is file.
>>
>> When I want to get sth from my database I get NULL, but I know that
>> there is
>> sth!
>> For example:
>>
>>> data$ID
>> NULL
>>> data$kod
>> NULL
>>
>> but command like below is always recognize by R
>>>
is
sth!
For example:
data$ID
NULL
data$kod
NULL
but command like below is always recognize by R
data[2,3]
[1] '082'
Tell is what happens when you enter:
str(data)
class(data)
Perhaps the third column is not named "ID" or "kod" or the object is
not a data
I have database write as .csv file.
When I want to get sth from my database I get NULL, but I know that there is
sth!
For example:
> data$ID
NULL
> data$kod
NULL
but command like below is always recognize by R
> data[2,3]
[1] '082'
In my opinion this problem is also connect
Also consider the "View" function for looking at the dataframe
On Wed, Apr 15, 2009 at 10:13 AM, Vladan Arsenijevic
wrote:
> Hi all,
>
> I am dealing with a big data frame. When printing something like
>
>> allData[[3]]
>
> 1 625.364 38.223 21.014 0.216 1.241411 V 1050o 58.38065 -0.06178768
bericht-
Van: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
Namens Vladan Arsenijevic
Verzonden: woensdag 15 april 2009 16:13
Aan: r-help@r-project.org
Onderwerp: [R] data frame display
Hi all,
I am dealing with a big data frame. When printing something like
> allData[[3]]
Hi all,
I am dealing with a big data frame. When printing something like
allData[[3]]
1 625.364 38.223 21.014 0.216 1.241411V 1050o 58.38065 -0.06178768
2 383.709 55.811 21.435 0.296 1.241411V 1050o 58.38308 -0.03328282
3 434.669 58.597 21.207 0.233 1.241411V 1050o 58.38334
Its not clear what it means for the sum of
the weeks in a month to equal the month
since the weeks don't evenly divide
a month but if we apportion them pro
rata then here is a possibility.
We create the input monthly series z and
then produce a series of Date class days
d covering the period. Mer
Hi R users,
I have a time series variable that is only available at a monthly level for
1 years that I need to decompose to a weekly time series level - can
anyone recommend a R function that I can use to decompose this series?
eg. if month1 = 1200 I would to decompose so that the sum of the
project.org
Subject: [R] Data manipulation - multiplicate cases
Hi listers,
I am trying to arrange my data and I didn't find any information how to do
it!
I have a data with 3 variables: X Y Z
1-I would like to multiplicate de information of X according to the number I
have for my Y vari
Is this what you are looking for:
> x
X Y Z
1 123 3 1
2 234 3 1
3 345 4 2
4 456 3 2
> new.x <- x[rep(seq(nrow(x)), times=x$Y),]
> new.x
X Y Z
1 123 3 1
1.1 123 3 1
1.2 123 3 1
2 234 3 1
2.1 234 3 1
2.2 234 3 1
3 345 4 2
3.1 345 4 2
3.2 345 4 2
3.3 345 4 2
4 456 3 2
4.1 456 3 2
4.
Hi listers,
I am trying to arrange my data and I didn't find any information how to do
it!
I have a data with 3 variables: X Y Z
1-I would like to multiplicate de information of X according to the number I
have for my Y variable...
2-Then I want to identify with a dicotomic variable by the number
The rows
of the data frame may be in arbitrary order.
Bill Venables
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Thomas S. Dye
Sent: Monday, 23 March 2009 7:36 AM
To: r-help@r-project.
Aloha all,
I have a data frame with 4 columns. The first three are factors (f1,
f2, f3) and the fourth is numeric. I'd like to explore these data
using median polish. To do that I plan to use medpolish() on the
matrix[f1,f2xf3], then medpolish on the resulting matrix[f2,f3]. This
appro
thx for ur fast responds.
but sorry for asking stupid, i am a turn beginner of R (just trying it out
<3 months, and i am taking my first course about it)
so, to tackle this questions,
i was told to use "nested design" method,
could you actually show me how would u attempt this problem?
(a) Determi
On Sat, Mar 21, 2009 at 5:13 PM, UBC wrote:
>
> so i am having this question
> what should i do if the give data file (.txt) has 4 columns, but different
> lengths?
