Displaying 20 results from an estimated 2000 matches similar to: "phantom NA/NaN/Inf in foreign function call (or something altogether different?)"
2012 Dec 04
2
computing marginal values based on multiple columns?
Hello all,
I have what feels like a simple problem, but I can't find an simple
answer. Consider this data frame:
> x <- data.frame(sample1=c(35,176,182,193,124),
sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156),
class=c('a','a','c','b','c'))
> x
sample1 sample2 sample3 class
1 35 198 12 a
2 176 176
2011 Aug 14
2
conditional filter resulting in 2 new dataframes
This is what I am starting with:
initial<- matrix(c(1,5,4,8,4,4,8,6,4,2,7,5,4,5,3,2,4,6), nrow=6,
ncol=3,dimnames=list(c("1900","1901","1902","1903","1904","1905"),
c("sample1","sample2","sample3")))
And I need to apply a filter (in this case, any value <5) to give me one
dataframe with only the
2010 Sep 07
1
average columns of data frame corresponding to replicates
Hi Group,
I have a data frame below. Within this data frame there are samples
(columns) that are measured more than once. Samples are indicated by
"idx". So "id1" is present in columns 1, 3, and 5. Not every id is
repeated. I would like to create a new data frame so that the repeated
ids are averaged. For example, in the new data frame, columns 1, 3,
and 5 of the original
2010 Nov 04
2
Converting Strings to Variable names
Hi all,
I am processing 24 samples data and combine them in single table called
CombinedSamples using following:
CombinedSamples<-rbind(Sample1,Sample2,Sample3)
Now variables Sample1, Sample2 and Sample3 have many different columns.
To make it more flexible for other samples I'm replacing above code with a
for loop:
#Sample is a string vector containing all 24 sample names
for (k in
2008 May 21
2
Problem in converting natural numbers to bits and others
Hi,
I just started using R for about one week and I have few problems.
i)I have a problem in finding right function to convert a table of natural
numbers to bitwise. For a simple example;
I have the below table:-
Column Col1 Col2 Col3
Sample1 5 7 10
Sample2 0 2 1
Sample3 4 0 0
Supposedly i wanted to convert to :-
Column Col1 Col2 Col3
2006 Jun 29
1
kmeans clustering
Hello R list members,
I'm a bio informatics student from the Leiden university
(netherlands). We were asked to make a program with different
clustering methods. The problem we are experiencing is the following.
we have a matrix with data like the following
research1 research2 research3 enz
sample1 0.5 0.2 0.4
sample2 0.4
2012 Apr 18
3
normal distribution assumption for multi-level modelling
Hello,
I'm analysing reaction time data from a linguistic experiment (a variant of
a lexical decision task). To ascertain that the data was normally
distributed, I used *shapiro.test *for each participant (see commands
below), but only one out of 21 returns a p value above p.0 05.
> f = function(dfr) return(shapiro.test(dfr$Target.RTinv)$p.value)
> p = as.vector(by(newdat,
2010 Jun 13
1
using latticeExtra plotting confidence intervals
I am wanting to plot a 95% confidence band using segplot, yet I am wanting
to have groups. For example if I have males and females, and then I have
them in different races, I want the racial groups in different panels. I
have this minor code, completely made up but gets at what I am wanting, 4
random samples and 4 samples of confidence, I know how to get A & B into one
panel and C&D in to
2010 May 21
2
Data reconstruction following PCA using Eigen function
Hi all,
As a molecular biologist by training, I'm fairly new to R (and statistics!),
and was hoping for some advice. First of all, I'd like to apologise if my
question is more methodological rather than relating to a specific R
function. I've done my best to search both in the forum and elsewhere but
can't seem to find an answer which works in practice.
