On 1/16/15 9:34 AM, Bert Gunter wrote:
Chee Hee's approach is both simpler and almost surely more efficient,
but I wanted to show another that walks the tree (i.e. the list)
directly using recursion at the R level to pull out the desired
components. This is in keeping with R's functional
On 8/30/2014 2:11 PM, Felipe Carrillo wrote:
library(plyr)
b - structure(list(SampleDate = structure(c(1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L), .Label = 5/8/1996, class = factor), TotalCount = c(1L,
2L, 1L, 1L, 4L, 3L, 1L, 10L, 3L), ForkLength = c(61L, 22L, NA,
NA, 72L, 34L, 100L, 23L, 25L),
On 8/25/2014 2:08 PM, Jonathon Love wrote:
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA512
hey,
i was able to solve the problem by explicitly putting
R_LIBS_USER=''
in the /etc/Renviron file (it isn't enough simply to comment the entry
out, because it has a default value that magically comes
On 7/20/2014 12:50 AM, Kevin Kunzmann wrote:
Hi,
I have developed a package and would like to switch documentation to
roxygen2 from manual :) However
roxygen2::roxygenize()
First time using roxygen2 4.0. Upgrading automatically...
Loading required package: nleqnslv
Error in
On 4/17/2014 5:44 AM, Michael Friendly wrote:
I know I can do that. My example was just a toy version of a more
complex graph I
generate on the logit scale, and save as gg.
I wanted to know if there was a way to transform it to the probability
scale by using
gg + coord_trans()
with some
On 3/7/2014 7:41 PM, Keith S Weintraub wrote:
Folks,
I have a data frame as follows:
foo-structure(list(name = c(A, B, C), num = c(3L, 2L, 1L)), .Names =
c(name,
num), row.names = c(NA, -3L), class = data.frame)
str(foo)
'data.frame': 3 obs. of 2 variables:
$ name: chr A B C
$
On 12/13/2013 1:14 PM, Silvano Cesar da Costa wrote:
Hi,
I'm using Sweave to create some tables. My code is:
label=Q1, echo=FALSE, results=tex=
tab1 = table(DISCIPLINA, Q1)
tab1.prop = round(addmargins(100*prop.table(tab1, 1),
FUN=list(Total=sum)), 2)
tab1.txt = xtable(tab1.prop,
On 11/17/2013 5:28 AM, David Winsemius wrote:
On Nov 17, 2013, at 3:21 AM, Chris89 wrote:
Hi everyone!
I am in the process of writing an R-package and while writing a summary
function, I have come across a problem. I am able to print a summary
table
(as in a standard glm() summary) by using
On 10/17/2013 11:54 AM, Stock Beaver wrote:
# Suppose I have a vector:
myvec = c(1,0,3,0,77,9,0,1,2,0)
# I want to randomly pick an element from myvec
# where element == 0
# and print the value of the corresponding index.
# So, for example I might randomly pick the 3rd 0
# and I would
On 9/26/2013 1:45 PM, Caitlin wrote:
Hi all.
I am attempting to graph data from three lab teams with milligrams of
maltose shown on the y-axis and 5 pH values (5 to 9) on the x-axis as
labels. Unfortunately, I can't seem to construct the graph in this manner
using the following code:
ph1 =
On 8/30/2013 3:36 AM, Uwe Ligges wrote:
On 30.08.2013 11:59, Christofer Bogaso wrote:
Hello again,
I have a string which I need to put in some legitimate date format.
My string is: MAY-14
And output format would be 05/01/2014, this should be of Date class,
so that I can make some
On 7/25/2013 8:13 AM, Juan Antonio Balbuena wrote:
Hello
I hope that there is a simple solution to this apparently complex problem.
Any help will be much appreciated:
I have a dataframe with Left and Right readings (that is, elements in each
row are paired). For instance,
On 7/25/2013 11:34 AM, Rui Barradas wrote:
Hello,
I'm not an expert in ggplot2 graphics but I can (partly) answer to your
first question. Inline.