> how can i read them in R?
> any idea for the following problem?
>
>
> Gas consumption (1000 cubic feet) was measured before and aft
This works with the example. If the real data is different it may not
work. To run the example below just copy and paste it into R.
To run with the real data replace textConnection(Lines) with
"insulation.txt" everywhere.
Lines <- "Before insulAfter insul.
tempgas tempgas
-0.8
If the input file has a separator other than a space (e.g., tabs or
commas) then you can read it is and the missing data will be NAs and
you can decide how to handle it. If it does not have a separator,
then maybe you can read it in with read.fwf. Otherwise when you read
it in, you can tell the s
so i am having this question
what should i do if the give data file (.txt) has 4 columns, but different
lengths?
how can i read them in R?
any idea for the following problem?
Gas consumption (1000 cubic feet) was measured before and after insulation
was put into
a house. We are interested in loo
_data_df<-test_data1_df
length_test_data<-length(grep('^Run',names(test_data_df)))
if(length_test_data==2)
{
reshaped_test_data<-reshape(test_data_df,
varying=list(c('Run.1')),
idvar='Location',direction='long')
} else if (length_test_d
look at package reshape there is a cool little function input that
once you get the hang of is handy.
On Mon, Mar 9, 2009 at 5:50 PM, Jason Rupert wrote:
> I think I am overlooking a call or concept in R to help me easily and quickly
> restructure my data.frame:
>
> Sometimes the data I receive
I think I am overlooking a call or concept in R to help me easily and quickly
restructure my data.frame:
Sometimes the data I receive looks like:
VariableName, Run1, Run2, Run3, Location
temp, 15.0, 16.0, 17.0, There
And other times it looks like:
VariableName, Run, Location
t
How do I do a data envelopment analysis in R...provide me with the step by
step procedure for that..thanks in advance...
Arup
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
Dear R-list members,
I have a data file with thousands of lines (cases), where each line
contains the values of several variables. I would like to separate
these lines in small groups, with each group followed by a blank
line, to ease the visual inspection of the data in some situations.
I am w
Hello all,
I have a *.csv file that looks like this (actual file is orders of magnitude
larger):
site taxa no.ind meadow LMA 2 meadow LCY 1 meadow MSA 2 forest LMA 1
forest LCY 1 forest MSA 1 forest MSX 1
I am interested in, but have failed to create, code that efficiently
converts it to a si
There are three different data editors, so you will need to start by
telling us your OS (and the other details asked for in the posting
guide).
But this is really an R-devel question, and that is where offers to
work on this (or to sponsor such work) should be posted.
As the author of one ve
Hi !
I am using "Tinn R" data editor.
This is wonderful and also thin one. Try this. I guess, yu will find what
you are looking for.
Regards,
Suresh
Simon Pickett-4 wrote:
>
> Hi all,
>
> I've used R for basic programming and data management for a few years now.
Hi all,
I've used R for basic programming and data management for a few years now. One
of the things that I think could be improved is the data editor.
Its a great feature and I use it alot by calling edit(data.frame); very useful
to see if what you tried to do actually worked.
However, one o
Hello Jim:
Yes, that's exactly what I needed!
Thank you!
Josip
- Original Message -
From: "jim holtman"
To: "Josip Dasovic"
Cc: r-help@r-project.org
Sent: Tuesday, January 27, 2009 4:45:31 PM GMT -08:00 US/Canada Pacific
Subject: Re: [R] Data Frame Manipula
Is the what you are after:
> df<-data.frame(cbind("country"=c(rep("Angola", 9), rep("Burundi", 7),
+ rep("Chad", 13)), "year"=c(1975:1983, 1989:1995, 1965:1977)),
+ "war"=c(rep(1,2), rep(0,5), rep(1,2), rep(1,2), rep(0,2), rep(1,3),
+ rep(1,4), rep(0,6), rep(1,3)))
> x <- split(df,
Dear R Helpers:
I have a data set where the unit of observation is country-year. I would like
to generate a new data set based on some inclusionary (exclusionary) criteria.
Here is an example of the type of data that I have.
df<-data.frame(cbind("country"=c(rep("Angola", 9), rep("Burundi", 7),
Although the message is rather unreadable, I guess you want to look at
?reshape.
