Hello,
struggling with the very basic needs... :( any help appreciated.
#using the package doBY
#who drinks how much beer per day and therefor cannot calculate rowise
maxvals
evaluation=data.frame(date=c(1,2,3,4,5,6,7,8,9),
name=c(Michael,Steve,Bob,
Michael,Steve,Bob,Michael,Steve,Bob),
Try this:
evaluation$maxVol - ave(evaluation$vol, evaluation$name, FUN = max)
or using SQL via sqldf like this:
library(sqldf)
sqldf(select * from evaluation join
(select name, max(vol) from evaluation group by name) using (name))
On 8/31/07, Calle [EMAIL PROTECTED] wrote:
Hello,
hello,
I wanna print something like this
Class Levels Values
Id_TrT1 1 2
Id_Geno764208 64209 64210 64211 64212 64213
64214
Id_Rep 2 12
Is it possible?
I have some problem I think taht
how about:
?str
ever considered reading an introductory text?
find some here:
http://cran.r-project.org/other-docs.html
Stefan
elyakhlifi mustapha wrote:
hello,
I wanna print something like this
Class Levels Values
Id_TrT1 1 2
Id_Geno
, new.info)
dat
It works. The R console result can be seen in the attachment.
CU, Corinna
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von elyakhlifi
mustapha
Gesendet: Montag, 23. April 2007 16:02
An: R-help@stat.math.ethz.ch
Betreff: [R] data
] data frame
hello,
I wanna print something like this
Class Levels Values
Id_TrT1 1 2
Id_Geno764208 64209 64210 64211 64212 64213
64214
Id_Rep 2 12
Is it possible?
I have some problem I
Its not usual to represent structures in this form in R but you
can do it if you really want:
data.frame(A = letters[1:3], B = 1:3, C = I(list(2, 1:6, 9)))
Note the I (capital i) to make sure the list gets passed in asis.
On 4/23/07, elyakhlifi mustapha [EMAIL PROTECTED] wrote:
hello,
I
This is a slightly different formulation of your problem, but you
might find it easier to work with:
Start by making a data frame):
df=data.frame(Father=c
(Fred,Fred,Fred,Barney,Barney),Mother=c
(Mary,Mary,Mary,Liz,Liz),Child.Age=c(4,7,9,3,5))
If you want to add a wedding date column
I have a data frame called df which has about 100 columns but thousands of
rows. I set D the index of df$D and M to be the index of df$M.
When I run the following loop as it is vs. df[,D] and df[,M] replaced with
df$D and df$M there is a real big time difference in completion (with the
latter
What are D and M? 'Index' here could be a number or a name.
In either case, df[[D]] would be the equivalent of df$D.
However, your computation does not need a loop at all, let alone two.
Try something like
tmp - with(df, paste(D, m))
dates - unique(tmp)
On Wed, 21 Feb 2007, Alp ATICI wrote:
Dear all,
Can anyone please shed some light onto how to do this?
This will give me all intensity columsn in my data frame:
intensityindeces - grep(^Intensity,names(dataframe),value=TRUE)
This will give me the maximum intensity for the first row:
intensityone - max(dataframe[1,intensityindeces])
Johannes Graumann wrote:
Dear all,
Can anyone please shed some light onto how to do this?
This will give me all intensity columsn in my data frame:
intensityindeces - grep(^Intensity,names(dataframe),value=TRUE)
This will give me the maximum intensity for the first row:
intensityone -
do.call(pmax, dataframe[,intensityindeces])
if I understand you aright.
On Mon, 19 Feb 2007, Johannes Graumann wrote:
Dear all,
Can anyone please shed some light onto how to do this?
This will give me all intensity columsn in my data frame:
intensityindeces -
On Monday 19 February 2007 11:53, Prof Brian Ripley wrote:
do.call(pmax, dataframe[,intensityindeces])
Thank you very much for your help!
Any idea why do.call(pmax,list(na.rm=TRUE),dataframe[,intensityindeces])
would give me
Error in if (quote) { : argument is not interpretable as logical
In
On Mon, 19 Feb 2007, Johannes Graumann wrote:
On Monday 19 February 2007 11:53, Prof Brian Ripley wrote:
do.call(pmax, dataframe[,intensityindeces])
Thank you very much for your help!
