[R] RODBC Excel sqlQuery insert into
I have searched the archives for using insert into to update spreadsheets
using RODBC and have come up short. So, first off, is it possible?
I have put together a dummy xls table (c:\foo.xls)for exploring
possibilities of RODBC. Ultimately, I am interested in replacing much of
our previous use of vba macros with R ( I'd prefer elimination, but will
take what I can get ). In order to achieve this, I will still have a need
to update spreadsheets directly through R scripts.
Simple queries seem to work fantastic! But, I am missing something when
it comes to writing.
sqlQuery( myChan,insert into [my customers$] ( CUST_ID, NAME, SOCIAL )
VALUES( 5,'robin',5678 ) )
[1] [RODBC] ERROR: Could not SQLExecDirect S1000 -3035
[Microsoft][ODBC Excel Driver] Operation must use an updateable query.
myChan = odbcConnectExcel(c:\\foo.xls)
myChan
RODB Connection 13
Details:
case=nochange
DBQ=c:\foo.xls
DefaultDir=c:\
Driver={Microsoft Excel Driver (*.xls)}
DriverId=790
MaxBufferSize=2048
PageTimeout=5
sqlTables(myChan)
TABLE_CAT TABLE_SCHEM TABLE_NAME TABLE_TYPE REMARKS
1 c:\\fooNA 'my customers orders$' TABLENA
2 c:\\fooNA'my customers$' TABLENA
3 c:\\fooNA 'my products$' TABLENA
4 c:\\fooNA 'poor table$' TABLENA
sqlQuery(myChan,select * from [my customers$] )
CUST_IDNAME SOCIALDOB
1 1superman 1234 1940-12-31
2 2 batman 2345 1960-01-01
3 3 wonderwoman 3456 1942-05-15
4 4 spiderman 4567 1982-09-30
sqlQuery(myChan,select * from [my customers orders$] )
ORDER_ID CUST_ID DATE PRODUCT_ID QUANTITY
11 3 1997-08-13 1 12
22 3 1998-07-23 7 24
33 1 1994-01-08 6 11
44 4 2001-11-13 5 32
55 2 1997-03-09 79
sqlQuery(myChan,select * from [my products$] )
PRODUCT_ID PRODUCT_NAME
1 1 cape
2 2 mask
3 3 tights
4 4boots
5 5 goggles
6 6 gloves
7 7 aspirin
sqlQuery(myChan,select * from [poor table$] )
a bunch of stuff F2 F3 F4 F5F6
1 NA NA NANA NANA
2 NA NA NANA NANA
3 NA NA NA data_i_like more data I likeNA
4 NA NA 100blue hot 94.16857
5 NA NA 200 red warm 35.85302
6 NA NA 300 green cold 232.09150
7 NA NA 400blue cold 45.40191
sqlQuery(myChan,select * from [my customers$] A, [my customers orders$] B
where A.CUST_ID = B.CUST_ID AND A.SOCIAL 3000, as.is=TRUE)
CUST_IDNAME SOCIAL DOB ORDER_ID CUST_ID DATE
PRODUCT_ID QUANTITY
1 3 wonderwoman 3456 1942-05-15 00:00:002 3
1998-07-23 00:00:00 7 24
2 3 wonderwoman 3456 1942-05-15 00:00:001 3
1997-08-13 00:00:00 1 12
3 4 spiderman 4567 1982-09-30 00:00:004 4
2001-11-13 00:00:00 5 32
$
f = sqlQuery( myChan,insert into [my customers$] ( CUST_ID, NAME, SOCIAL
) VALUES( 5,'robin',5678 ) )
[1] [RODBC] ERROR: Could not SQLExecDirect S1000 -3035
[Microsoft][ODBC Excel Driver] Operation must use an updateable query.
Additional reference:
http://www.microsoft.com/technet/scriptcenter/resources/officetips/jun05/tips0607.mspx
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Re: [R] RODBC Excel sqlQuery insert into
I do recall seeing posts on RDOM flying around from time to time and today
have stumbled across as well through my RODBC searches. I have also been
reminded from another reply of RCOM being a potential solution. I had
hoped to stick to more of a pure sql based process, but certainly R makes
it easy to change course when necessary. I shall do some digging into
each of these packages now as well. Thanks for the response!
--Cheers.