Uwe Ligges
oscar linares wrote:
Dear Rxperts,
I would like to convert the following:
StudyStudy.NameParameterDestSrcFormValueMin
MaxFSD
1NT_1-0BFK(03)03A0.
Dear Rxperts,
I would like to convert the following:
StudyStudy.NameParameterDestSrcFormValueMin
MaxFSD
1NT_1-0BFK(03)03A0.128510.0E+001.
0.41670E-01
1NT_1-0BFL(00,03)0003D0.36577
1NT_1-0BFL(00
? gsub
>
> gsub("\\(|\\)", "", var)
You can then read.table on a textConnection.
> read.table(textConnection(gsub("\\(|\\)", "", var) ))
V1 V2
1 p1 10
2 p1 3
3 p1 4
4 p2 20
5 p2 30
6 p2 40
7 p3 4
8 p3 1
9 p1 2
On Jan 18, 2009, at 12:13 PM, oscar linares wrote:
Dear Rxperts,
I have a
Does this give you what you want:
> x <- read.table(textConnection("p(1) 10
+ p(1) 3
+ p(1) 4
+ p(2) 20
+ p(2) 30
+ p(2) 40
+ p(3) 4
+ p(3) 1
+ p(1) 2"), as.is=TRUE)
> # remove parenthesis
> x$V1 <- gsub("[()]", "", x$V1)
>
>
> x
V1 V2
1 p1 10
2 p1 3
3 p1 4
4 p2 20
5 p2 30
6 p2 40
7 p3 4
8 p3
Dear Rxperts,
I have a varaibles data file that looks like this
p(1) 10
p(1) 3
p(1) 4
p(2) 20
p(2) 30
p(2) 40
p(3) 4
p(3) 1
p(1) 2
I cannot process these data with R because it does not like the parentheses.
How can I get these to look like:
p1 10
p1 3
p1 4
p2 20
p2 30
p2 40
p3 4
p3 1
p3 2
The
Hi,
On my system (see below), it works fine (inputing the code below at
the R prompt). Make sure that the encoding of the input file is
encoded UTF-8.
Rgds,
Ivan
> sessionInfo()
R version 2.8.1 Patched (2009-01-14 r47602)
i386-apple-darwin9.6.0
locale:
en_GB.UTF-8/en_GB.UTF-8/C/C/en_GB.
Hi,
I ran into this issue previously and managed to solve it, but I've
forgotten how and am getting frustrated...
I have a data frame (see below) with scandinavian characters in R
(2.7.1) running on a Win Xp-computer. I save the data frame in an
RData-file on a usb stick, and load() it in R (2.8.0
Use read.zoo and aggregate.zoo from zoo and
months, hours and as.chron from chron.
Note that we must read in col 1 as character to ensure leading
zeros don't get dropped. There are two mph columns and it is
assumed you want both:
Lines <- "LST inch mphDeg DegF DegF%volts D
Does this do it for you:
> # quick and dirty -- remove the 'mm' from the data and then aggregate
> x$hours <- (x$LST %/% 100) * 100
> aggregate(x$mph, list(x$hours), mean)
Group.1x
1 50601 13.82500
2 506010100 17.55000
3 506010200 23.04167
4 506010300 22.0
>
>
You can also 'fi
Dear all-
I have a dataset (see a sample below - but the whole dataset is June
2005 - June 2008). The "LST" format is "YYMMDDHHmm" and I would like to
get the hourly average of the "mph" for the summer months (spanning all
years). I have been trying to use "aggregate" but am not having much
On Mon, Dec 08, 2008 at 09:34:35PM -0800, Feanor22 wrote:
>
> Hi experts of R,
>
> Are there any functions in R to test a univariate series for long memory
> effects, structural breaks and time reversability?
> I've found for ARCH effects(ArchTest), for normal (Shapiro.test,
> KS.test(comparing w
Hi experts of R,
Are there any functions in R to test a univariate series for long memory
effects, structural breaks and time reversability?
I've found for ARCH effects(ArchTest), for normal (Shapiro.test,
KS.test(comparing with randn) and lillie.test) but not for the above
mentioned.
