Hello,
This is definitely possible with R, there a lots of package to make
good graphics.
However, the easiest way for us to help you is if you give us a small
reproducible example, as you started to with your initial post.
If you have an object in your R session that you'd like help with
you
Greetings Gabor,
is it possible to open a channel to a data frame in the default environment?
there are cases when using a sql update statement is the simplest
alternative, so instead of dumping the df and then updating and then
reimporting it I would like to update the df directly in R.
Thank
On Thu, Aug 26, 2010 at 1:18 PM, stephenb stephen.b...@cibc.com wrote:
is it possible to open a channel to a data frame in the default environment?
there are cases when using a sql update statement is the simplest
alternative, so instead of dumping the df and then updating and then
reimporting
Dear all,
I have an xts object , t.xts with 4 columns: v1 DD1 v2 DD2 and
created a data frame :
t - as.data.frame(t.xts)
I would like to extract data and create a new data frame for when the values
in column DD1 falls between 0 and 30 and extract the corresponding v1 value.
How can I do
On Aug 16, 2010, at 5:53 AM, Lily_stats wrote:
Dear all,
I have an xts object , t.xts with 4 columns: v1 DD1 v2 DD2 and
created a data frame :
t - as.data.frame(t.xts)
t is not the best choice of names for an object because it is the
name of a commonly used function
Let's instead
Hi listers,
I made some search, but i didn`t find in the forum.
I have a data set.
I would like to make a search (conditon) on my data set.
x-c(1,2,3,4,5,6,7,8,9,10)
count-0
if (CONDITON){count-1}else{count-0}
My CONDITION would be: is there number 5 in my data set?
Thanks in advance,
Marcio
?match, look at the %in% operator.
Mestat wrote:
Hi listers,
I made some search, but i didn`t find in the forum.
I have a data set.
I would like to make a search (conditon) on my data set.
x-c(1,2,3,4,5,6,7,8,9,10)
count-0
if (CONDITON){count-1}else{count-0}
My CONDITION would be: is there
In the real data the months are all complete, but the years can be missing.
So years can be missing up front, in the middle, at the end. but if a year
is present than every month has a value or NA.
To create regular R ts I had to plow through the data frame, collect a year
caluculate an index to
On Sun, Aug 8, 2010 at 2:01 AM, steven mosher mosherste...@gmail.com wrote:
In the real data the months are all complete, but the years can be missing.
So years can be missing up front, in the middle, at the end. but if a year
is present than every month has a value or NA.
To create regular R
Ok,
I'm a bit confused by what you mean by regularly spaced
After I do the do.call I do get a data structure with all the times present
and every time has a NA or a data value.
Steve
On Sun, Aug 8, 2010 at 2:46 AM, Gabor Grothendieck
ggrothendi...@gmail.comwrote:
On Sun, Aug 8, 2010 at 2:01
On Sun, Aug 8, 2010 at 11:21 AM, steven mosher mosherste...@gmail.com wrote:
Ok,
I'm a bit confused by what you mean by regularly spaced
After I do the do.call I do get a data structure with all the times present
and every time has a NA or a data value.
Steve
regularly spaced means that
On Sun, Aug 8, 2010 at 11:55 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Sun, Aug 8, 2010 at 11:21 AM, steven mosher mosherste...@gmail.com wrote:
Ok,
I'm a bit confused by what you mean by regularly spaced
After I do the do.call I do get a data structure with all the times
Thanks again,
They worked for me as well. I did a simpler example with fewer years just to
show that it worked...( shorted here for display)
f - function(x) {
+dat - x[-(1:2)]
+tim - as.yearmon(outer(x$Year, seq(0, length = ncol(dat))/12,
+))
+zoo(c(as.matrix(dat)), tim)
On Sun, Aug 8, 2010 at 5:54 PM, steven mosher mosherste...@gmail.com wrote:
z-as.zooreg(as.ts(g))
z
X12345 X34567 X56789
1989(1) NA 3 6
1989(2) NA 3 6
1989(3) NA 3 6
1989(4) NA 3 6
1989(5) NA 3 6
1989(6)
Given a data frame, or it could be a matrix if I choose to.
The data consists of an ID, a year, and data for all 12 months.
Missing values are a factor AND missing years.
Id-c(rep(67543,4),rep(12345,3),rep(89765,5))
Years-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1))
On Sat, Aug 7, 2010 at 4:49 PM, steven mosher mosherste...@gmail.com wrote:
Given a data frame, or it could be a matrix if I choose to.