James W. MacDonald [EMAIL PROTECTED]
03/13/2007 02:05 PM
To
[EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] RODBC Excel sqlQuery insert into
Hi Toby,
[EMAIL PROTECTED] wrote:
I have searched the archives for using insert into to update
spreadsheets
using RODBC and have come up short. So, first off, is it possible?
I don't think so. Writing to an Excel spreadsheet is probably easier
done using the RDCOMClient package. There are some good examples of how
to do things that come with the package. Searching the R-help archives
should also come up with some good examples.
Best,
Jim
--
James W. MacDonald, M.S.
Biostatistician
Affymetrix and cDNA Microarray Core
University of Michigan Cancer Center
1500 E. Medical Center Drive
7410 CCGC
Ann Arbor MI 48109
734-647-5623
**
Electronic Mail is not secure, may not be read every day, and should not
be used for urgent or sensitive issues.
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[R] Matrix multiplication using apply() or lappy() ?
I am trying to divide the columns of a matrix by the first row in the
matrix.
I have tried to get this using apply and I seem to be missing a concept
regarding the apply w/o calling a function but rather command args %*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of approach. Does
anybody have a snippet of a call to apply() that would accomplish this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
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Re: [R] Matrix multiplication using apply() or lappy() ?
The apply was exactly what I was after. And, I will check out the others
as well. great tips!
Gabor Grothendieck [EMAIL PROTECTED]
09/06/2006 11:11 AM
To
[EMAIL PROTECTED] [EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] Matrix multiplication using apply() or lappy() ?
This last one could also be written slightly shorter as:
t(apply(a, 1, /, a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
And here is one more:
t(apply(a, 1, function(x) x/a[1,]))
On 9/6/06, Gabor Grothendieck [EMAIL PROTECTED] wrote:
Here are a few possibilities:
a - matrix(1:24, 4) # test data
a / rep(a[1,], each = 4)
a / outer(rep(1, nrow(a)), a[1,])
a %*% diag(1/a[1,])
sweep(a, 2, a[1,], /)
On 9/6/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to divide the columns of a matrix by the first row in
the
matrix.
I have tried to get this using apply and I seem to be missing a
concept
regarding the apply w/o calling a function but rather command args
%*% /
etc. Would using apply be more efficient than this approach?
I have observed examples in the archives using this type of
approach. Does
anybody have a snippet of a call to apply() that would accomplish
this as
well?
Thanks!
seed=50
$a = array(rnorm(20),dim=c(4,5))
$b = matrix(a[1,],dim(a)[1],dim(a)[2],byrow=T)
$a
[,1] [,2] [,3][,4] [,5]
[1,] -1.3682810 -0.4314462 1.57572752 0.67928882 -0.3672346
[2,] 0.4328180 0.6556479 0.64289931 0.08983289 0.1852306
[3,] -0.8113932 0.3219253 0.08976065 -2.99309008 0.5818237
[4,] 1.4441013 -0.7838389 0.27655075 0.28488295 1.3997368
$a/b
[,1] [,2] [,3] [,4] [,5]
[1,] 1.000 1.000 1. 1.000 1.00
[2,] -0.3163225 -1.5196515 0.40800157 0.1322455 -0.504393
[3,] 0.5930018 -0.7461539 0.05696457 -4.4062113 -1.584338
[4,] -1.0554128 1.8167710 0.17550671 0.4193841 -3.811560
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and provide commented, minimal, self-contained, reproducible code.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] cumulative growth rates indexed to a common starting point over n series of observations
Perfect, the apply was exactly what I was after, just wasn't clicking
with me, and I overlooked the cumprod ... sweet !
Thanks to all for pushing me down the right path!
Dirk Eddelbuettel [EMAIL PROTECTED]
08/31/2006 05:03 PM
To
[EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] cumulative growth rates indexed to a common starting point over n
series of observations
On 31 August 2006 at 15:22, [EMAIL PROTECTED] wrote:
| What is the R way of computing cumulative growth rates given a series of
| discrete values indexed .
|
| For instance, given a matrix of 20 observations for each of 5 series
(zz),
| what is the most straight forward technique in R for computing
cumulative
| growth (zzcum) ?
| It seems for the solution I'm after might be imbedding the following cum
| growth rate calc as a function into a function call to apply, mapply,
...?
|
|
|
| zz = rnorm(100)
| dim(zz) = c(20,5)
| zz
| zzcum=matrix(nrow=20,ncol=5)
| zzcum
| zzcum[1,]=100*(1+zz[1,]/100)
| zzcum
| for(i in 2:20){zzcum[i,] = zzcum[i-1,]*(1+zz[i,]/100)}
| zzcum
How about cumprod() inside apply() ?
zzcum - matrix(rnorm(100), ncol=5)
apply(zzcum/100 + 1, 2, cumprod)
Hth, Dirk
--
Hell, there are no rules here - we're trying to accomplish something.