Where can I
hi there
I have a dataframe
abc 123 234
abc 234 456
def 567 234
elm 123 456
klm 234 678
klm 465 678
I want the unique of first colum along with the values in colum 2 and 3.I By
default it will select the first element for the unique so my out put should
be
abc 123 234
def 567 234
elm 123 456
k
This was the scenario that Phoebe posted about. Her data was:
2.93290e-06 1.17772e-06 -0.645205 rs2282755
3.07521e-06 3.14000e-04 0.412997 rs1336838
4.84017e-06 2.18311e-01 0.188669 rs2660664 rs967785
9.77861e-06 7.04740e-02 0.294653 rs2660664
1.22767e-05 1.56325e-05 0.569826 rs6870519
2.
I have the same problem. If the first row has more columns than later rows,
the fill=TRUE works; however when the first row has LESS columns than later
rows, R won't read in more columns than the length of the first row. Anyone
has solutions?
Yanni
Marc Schwartz wrote:
>
> on 11/11/2008 03:39
on 11/11/2008 03:39 PM phoebe kong wrote:
> Hi all,
>
> I have problem reading in a text file as follow.
>
> The following data has not column header. I tried the following
> command but failed,
>
> temp<-read.table("data.txt",header=F)
>
> An error message stated that line 3 did not have 4 ele
Hi all,
I have problem reading in a text file as follow.
The following data has not column header. I tried the following
command but failed,
temp<-read.table("data.txt",header=F)
An error message stated that line 3 did not have 4 elements.
0.293290E-05 0.117772E-05 -0.645205 rs22
Hi Keun-Hyung,
Can you send the data (off list) to me or at least show what
str(gh1)
produces, and show us the output from
require(vegan)
sessionInfo()
Without that it is difficult to help.
G
On Tue, 2008-11-11 at 17:19 +0900, [EMAIL PROTECTED] wrote:
> Dear all,
>
> I'm using R2.8 version,
Dear all,
I'm using R2.8 version, and am trying to do NMDS and calculate other
diversity indices in vegan package.
The problem is that it works with a small set of data (43 X 23; row by
column), but the following error message comes up with a larger data set (43
X 104) (it seems not large to me a
See ?ave and ?seq_along
DF <- data.frame(Name = c("Mary", "Mary", "Mary", "Sam", "Sam",
"John", "John", "John", "John"), stringsAsFactors = FALSE)
DF$index <- ave(1:nrow(DF), DF$Name, FUN = seq_along)
On Sat, Nov 8, 2008 at 5:43 AM, jie feng <[EMAIL PROTECTED]> wrote:
> Hi, there,
>
> I have
one way is with ave(), e.g.,
dat <- data.frame(name = rep(c("Mary", "Sam", "John"), c(3,2,4)))
dat$freq <- ave(seq_along(dat$name), dat$name, FUN = seq_along)
dat
I hope it helps.
Best,
Dimitris
jie feng wrote:
Hi, there,
I have a simple data manipulation question for you. Thank you for yo
Hi, there,
I have a simple data manipulation question for you. Thank you for your help!
Suppose that I have this data about people appearing in a class
Mary
Mary
Mary
Sam
Sam
John
John
John
John
Then I want to find out what exact time(s) the student appears at the
moment such as
Mary 1
Mary
dles some special cases outside my small example dataset.
Thank you again!
Peter.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of bartjoosen
Sent: 6. november 2008 11:31
To: r-help@r-project.org
Subject: Re: [R] Data manipulation question
How about:
id
How about:
id <- c(rep("a",4),rep("b",2), rep("c",5), rep("d",1))
start <- c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
stop <- c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
data <- data.frame(id,start,stop)
f <- function(data){
m <- match(data$start,data$stop) + 1
if (length
On Thu, Nov 6, 2008 at 4:23 PM, Peter Jepsen <[EMAIL PROTECTED]> wrote:
>
> Here is an example:
>
> id <- c(rep("a",4),rep("b",2), rep("c",5), rep("d",1))
> start <- c(c(0,6,17,20),c(0,1),c(0,5,10,11,50),c(0))
> stop <- c(c(6,12,20,30),c(1,10),c(3,10,11,30,55),c(6))
> data <- as.data.frame(cbind(id
Dear R-listers,
I am a relatively inexperienced R-user currently migrating from Stata. I
am deeply frustrated by this data manipulation question: I know how I
could do it in Stata, but I cannot make it work in R.