Em 25-07-2013 18:30, Gavin Rudge escreveu:
Further to my recent post on this topic and thanks to help received
already (thanks BTW), I've got
On 7/19/2013 12:54 PM, Pete Brecknock wrote:
Hi
I am trying to add the contents of the list myList to a new column z in
the data frame myDataframe
myList - list(c(A1,B1), c(A2,B2,C2), c(A3,B3))
myDataframe - data.frame(x=c(1,2,3), y=c(R,S,T))
Would like to produce
x y z
1 R A1,B1
On 6/27/2013 12:34 AM, Suparna Mitra wrote:
Hello R experts,
I am having a problem to edit legend in ggplot using four variables.
My data structure is :
str(df)
'data.frame': 10 obs. of 6 variables:
$ id: Factor w/ 2 levels 639A,640: 1 1 1 1 1 2
2 2 2 2
$
On 1/25/2013 2:29 PM, emorway wrote:
I played around with your example on the smaller dataset, and it seemed like
it was doing what I wanted. However, applying it to the larger problem, I
didn't get a resized 2D dataset that preserved the order I was hoping for.
Hopefully the following
On 1/23/2013 6:40 PM, C W wrote:
Hey Mark,
I am aware of this package. The situation is,
1. I am emailing my code from my machine to a public machine. Installation
is a hassle.
2. Copy pasting for a website.
For your second use case, there is the Pretty R syntax highlighter at
On 11/13/2012 11:19 AM, Duncan Murdoch wrote:
On 13/11/2012 1:33 PM, Jamie Olson wrote:
I was surprised to notice that statements like:
h = function(...){list(...)}(x=4)
do not throw syntax errors. R claims that 'h' is now a function, but I
can't seem to call it.
h =
On 11/6/2012 11:11 PM, ChangLH wrote:
Hi,
Color of my step plot is now by default. Now I'd like to change the color as
the grey scale I specified. I don't know why I got three black plot. Here I
attach two version of codes. The first one produces a step plot with color
by default. The second
On 11/2/2012 11:11 AM, Ni, Shenghua wrote:
r-c(1,1,9,1,1,1)
col_no-cut(r,c(0,2,3,6,8,10,100))
levels(col_no)-c(2%,2-4%,4-6%,6-8%,8-10%,10%)
col_no
[1] 2% 2% 8-10% 2% 2% 2%
Levels: 2% 2-4% 4-6% 6-8% 8-10% 10%
Yes.
(Or, I get the same output from this code. What is the question?)
On 11/1/2012 1:06 PM, mdvaan wrote:
I should have been more specific:
y - list()
a - c(A, K)
b - c(B, L)
c - c(C, M)
d - c(D, N)
e - c(E, O)
y[[1]] - a
y[[2]] - b
y[[3]] - c
y[[4]] - d
y[[5]] - e
y
[[1]]
[1] A K
[[2]]
[1] B L
[[3]]
[1] C M
[[4]]
[1] D N
[[5]]
[1] E O
How do I get a list
On 11/3/2012 5:47 PM, Jim Lemon wrote:
On 11/04/2012 06:27 AM, Nathan Miller wrote:
Hi,
I'm trying to create a plot showing the density distribution of some
shipping data. I like the look of violin plots, but my data is not
continuous but rather binned and I want to make sure its binned nature
On 10/23/2012 5:59 AM, PIKAL Petr wrote:
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
On 10/19/2012 11:42 AM, autumnlin wrote:
http://r.789695.n4.nabble.com/file/n4646789/1.bmp
hi is it possible to draw a graph like the attached one using ggplot?
Yes
how should i code it in r?
Use a geom_point() where the shape aesthetic is mapped to the grouping
variable and also add a
On 10/2/2012 10:01 AM, Starkweather, Jonathan wrote:
I'm relatively new to R and would first like to sincerely thank all
those who contribute to its development. Thank you.
