Displaying 20 results from an estimated 1000 matches similar to: "Finding a vector position using names rather than values"
2010 Jul 09
2
select columns from vector of column names
Hi
I want to extract columns from a data frame using a vector with the desired
column names.
This short example uses the select argument in the subset function to
accomplish what I am trying to do. Is there a better solution?
#names of desired columns
colnames <- c("col1","col3")
#my data
data <-
2010 Jul 09
1
select columns from dataframe
Hi,
I would like to extract columns from a dataframe using a vector of desired
column names.
The following working example uses the select argument in the subset
function to accomplish what I am trying to do. Is there a better solution?
Thanks.
#my data
data <- data.frame("col1"=c(1,2,3),"col2"=c("A","B","C"),"col3"=c(4,5,6))
2008 Nov 25
1
Efficient passing through big data.frame and modifying select
> -----Original Message-----
> From: William Dunlap
> Sent: Tuesday, November 25, 2008 9:16 AM
> To: 'johannes_graumann at web.de'
> Subject: Re: [R] Efficient passing through big data.frame and
> modifying select fields
>
> > Johannes Graumann johannes_graumann at web.de
> > Tue Nov 25 15:16:01 CET 2008
> >
> > Hi all,
> >
> >
2008 Dec 23
1
quotation problem/dataframe names as function input argument.
Dear R friends:
Can someone help me with the following problem? Many thanks in advance.
# Problem Description:
# I want to write functions which take a (character) vector of dataframe
names as input argument.
# For example, I want to extract the number of observations from a number of
dataframes.
# I tried the following:
nobs.fun <- function (dframe.vec)
{
nobs.vec <-
2012 Mar 30
1
How to use access results of gregexpr in data frames
Hello,
I'm trying to figure out how to find the index of the second occurrence of "/" in a string (which happens to represent a date) within a data frame column.
I've used the following code successfully to find the first instance of "/".
dframe <- data.frame(date=c("5/14/2011", "4/7/2011"))
dframe$x1 <- regexpr("/", dframe[, 1])
2011 Jun 09
3
How to subset based on column name that is a number ?
Hi,
I have a data frame with column names "1", "2", "3", ... and I'd like to extract
a subset based on the values in the first column. None of the methods I tried
worked (below).
x <- subset(dframe, 1 = = "My Text")
x <- subset(dframe, "1" = = "My Text")
x <- subset(dframe, names(dframe)[1] = = "My Text")
Q
2005 Nov 09
2
error in NORM lib
Dear alltogether,
I experience very strange behavior of imputation of NA's with the NORM
library. I use R 2.2.0, win32.
The code is below and the same dataset was also tried with MICE and
aregImpute() from HMISC _without_ any problem.
The problem is as follows:
(1) using the whole dataset results in very strange imputations - values
far beyond the maximum of the respective column, >
2010 Oct 13
4
Change global env variables from within a function
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column names in a data frame, I ran into the following problem: When I
rename the columns within my function, I can't seem to get it to
2006 Jan 20
3
Selecting data frame components by name - do you know a shorter way?
Hi! I suspect there must be an easy way to access components of a data frame by name, i.e. the input should look like "name1 name2 name3 ..." and the output be a data frame of those components with the corresponding names. I ´ve been trying for hours, but only found the long way to do it (which is not feasible, since I have lots of components to select):
2011 Jul 08
1
Referencing a vector of data labels in ggplot function
Hi,
I really feel I've looked everywhere, although I know this can't be a hard
problem. I'd like to be able to call the graph below as a function, but I
can't get the function to recognize variables beyond 'dframe'. I've read
through many papers on writing functions in R, but I can't get this to work.
data <- data.frame('date' = as.Date(rep(c(15101,
1999 Oct 19
2
Summary bug?
Hi,
It seems that there's a bug in summary, in the max. output... but max() alone
works fine.
> hw04.dframe$area
...
[41] 1790 1380 1296 2745 798 2306 438649 1481 1559 2450
...