I have a data frame of hospitalization data where each row represents an
admission.
--
> View this message in context:
> http://www.nabble.com/how-to-cite-R-data-sets-tp20325205p20325205.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat
o cite this data set?
Thanks,
~D
--
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.c
lee
Sent: Friday, September 26, 2008 3:10 PM
To: r-help@r-project.org
Subject: [R] data frame column name as a function argument
Hello,
I'd like to pass a column name as the argument for a function, but I'm
getting "NULL" as a return value. Any suggestions? Thanks.
> d <
Hello,
I'd like to pass a column name as the argument for a function, but I'm
getting "NULL" as a return value. Any suggestions? Thanks.
> d <- data.frame(cbind(x=1, y=1:10))
> d
x y
1 1 1
2 1 2
3 1 3
4 1 4
5 1 5
6 1 6
7 1 7
8 1 8
9 1 9
10 1 10
> testing <- function(var) {
Hello,
maybe it is better if you copy an extract of your dataset file in the
message because the attached file did'nt seem to get through.
Margherita
2008/9/14 Ndoh Innocent (Holy) <[EMAIL PROTECTED]>
> Greetings dear friends.
> Please, I really find problems having the program read my datasets
Greetings dear friends.
Please, I really find problems having the program read my datasets (here
attached).
Have converted datasets to csv, imported but always not reaching the target.
Would be very happy if some one out can help me on time.
Thanks
Ndoh Mbue Innocent
International corporation of
You can use a "connection" and read a portion of the data in at a time
and process it. Do you need all the data at once? If so, I would
agree that you either need more memory (and possibly a 64-bit version
of the system), or you come up with a different approach to your
processing. You have not i
e 2 million records? Chapter 4
of the R Data Import/Export Manual may help.
2. Failing that, buy more memory for your PC.
Regards,
Richie.
Mathematical Sciences Unit
HSL
ATTENTION:
This message contains privileged and
Dear All,
I have a data set containing 2,122,164 records and 38198952 fields.
I can not import this data due to "momory problem".
Is there a way to solve this problem?
Thanks
[[alternative HTML version deleted]]
__
R-help@r-project.org ma
I have a data frame containing sequences and I am interested in changing a few
sequences in a window and the swapping the original sequence back after I have
completed my analysis.
My temporary data frame that I am creating seq.in.window does not like the way
I am making me assignment. The vari
Amanda1988 <[EMAIL PROTECTED]> wrote:
>
> I was having a problem with a little simple function I wrote in R and I think
> the problem was that R is representing fractional numbers in binary floating
> point and not decimal notation, so sometimes I was having extra data points
> counted. Is there
faq 7.31
On Fri, Aug 15, 2008 at 9:16 AM, Amanda1988 <[EMAIL PROTECTED]> wrote:
>
> I was having a problem with a little simple function I wrote in R and I think
> the problem was that R is representing fractional numbers in binary floating
> point and not decimal notation, so sometimes I was havi
I was having a problem with a little simple function I wrote in R and I think
the problem was that R is representing fractional numbers in binary floating
point and not decimal notation, so sometimes I was having extra data points
counted. Is there a way to cast a number stored in a variable as an
to view matrices I am working with in a clean, easy to read,
>> separate window.
>>
>> A friend showed me how to do something like I want with edit(). I can
>> view
>> the matrix in the 'R Data Editor':
>>
>> For a sample matrix:
>>
>>
ug 1, 2008, at 7:29 PM, Rachel Schwartz wrote:
Hi,
I would like to view matrices I am working with in a clean, easy to read,
separate window.
A friend showed me how to do something like I want with edit(). I can
view
the matrix in the 'R Data Editor':
For a sample matrix:
mat=mat
working with in a clean, easy to
read,
separate window.