I would humbly like to propose a rule which creates a standard (i.e.,
strongly encouraged, mandatory, etc.) for authors to
On 9/30/2012 11:38 AM, Jonas Stein wrote:
Hi,
how can i adjust the font in a ggplot2 qplot so that it will look
similar to the LaTeX font?
Computer Modern Sans Serif in the same size would be nice.
My output device is
ggsave(filename=test.pdf, width=5.5, height=3, dpi=300)
and i will include
Light and\nheavy good\nvehicles\n(diesel) - GVX
On 8/7/2012 8:17 AM, Brian Diggs wrote:
On 8/6/2012 9:07 PM, vd3000 wrote:
Hi, all
I am trying to use the label_wrap_gen function in this website.
https://github.com/hadley/ggplot2/wiki/labeller
I tried to make a long name like this
Light
On 8/6/2012 9:07 PM, vd3000 wrote:
Hi, all
I am trying to use the label_wrap_gen function in this website.
https://github.com/hadley/ggplot2/wiki/labeller
I tried to make a long name like this
Light and heavy good vehicles (diesel) -\nGVX
f2 = facet_grid(vehicle ~ .,
A few comments in-line below
On 8/7/2012 1:53 PM, Jennifer Sabatier wrote:
PROV.PM.FBCTS - c(0.00 ,0.00, 33205.19, 25994.56, 23351.37, 26959.56
,27632.58, 26076.24, 0.00, 0.00 , 6741.42, 18665.09 ,18129.59 ,21468.39
,21294.60 ,22764.82, 26076.73)
FBCTS.INV.TOT - c(0 , 0, 958612,
On 7/13/2012 8:37 PM, darnold wrote:
Hi,
I hope that folks can give me some simple approaches to taking the data set
below, which is accumulated in two columns called long and group, then
arrange the data is the long column into a data frame containing five
variables: Group 1, Group 2, Group 3,
On 7/10/2012 7:53 AM, Peter Ehlers wrote:
On 2012-07-10 06:57, Rui Barradas wrote:
Hello,
If you write a function, it becomes less convoluted...
empty - function(x){
if(NROW(x) == 0){
y - rep(NA, NCOL(x))
names(y) - names(x)
y
}else x
}
(.xb - iris[
On 6/4/2012 12:12 PM, Marc Schwartz wrote:
To jump into the fray, he really needs to read the Details section of
?read.table and arguably, the source code for read.table().
It is not that the resultant data frame has row names, but that an
additional first *column name* called 'row.names' is
On 4/23/2012 9:24 AM, Matthias Rieber wrote:
Hello,
I've some problem with the ggplot2. Here's a small example:
--8--
library(ggplot2)
molten- data.frame(date=c('01','01','01','01',
'02','02','02','02'),
channel=c('red','red','blue','blue',
On 4/19/2012 5:20 AM, Brian Smith wrote:
Thanks Brian. That worked. I also wanted to increase the size of the
'points' on the graph. Is there any way I can get rid of the 'legend' (in
this case '3') appearing on the plot?
=== code
library(ggplot2)
leaves- letters[1:8]
On 4/20/2012 12:11 PM, Bush, Daniel P. DPI wrote:
I'm designing a set of plots intended for a general audience; here's
the code for one of them, using the latest version of ggplot:
plot.enr.all-
ggplot(data=df1, aes(x=HS_GRAD_YEAR, y=Percentage, group=Enrolled_by,
On 4/16/2012 7:31 AM, Brian Smith wrote:
Hi,
I was trying to replicate one of the graphs given on the ggplot2 website. I
have given a sample code below. I would like to combine the legends, since
each color is uniquely mapped to a shape.
###
library(ggplot2)
leaves- letters[1:8]
On 4/16/2012 1:23 PM, Jeff wrote:
I'm new to R.
