correctly then. I would like to set the
number of decimal by default, for the whole R session, like I do with
options(digits=6). Except that digits sets up the number of digits
(including what is befor
On Jan 28, 2010, at 10:04 AM, Oliver wrote:
OK, now it works... just using [ and ] or [[ and ]] works.
I thought have tried it before... why does it workj now and not
before?
Provide your console session and someone can tell you. Failing that,
you are asking us to read your mind.
OK, now it works... just using [ and ] or [[ and ]] works.
I thought have tried it before... why does it workj now and not before?
hmhh
sorry for the traffic
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PLEASE do
Hello,
say I have a dataframe x
and it contains rows like "ch_01", "ch_02" and so on.
How can I select those channels iundirectly, by name?
I tried to select the data with get() but get() seems only to work
on simple variables?
Or how to do it?
I need something like that:
name1 <- "ch_01"
Hi,
I have a list of dates like this:
date
2009-12-03
2009-12-11
2009-10-07
2010-01-25
2010-01-05
2009-09-09
2010-01-19
2010-01-25
2009-02-05
2010-01-25
2010-01-27
2010-01-27
...
and am creating a histogram like this
t <- read.table("test.dat",header=TRUE)
hist(
I have beautiful box and whisker charts formatted with lattice, which is
obviously calculating summary statistics internally in order to draw the
charts. Is there a way to dump the associated summary tables that are being
used to generate the charts? Realize I could use tapply or such to get
somet
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