-- Thomas A. Edison
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] cumulative growth rates indexed to a common starting point over n series of observations
The apply with the cumprod was exactly what I was after. The apply just
wasn't clicking with me, and I had overlooked the cumprod. Thanks to all
for pushing me down the right path!
Actually, what I am ultimately after is a way to link this series, without
having to use a for loop ( the only way I can think of ... ). But, would
like to see if it can be linked using mapply or apply against the rows
and to compute the linked results.
zz = matrix(rnorm(20), ncol=2)
zzcum = apply(zz/100 + 1, 2, cumprod)
zzlinkcum = 100*zzcum
for(i in 2:length(zz[,1])){ zzlinkcum[i,]=zzlinkcum[i-1,]*zzcum[i,]} ###
Is there a better way here ?
zz
[,1] [,2]
[1,] 0.8563323 -0.3895789
[2,] 0.8311070 -0.7483010
[3,] 0.3526344 -1.1702419
[4,] 1.1105516 0.5238831
[5,] 0.6471324 -0.1419063
[6,] -0.9821008 0.8378471
[7,] 1.6799099 1.7025973
[8,] -0.1904968 -2.9203921
[9,] 0.4218418 0.4799355
[10,] -0.5989616 -1.2121375
zzcum
[,1] [,2]
[1,] 1.007118 1.001036
[2,] 1.011022 1.011093
[3,] 1.007719 1.023287
[4,] 1.026238 1.030673
[5,] 1.031242 1.023557
[6,] 1.029199 1.021431
[7,] 1.044428 1.017244
[8,] 1.034568 1.026164
[9,] 1.032941 1.024926
[10,] 1.032490 1.026172
zzlinkcum
[,1] [,2]
[1,] 100.7118 100.1036
[2,] 101.8219 101.2140
[3,] 102.6078 103.5711
[4,] 105.3000 106.7479
[5,] 108.5898 109.2626
[6,] 111.7605 111.6042
[7,] 116.7259 113.5287
[8,] 120.7608 116.4990
[9,] 124.7388 119.4028
[10,] 128.7916 122.5278
Dirk Eddelbuettel [EMAIL PROTECTED]
08/31/2006 05:03 PM
To
[EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] cumulative growth rates indexed to a common starting point over n
series of observations
On 31 August 2006 at 15:22, [EMAIL PROTECTED] wrote:
| What is the R way of computing cumulative growth rates given a series of
| discrete values indexed .
|
| For instance, given a matrix of 20 observations for each of 5 series
(zz),
| what is the most straight forward technique in R for computing
cumulative
| growth (zzcum) ?
| It seems for the solution I'm after might be imbedding the following cum
| growth rate calc as a function into a function call to apply, mapply,
...?
|
|
|
| zz = rnorm(100)
| dim(zz) = c(20,5)
| zz
| zzcum=matrix(nrow=20,ncol=5)
| zzcum
| zzcum[1,]=100*(1+zz[1,]/100)
| zzcum
| for(i in 2:20){zzcum[i,] = zzcum[i-1,]*(1+zz[i,]/100)}
| zzcum
How about cumprod() inside apply() ?
zzcum - matrix(rnorm(100), ncol=5)
apply(zzcum/100 + 1, 2, cumprod)
Hth, Dirk
--
Hell, there are no rules here - we're trying to accomplish something.
-- Thomas A. Edison
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] cumulative growth rates indexed to a common starting point over n series of observations
That is choice !I'm starting to love this language. Thank-you !
Dirk Eddelbuettel [EMAIL PROTECTED]
09/01/2006 02:59 PM
To
[EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] cumulative growth rates indexed to a common starting point over n
series of observations
On 1 September 2006 at 14:30, [EMAIL PROTECTED] wrote:
| The apply with the cumprod was exactly what I was after. The apply just
| wasn't clicking with me, and I had overlooked the cumprod. Thanks to
all
| for pushing me down the right path!
|
| Actually, what I am ultimately after is a way to link this series,
without
| having to use a for loop ( the only way I can think of ... ). But,
would
| like to see if it can be linked using mapply or apply against the rows
| and to compute the linked results.
|
| zz = matrix(rnorm(20), ncol=2)
| zzcum = apply(zz/100 + 1, 2, cumprod)
| zzlinkcum = 100*zzcum
| for(i in 2:length(zz[,1])){ zzlinkcum[i,]=zzlinkcum[i-1,]*zzcum[i,]} ###
| Is there a better way here ?