> summary(hw04.dframe)
area
Min. : 798
1st Qu.: 1349
Median : 1690
Mean : 6962
3rd Qu.: 2306
Max. :438600 ### should read 438649
or, to the point,
2006 Jun 09
3
sqlSave() and rownames=TRUE makes my Rgui crash
Hello,
I created a table in MySQL with this command
CREATE TABLE example (pk INT NOT NULL AUTO_INCREMENT,PRIMARY KEY(pk),
id VARCHAR(30),col1 VARCHAR(30),col2 VARCHAR(30))
### In R, I can connect to this table:
library(DBI)
library(RODBC)
chan <- odbcConnect("MySQL51", uid="root", pwd="xxx")
first <- sqlQuery(chan, "select * from example")
2000 Jun 25
1
renaming columns
I frequently get data sets with cryptically-named variables. The datasets
are more useful to me with informative variable names. I know that I can
rename variables using the following command:
dimname(dataset[[2]][index.of.variable.to.be.renamed]<-new.variable.name
If I want to do this inside a function (say something I call RenameCol)
what is the best way to communicate the
2012 Aug 24
1
help with a special variant of balloonplot
Hi,
I am interested in implementing a special variant of
balloonplot. Let me
explain with an example dataset from the reference manual :
library(gplots)
data(Titanic)
dframe<-as.data.frame(Titanic)
survived<-dframe[dframe$Survived=="Yes",]
attach(survived)
balloonplot(x=Class,y=list(Age,Sex),z=Freq,sort=TRUE,show.zeros=TRUE,cum.margins=FALSE,
main="BalloonPlot :
2011 Jul 22
2
averaging rows based on string¿?
Hi Folks,
Ran into something I'd really like to do in R simply/elegantly, but my R -
coding skills seem surpassed. This is the thing. Imagine the following data:
labs<-c("abcdef","abcgg","tgthefdk","tgtijuel","tgtnjmoi","gbnt","dlift")
dat<-c(0.5,0.25,1,2,16,0.250,4)
dframe<-data.frame(labs,dat)
I would like to
2007 Oct 13
1
geom_errorbar with ggplot2
Dear All,
I am trying to build an error bar plot with ggplot2. However, even if
the code seems to work, the plot is rather odd, with what seems to be
a continuous line linking the different bars.
I attach below the code I used, as well as the data.
================================================================
DFrame<-data.frame(x=1:10,y=0.5*(1:10),Upper=(1:10)+0.5, Lower=(1:10)-0.5)
p
2012 Nov 15
1
Can't see what i did wrong..
with
pred.pca<-predict(splits[[i]]$pca,trainingData at samples)[,1:nPCs]
dframe<-as.data.frame(cbind(pred.pca,class=isExplosive(trainingData,2)));
results[[i]]$classifier<-ksvm(class~.,data=dframe,scaled=T,kernel="polydot",type="C-svc",
C=C,kpar=list(degree=degree,scale=scale,offset=offset),prob.model=T)
and a degree of 5 i get an error of 0 reported by the ksvm
2000 Apr 03
1
cbind.ts(),ts.union() (PR#508)
Full_Name: Wolfgang Koller
Version: 1.0.0
OS: linux
Submission from: (NULL) (137.208.7.48)
> data(UKLungDeaths)
> ts.union(mdeaths, fdeaths,dframe=TRUE)
Error in names<-.default(*tmp*, value = nmsers) :
names attribute must be the same length as the vector
> cbind(mdeaths, fdeaths,dframe=TRUE)
Error in names<-.default(*tmp*, value = nmsers) :
names attribute
2010 Feb 05
1
Finding ranges on a per-row basis from several objects
Dear all,
I'm attemping to find the overall range of values from column 5 of a series of data frames, on a per-row basis, and assign the results to a new object. At present, I'm only able to receive the overall range of all values, whereas I'm intending to get the results of the range for each corresponding row. This is what I have so far:
for (i in seq(nrow(dframe[1])))
???
2016 Feb 09
2
problem plotting "ts" in a data.frame
Hello:
I'm having trouble plotting an object of class "ts" that is in a
data.frame. I can do it with(data.frame, plot(...)) but not with
plot(..., data.frame); see the example below.
This work around gets me past this problem. However, I thought
the R Core team might want to know about this if they don't already.
Thanks for all your work in making