The data consists of an ID, a year, and data for all 12 months.
Missing values are a factor AND missing years.
Id-c(rep(67543,4),rep(12345,3),rep(89765,5))
Thanks Gabor, I probably should have done an example with fewer columns.
i will rework the example and post it up so the next guys who has this issue
can have a
clear example with a solution.
On Sat, Aug 7, 2010 at 5:04 PM, Gabor Grothendieck
ggrothendi...@gmail.comwrote:
On Sat, Aug 7, 2010
Very Slick.
Gabor this is a Huge speed up for me. Thanks. ha, Now I want to rewrite a
bunch of working code.
Id-c(rep(67543,4),rep(12345,3),rep(89765,5))
Years-c(seq(1989,1992,by =1),1991,1993,1994,seq(1991,1995,by=1))
Values2-c(12,NA,34,21,NA,65,23,NA,13,NA,13,14)
On Sat, Aug 7, 2010 at 9:18 PM, steven mosher mosherste...@gmail.com wrote:
Very Slick.
Gabor this is a Huge speed up for me. Thanks. ha, Now I want to rewrite a
bunch of working code.
Id-c(rep(67543,4),rep(12345,3),rep(89765,5))
Years-c(seq(1989,1992,by
Hi,
I have managed to convert my data frames into xts such as :
str(z)
An âxtsâ object from 1983-01-03 19:00:00 to 2006-01-01 22:00:00 containing:
Data: num [1:182959, 1:2] 12.6 11.3 12.7 12.8 10.9 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:2] v DD
Indexed by
Hi,
I am trying to convert my dataset into xts. I have tried the following :
data1-read.table(data1.txt,header=F)
data2-read.table(data2.txt,header=F)
data1.xtsas.xts(data1,descr=my new xts object)
However, I get an error :
Error in as.POSIXlt.character(x, tz, ...) :
character string is
On Fri, Jul 30, 2010 at 9:02 AM, Lily_stats sund...@gmail.com wrote:
Hi,
I am trying to convert my dataset into xts. I have tried the following :
data1-read.table(data1.txt,header=F)
data2-read.table(data2.txt,header=F)
data1.xtsas.xts(data1,descr=my new xts object)
However, I get an
Hi,
I am very new to R so these questions may seem simple!
I have a huge 2 sets of data(matrix 5x2++) in the following formats ,
for example data.txt and data2.txt:
Date Time X Y
03/03/1983 20:00 0.1 990
I would like to recreate
Convert your datasets into xts objects and then do a cbind ordering by the
column you want. Do a ?cbind.
HTH
Raghu
On Fri, Jul 30, 2010 at 10:33 AM, Lily_stats [via R]
ml-node+2307770-1033893256-309...@n4.nabble.comml-node%2b2307770-1033893256-309...@n4.nabble.com
wrote:
Hi,
I am very new
Please try:
data - xts(data[,2:n], order.by=as.POSIXct(strptime(data[,1],
%d/%m/%Y)))
Use similar strptime for hours also.n=number of columns.
Good Luck
Raghu
On Fri, Jul 30, 2010 at 2:02 PM, Lily_stats [via R]
Hi
I am trying to modify a data frame D with lists x and y in such a way that if a
value in x==0 then it should replace that value with the last not zero value in
x. I.e.
for loop over i{
if(D$x[i]==0)
D$x[i]=D$x[i-1]
}
The data frame is quite large in size ~ 43000 rows. This operation
-Oorspronkelijk bericht-
Van: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] Namens jd6688
Verzonden: maandag 26 juli 2010 7:07
Aan: r-help@r-project.org
Onderwerp: [R] data arranged by p-values
Idcat1locationitem_values p-values
sequence
Id cat1locationitem_values p-valuessequence
a1111 3002737 0.196504377 0.011
a1121 3017821 0.196504377 0.052
a1131 3027730 0.196504377 0.023
a1141 3036220 0.196504377 0.04
Dear All,
I have a raster map of the class 'SpatialPointsDataFrame' and coordinates
of the class 'SpatialPoints'. I would like to retrieve the values that are
contained in the raster map at the specific locations given by the
coordinates.
Can anyone help me out?