Sure, why not call apply again?
set.seed(42); zz - matrix(rnorm(20), ncol=2)
zzcum - apply(1+zz/100, 2, cumprod)
apply(rbind(c(1,1), zzcum), 2, cumprod)*100
[,1] [,2]
[1,] 100. 100.
[2,] 101.3710 101.3049
[3,] 102.1804 104.9735
[4,] 103.3704 107.2642
[5,] 105.2360 109.2994
[6,] 107.5684 111.2246
[7,] 109.8358 113.9036
[8,] 113.8461 116.3156
[9,] 117.8912 115.6233
[10,] 124.5442 112.1301
[11,] 131.4900 110.1781
Hth, Dirk
--
Hell, there are no rules here - we're trying to accomplish something.
-- Thomas A. Edison
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Re: [R] cumulative growth rates indexed to a common starting point over n series of observations
Yes I do. I was still scratching my head still trying to figure out why
I couldn't get my actual #s to match the example from the random series.
This now matches my brute force calcs.
btw:
Also had a former colleague and good friend suggest this could be done
using ranges. Which opened my eyes as well, although I haven't tested
yet. But I do think the apply approach is more elegant.
from my original post...
zz = matrix(rnorm(20), ncol=2)
zzcum = apply(zz/100 + 1, 2, cumprod)
zzlinkcum = 100*zzcum
zzlinkcum[2:length(zz[,1]),] -
zzlinkcum[1:(length(zz[,1])-1),]*zzcum[2:length(zz[,1]),]
Thanks so much for al the help !
Dirk Eddelbuettel [EMAIL PROTECTED]
09/01/2006 02:59 PM
To
[EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] cumulative growth rates indexed to a common starting point over n
series of observations
On 1 September 2006 at 14:30, [EMAIL PROTECTED] wrote:
| The apply with the cumprod was exactly what I was after. The apply just
| wasn't clicking with me, and I had overlooked the cumprod. Thanks to
all
| for pushing me down the right path!
|
| Actually, what I am ultimately after is a way to link this series,
without
| having to use a for loop ( the only way I can think of ... ). But,
would
| like to see if it can be linked using mapply or apply against the rows
| and to compute the linked results.
|
| zz = matrix(rnorm(20), ncol=2)
| zzcum = apply(zz/100 + 1, 2, cumprod)
| zzlinkcum = 100*zzcum
| for(i in 2:length(zz[,1])){ zzlinkcum[i,]=zzlinkcum[i-1,]*zzcum[i,]} ###
| Is there a better way here ?
Sure, why not call apply again?
set.seed(42); zz - matrix(rnorm(20), ncol=2)
zzcum - apply(1+zz/100, 2, cumprod)
apply(rbind(c(1,1), zzcum), 2, cumprod)*100
[,1] [,2]
[1,] 100. 100.
[2,] 101.3710 101.3049
[3,] 102.1804 104.9735
[4,] 103.3704 107.2642
[5,] 105.2360 109.2994
[6,] 107.5684 111.2246
[7,] 109.8358 113.9036
[8,] 113.8461 116.3156
[9,] 117.8912 115.6233
[10,] 124.5442 112.1301
[11,] 131.4900 110.1781
Hth, Dirk
--
Hell, there are no rules here - we're trying to accomplish something.
-- Thomas A. Edison
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and provide commented, minimal, self-contained, reproducible code.
[R] cumulative growth rates indexed to a common starting point over n series of observations
What is the R way of computing cumulative growth rates given a series of
discrete values indexed .
For instance, given a matrix of 20 observations for each of 5 series (zz),
what is the most straight forward technique in R for computing cumulative
growth (zzcum) ?
It seems for the solution I'm after might be imbedding the following cum
growth rate calc as a function into a function call to apply, mapply, ...?
zz = rnorm(100)
dim(zz) = c(20,5)
zz
zzcum=matrix(nrow=20,ncol=5)
zzcum
zzcum[1,]=100*(1+zz[1,]/100)
zzcum
for(i in 2:20){zzcum[i,] = zzcum[i-1,]*(1+zz[i,]/100)}
zzcum
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and provide commented, minimal, self-contained, reproducible code.