Any idea why do.call(pmax,list(na.rm=TRUE),dataframe[,intensityindeces])
You want something like
Try do.call(pmax,c(dataframe[,intensityindices],na.rm=TRUE))
This is like the second example in the help page for do.call
On 19/02/07, Johannes Graumann [EMAIL PROTECTED] wrote:
On Monday 19 February 2007 11:53, Prof Brian Ripley wrote:
do.call(pmax, dataframe[,intensityindeces])
Thank you
Thanks to you and Brian Ripley. Quite confusing all this ...
Thanks again.
Joh
On Monday 19 February 2007 13:42, David Barron wrote:
Try do.call(pmax,c(dataframe[,intensityindices],na.rm=TRUE))
This is like the second example in the help page for do.call
On 19/02/07, Johannes Graumann
Dear R users,
When you do:
x - rnorm(10)
y - rnorm(10)
z - rnorm(10)
a - data.frame(x,y,z)
a$x
[1] 1.37821893 0.21152756 -0.55453182 -2.10426048 -0.08967880 0.03712110
[7] -0.80592149 0.07413450 0.15557671 1.22165341
Why does this not work:
a[a$y0.5,y] -1
Error in
When specifying a column name with [ the name must be quoted (unlike
when using it with $):
a[a$y 0.5, y] - 1
On 8/4/06, Sander Oom [EMAIL PROTECTED] wrote:
Dear R users,
When you do:
x - rnorm(10)
y - rnorm(10)
z - rnorm(10)
a - data.frame(x,y,z)
a$x
[1] 1.37821893
/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Sander Oom [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Friday, August 04, 2006 1:48 PM
Subject: [R] Data frame referencing
://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Sander Oom [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Friday, August 04, 2006 1:48 PM
Subject: [R] Data frame referencing?
Dear R users,
When you do:
x - rnorm(10)
y - rnorm(10)
z - rnorm(10
Hi,
I want to make a data frame which contains the positions of some searched
values in another data frame.
Like:
Dataframe 1:
1 2 3 4 1 2 3 4
2 3 4 1 2 3 4 2
4 1 2 3 2 3 4 1
Let's say I searched on 4, then Dataframe 2 should contain:
x y
1 4
1 8
2 3
2 7
3 1
3 7
I
Try this:
which(DF1 == 4, arr.ind = TRUE)
On 6/24/06, Bart Joosen [EMAIL PROTECTED] wrote:
Hi,
I want to make a data frame which contains the positions of some searched
values in another data frame.
Like:
Dataframe 1:
1 2 3 4 1 2 3 4
2 3 4 1 2 3 4 2
4 1 2 3 2 3 4
I'll suggest an algorithm with a little code but haven't actually tried it.
R loves vector/matrix operations (and, incidentally, you probably ought to
be using matrices here and not data frames, particularly if the images are
large).
Here is your matrix from your example (I'll call this x)
1 2
Thanks, couldn't find this function,
Best regards
Bart
- Original Message -
From: Gabor Grothendieck [EMAIL PROTECTED]
To: Bart Joosen [EMAIL PROTECTED]
Cc: r-help@stat.math.ethz.ch
Sent: Saturday, June 24, 2006 1:15 PM
Subject: Re: [R] data frame search
Try this:
which(DF1 == 4
: Saturday, June 24, 2006 1:15 PM
Subject: Re: [R] data frame search
Try this:
which(DF1 == 4, arr.ind = TRUE)
On 6/24/06, Bart Joosen [EMAIL PROTECTED] wrote:
Hi,
I want to make a data frame which contains the positions of some searched
values in another data frame.
Like
I guess you are looking for
print(summary(lm(y ~ ., data=x.d)))
since '.' refers to all the columns of 'data' except perhaps the response.
On Sat, 25 Mar 2006, ivo welch wrote:
Dear R wizards: This must have an obvious solution, but I am stumped.
I can run a linear regression giving
Dear R wizards: This must have an obvious solution, but I am stumped.
I can run a linear regression giving a matrix as the independent set
of variables, but if I give a data frame (which I would like to give,
because it should tell the linear model the names of the variables), R
does not like
Try using dot notation:
lm(y ~., x.d)
On 3/25/06, ivo welch [EMAIL PROTECTED] wrote:
Dear R wizards: This must have an obvious solution, but I am stumped.
I can run a linear regression giving a matrix as the independent set
of variables, but if I give a data frame (which I would like to
try:
DF2 - as.data.frame(matrix(vec, nr=nrow(DF),nc=ncol(DF))==
matrix(1:ncol(DF),nr=nrow(DF),nc=ncol(DF),byrow=T))
DF3 - data.frame(mapply(function(z,x,y) { x[y] - 0 ; x },
names(DF), DF, DF2, SIMPLIFY=F))
but there must be an easier way...