A friend showed me how to do something like I want with edit(). I
can view
the matrix in the 'R Data Editor':
For a sample matrix:
mat=matrix(1:15,ncol=3)
mat
[,1] [,2] [,3]
[1,]16 11
[2,]27 12
[3,]3
edit(). I can view
> the matrix in the 'R Data Editor':
>
> For a sample matrix:
>
>> mat=matrix(1:15,ncol=3)
>> mat
> [,1] [,2] [,3]
> [1,]16 11
> [2,]27 12
> [3,]38 13
> [4,]49 14
> [5,]5 10 15
&
, Aug 1, 2008 at 10:29 AM, Rachel Schwartz <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I would like to view matrices I am working with in a clean, easy to read,
> separate window.
>
> A friend showed me how to do something like I want with edit(). I can view
> the matrix in the
Rachel Schwartz wrote:
Hi,
I would like to view matrices I am working with in a clean, easy
to read,
separate window.
A friend showed me how to do something like I want with edit().
I can view
the matrix in the 'R Data Editor':
showed me how to do something like I want with edit(). I can
>> view
>> the matrix in the 'R Data Editor':
>>
>> For a sample matrix:
>>
>> mat=matrix(1:15,ncol=3)
>>> mat
>>>
>> [,1] [,2] [,3]
>> [1,]16 11
with edit(). I can view
the matrix in the 'R Data Editor':
For a sample matrix:
mat=matrix(1:15,ncol=3)
mat
[,1] [,2] [,3]
[1,]16 11
[2,]27 12
[3,]38 13
[4,]49 14
[5,]5 10 15
look=function(x) invisible(edit(x))
look(mat)
That
Hi,
I would like to view matrices I am working with in a clean, easy to read,
separate window.
A friend showed me how to do something like I want with edit(). I can view
the matrix in the 'R Data Editor':
For a sample matrix:
> mat=matrix(1:15,ncol=3)
> mat
[,1] [,2] [,3
I think the
merge()
function should be adequate for this task. Here is an example.
A <- data.frame(day=1:5, x=runif(5))
B <- data.frame(day=3:7, x=runif(5))
A
day x
1 1 0.9764534
2 2 0.9693998
3 3 0.1324933
4 4 0.8311153
5 5 0.3264465
B <- data.frame(day=3:8, x=run
Sorry, the last one should be:
ix <- match(B$DayOfYear, A$DayOfYear)
A[ix, "x"] <- A[ix, "x"] - B$x
Again we are assuming B's days are a subset of A's.
On Sat, Jul 26, 2008 at 6:08 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Here is a third solution.
>
> A[B$DayOfYear, "x"] <- A[B$DayOfY
Here is a third solution.
A[B$DayOfYear, "x"] <- A[B$DayOfYear, "x"] - B$x
Its assumes B's days are a subset of A's but if that's not the case then
you would need to intersect them first: ?intersect
On Sat, Jul 26, 2008 at 5:26 PM, <[EMAIL PROTECTED]> wrote:
> I have two vectos (list) that repr
Here is a second solution. This one uses sqldf instead of zoo:
library(zoo)
sqldf("select A.x - ifnull(B.x, 0) from A left join B using(DayOfYear)")
See
http://sqldf.googlecode.com
On Sat, Jul 26, 2008 at 5:26 PM, <[EMAIL PROTECTED]> wrote:
> I have two vectos (list) that represent a years of
For the last statement we may prefer this
so it stays a zoo object:
> m <- merge(Az, Bz, fill = 0)
> m[,1] - m[,2]
12345
14290 30490 2219
On Sat, Jul 26, 2008 at 5:53 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> Look at merge.zoo
>
>> library(zoo)
>> dput(A)
> st
Look at merge.zoo
> library(zoo)
> dput(A)
structure(list(DayOfYear = 1:5, x = c(1429L, 3952L, 3049L, 2844L,
2219L)), .Names = c("DayOfYear", "x"), class = "data.frame", row.names = c("1",
"2", "3", "4", "5"))
> B <- A[c(2,4),]
> Az <- zoo(A$x, A$DayOfYear)
> Bz <- zoo(B$x, B$DayOfYear)
> merge(Az
I have two vectos (list) that represent a years of data. Each "row" is
represented by the day of year and the quantity that was sold for that day. I
would like to form a new vector that is the difference between the two years of
data. A sample of A (and similarly B) looks like:
> A[1:5,]
Day
on 07/22/2008 11:24 AM Christian Hof wrote:
Dear all,
how can I, with R, transform a presence-only table (with the names of
the species (1st column), the lat information of the sites (2nd column)
and the lon information of the sites (3rd column)) into a
presence-absence (0/1) matrix of specie
Dear all,
how can I, with R, transform a presence-only table (with the names of
the species (1st column), the lat information of the sites (2nd column)
and the lon information of the sites (3rd column)) into a
presence-absence (0/1) matrix of species occurrences across sites, as
given in the
one. In SQL, we can just simply use the join comment.