I have two data frames I need to merge. One has an ID column the other
does not, but both have the same number of rows that are ordered in the
same way - e.g., row 1 is the same person in both data frames. For this
reason, there is no need to
corresponding non-NA element in the
##' given vectors in the order they are specified
##' @author Brian Diggs
coalesce - function(...) {
dots - list(...)
ret - Reduce(function (x,y) ifelse(!is.na(x),x,y), dots)
class(ret) - class(dots[[1]])
ret
}
And using your example data
On 3/15/2012 4:11 PM, tibaker wrote:
Hi
When I run my script using ggplot and geom_smooth I get messages that I
would like to suppress:
p- ggplot(dataSubset)
p- p + aes(x = as.Date(factor(key),format=%Y%m%d)) + geom_line()
p- p + geom_smooth(span=0.2,se=FALSE,size=0.7)
The messages look like
On 3/16/2012 10:29 AM, mdvaan wrote:
Hi,
This is probably an easy one, but I am new to ggplot2 and cannot find an
answer online.
Just so you know, there is a mailing list devoted to ggplot2. You can
subscribe and view messages at
https://groups.google.com/forum/?fromgroups#!forum/ggplot2
On 3/12/2012 10:47 AM, J Toll wrote:
Hi,
I have a problem that I'm finding a bit tricky. I'm trying to use
mapply and assign to generate curried functions. For example, if I
have the function divide
divide- function(x, y) {
x / y
}
And I want the end result to be functionally equivalent
On 2/22/2012 4:53 PM, Vinny Moriarty wrote:
Hello,
I have current data from a nortek ADP, which is basically current speed and
direction data in a 3 dimensional X Y Z format
http://www.nortekusa.com/usa/products/current-profilers/aquadopp-profiler-1
The instrument logs data in a complex
See comments inline
On 2/23/2012 3:27 PM, Kara Przeczek wrote:
Sorry. I forgot to note that I am using R version 2.8.0.
That's a rather old version; 2.8.0 came out in October 2008; maybe you
don't have control over that, though.
From:
Please include the context of the discussion in your responses. See
inline below.
On 1/24/2012 11:33 PM, Ajay Askoolum wrote:
Thanks you,
I can get the length of aa with length(unlist(aa)). If aa has 4
dimensions, I imagine I'd need to do
max(sapply(aa,sapply,sapply,length)
Correct.
How
On 1/25/2012 10:09 AM, patzoul wrote:
I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1] = b[1].
How can I do that without using a loop?
A combination of Reduce and Map will work. Map to stitch together the a
and b vectors, Reduce
On 1/26/2012 10:33 AM, Berend Hasselman wrote:
On 26-01-2012, at 19:10, Berend Hasselman wrote:
On 26-01-2012, at 17:58, Brian Diggs wrote:
On 1/25/2012 10:09 AM, patzoul wrote:
I have 2 series of data a,b and I would like to calculate a new series which
is z[t] = z[t-1]*a[t] + b[t] , z[1
On 1/24/2012 2:47 PM, Ajay Askoolum wrote:
Given a variable aa in the workspace, some of its attributes are:
typeof(aa)
[1] list
mode(aa)
[1] list
length(aa)
[1] 2
How do I retrieve the maximum indices, in this case 2,3,4? The variable itself
is:
aa
[[1]]
[[1]][[1]]
[[1]][[1]][[1]]
On 1/20/2012 9:12 AM, cameron wrote:
Can anyone please help me with this?
I have a list of business dates. What I want is to have last day of last
month and paste them on next month.
What i haveWhat i want
5725 2011-09-22
5726 2011-09-23
5727 2011-09-26
5728 2011-09-27
On 1/13/2012 2:26 PM, MacQueen, Don wrote:
It's a nice idea, but I wouldn't be optimistic about it happening:
Each of these public databases no doubt has its own more or less unique
API, and the people likely to know the API well enough to write R code to
access any particular database will be
On 11/18/2011 1:22 AM, ren...@vannieuwkoop.ch wrote:
\documentclass[11pt,a4paper]{article}
\usepackage{Sweave}
\begin{document}
=
x = runif(100, 1, 10)
y = 2 + 3 * x + rnorm(100)
@
echo=FALSE,results=tex=
library(xtable)
print(xtable(summary(lm(y~x)),
align=r,
caption=Summary
On 11/10/2011 11:32 AM, Jacob Wegelin wrote:
On my MacBook Pro (OS 10.6.8), after updating to R version 2.14.0
(2011-10-31) and reinstalling the Hmisc package, I am unable to load the
Hmisc library.
