Displaying 20 results from an estimated 8000 matches similar to: "Function to check if a vector contains a given value?"
2009 Oct 13
4
replacing period with a space
Dear R-ers!
I have x as a variable in a data frame x.
x<-data.frame(x=c("aa.bb","cc.dd.ee"))
x$x<-as.character(x$x)
x
I am sorry for such a simple question - but how can I replace all
periods in x$x with spaces?
sub('.', ' ', x$x) - removes all letters to the left of each period...
Thanks a lot for your advice!
--
Dimitri Liakhovitski
Ninah.com
2010 Mar 04
4
Analogue to SPSS regression commands ENTER and REMOVE in R?
I am not sure if this question has been asked before - but is there a
procedure in R (in lm or glm?) that is equivalent to ENTER and REMOVE
regression commands in SPSS?
Thanks a lot!
--
Dimitri Liakhovitski
Ninah.com
Dimitri.Liakhovitski at ninah.com
2010 Mar 18
1
R takes long time to open
Hello.
Until today I've been using R2.9 and since today R2.10 (on a PC).
In both of them it takes about 20 sec for the prompt to appear IN R
console after I start R. And every time it says: "Previous saved work
space restored" - even if I have not saved any workspace or, in case
of R2.10 - even though I have not used it once.
In the older versions - R would start within 2-3 sec.
Is
2009 Sep 04
2
transforming a badly organized data base into a list of data frames
Dear R-ers!
I have a badly organized data base in Excel. Once I read it into R it
looks like this (all variables become factors because of many spaces
and other characters in Excel):
2009 Sep 17
2
referring to a row number and to a row condition, and to columns simultaneously
Hello, dear R-ers!
I have a data frame:
x<-data.frame(a=c(4,2,4,1,3,4),b=c(1,3,4,1,5,0),c=c(NA,2,5,3,4,NA),d=rep(NA,6),e=rep(NA,6))
x
When x$a==1, I would like to replace NAs in columns d and e with 8 and
9, respectively
When x$a != 1, I would like to replace NAs in columns d and e 101 and
1022, respectively.
However, I only want to do it for rows 2:5 - while ignoring what's
happening in
2010 Mar 30
1
Efficiency question: replacing all NAs with a zero
Dear R'ers,
I have a very large data frame (over 4000 rows and 2,500 columns). My
task is very simple - I have to replace all NAs with a zero. My code
works fine on smaller data frames - but I have to deal with a huge one
and there are many NAs in each column.
R runs out of memory on me ("Reached total allocation of 1535Mb: see
help(memory.size)"). Is there any other, more efficient
2010 Mar 25
1
Precision level
Hello!
I am wondering at what point does R consider a numeric value to be
equal to zero - for statements of the type x==0 and x %in% 0.
Thank you very much!
--
Dimitri Liakhovitski
Ninah.com
Dimitri.Liakhovitski at ninah.com
2009 Sep 23
0
R: Function to check if a vector contains a given value?
?any
any(x==2)
Stefano
-----Messaggio originale-----
Da: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org]Per conto di Dimitri Liakhovitski
Inviato: mercoled? 23 settembre 2009 17.38
A: R-Help List
Oggetto: [R] Function to check if a vector contains a given value?
Dear R'rs,
is there a function that checks if a given vector contains a certain value.
E.g.,
2010 Mar 30
4
Code is too slow: mean-centering variables in a data frame by subgroup
Dear R-ers,
I have a large data frame (several thousands of rows and about 2.5
thousand columns). One variable ("group") is a grouping variable with
over 30 levels. And I have a lot of NAs.
For each variable, I need to divide each value by variable mean - by
subgroup. I have the code but it's way too slow - takes me about 1.5
hours.
Below is a data example and my code that is too
2010 Mar 09
2
looping through predictors
Dear R-ers,
I have a data frame data with predictors x1 through x5 and the
response variable y.