Kenneth Cabrera a écrit :
Hi, R
wrote:
Date sent: Wed, 18 Jan 2006 02:35:35 -0500
From: Kenneth Cabrera [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject:[R] Data frame index?
Hi, R users:
I have a data.frame (not a matrix), I got a vector with the same
/dimitris.htm
- Original Message -
From: Kenneth Cabrera [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Sent: Wednesday, January 18, 2006 8:35 AM
Subject: [R] Data frame index?
Hi, R users:
I have a data.frame (not a matrix), I got a vector with the same
length
as the
number of records (rows
On 1/18/2006 2:35 AM, Kenneth Cabrera wrote:
Hi, R users:
I have a data.frame (not a matrix), I got a vector with the same length
as the
number of records (rows) of the data frame, and each element of
that vector is the column number (in a specific range of columns) of the
corresponding
It's worth noting that there are quite a few for loops inside the code
used by matrix indexing of data frames.
I think a single for-loop over the columns is as good as any, something
like
DF - data.frame(x=1, y=rep(a, 4), z = 3)
ind - c(1,3,3,1) # only numeric cols
for(i in unique(ind))
Hi, R users:
I have a data.frame (not a matrix), I got a vector with the same length
as the
number of records (rows) of the data frame, and each element of
that vector is the column number (in a specific range of columns) of the
corresponding
record that I must set to zero.
How can I do
R-help@stat.math.ethz.ch
Rhett Eckstein [EMAIL PROTECTED] a écrit : Dear R users:
s4 - seq(length=10, from=1, by=5)
s-data.frame(s4,s4,s4)
I would like to do some modification to s.
And I want the form like the following,if it is possible, how should I do?
The last column is the sum of
Dear R users:
s4 - seq(length=10, from=1, by=5)
s-data.frame(s4,s4,s4)
I would like to do some modification to s.
And I want the form like the following,if it is possible, how should I do?
The last column is the sum of previous three column.
s4 s4.1 s4.2sum
1 11
2
Here is one way. You can change depending on what you want the offsets to
be:
s4 - seq(length=10, from=1, by=5)
s4
[1] 1 6 11 16 21 26 31 36 41 46
f.x - function(vec, n) c(rep(0,n), vec)[1:length(vec)]
f.x(s4,2)
[1] 0 0 1 6 11 16 21 26 31 36
df - data.frame(s4=s4, s4.1=f.x(s4,2),
On 12/22/05 9:11 AM, Rhett Eckstein [EMAIL PROTECTED] wrote:
Dear R users:
s4 - seq(length=10, from=1, by=5)
s-data.frame(s4,s4,s4)
I would like to do some modification to s.
And I want the form like the following,if it is possible, how should I do?
The last column is the sum of
I know this is very basic--I'm wondering if there is a way to write data
frames as outputs from a loop.
In other words, take this simple example:
a - data.frame(x = c(1,2,3,4), y = c(1,2,1,2))
Given a, how would you write a loop that creates two data frames, x and y,
where the first column
No need for computer-intensive loops. Try the following:
a - data.frame(x = c(1,2,3,4), y = c(1,2,1,2))
xx=data.frame(x.1=a[,1],x.2=2*a[,1])
yy=data.frame(y.1=a[,2],y.2=2*a[,2])
Cheers
Francisco
From: Apoian, Zack [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Subject: [R] data frame output
Hello,
is't possible to get excerptions of data frame using some contstraints,
something like q1 = q[q$V31] ?
Regards,
Oleg
_
Oleg Bartunov, sci.researcher, hostmaster of AstroNet,
Sternberg Astronomical Institute,
PM
Subject: [R] data frame excerption
Hello,
is't possible to get excerptions of data frame using some contstraints,
something like q1 = q[q$V31] ?
Regards,
Oleg
_
Oleg Bartunov, sci.researcher, hostmaster of AstroNet,
Sternberg
Oleg Bartunov wrote:
Hello,
is't possible to get excerptions of data frame using some contstraints,
something like q1 = q[q$V31] ?
Yes, but you have to use the indexing matrix-like (see the manuals):
q1 - q[q$V3 1, ]
Uwe Ligges
Regards,
Oleg
Hi
I have used merge() to merge two data frames, very much like performing
a SQL join. Now I want to do a few different SQL-style things and I
wondered if there were functions to do it...
Is there a group by style function? For example if I merge() two data
frames and end up with multiple
Hi Michael
On 1 Dec 2004 at 11:50, michael watson (IAH-C) wrote:
Hi
I have used merge() to merge two data frames, very much like
performing a SQL join. Now I want to do a few different SQL-style
things and I wondered if there were functions to do it...
