Displaying 20 results from an estimated 5000 matches similar to: "Using apply for two datasets"
2011 Feb 25
1
speed up process
Dear users,
I have a double for loop that does exactly what I want, but is quite
slow. It is not so much with this simplified example, but IRL it is slow.
Can anyone help me improve it?
The data and code for foo_reg() are available at the end of the email; I
preferred going directly into the problematic part.
Here is the code (I tried to simplify it but I cannot do it too much or
else it
2006 Feb 20
3
Boxplot Help for Neophyte
R helpers
I am getting to grips with R but came across a small problem today that I
could not fix by myself.
I have 3 text files, each with a single column of data. I read them in
using:
myData1<-scan("C:/Program Files/R/myData1.txt")
myData2<-scan("C:/Program Files/R/myData2.txt")
myData3<-scan("C:/Program Files/R/myData3.txt")
I wanted to produce a
2012 Jul 03
1
insert missing dates
Hello
I have dataframes.
mydata1 <-data.frame(value=c(15,20,25,30,45,50),dates=c("2005-05-25 07:00:00
","2005-05-25 19:00:00","2005-06-25 07:00:00","2005-06-25 19:00:00
","2005-07-25 07:00:00","2005-8-25 19:00:00"))
or
mydata2 <-data.frame(value=c(15,20,25,30,45,50),dates=c("2005-05-25 00:00:00
","2005-05-25
2008 Apr 15
2
How can I import user-defined missings from Spss?
Hi,
It works for me to import spss datasets via library(foreign) with read.spss or via library Hmisc by (spss.get).
But no matter which way I do import the data, user-defined missings from Spss are always lost.
(it makes no difference if there are a single value, a range, or any combination of them. They are always ignored).
Is there any way in R to find out if any value was user-defined missing
2010 Nov 11
2
Kolmogorov Smirnov Test
I'm using ks.test (mydata, dnorm) on my data. I know some of my
different variable samples (mydata1, mydata2, etc) must be normally
distributed but the p value is always < 2.0^-16 (the 2.0 can change
but not the exponent).
I want to test mydata against a normal distribution. What could I be
doing wrong?
I tried instead using rnorm to create a normal distribution: y = rnorm
2011 Feb 28
0
Fwd: Re: speed up process
Dear Jim,
Here is again exactly what I did and with the output of Rprof (with this
reduced dataset and with a simpler function, it is here much faster than
in real life).
Thanks you again for your help!
## CODE ##
mydata1<- structure(list(species = structure(1:8, .Label =
c("alsen","gogor", "loalb", "mafas", "pacyn", "patro",
2010 Jan 22
1
confidence intervals for mean (GLM)
Dear useRs,
How could I obtain the confidence intervals for the means of my treatments, when my data was fitted to a GLM?
I need the CI's for the Poisson and Negative Binomial distributions.
Here's what I have:
mydata1 <- data.frame('treatments'=gl(4,20), 'value'=rpois(80, 1))
model1 <- glm(value ~ treatments, data=mydata1, family=poisson)
means1 <-
2007 Mar 08
1
R: Searching and deleting elements of list
you could try mapply
mydata2<-mapply("[", mydata, lapply(mydata, function(x) !x %in% A))
mydata2[[1]]<-A #to replace the obviously deleted elements of "A"
mydata2
mydata2[[1]]
mydata2[[2]]
mydata2[[3]]
mydata2[[4]]
Stefano
-----Messaggio originale-----
Da: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch]Per conto di jastar
2017 Jul 06
3
Efficient swapping
Thanks a lot, Ista! I really appreciate it.
How about a slightly different case as the following:
set.seed(1)
(tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace =
TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE)))
x R1 R2
1 C B
2 B B
3 C E
4 E C
5 E B
6 D E
7 E E
8 D F
9 C D
10 A E
Notice that the factor levels between
2006 May 16
2
retaining character matrices when combining into data frames
Hello everyone,
If I want to convert or combine a (large) character matrix into a data
frame without having any of its columns convert into a factor class,
is there a simple solution? I() says it will operate on 'an
object' but it seems that unless the object is a vector, the results
are not what I expect.