I am carrying out
2011 Jan 20
1
Problems with ecodist
Dear Dr.Goslee and anyone may intrested in matrix manipulate,
I am using your ecodist to do mantel and partial mantel test, I have
locality data and shape variation data, and the two distance matrixs are
given as belowings. When I run the analysis, it is always report that the
matrix is not square, but I didn't know what's wrong with my data. Would you
please help me on this. I am quite
2011 Sep 26
1
How to Store the executed values in a dataframe & rle function
Hi group,
This is how my test file looks like:
Chr start end sample1 sample2
chr2 9896633 9896683 0 0
chr2 9896639 9896690 0 0
chr2 14314039 14314098 0 -0.35
chr2 14404467 14404502 0 -0.35
chr2 14421718 14421777 -0.43 -0.35
chr2 16031710 16031769 -0.43 -0.35
chr2 16036178 16036237 -0.43 -0.35
chr2 16048665 16048724 -0.43 -0.35
chr2 37491676 37491735 0 0
chr2 37702947 37703009 0 0
2009 Sep 20
4
correlation help
Dear group,
I have a matrix like the following:
Name Sample1 sample2 sample3 sample4 ..... sample(n)
nm1 10.5 13.5 30 31
nm2 8 11 34 29
nm3 9 10.3 27.8 35
nm(j)
I want to be able to calculate correlation between all pairs of names.
For example (nm1,nm2),
2009 Feb 02
1
A question regarding bootstrap
Dear List Members,
I have two small samples (n=20), the distributions are highly skewed. Does
it make any sense to do a boostrap test to check for difference in means?
And if so, could this be done like this:
x <- numeric(10000)
for(i in 1:10000) {
x[i] <- mean(sample(sample1,replace=TRUE)) -
mean(sample(sample2,replace=TRUE))
}
(mean(sample1)-mean(sample2))/sd(x)
Regards,
Erika
2010 Feb 25
1
taking the median across similar data
Dear All,
I am analyzing the miRNA data set in which I have 817 unique probes for each
they have 20 features each . I have to group the similar features and take
the median across them so that I have a data with no repeats to perform
invariant analysis .
My data looks something similar format
probename sample1 sample2 sample3
A 2.3 2.4 2.5
A
2012 Mar 29
1
Random sample from a data frame where ID column values don't match the values in an ID column in a second data frame
Hello,
Let's say I've drawn a random sample (sample1.df) from a large data frame
(main.df), and I want to create a second random sample (sample2.df) where
the values in its ID column *are not* in the equivalent ID column in the
first sample (sample1.df). How would I go about doing this?
In other words:
The values in sample2.df$ID *are not found* in sample1.df$ID, and both
samples are
2007 May 10
1
how to pass "arguments" to a function within a function?
I have searched the r-help files but have not been able to find an answer to this question. I apologize if this questions has been asked previously.
(Please excuse the ludicrousness of this example, as I have simplified my task for the purposes of this help inquiry. Please trust me that something like this will in fact be useful what I am trying to accomplish. I am using R 2.4.1 in Windows XP.)
2008 May 14
3
[LLVMdev] Help needed after hiatus
Hi,
I've restarted my Elsa/LLVM project after three months of having real
life intrude. I upgraded my LLVM source to the current trunk. I had to
make a few changes to my source, e.g. LLVMFoldingBuilder became
IRBuilder and several instances of "new" became "Create".
Now, a test case that previously succeeded fails. I run the following
script:
#!/bin/sh
if [ 1 -ne 0 ]
2008 Feb 07
1
How to split a factor (unique identifier) into several others?
Hello,
I have a data frame with a factor column, which uniquely identifies
the observations in the data frame and it looks like this:
sample1_condition1_place1
sample2_condition1_place1
sample3_condition1_place1
.
.
.
sample3_condition3_place3
I want to turn it into three separate factor columns "sample",
"condition" and "place".
This is what I did so far:
#
2011 Sep 28
0
Rle function to expand for many samples
Dear R experts,
code:
>m<-read.table("test.txt",sep='\t',header=TRUE,colClasses=c('character','integer','integer','rep('numeric',150))
> s<-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]]))
> names(s)=c("Values","Probes")
2011 Mar 26
2
simple if question
Hi everyone,
I have just got different samples from a dataframe (independent and
exclusive, there aren't common elements among them). I want to create a
variable that indicate the sampling selection of the elements in the
original dataframe (for example, 0 = no selected, 1= sample 1, 2=sample
2, etc.).
I have tried to do it with ifelse command, but the problem is that the
second line