Is there a group by style
://www.student.kuleuven.ac.be/~m0390867/dimitris.htm
- Original Message -
From: michael watson (IAH-C) [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, December 01, 2004 12:50 PM
Subject: [R] Data Frame Manipulations
Hi
I have used merge() to merge two data frames, very much like
performing
a SQL
On Thu, 18 Mar 2004, Randy Zelick wrote:
Is there a way to *not* supress leading zeros when printing (to the
console window or to a file) a dataframe?
If you mean via print() or autoprinting, no.
I am not sure why you would want to do this, but it seems that using
format() and then gsub should
Randy == Randy Zelick [EMAIL PROTECTED]
on Thu, 18 Mar 2004 21:00:08 -0800 (PST) writes:
Randy Hello list,
Randy Is there a way to *not* supress leading zeros when printing (to the
Randy console window or to a file) a dataframe?
Yes, e.g. use
formatC(..., format = f)
to
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
On Fri, 19 Mar 2004, Randy Zelick wrote:
I got three responses for help on the leading zero problem. Thank you.
Well, it seems that you didn't tell us what the actual problem was: please
consult the posting guide and its references and learn to ask the right
question.
Alas I still don't have
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]
Department of Biology voice: 503-725-3086
Portland State University
Hi, R-help:
I am a new user of R and am very pleased with R's features. Here I have
one question regarding data frame manipulation. I have a data frame look
like this:
Fruit Condition
1 Orange Good
2 OrangeBad
3 Orange Good
4 Orange Good
5 OrangeBad
6 Apple Good
7
is returned since there
is no else leg to the if. rbind then binds the groups back into
a data frame.
---
Date: Wed, 10 Mar 2004 17:09:23 -0500
From: Sean Liang [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject: [R] data frame filtration
Hi, R-help:
I am a new user of R and am very pleased
Hi All,
I want to ask if there is a transpose function for data frame like the
procedure of transpose in SAS? Because I want to partially transpose a
data frame which contains 5 columns (siteid, date, time, obs, mod), what
I want to do is to put time as the column variables along with siteid,
On Mon, 29 Sep 2003 [EMAIL PROTECTED] wrote:
Hi All,
I want to ask if there is a transpose function for data frame like the
procedure of transpose in SAS? Because I want to partially transpose a
data frame which contains 5 columns (siteid, date, time, obs, mod), what
I want to do is to put
Dave -
I'm not sure whether there is already a function which does
exactly what you want, because this is kind of a special case.
The functions I wold look at are: by, aggregate, tapply,
mapply, and, in the package nlme one I didn't know about
before called gapply.
But, in your case, the part
]: [EMAIL PROTECTED]: Mon, 29 Sep 2003 13:15:36
-0400Subject: [R] Data frame transposeHi All,I want to ask if there is a transpose
function for data frame like theprocedure of transpose in SAS? Because I want to
partially transpose adata frame which contains 5 columns (siteid, date, time, obs,
mod
$siteid,m$date),]
--- On Mon 09/29, [EMAIL PROTECTED] wrote:
From: [mailto: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Date: Mon, 29 Sep 2003 13:15:36 -0400
Subject: [R] Data frame transpose
brbrbrbrHi All,brbrI want to ask if there is a transpose function for data
frame like thebrprocedure
This seems to be a simple problem, and I feel that there ought to be a
simple answer, but I can't seem to find it.
I have a function that returns a number of values as a heterogeneous list -
always the same length and same names(), but a number of different data
types, including character. I
22, 2003 5:15 AM
To: [EMAIL PROTECTED]
Subject: [R] Data frame from list of lists
This seems to be a simple problem, and I feel that there
ought to be a simple answer, but I can't seem to find it.
I have a function that returns a number of values as a
heterogeneous list - always
part (for me) is how to get
that last column back to POSIXct. I have not dealt with date/time in R
before.
HTH,
Andy
-Original Message-
From: Gregory Jefferis [mailto:[EMAIL PROTECTED]
Sent: Monday, September 22, 2003 5:15 AM
To: [EMAIL PROTECTED]
Subject: [R] Data frame from
Scot,
Thanks for the additional information. On further
reflection... whether one uses SAS PROC EXPORT or uses
a SAS LIBNAME yourfile XPORT 'yourpathname';
statement, an intermediate file is created in either
case. As far as experience tells me now, PROC EXPORT
is a far superior choice, because
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