> What should we do in R? Is there any package that allows us to run SQL
> statements with R data?
>
> Thank you in advance for your help,
> Willa
>
> This message contains confidential information and is in...{{dropped
n this two
data frames into one. In SQL, we can just simply use the join comment.
What should we do in R? Is there any package that allows us to run SQL
statements with R data?
Thank you in advance for your help,
Willa
This message contains confidential information and is i
See sqldf home page:
http://sqldf.googlecode.com
e.g.
library(sqldf)
set.seed(1)
pti <-rnorm(7,10)
fid <- rnorm(7,100)
finc <- rnorm(7,1000)
# set is a reserved word in SQL so use sset
sset <- data.frame(fid,pti,finc)
system.time(out <- sqldf("select fid, sum(pti) from sset group by
On Sat, Jul 12, 2008 at 7:44 AM, sj <[EMAIL PROTECTED]> wrote:
> Hello,
>
> I am trying to do some fairly straightforward data summarization, i.e., the
> kind you would do with a pivot table in excel or by using SQL queires. I
> have a moderately sized data set of ~70,000 records and I am trying to
-
-Ursprüngliche Nachricht-
Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im
Auftrag von Daniel Malter
Gesendet: Friday, July 11, 2008 7:53 PM
An: 'sj'; 'r-help'
Betreff: Re: [R] data summarization etc...
The problem is that you do not really have categories. Yo
Betreff: [R] data summarization etc...
Hello,
I am trying to do some fairly straightforward data summarization, i.e., the
kind you would do with a pivot table in excel or by using SQL queires. I
have a moderately sized data set of ~70,000 records and I am trying to
compute some group averages and
Hello,
Have you tried using the GUI Rattle from www.rattle.togaware.com . It
works pretty well for summarization.
Regards,
Ajay
www.decisionstats.com
On Sat, Jul 12, 2008 at 4:14 AM, sj <[EMAIL PROTECTED]> wrote:
>
> Hello,
>
> I am trying to do some fairly straightforward data summarization, i
Hello,
I am trying to do some fairly straightforward data summarization, i.e., the
kind you would do with a pivot table in excel or by using SQL queires. I
have a moderately sized data set of ~70,000 records and I am trying to
compute some group averages and sum values within groups. the code exam
Hello,
I am trying to do some fairly straightforward data summarization, i.e., the
kind you would do with a pivot table in excel or by using SQL queires. I
have a moderately sized data set of ~70,000 records and I am trying to
compute some group averages and sum values within groups. the code exam
hu, 7/3/08, arghya ganguli <[EMAIL PROTECTED]> wrote:
> From: arghya ganguli <[EMAIL PROTECTED]>
> Subject: [R] Data size
> To: r-help@r-project.org
> Received: Thursday, July 3, 2008, 7:41 AM
> Can somebody please let me know what is the maximum number
> of rows a
Can somebody please let me know what is the maximum number of rows and
columns that R can handle in a datafile?
Thanks & Regards,
Arghya Ganguli
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the postin
mat <- outer( 0:9, 0:(1024-1), function(x,y) y %/% (2^x) %% 2 )
On Thu, 26 Jun 2008, Daniel Folkinshteyn wrote:
this is probably a cludge, and there may be a "neater" way to do this, but...
here's one:
a = 0:1
for (i in 1:9){ a= merge(unname(a), 0:1) }
a = t(a)
after the for loop, 'a'
Try this also:
t(expand.grid(rep(list(0:1), 10)))
On Thu, Jun 26, 2008 at 3:18 PM, SARAH A DEPAOLI <[EMAIL PROTECTED]> wrote:
> I am looking for a way to generate a data matrix that contains all possible
> response patterns for 10 binary items. This should produce a matrix with 10
> rows (repre
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