[R] Help needed understanding eval,quote,expression
I am trying to build up a quoted or character expression representing a
component in a list in order to reference it indirectly.
For instance, I have a list that has data I want to pull, and another list
that has character vectors and/or lists of characters containing the names
of the components in the first list.
It seems that the way to do this is as evaluating expressions, but I seem
to be missing something. The concept should be similar to the snippet
below:
For instance:
$x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
$y = quote(x$y$y1)
$eval(y)
[1] hello
but, I'm trying to accomplish this by building up y as a character and
then evaluating it, and having no success.
$y1=paste(x$y$,y1,sep=)
$y1
[1] x$y$y1
How can I evaluate y1 as I did with y previously? or can I?
Much Thanks !
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Re: [R] Help needed understanding eval,quote,expression
Great help, thanks to both of you. I actually think I get it.
I think I was locked into eval(expression ... ) as the solution. I did
search the archives for this question, but it must not have clicked with
me.The Thomas Lumley R-help (February 2005) was on the money. I was
missing the power flexibility of the [[ operation. It is certainly more
direct.
I do believe this pattern will satisfy my original problem.
x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
ids = list( zIds=c(z1,z2),yIds=c(y1,y2))
str=y1
x$y[[str]]
[1] hello
btw: I set my prompt to $, so the first post, the $ at the beginning of
the line was the prompt. apologies for the confusion.
cheers.
Prof Brian Ripley [EMAIL PROTECTED]
06/29/2006 04:06 AM
To
Joerg van den Hoff [EMAIL PROTECTED]
cc
[EMAIL PROTECTED], r-help@stat.math.ethz.ch
Subject
Re: [R] Help needed understanding eval,quote,expression
On Thu, 29 Jun 2006, Joerg van den Hoff wrote:
Prof Brian Ripley wrote:
You are missing eval(parse(text=)). E.g.
x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
(what do you mean by the $ at the start of these lines?)
eval(parse(text=x$y$y1))
[1] hello
However, bear in mind
fortune(parse)
If the answer is parse() you should usually rethink the question.
-- Thomas Lumley
R-help (February 2005)
In your indicated example you could probably use substitute() as
effectively.
On Wed, 28 Jun 2006, [EMAIL PROTECTED] wrote:
I am trying to build up a quoted or character expression representing
a
component in a list in order to reference it indirectly.
For instance, I have a list that has data I want to pull, and another
list
that has character vectors and/or lists of characters containing the
names
of the components in the first list.
It seems that the way to do this is as evaluating expressions, but I
seem
to be missing something. The concept should be similar to the snippet
below:
For instance:
$x = list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
$y = quote(x$y$y1)
$eval(y)
[1] hello
but, I'm trying to accomplish this by building up y as a character and
then evaluating it, and having no success.
$y1=paste(x$y$,y1,sep=)
$y1
[1] x$y$y1
How can I evaluate y1 as I did with y previously? or can I?
Much Thanks !
if I understand you correctly you can achieve your goal much easier than
with
eval, parse, substitute and the like:
x - list(y=list(y1=hello,y2=world),z=list(z1=foo,z2=bar))
s1 - 'y'
s2 - 'y1'
x[[s1]][[s2]]
i.e. using `[[' instead of `$' for list component extraction allows to
use
characters for indexing (in other words: x$y == x[['y']])
But what he actually asked for was
I am trying to build up a quoted or character expression representing
a
component in a list in order to reference it indirectly.
You just typed in x[[s1]][[s2]], not 'built [it] up'. Suppose the
specification had been
r - x
s - c(y, y1)
and s was of variable length? Then you need to construct a call similar
to x[[y]][[y1]] from r and s.
[There was another reason for sticking with $ rather than using [[: the
latter makes unnecessary copies in released versions of R.]
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] Help needed using lattice for area plots lpolygon, xyplot.
I am trying to learn how to use the graphics from the lattice package (
and am very new to R).
I am trying to replicate the example plot referenced below, by using the
lattice xyplot lpolygon to create panels. I get what appears to be the
correct shape of the filled region, but cannot get the position to overlay
properly. I have attempted with various settings of position. ( i.e.
position = c(0,0,1,1) etc settings as well. I am not understanding
something about the positioning panels. I am missing some subtle
difference between polygon lpolygon, or am missing something about panel
overlays /or panel postions.