Hmisc was working *before* I updated R.
Any idea what's wrong?
Likely the same problem as
On 11/3/2011 12:37 AM, Rolf Turner wrote:
I have just installed R version 2.14.0 and tried to re-build and
re-check some
of the packages that I maintain.
I'm getting a warning (in the process of running R CMD check on my deldir
package):
* checking PDF version of manual ... WARNING
LaTeX
to the list. The
gist of the
off-line replies was ``you need to install some fonts'', and (unusually,
and very
kindly! :-) ) the responders told me *how* to do so.
SNIP (descriptions of how to install on various Linux distributions)
The third message (which *was* posted to the list) was from Brian
On 11/3/2011 3:30 PM, Brian Diggs wrote:
SNIP
The error on R CMD check I get is:
* checking PDF version of manual ... WARNING
LaTeX errors when creating PDF version.
This typically indicates Rd problems.
LaTeX errors found:
!pdfTeX error: pdflatex.EXE (file ec-inconsolata): Font ec-inconsolata
I'm going to put on my fire suit and wade in (see inline)
On 10/4/2011 8:11 AM, Bert Gunter wrote:
On Tue, Oct 4, 2011 at 7:42 AM, Jeanne M. Spicerxn8spi...@gmail.comwrote:
I'm not sure how returning an incorrect result is ever a 'positive' feature
It is **not** incorrect; perhaps
On 10/1/2011 3:03 AM, jim holtman wrote:
try this:
vec1-
c(4,5,6,7,8,9,10,11,12,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81)
vec2- c
On 9/29/2011 8:11 PM, wirichada wrote:
Dear R users,
I am trying to do the forest plot follow the function given on web. However,
the order of the tests has been sorted alphabetically. I would prefer
keeping the order as data frame input so that I can group and compare (from
the graph) the
On 9/29/2011 8:14 PM, Jp zhu wrote:
I am a new R user. I trying to install a package all Matching but failed.
Here is the error msg.
install.packages(Matching, dependencies=TRUE)
Installing package(s) into ‘C:/Users/jzhu/Documents/R/win-library/2.13’
(as ‘lib’ is unspecified)
--- Please select
On 9/30/2011 8:31 AM, Durant, James T. (ATSDR/DTEM/PRMSB) wrote:
Happy Friday fellow R users.
I need some help - I am trying to make a graph using ggplot 2 of some lead isotope
ratios. Normally, the isotope mass number appears as a superscript before the chemical
symbol. However, I cannot
On 9/6/2011 4:01 AM, eldor ado wrote:
I have a related question:
dataframe df contains values like
df
.. \\textbf{ 0.644 } ..
and the line
print( xtable(df , sanitize.text.function = function(x){x}))
sanitize.text.function is an argument of print.xtable, not xtable. Try
print(
On 8/17/2011 11:13 AM, Uwe Ligges wrote:
Actually require() is a wrapper around library() with more error
handling to be used inside other functions. Just type require(), you can
read the few lines of code quickly.
I think the unstated corollary is that library() is preferred when not
inside
On 8/17/2011 11:29 AM, mkzo...@comcast.net wrote:
I'm trying to create a dotplot with some grouping.