I am running a simple regression:
reg<-lm(y~x1, data=data)
I would like to loop through all predictors. Something like:
predictors<-c("x1","x2",... "x10)
for(i in predictors){
reg<-lm(y~i)
etc.
}
But it's not working. I am getting an error:
Error in
2009 Sep 22
1
any way to make it work faster (deleting rows that contain certain values)
Hello, dear R'ers,
index<-expand.grid(1:7,1:4,1:4,1:4,1:4,1:4,1:4,1:4,1:4,1:4,1:4)
In this case, dim(index) is 7,340,032 (!) and 11.
I realize it's huge.
Then, I am trying to get rid of the undesired combinations of columns.
They should not contain identical values in any 2 columns.
Also if column 1 has a value of 5, there should be no 2 in any other column,
if column 1 has a value
2010 Oct 01
3
Suppressing printing in the function
Hello!
I wrote a function that returns a data frame. Nowhere in the function
do I say print(my.data.frame), but when I run the function - the data
frame is printed on the console.
Is there any way to suppress it?
Thank you!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
2010 Jan 20
5
standardizing one variable by dividing each value by the mean - but within levels of a factor
Hello!
I have a data frame with a factor and a numeric variable:
x<-data.frame(factor=c("b","b","d","d","e","e"),values=c(1,2,10,20,100,200))
For each level of "factor" - I would like to divide each value of
"values" by the mean of "values" that corresponds to the level of
"factor"
In other
2010 Jun 29
2
transposing a data frame from horizontal to vertical (stacking)
Hello, everyone!
I have a very simple task - I have a data frame (see MyData below) and
I need to stack the data (see result below).
I wrote the syntax below - it's very basic and it does what I need.
But I am sure what I am trying to do is a very typical task and there
must be a much shorter/more elegant way of doing it.
Any advice?
Thank you very much!
2010 May 08
1
Increasing the font size on axes in trellis
Hello,
the code below gives me the picture I need - but there is on small
thing I can't figure out.
The plot has very small tick mark labels for both axes. I don't mean
the axis labels - they are both good, but what is shown near the tick
marks.
Please help me figure out what parameter I should add to make those
larger. I tried sticking cex.lab=1.3 in different places but it didn't
2010 Oct 25
3
finding the year of a date
I know that I can use as.yearmon in the package "zoo" to find the year
and the month of a date.
I can use as. yearqtr to find the year and the quarter.
But how can one find just the year of a date?
Thanks a lot!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
2011 Feb 25
6
preventing repeat in "paste"
Hello!
s<-"start"; e<-"end"
middle<-as.character(c(1,2,3))
I would like to get the following result:
"start 123 end" or "start 1 2 3 end" or "start 1,2,3 end"
How can I avoide this (undesired) result:
paste(s,middle,e,sep=" ")
Thank you!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
2010 Aug 13
3
transforming dates into years
Hello!
If I have in my data frame MyFrame a variable saved as a Date and want
to translate it into years, I currently do it like this using "zoo":
library(zoo)
as.year <- function(x) as.numeric(floor(as.yearmon(x)))
myFrame$year<-as.year(myFrame$date)
Is there a function that would do it directly - like "as.yearmon" -
but for years?
Thank you!
--
Dimitri
2011 Feb 07
1
Question about checkTmvArgs function in rtmvnorm (package tmvtnorm)
Hello!
I was wondering if it's possible to see the actual code of
checkTmvArgs function that is part of the code for rtmvnorm (which is
below - I just typed "rtmvnorm" on the prompt). I get an error:
Error in checkTmvArgs(mean, sigma, lower, upper) :
sigma must be a symmetric matrix
At the same time I am pretty sure that the matrix I am passing as
sigma is a var-covar matrix
2010 May 05
1
What is the default nPerm for regression in randomForest?
Could not find it in ?randomForest.
Thank you for your help!
--
Dimitri Liakhovitski
Ninah.com
Dimitri.Liakhovitski at ninah.com