Kind regards,
Katrin Fleischer
-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of kfl...@falw.vu.nl
Sent: Tuesday, 20 July 2010 9:42 PM
To: r-help@r-project.org
Subject: [R] data from SpatialGridDataFrame
Dear All,
I have a raster map of the class 'SpatialPointsDataFrame' and coordinates
of the class 'SpatialPoints'. I
see ?sp::overlay and section 5.2 of Applied Spatial Data Analysis with R
I see there is now also raster::overlay, but I can't claim experience
with that funciton (however my impression is that the raster package
is a powerful tool for working with potentially very large rasters in
R).
hth,
Hello!
So, this is going to seem like a very simple question - I am quite new to R
and am having some trouble figuring out little nuances:
I am running a loop that goes through my data and performs a nls parameter
estimation on each data set. At the end of the loop, I would like to collect
the
Not necessarily the best way if your dataframe will get large, but it
should work:
parameters - NULL # where you will collect the result
for(j in 1:dim(r)[2]){
indiv=r[,j][which(r[,j]-1)] #removes -1 growth data
age.1=age[1:length(indiv)]
length.ind=data.frame(age.1,indiv,
hello,
I am trying to develop a triangle plot but am having difficultly assigning
the row.names to the 3 columns in the data.frame
Here is what I've done,
attach(SoilVegHydro)
dim(SoilVegHydro)
129239
# now take 3 variables from main data.frame for plotting
dat - cbind.data.frame(TP,
On Jul 8, 2010, at 4:41 PM, steve_fried...@nps.gov wrote:
hello,
I am trying to develop a triangle plot but am having difficultly
assigning
the row.names to the 3 columns in the data.frame
Here is what I've done,
attach(SoilVegHydro)
dim(SoilVegHydro)
129239
# now take 3
Hi,
Thanks a lot.
The Vectorize method worked and its much faster than looping through the
data frame.
Regards,
Harsh Yadav
On Thu, Jul 8, 2010 at 11:06 PM, David Winsemius dwinsem...@comcast.netwrote:
On Jul 8, 2010, at 10:33 PM, Erik Iverson wrote:
I have a data frame:
id url
Really? I don't usually think of Vectorize as a performance
enhancement, probably because my use of with a complex function then
gets applied to 4.5 million records. I need to go out, get a cup of
coffee, and leave it alone for about half an hour. I tried recently
to figure out how I can
Hi all,
I have a data frame for which I want to limit the output by checking whether
row values for specific column meets particular conditions.
Here are the more specific details:
I have a function that checks whether an input string exists in a defined
list:-
checkBaseLine - function(s){
It will be a lot easier to help you if you follow the posting guide and PLEASE
do read the posting guide and provide commented, minimal, self-contained,
reproducible code.
You gave your function definition, which is good. Use ?dput to give us a small
data.frame that can accurately show what
Hi,
Here is a somewhat detailed explanation of what I want to achieve:
I have a data frame:
id url
urlType
1 1 www.yahoo.com1
2 2 www.google.com/?search= 2
3 3 www.google.com
On Jul 8, 2010, at 10:09 PM, harsh yadav wrote:
Hi,
Here is a somewhat detailed explanation of what I want to achieve:
I have a data frame:
id url
urlType
1 1 www.yahoo.com1
2 2 www.google.com/?search= 2
3
I have a data frame:
id url
urlType
1 1 www.yahoo.com http://www.yahoo.com
1
2 2 www.google.com/?search= http://www.google.com/?search=
2
3 3
On Jul 8, 2010, at 10:33 PM, Erik Iverson wrote:
I have a data frame:
id
url urlType
1 1 www.yahoo.com http://
www.yahoo.com1
2 2 www.google.com/?search=
, 2010 9:44 PM
To: r-help@r-project.org
Subject: [R] Data Labels in a barchart (Lattice or otherwise)
Hi,
Can anyone please help me with how I could add labels with the value
for
each bar in a barchart? (similar to how data labels can be added in
Excel) I
have done a lot of searching
Hi,
Can anyone please help me with how I could add labels with the value for
each bar in a barchart? (similar to how data labels can be added in Excel) I
have done a lot of searching but havent been lucky.
Thanks,
Raoul
--
View this message in context:
On Jul 4, 2010, at 11:43 PM, RaoulD wrote:
Hi,
Can anyone please help me with how I could add labels with the value
for
each bar in a barchart? (similar to how data labels can be added in
Excel) I
have done a lot of searching but havent been lucky.
This is generally pretty easy with
Thank You David. Yes, I am using the lattice barchart and have managed to add
data labels, however, they tend to be on the tip of each bar and are
difficult to read as they are partially on the bar. Any help would be
greatly appreciated.
