Re: [R] linux gplots install unhappy

2007-02-21 Thread Randy Zelick

Thanks to the many folks who responded, and apologies for being so dense!

The different behavior between windows and linux relative to install 
defaults tripped me up. I had not thought about the dependencies switch.

All working fine now.

Cheers,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] linux gplots install unhappy

2007-02-20 Thread Randy Zelick
Hello all,

I use R on both windows and a mainframe linux installation (RedHat 
enterprise 3.0, which they tell me is soon to be upgraded to 4.0). On 
windows I installed the package gplots without trouble, and it works fine. 
When I attempted to do the same on the unix computer, the following error 
message was forthcoming:




downloaded 216Kb

* Installing *source* package 'gplots' ...
** R
** data
** inst
** preparing package for lazy loading
Loading required package: gtools
Warning in library(pkg, character.only = TRUE, logical = TRUE, lib.loc = 
lib.loc) :
  there is no package called 'gtools'
Error: package 'gtools' could not be loaded
Execution halted
ERROR: lazy loading failed for package 'gplots'
** Removing '/n/fs/disk/resuser02/u/zelickr/R/library/gplots'

The downloaded packages are in
 /tmp/RtmpikM2JW/downloaded_packages
Warning messages:
1: installation of package 'gplots' had non-zero exit status in: 
install.packages(gplots, lib = ~/R/library)
2: cannot create HTML package index in: 
tools:::unix.packages.html(.Library)



Can someone provide the bit of information I need to progress with this?

Thanks very much,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] fit sine?

2006-12-19 Thread Randy Zelick
Hello list,

I am making scatterplots of data that vary sinusoidally over 24 hours. 
With a bit of previous help from this list, I now can get an x-axis with 
time tics from 00:00 on the left edge to 12:00 in the middle and 00:00 on 
the right edge, i.e., just 24 hours.

Now I would like to fit a sin function to the plot. I've looked all over 
CRAN and the web in general, but have not found a method or package for 
this.

Can anyone make a suggestion?

Thanks,

=Randy=


Using R 2.4.0 on Solaris (unix).


R. Zelick   email: [EMAIL PROTECTED]
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[R] samples to dataframes

2006-10-11 Thread randy zelick
Hello all,

I'll bet this is not complicated, but I really did hunt for information 
on it, and tried some ideas without success.

Running Windows XP and R 2.3.1

I am using the Sound package to read in .wav files, and this works fine. 
The R object produced is of class sample. I want to use some of the 
functions in the package Seewave to do further analysis, but this 
package wants data objects to be dataframes. So I tried various ways to 
coerce samples to dataframes, but was not successful.

Does anyone have a hint/solution?

Thanks much,

=Randy=

-- 
Randy Zelickemail: [EMAIL PROTECTED]
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[R] file.choose but for folder/directory?

2006-07-10 Thread Randy Zelick

Hello there,

This question is relative to WindowsXP, using R 2.2.1:

I am looking for a function that allows a user to interactively choose a 
directory so I can use list.files to process all the files in that 
directory. I've looked at getwd, but this is not interactive. The 
functions file.choose and choose.files are the right idea, but these only 
permit selection of a file within a folder, not the folder itself.

Thanks for any and all suggestions,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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Re: [R] file.choose but for folder/directory -- THANKS!

2006-07-10 Thread Randy Zelick

Thanks to all who responded. I upgraded to 2.3.1 straight away, and 
choose.dir does exactly what I need.

Cheers,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] x y averaging thanks

2006-06-07 Thread Randy Zelick

I have three completely different(!) but workable solutions. Thanks much 
for your help.

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] x y averaging

2006-06-06 Thread Randy Zelick
Hello,

I am trying to average a number of data sets where the x vector contains 
times and the Y data are instrument readings. The vectors all have 
different numbers of values, but many X time values match. For example:

A fragment of the first data set:

x vectvalue
14:56:10  0.325
14:62:11  0.111
14:68:11  0.214
.
.
.
this can go on for 100's of values, spaced by 6 minutes, but there can be 
gaps too, like:

16:05:18  0.245
16:11:09  0.266
16:17:05  0.271
16:33:00  0.304
16:39:05  0.300


A fragment of the second data set:

15:59:08  0.255
16:05:44  0.281
16:11:25  0.249
16:17:39  0.238
16:23:51  0.288


...and the result I am looking for is a new vector that looks like this:

14:56:10  0.325
14:62:11  0.111
14:68:11  0.214
.
.
.
15:59:08  0.255
16:05:18  0.2630  * matches, so average
16:11:09  0.2575  * matches, so average
16:17:05  0.2545  * matches, so average
16:33:00  0.304
16:39:05  0.300


The times values in the new X vector are properly interleaved taking data 
across the datasets, but when there is a match the Y values are averaged. 
Note that the seconds don't matter.