I've been able to create what I want using dotchart (basic
graphics), but can't quite get it using dotplot (lattice). I prefer
to use lattice (or ggplot2) because I think it's a bit easier to
On 8/18/2011 1:40 PM, Alexander Schwall wrote:
Hi R community,
I have been trying to figure out why R is reversing the order of rows after
I run data.matrix()
Here is my data:
df-structure(list(itmID = c(1L, 2L, 1L, 2L, 1L, 2L), variable =
structure(c(1L,
1L, 2L, 2L, 3L, 3L), .Label = c(3,
On 8/9/2011 10:32 AM, Fernando Andreacci wrote:
Hi, I got what I want using
lines(varmeasure ~ vardates)
but, as I'm using as.Date(vardates) in my boxplot code
boxplot(varmeasure ~ as.Date(vardate))
I want to change how vardate is displayed
I used strftime(as.Date(vardate), format=%m/%Y)
On 8/6/2011 9:21 AM, zhenjiang xu wrote:
Unfortunately the list names of my real data are irregular with mixed
digit and letters at the end. This is good idea though. It inspired me
to give another solution based on that:
x- list(A=c(d, e, f), B=c(d, e), C=c(d,g))
tmp- unlist(x, use.names=F)
a
On 8/3/2011 6:07 AM, wwreith wrote:
So I take it 3D pie charts are out?
At least with ggplot, yes. 2D pie charts are somewhat tricky with
ggplot, even. They can be gone with stacked, normalized bar charts
projected into polar coordinates, if I recall properly.
Not limited to ggplot,
On 8/2/2011 11:39 AM, wwreith wrote:
Does anyone know how to create a 3D Bargraph using ggplot2/qplot. I don't
mean 3D as in x,y,z coordinates. Just a 2D bar graph with a 3D shaped bard.
See attached excel file for an example.
It is not possible.
Before anyone asks I know that 3D looking
On 7/31/2011 6:24 PM, Alexandre Aguiar wrote:
Em Domingo 31 Julho 2011, você escreveu:
My memory is that this question gets asked every few months and one of
the stock answers is to use the function 'package.skeleton' in the
utils package as a starting point.
Got that from docs. And actually
On 7/25/2011 8:27 PM, Sigrid wrote:
Thank you Brian.
Sorry for being such a noob. I am not a programmer and just learning R by
myself. This is was I typed, but ended up with a couple error messages.
df-structure(list(year = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
+ 1L, 1L, 1L, 1L, 1L, 1L,
On 7/18/2011 9:23 PM, Sigrid wrote:
Hi
I apologize for not providing reproducible codes more clearly, and I hope
this will be more understandable.
I have 14 lines (7 per facet that I would like to add). I will provide you
with six of the lines from the data as that should enough data to work
On 7/7/2011 3:23 PM, elephann wrote:
Hi everyone!
I have a data frame with 1112 time series and I am going to randomly
sampling r samples for z times to compose different portfolio size(r
securities portfolio). As for r=2 and z=1,that's:
z=1
A=seq(1:1112)
x1=sample(A,z,replace =TRUE)
Merging two posts (data and questions); see inline below.
On 7/11/2011 7:55 PM, Sigrid wrote:
Thank you, Dennis.
This is my regenerated dput codes. They should be correct as I closed off R
and re-ran them based on the dput output.
NB, this is the test dataset used later
See inline below,
On 7/4/2011 8:00 PM, Ungku Akashah wrote:
Hello.
My name is Akashah. i work at metabolic laboratory. From my study, i
found that volcano plot can help a lot in my section. i already
studied about the volcano plot and get the coding to run in R
software, unfortunately, there
On 6/22/2011 2:28 PM, Alexander Shenkin wrote:
On 6/22/2011 4:09 PM, Brian Diggs wrote:
On 6/22/2011 1:37 PM, Alexander Shenkin wrote:
On 6/22/2011 3:34 PM, Brian Diggs wrote:
On 6/22/2011 12:09 PM, Luke Miller wrote:
For what it's worth, I cannot reproduce this problem under a nearly
On 6/22/2011 11:02 PM, Idris Raja wrote:
Brian,
I'm a bit confused about how the following line works, specifically, what is
happening in freq=length(x)? Is it just taking the length of x after it has
been summarized by different combinations x y? I guess that must be the
case, because that
On 6/22/2011 12:09 PM, Luke Miller wrote:
For what it's worth, I cannot reproduce this problem under a nearly
identical instance of R (R 2.12.1, Win 7 Pro 64-bit). I also can't
reproduce the problem with R 2.13.0. You've got something truly weird
going on with your particular instance of R.