This is the code I am using:
On Jul 5, 2010, at 1:14 PM, RaoulD wrote:
Thank You David. Yes, I am using the lattice barchart and have
managed to add
data labels, however, they tend to be on the tip of each bar and are
difficult to read as they are partially on the bar. Any help would be
greatly appreciated.
This is
Hello,
I am trying to calculate the mean value of each row in a data frame (d),
I am having troubles and getting errors using the code I have written.
Below is a brief example of the code, any thought or suggestions would
be great.
Thank you for your time,
Doug
# Example Code:
d -
Doug -
Try
d$avg = apply(d,1,mean,na.rm=TRUE)
d
st1 st2 st3 st4 avg
1 1 2 5 6 3.50
2 2 5 5 5 4.25
3 3 6 NA 7 5.33
4 4 7 7 8 6.50
(If you must use a loop, calculate
mean(as.numeric(d[i,1:4]))
Take a look at mean(d[1,1:4]) to
Douglas M. Hultstrand wrote:
Hello,
I am trying to calculate the mean value of each row in a data frame (d),
I am having troubles and getting errors using the code I have written.
Below is a brief example of the code, any thought or suggestions would
be great.
Thank you for your time,
Hello Doug,
I just wanted to add that a faster way to initialize a vector is:
avg - vector(numeric, nrow(d))
Also you might like nrow(d) over length(d[ , 1]) if the number of rows
is what you are after. Its sister function is ncol() .
Best regards,
Josh
On Mon, Jun 28, 2010 at 11:37 AM,
Dear list,
I have the following problem. I have a data frame like this
CLUSTERYEAR variableDelta R_pivot
M1 2005 EC01 NA NA
M1 2006 EC012
You've posted this repeatedly, and yet received no answer. Perhaps
that is because you haven't read the posting guide! You didn't
provide a reproducible example, you didn't tell us where ddply came
from, you didn't tell us what was wrong with the code you suggested.
It would also be rather easier
Dear list,
I have the following problem. I have a data frame like this
CLUSTERYEAR variableDelta R_pivot
M1 2005 EC01 NA NA
M1 2006 EC012
Dear list,
I have the following problem. I have a data frame like this
CLUSTERYEAR variableDelta R_pivot
M1 2005 EC01 NA NA
M1 2006 EC012
Dear R´ers..
In this mock dataset how can I generate a logical variable based on whether
just tes or tes3 are NA in each row??
test-sample(c(A,NA,B),100,replace=T)
test2-sample(c(A,NA,B),100,replace=T)
test3-sample(c(A,NA,B),100,replace=T)
tes-cbind(test,test2,test3)
sam-c(test,test3)
?any
Not really a reproducible answer, but I think you're looking
for
apply(tes[,sam],1,function(x)any(is.na(x)))
- Phil Spector
Statistical Computing Facility
Department
Hi there,
One option would be
apply(tes, 1, function(.row) any(is.na(.row[c(1,3)])))
See ?any, ?is.na and ?apply for more information.
HTH,
Jorge
On Thu, Jun 3, 2010 at 3:20 PM, moleps wrote:
Dear R´ers..
In this mock dataset how can I generate a logical variable based on whether
just
On Jun 3, 2010, at 2:20 PM, moleps wrote:
Dear R´ers..
In this mock dataset how can I generate a logical variable based on whether
just tes or tes3 are NA in each row??
test-sample(c(A,NA,B),100,replace=T)
test2-sample(c(A,NA,B),100,replace=T)
test3-sample(c(A,NA,B),100,replace=T)
you probably want to use the apply function:
d=sample(1000,500);
d[sample(500,50)]-NA; #put 50 NAs into the data
d=data.frame(matrix(d,ncol=50));
names(d)=paste('var',1:50,sep='.')
d
apply(d,1,sum) #are any of the row values NA ?
apply(d,2,function(x)sum(is.na(x))) #how many values for each of
-Any- was my fix... Appreciate it.
//M
On 3. juni 2010, at 21.33, Phil Spector wrote:
?any
Not really a reproducible answer, but I think you're looking
for
apply(tes[,sam],1,function(x)any(is.na(x)))
- Phil Spector
!!
if (nrow(futures)==0) futures-data.frame(...)