I am using R Version 2.2.1 on a PC

Thanks for thinking about it,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] trouble with plotrix package

2006-05-19 Thread Randy Zelick
Hello list,

I wrote a simple program to plot data on polar axes, taking advantage of 
the plotrix package and its function radial.plot. The basic plot works 
fine, but I am having difficulties with the formatting. There are three 
problems, but I thought I would attack them one at a time. Here is the 
first:

If my data set contains values with all vector lengths between 0 and 100 
(and various angles), and I set rp.type=s, I get a nice bullseye type 
plot with the data shown on a background of concentric circles labeled 
appropriately. On the other hand, if my data only contain vector length 
values from, say, 80 to 90 then the first concentric ring is at 80, and 
five more rings are scrunched between 80 and 100.

It looks like there is an autoscaling feature turned on that says make a 
fixed number of rings starting at a nice value below the user's lowest 
data value and extending to a nice value above.

There is a switch to set the upper bound on rings (radial.lim) but I don't 
see a way to specify the lower bound. What I want is a bullseye plot 
that goes from my start value (first ring at 10) to my end value (last 
ring at 100) independent of the data range.

Thanks!!

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] extract a row

2004-08-16 Thread Randy Zelick
Hello there,

Using 1.9.0 on WinXP...

I have a data frame, one column of which is named rate. The column has
text entries like fast, medium, slow, very slow, and so forth. I
have not tried to make them factors, but maybe R did this automatically.

Anyway, I would like to display on the console rows that meet a rate
criterion.

So I want to type something like:

dataframe[rate==slow]

and get back this sort of output:

test  subject  trial  rate  score

 34B27  3 slow   27
 55B55  4 slow   34

where test, subject, trial, rate, and score are all the
dataframe's columns.

How do I do that??

Thanks,

=Randy=


R. Zelick   email: [EMAIL PROTECTED]
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[R] swapping with data.frame

2004-06-01 Thread Randy Zelick
Hi there,

I have some data which are convenient to enter as lists. For example:

t1-list(fname=animal1,testname=hyla,dspkr=left,res1=39.7,res2=15.0)
t2-list(fname=animal1,testname=bufo,dspkr=left,res1=14.4,res2=56.1)
t3-list(fname=animal2,testname=hyla,dspkr=right,res1=22.6,res2=11.8)

I would like to generate a dataframe, but *not* the way this approach
works...

fdf-data.frame(t1,t2,t3)

fdf

fname testname dspkr res1 res2   fname testname dspkr res1 res2 ...
1 animal1 hyla  left 39.7   15 animal1 bufo  left 14.4 56.1 ...


Instead, what I would like is:

 fname   testname   dspkr   res1res2

t1 animal1   hylaleft   39.715.0
t2 animal1   bufoleft   14.456.1
t3 animal2   hyla   right   22.611.8


or this would be fine too...

x   fname   testname   dspkr   res1res2

1  t1 animal1   hylaleft   39.715.0
2  t2 animal1   bufoleft   14.456.1
3  t3 animal2   hyla   right   22.611.8

Is there a practical (hopefully simple) way to do this?

Thanks,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] formatted table to disk

2004-04-21 Thread Randy Zelick
Hello there,

Using R 1.8.1 and WinXP...

I would like to write a disk file (ascii text) of an R data frame. When I
type the name of the object in the console window, it is sensibly and
nicely formatted. When I use write.table, it is *not*, because the
separation character is just a space.

I looked through the introductory material (my level) on the R website,
but did not find a reference to this issue.

At a minimum I suppose there is a way to insert a tab character in the
sep=  field, but I don't know how to do that. Alternatively, I could
deal with a variable specifying fixed-field widths for each of the
variables in the frame. Didn't find that either.

Even so, you would loose the right-justification and the fact that R seems
to take into account the width of the largest value in a column.

Of course the really slick deal would be to have paginated output where
the column titles repeat at the top of the page (having entered a
parameter for the number of lines on your page after margins and so
forth).

A somewhat brutal solution would be to create all the text strings for
making a LaTex document/table environment and write all that stuff out to
the text file. Please don't tell me I have to do that!

So is there a function for producing a file on disk with the same
formatting as the screen?

Thanks,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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Re: [R] formatted table to disk

2004-04-21 Thread Randy Zelick
Hello,

Thanks for the help, all who made suggestions.

Sink worked -- I don't know why I can't find these things.

I'll look at the xtable package as suggested also.