On 6/22/2011 1:23 PM, William Dunlap wrote:
The isdst value -1 doesn't seem right. Shouldn't it
be either 0 (not daylight savings time) or 1 (yes dst)?
I've only seen isdst==-1 when all the other entries
were NA (that happens when the string doesn't match
the format).
A isdst of -1 indicates
On 6/22/2011 1:37 PM, Alexander Shenkin wrote:
On 6/22/2011 3:34 PM, Brian Diggs wrote:
On 6/22/2011 12:09 PM, Luke Miller wrote:
For what it's worth, I cannot reproduce this problem under a nearly
identical instance of R (R 2.12.1, Win 7 Pro 64-bit). I also can't
reproduce the problem with R
On 6/21/2011 11:30 AM, Idris Raja wrote:
I have a dataframe df with two columns x and y. I want to count the number
of times a unique x, y combination occurs.
For example
x- c(1,2,3,4,5,1,2,3,4)
y- c(1,2,3,4,5,1,2,4,1)
df-as.data.frame(cbind(x, y))
#what is the correct way to use ddply for
On 6/20/2011 11:25 AM, Vickie S wrote:
Hi Dennis,
It looks like something is wrong about configuration of ggplot dependency with
plyr.Since i saw some threads about this particular error message.
I tried several times by installing different versions of plyr but it did not
work.
On 6/20/2011 12:23 PM, ivan wrote:
Hi,
I have two datasets, x and y. Simplified x and y denote:
X
Y
A B C A B C . . . . . . . . . . . . . . . . . .
I want to implement all possible models such as lm(X$A~Y$A), lm(X$B~Y$B),
lm(X$C~Y$C)... I have tried the following:
fun- function(x,y){
On 6/17/2011 2:24 PM, (Ted Harding) wrote:
And the extra twist in the tale is exemplified by this
mini-version of Albert-Jan's first example:
DF- data.frame(A=c(1,2,3))
DF$B- c(4,5,6)
DF$C- c(7,8,9)
DF
# A B C
# 1 1 4 7
# 2 2 5 8
# 3 3 6 9
DF$D- DF[A]/DF[B]
DF
On 6/9/2011 12:27 PM, Abraham Mathew wrote:
I have a repetative task in R and i'm trying to find a more efficient way to
perform
the following task.
lst- list(roots = c(car insurance, auto insurance),
roots2 = c(insurance), prefix = c(cheap, budget),
prefix2 = c(low
On 6/7/2011 8:08 AM, wwreith wrote:
I am learning ggplot2 commands and I have figured out how to create
histograms and density curves but I am not sure how to add a density curve
on top of a histogram.
Here are the two graphs that I created.
## Histogram
t-rnorm(500)
w-qplot(t, main=Normal
On 5/26/2011 11:54 PM, Marius Hofert wrote:
Dear Prof. Ripley,
many thanks for your quick reply.
A character matrix (although clearly not very elegant) would be no problem,
xtable deals
with that.