#http://addictedtor.free.fr/graphiques/graphcode.php?graph=7.
par(bg=white)
n - 100
set.seed(43214) #just so we have the same exact graph
x - c(0,cumsum(rnorm(n)))
y - c(0,cumsum(rnorm(n)))
xx - c(0:n, n:0)
yy - c(x, rev(y))
plot(xx, yy, type=n, xlab=Time, ylab=Distance)
polygon(xx, yy, col=gray)
title(Distance Between Brownian Motions)
# using lattice.
p1 - xyplot( yy~xx,type='l');
p2 - lpolygon(xx,yy,col='blue');
print(p1,position=c(0,0,1,1), more=TRUE);
print(p2,position=c(0,0,1,1));
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[R] Fw: Help needed using lattice for area plots lpolygon, xyplot.
I am trying to learn how to use the graphics from the lattice package (
and am very new to R).
I am trying to replicate the example plot referenced below, by using the
lattice xyplot lpolygon to create panels. I get what appears to be the
correct shape of the filled region, but cannot get the position to overlay
properly. I have attempted with various settings of position. ( i.e.
position = c(0,0,1,1) etc settings as well. I am not understanding
something about the positioning panels. I am missing some subtle
difference between polygon lpolygon, or am missing something about panel
overlays /or panel postions.
#http://addictedtor.free.fr/graphiques/graphcode.php?graph=7.
par(bg=white)
n - 100
set.seed(43214) #just so we have the same exact graph
x - c(0,cumsum(rnorm(n)))
y - c(0,cumsum(rnorm(n)))
xx - c(0:n, n:0)
yy - c(x, rev(y))
plot(xx, yy, type=n, xlab=Time, ylab=Distance)
polygon(xx, yy, col=gray)
title(Distance Between Brownian Motions)
# using lattice.
p1 - xyplot( yy~xx,type='l');
p2 - lpolygon(xx,yy,col='blue');
print(p1,position=c(0,0,1,1), more=TRUE);
print(p2,position=c(0,0,1,1));
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Re: [R] Fw: Help needed using lattice for area plots lpolygon, xyplot.
Perfect. Thank-you!
Deepayan Sarkar [EMAIL PROTECTED]
06/07/2006 11:19 AM
To
[EMAIL PROTECTED] [EMAIL PROTECTED]
cc
r-help@stat.math.ethz.ch
Subject
Re: [R] Fw: Help needed using lattice for area plots lpolygon, xyplot.
On 6/7/06, [EMAIL PROTECTED]
[EMAIL PROTECTED] wrote:
I am trying to learn how to use the graphics from the lattice package (
and am very new to R).
I am trying to replicate the example plot referenced below, by using the
lattice xyplot lpolygon to create panels. I get what appears to be
the
correct shape of the filled region, but cannot get the position to
overlay
properly. I have attempted with various settings of position. ( i.e.
position = c(0,0,1,1) etc settings as well. I am not understanding
something about the positioning panels. I am missing some subtle
difference between polygon lpolygon, or am missing something about
panel
overlays /or panel postions.
#http://addictedtor.free.fr/graphiques/graphcode.php?graph=7.
par(bg=white)
n - 100
set.seed(43214) #just so we have the same exact graph
x - c(0,cumsum(rnorm(n)))
y - c(0,cumsum(rnorm(n)))
xx - c(0:n, n:0)
yy - c(x, rev(y))
plot(xx, yy, type=n, xlab=Time, ylab=Distance)
polygon(xx, yy, col=gray)
title(Distance Between Brownian Motions)
# using lattice.
p1 - xyplot( yy~xx,type='l');
p2 - lpolygon(xx,yy,col='blue');
print(p1,position=c(0,0,1,1), more=TRUE);
print(p2,position=c(0,0,1,1));
You are missing a fundamental concept in lattice, namely that of panel
functions. A literal translation of that example would be
xyplot(yy ~ xx, panel = lpolygon, col = gray)
which is more or less equivalent to
xyplot(yy ~ xx,
panel = function(x, y, ...) {
# panel.xyplot(x, y, ...) # unnecessary
lpolygon(x, y, col = gray)
})
The line commented out is the equivalent of plot(...type='n'), but is
unnecessary here.
Deepayan
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