For instance, if g is a 2x2 character matrix, as.data.frame(I(g)) will
return an
2011 Mar 10
3
A question about data frame
A very simple question. With a data frame like this:
> n = c(2, 3, 5)
> s = c("aa", "bb", "cc")
> df = data.frame(n, s)
I want df$s[1] or df[1,2], but how can I get rid of the extra line in
the output about the factor levels:
> df$s[1]
[1] aa
Levels: aa bb cc
Thanks,
Gang
2012 Feb 07
6
Setting up infile for R CMD BATCH
Suppose I create an R program called myTest.R with only one line like
the following:
type <- as.integer(readline("input type (1: type1; 2: type2)? "))
Then I'd like to run myTest.R in batch mode by constructing an input
file called answers.R with the following:
source("myTest.R")
1
When I ran the following at the terminal:
R CMD BATCH answer.R output.Rout
it failed
2013 May 02
0
Data in packages: save or write.table?
Hi all,
I am trying to understand Writing R Extension...
Section 1.1.5, data: I include two datasets in a package, one using 'save',
the other using 'write.table':
--- 8< ----
myData1 <- data.frame(x=1:10)
write.table(myData1,file="myData1.txt")
myData2 <- data.frame(x=2:10)
save(myData2,file="myData2.Rdata")
--- 8< ----
Then R CMD check aks me to
2010 Mar 27
3
Calculate variance/covariance with complex numbers
Anybody knows what functions can be used to calculate
variance/covariance with complex numbers? var and cov don't seem to
work:
> a
1
V1 0.00810014+0.00169366i
V2 0.00813054+0.00158251i
V3 0.00805489+0.00163295i
V4 0.00809141+0.00159533i
V5 0.00813976+0.00161850i
> var(a)
1
1 1.141556e-09
Warning message:
In var(a) : imaginary parts discarded in
2013 Dec 14
2
Change factor levels
Suppose I have a dataframe 'd' defined as
L3 <- LETTERS[1:3]
d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace
= TRUE))
(d <- d0[d0$fac %in% c('A', 'B'),])
x y fac
2 1 2 B
3 1 3 A
4 1 4 A
5 1 5 A
6 1 6 B
8 1 8 A
Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the
birthmark
2012 May 15
4
reading data into R
Hi I am really new using R, so this is really a beginner stuff! I
created a very small data set on excel and then converted it to .csv
file. I am able to open the data on R using the command "read.table
("mydata1.csv", sep=",", header=T)" and it just works fine. But when I
want to work on the data (e.g. calculate the mean of variable "X") R
says
2011 Sep 07
1
randomForest memory footprint
Hello, I am attempting to train a random forest model using the
randomForest package on 500,000 rows and 8 columns (7 predictors, 1
response). The data set is the first block of data from the UCI
Machine Learning Repo dataset "Record Linkage Comparison Patterns"
with the slight modification that I dropped two columns with lots of
NA's and I used knn imputation to fill in other gaps.
2012 Oct 20
2
Help with programming a tricky algorithm
Hi All,
I'm a little stumped by the following problem. I've got a dataset with
the following structure:
idxy ix iy country (other variables)
1 1 1 c1 x1
2 1 2 c1 x2
3 1 3 c1 x3
. . . . .
3739 55 67 c7 x3739
3740 55 68 c7 x3740
where ix and
2017 Jul 06
0
Efficient swapping
Untested, but I expect that setting the levels to be the same across the
two factors
levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6]
and proceeding as before should be fine.
Best,
Ista
On Jul 6, 2017 6:54 PM, "Gang Chen" <gangchen6 at gmail.com> wrote:
Thanks a lot, Ista! I really appreciate it.
How about a slightly different case as the following:
set.seed(1)
(tmp
2010 Feb 05
3
metafor package: effect sizes are not fully independent
In a classical meta analysis model y_i = X_i * beta_i + e_i, data
{y_i} are assumed to be independent effect sizes. However, I'm
encountering the following two scenarios:
(1) Each source has multiple effect sizes, thus {y_i} are not fully
independent with each other.
(2) Each source has multiple effect sizes, each of the effect size
from a source can be categorized as one of a factor levels