=Randy=

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[R] histogram y-scaling

2004-04-19 Thread Randy Zelick
Hello all,

Relative to WinXP  R1.8

I have two histograms to plot, and for comparison purposes I want them to
have the same Y-scaling. I tried to find the size of the bin with the
maximum count before generating the histogram, but this did not work (see
below). What is a better way?

par(mfrow=c(2,1)) # set up for plotting in 2 rows and 1 column

x1-seq(-0.5,58.5,1) # make a range of x values for histogram

I thought the following lines would allow me to capture the results of the
hist function and determine the max bin count for scaling *before* making
the plot, but R cleverly saw around my method and plots it anyway. With
this code I get two plots.

q=hist(mt1,x1)   # stick results in a variable... alas also plots
cts=q$counts # get the bin counts
mct1=max(cts)# how many values in the bin with the most values
hist(mt1,x1) # generate histogram plot

# go on with histogram #2...


Thanks,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] basic programming question

2004-04-15 Thread Randy Zelick

Hello list,

I am just starting to write some R functions (R 1.8 and Windows XP) and
got stuck as described below:

We have a bunch of time series data files, each is about 10,000 values.
There is a 12-line header in each one. I can read them and plot them
easily with R, but I want to make an automated system to read all the
files in a folder and have them appear as R objects that I can assemble in
different plots, or analyze together.

Perhaps this would be best done reading into a data table, where each row
is one of the files and each row extends for 10,000 columns? Or perhaps as
a matrix? I could use advice on that. My first notion, though, was just to
have them be individual files.

I started as follows:

#
# build a path from its parts:
#
p1=C:
p2=Work-PSU
p3=Lab-People-and-Projects
p4=Dane
p5=2004-04-06
pt=file.path(p1,p2,p3,p4,p5,fsep=/)
#
lf-list.files(pt)
ff=files
sayfiles=paste(length(lf),ff,sep= )
#
# at this point if I type sayfiles I get 30 files printed at the
console. OK, this is progress.
#
# Now I want to cycle through all the files and read in each one. The
# code here almost works, but I don't know how to do in R the simple task
# of substituting the read file name as the data object name.
#
d=dir(pt)
for (i in d){d[i]=scan(paste(pt,i,sep=/),skip=12)} # this almost works
#

I know this is wrong, I just want to show my intention -- that is on each
iteration of the loop have a new R object made called d[i], the name of
the current file. This new object is then the result of the scan operation
on the file of that same name.

Thanks,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
Department of Biology   voice: 503-725-3086
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mailing:
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R. Zelick   email: [EMAIL PROTECTED]
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[R] Re: [R-gui] Editor in R

2004-04-07 Thread Randy Zelick
Hello there,

I have been using WinEdt as a text editor for a while. I use it in a
simple general purpose way, and take advantage of just a few of the 100's
(1000s?) of features. I would say that used this way it is certainly not
more difficult than R. So it would seem that if you are successful
teaching R to biologists, additionally learning WinEdt should not push
anyone over the edge.

Still, another alternative is Pico, which you can get for free for
Windoze.

By the way, although Notepad works for R scripts, when I tried Wordpad it
did not. Probably puts whacko crapola at the end of each line.

Cheers,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] x-only zoom and pan?

2004-04-04 Thread Randy Zelick
Hello list,

Could the following be done without too much grief...?

Lets say I have two or three time series objects that I want to inspect
visually. Each I would plot with a y-offset so they stack up. They share
the same X scaling. The problem is that each is perhaps 100K values. Due
to the large number of values, features of the data sets cannot be seen
when all values are plotted.

What would be nice is to plot a fraction of the X range (say 10%). This
would be equivalent to zooming in the X direction. Then using a key
(ideally an arrow key), shift the viewing frame right or left to
effectively scroll through the data. So first you view 0-10%, then 10-20%
and so forth.

If necessary I can fabricate a vector with X values in it and plot(x,y)
instead of as time series, if this makes it any easier.

I am using a Windows version of R.

Thanks,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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[R] data frame output almost

2004-03-19 Thread Randy Zelick
Hello again,

I got three responses for help on the leading zero problem. Thank you.
Alas I still don't have it working. Here are more specifics:

I read in a data file like this:

participants-read.table(C:/Work/blah-blah)

The data file consists of the fields last name, first name, social
security number, response score 1, response score 2 and so forth.

If in the console window I type participants I get something like:

 Jones  Norman  786123344 98.2 16.3
FlintstoneFred  111457654 10.1  8.8
  UglyButt   89733456 66.7 32.0

The problem is that the 3rd social security number is really 089733456
and it needs to look like that.

None of the methods suggested seemed to work. I could make the social
security object alone print with leading zeros, but not as part of the
data frame.

Thanks again,

=Randy=

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[R] data frame output

2004-03-18 Thread Randy Zelick
Hello list,

Is there a way to *not* supress leading zeros when printing (to the
console window or to a file) a dataframe?

Thanks,

=Randy=

R. Zelick   email: [EMAIL PROTECTED]
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