-Original Message-
From: Peter Ehlers [mailto:ehl...@ucalgary.ca]
Sent: Tuesday, June 01, 2010 12:07 PM
To: arnaud Gaboury
Cc: 'Prof Brian Ripley'; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On 2010
Subject: Re: [R] data frame manipulation with zero rows
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something
else than
ddply() if I want to avoid any error each time my df has zero
rows?
Am I correct?
You could
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures -
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10, SUGAR
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures -
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10
Dear group,
Here is my data frame:
futures -
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10, CORN Jul/10, CORN Jul/10, LIVE CATTLE Aug/10,
LIVE CATTLE Aug/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10,
SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10, SUGAR NO.11 Jul/10
),
01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010, arnaud Gaboury wrote:
Dear group,
Here is the kind of data.frame I obtain every day with my function :
futures-
structure(list(DESCRIPTION = c(CORN Jul/10, CORN Jul/10,
CORN Jul/10
...@uni-bremen.de]
Sent: Tuesday, June 01, 2010 11:38 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation ddply
Hi Arnaud,
maybe aggregate can help:
PosFut - aggregate(futures$QUANTITY, list(DESCRIPTION =
futures$DESCRIPTION,
SETTLEMENT
of zero rows: it may or may not be the one in package
plyr.
-Peter Ehlers
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 9:47 AM
To: arnaud Gaboury
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1 Jun 2010
It is indeed ddply() from package plyr.
-Original Message-
From: Prof Brian Ripley [mailto:rip...@stats.ox.ac.uk]
Sent: Tuesday, June 01, 2010 12:24 PM
To: Peter Ehlers
Cc: arnaud Gaboury; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On Tue, 1
Brian Ripley'; r-help@r-project.org
Subject: Re: [R] data frame manipulation with zero rows
On 2010-06-01 1:53, arnaud Gaboury wrote:
Brian,
If I do understand correctly, I must use in my function something
else than
ddply() if I want to avoid any error each time my df has zero rows?
Am
I'm interested in using a data frame as if it were a hash table. For
instance if I had the following,
(d - data.frame(key=seq(0.5, 3, 0.5), value=rnorm(6)))
keyvalue
1 0.5 -1.118665122
2 1.0 0.465122921
3 1.5 -0.529239211
4 2.0 -0.147324638
5 2.5 -1.531503795
6 3.0 -0.002720434
Then
On Sun, May 30, 2010 at 9:03 AM, Alan Lue alan@gmail.com wrote:
I'm interested in using a data frame as if it were a hash table. For
instance if I had the following,
(d - data.frame(key=seq(0.5, 3, 0.5), value=rnorm(6)))
key value
1 0.5 -1.118665122
2 1.0 0.465122921
3 1.5
You might want to investigate the 'data.table'
package.
On 30/05/2010 09:03, Alan Lue wrote:
I'm interested in using a data frame as if it were a hash table. For
instance if I had the following,
(d- data.frame(key=seq(0.5, 3, 0.5), value=rnorm(6)))
keyvalue
1 0.5 -1.118665122
2
To:r-help@r-project.org,alan@gmail.com
Subject: Re: [R] Data Frame as Hash Table
Message-ID:4c0220b6.7090...@pburns.seanet.com
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
You might want to investigate the 'data.table'
package.
On 30/05/2010 09:03, Alan Lue wrote:
I'm
:
Message: 40
Date: Sun, 30 May 2010 09:24:22 +0100
From: Patrick Burnspbu...@pburns.seanet.com
To:r-help@r-project.org,alan@gmail.com
Subject: Re: [R] Data Frame as Hash Table
Message-ID:4c0220b6.7090...@pburns.seanet.com
Content-Type: text/plain; charset=ISO-8859-1; format=flowed
You
Hi there,
I am glad it helped.
I used mean as something to use, not because I had an understanding that
this is what you need - so if you believe sum is what you where after - go
with it :)
Regarding loving R, and time spending - everyone on this list probably know
how you feel. We all spent
Hi there,
The tool to learn for this is the cast function using the reshape package.
In your example you have more then one value for RTL, which you should think
of how to account for.
But basically, here is a solution to what you asked for (assuming I
understood you correctly)
require(reshape)
Great, these are valuable tips. Thanks both of you. I appreciate it. :)
Timothy
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Tal,
Wow, i cant believe how many different manipulations i went through trying
to coerce it into the format i wanted. The below works nearly perfectly, i had
to change the mean call to sum. Im curious why you used mean? Other than
that thank you very much, i feel a little foolish about
,
something like :
if(trade$Trade.Status==DEL)switch(.)