I tried as.data.frame() before, but if one wants to have the same rows
as in ft, one has to use
On 5/26/2011 12:29 PM, Julian TszKin Chan wrote:
Hi all,
Is there any way for me to to string in the argument of qplot or ggplot? for
example
qplot(x='carat',y='price',data=diamonds,geom=c('point','smooth'))
instead of
qplot(x=carat,y=price,data=diamonds,geom=c('point','smooth'))
I don't
-help-boun...@r-project.org] On Behalf Of Brian Diggs
Sent: Thursday, May 26, 2011 2:07 PM
To: Julian TszKin Chan
Cc: r-help@r-project.org
Subject: Re: [R] Question about ggplot2
On 5/26/2011 12:29 PM, Julian TszKin Chan wrote:
Hi all,
Is there any way for me to to string in the argument of
qplot
On 4/25/2011 10:19 AM, Christoph Jäckel wrote:
Hi Together,
I have a problem with the plyr package - more precisely with the ddply
function - and would be very grateful for any help. I hope the example
here is precise enough for someone to identify the problem. Basically,
in this step I want to
On 4/25/2011 11:55 AM, William Dunlap wrote:
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Brian Diggs
Sent: Monday, April 25, 2011 11:05 AM
To: christoph.jaec
On 4/25/2011 1:07 PM, Hadley Wickham wrote:
If you need plyr for other tasks you ought to use a different
class for your date data (or wait until plyr can deal with
POSIXlt objects).
How do you get POSIXlt objects into a data frame?
df- data.frame(x = as.POSIXlt(as.Date(c(2008-01-01
On 4/22/2011 12:42 PM, John Dennison wrote:
Greetings All,
I am looking to write a parametrized Rscript that will accept a variable
name(that also is the name of the flat file), transform the data into a data
frame and preform various modeling on the structure and save the output and
plot of
On 4/14/2011 2:04 PM, Cliff Clive wrote:
I have a vector of character strings that I would like to split in two, and
place in columns of a dataframe.
So for example, I start with this:
beatles- c(John Lennon, Paul McCartney, George Harrison, Ringo
Starr)
and I want to end up with a data frame
On 4/11/2011 9:33 AM, Simon Hayward wrote:
Hi all,
I am practising a bit with ggplot2 but I have a problem when I try to
use facet_grid.
The following code:-
p- ggplot(diamonds, aes(carat, ..density..)) +
+ geom_histogram(binwidth = 1)
p + facet_grid(cut ~ clarity, margins=TRUE)
produce the
On 4/6/2011 2:17 PM, dirknbr wrote:
I am aware this has been asked before but I could not find a resolution.
I am doing a logit
lg- glm(y[1:200] ~ x[1:200,1],family=binomial)
glm (and most modeling functions) are designed to work with data frames,
not raw vectors.
Then I want to predict
On 3/29/2011 2:48 AM, meytar wrote:
Hello
I am trying to make a graph of 10 different lines built each from 4
different
segments and to add a darker line that will represent the average of all
graphs
- all in the same plot.Actually each line is a ROC plot
The code I'm using for plotting one line
On 3/25/2011 3:13 AM, Denis Kazakiewicz wrote:
Hello
I simply want to plot two variables against one 'year' variable in
qplot.
Is any way of doing this without reshaping data in long format and using
facet function afterwards?
Yes, its is possible. But it is harder and more convoluted than
On 3/20/2011 12:19 PM, David.Epstein wrote:
I like Sweave, which I consider to be a great contribution. I have just
written a .Rnw document that comes to about 6 pages of mixed code and
mathematical explanation. Now I want to turn the R code into a function. My
R code currently contains
Taby,
First, it is better to reply to the whole list (which I have included on
this reply); there is a better chance of someone helping you. Just
because I could help with one aspect does not mean I necessarily can (or
have the time to) help with more.
Further comments are inline below.
On 3/16/2011 8:04 AM, taby gathoni wrote:
data-data.frame(id=1:(165+42),main_samp$SCORE,
x=rep(c(BAD,GOOD),c(42,165)))
f-function(x) {
+ str.sample-list()
+ for (i in 1:length(levels(x$x)))
+ {
+ str.sample[[i]]-x[x$x==levels(x$x)[i]
,][sample(tapply(x$x,x$x,length)[i],20,rep=T),]
+ }
+
On 3/13/2011 6:46 PM, David Winsemius wrote:
On Mar 13, 2011, at 8:51 PM, Mark Linderman wrote:
David, thank you for your quick reply. I spent a few minutes getting your
command to work with some sparse synthetic data, and then spent several
hours trying to figure out why my data didn't work
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