I would like to avoid the loop .
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Wednesday, May 26, 2010 9:15 PM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy
.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement
..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due to the switch statement
NULL
$Net.Charges..sum.
NULL
That's certainly not what I want.
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 8:43 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
The loop is due
Hi,
I'm very confused about R structures and the methods to go with them. I'm
using R for microarray analysis with Bioconductors. Suppose without reading
the documentations, what's the best way to explore a data structure when you
know nothing about it?
I am currently using is() / class() to see
want !!
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 10:38 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
Off course. You put in a matrix to sapply, but sapply is for vectors. You
want
, Instrument.Long.Name,
Delivery.Prompt.Date, Buy.Sell..Cleared., Volume, Price,
Net.Charges..sum.), row.names = c(NA, 3L), class = data.frame)
From: Joris Meys [mailto:jorism...@gmail.com]
Sent: Thursday, May 27, 2010 10:38 AM
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame
To: arnaud Gaboury
Cc: r-help@r-project.org
Subject: Re: [R] data frame manipulation change elements meeting criteria
Off course. You put in a matrix to sapply, but sapply is for vectors. You
want to apply the switch command on every entry of the vector
trades$Buy.Sell..Cleared for which trades
On 05/27/2010 02:13 AM, Timothy Wu wrote:
Hi,
I'm very confused about R structures and the methods to go with them.
I'm using R for microarray analysis with Bioconductors. Suppose
without reading the documentations, what's the best way to explore a
data structure when you know nothing about
: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Timothy Wu
Sent: Thursday, May 27, 2010 3:14 AM
To: r-help@r-project.org
Subject: [R] Methods to explore R data structures
Hi,
I'm very confused about R structures and the methods to go with them.
I'm
Hello All,
Please consider the following:
TotEmp-c(19,6,1,1,8,44,2,33,48,1)
ClusterType-c(AGF,CNS,OSV,RTL,RTL,TRN,REL,ACC_CLUST,RTL,WHL)
Taz-c(0,0,0,100,100,100,101,101,102,103)
AllCtTypes_-c(AGF,CNS,OSV,RTL,TRN,REL,ACC_CLUST,WHL,ADM_CLUST,
Dear group,
Here is my df :
trade -
structure(list(Trade.Status = c(DEL, INS, INS), Instrument.Long.Name =
c(SUGAR NO.11,
CORN, CORN), Delivery.Prompt.Date = c(Jul/10, Jul/10,
Jul/10), Buy.Sell..Cleared. = c(Sell, Buy, Buy), Volume = c(1L,
2L, 1L), Price = c(15.2500, 368., 368.5000),
see ?switch
X- rep(c(Buy,Sell,something else),each=5)
Y- rep(c(DEL,INS,DEL),5)
new.vect - X
for (i in which(Y==DEL)){
new.vect[i]-switch(
EXPR = X[i],
Sell=Buy,
Buy=Sell,
X[i])
}
cbind(new.vect,X,Y)
On Wed, May 26, 2010 at 7:43 PM, arnaud Gaboury
I understand that everything passed to an R function is passed by
value. This would seem to include data frames, which my current
application uses heavily, both for storing program inputs, and holding
intermediate and final results. In trying to get greater performance
out of my R code, I am
If you don't modify the data frame in your function it won't
physically make a new copy.
On Mon, May 24, 2010 at 10:29 AM, gschu...@scriptpro.com wrote:
I understand that everything passed to an R function is passed by
value. This would seem to include data frames, which my current
R is pretty smart about duplicating only when necessary. That is,
arguments passed to a function are copy-on-write. Also, I think (someone
more knowledgeable please correct if I'm wrong) it may be better to use
the data frame, which is just a list internally, because if you only
modify one column,
Hi Thomas,
Thanks very much for your reply. I used svd and it worked perfectly for my
purposes!
Thanks again,
Julia
--
View this message in context:
http://r.789695.n4.nabble.com/Data-reconstruction-following-PCA-using-Eigen-function-tp2226535p2229191.html
Sent from the R help mailing list
Looks like you have some numerical precision issues. Why not use the svd
function directly? (See below.)
-tgs
x - read.table(
textConnection(
Sample1 0.7329881 0.76912670 2.45906143 -0.06411602 1.2427801
0.3785717 2.34508664 1.1043552 -0.1883830 0.6503095
Sample2 -2.0446131
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