[Genabel-commits] r1138 - tutorials/GenABEL_general
noreply at r-forge.r-project.org
noreply at r-forge.r-project.org
Wed Mar 13 20:56:55 CET 2013
Author: lckarssen
Date: 2013-03-13 20:56:54 +0100 (Wed, 13 Mar 2013)
New Revision: 1138
Modified:
tutorials/GenABEL_general/ImputedDataAnalysis.Rnw
Log:
- A few small spelling fixes
- Workaround for ProbABEL bug #2575 (weird output file names)
Modified: tutorials/GenABEL_general/ImputedDataAnalysis.Rnw
===================================================================
--- tutorials/GenABEL_general/ImputedDataAnalysis.Rnw 2013-03-13 17:35:20 UTC (rev 1137)
+++ tutorials/GenABEL_general/ImputedDataAnalysis.Rnw 2013-03-13 19:56:54 UTC (rev 1138)
@@ -33,9 +33,9 @@
Let us start with analysis of directly typed SNPs. For that,
we will use \texttt{mlreg} function of \GA{}. This function
-implements ML-regression and Wald test of siginificance\footnote{
-In general the score test may be preferred becuse it is faster and
-more robust
+implements ML-regression and Wald test of significance\footnote{
+In general the score test may be preferred because it is faster and
+more robust.
}.
This will later on allow us direct comparison with the results of \PA{},
which implements the same testing procedure.
@@ -79,19 +79,19 @@
@
\begin{Exercise}
-It is known that rare variation in presence of
+It is known that rare variation in the presence of
outliers can generate spurious associations.
-Do you believe this is true association in this
+Do you believe this is a true association in this
particular case? What you can do to check whether
-this is true association or not?
+this is a true association or not?
\end{Exercise}
\begin{Answer}
Firstly, you can check (by producing a cross-plot
-of genotype vs. phenotype) if association is indeed
+of genotype vs.~phenotype) if association is indeed
due to extreme phenotypic outliers. A related question is
whether the distribution is skewed. Additionally,
a permutation-based test can help establishing
-correct p-value, taking into account the nature
+correct $p$-value, taking into account the nature
of the data in question.
However, to give an ultimate answer, a replication
@@ -102,7 +102,7 @@
\section{Analysis of imputed data with \PA{}}
-Here, you will analyse imputed data. In \texttt{RData}
+Here, you will analyse imputed data. In the \texttt{RData}
directory, you will find the necessary files:
\texttt{mach1.mldose.fvi} and \texttt{mach1.mldose.fvd}
(these files represent \texttt{mldose} data produced by
@@ -113,10 +113,10 @@
We will start with producing a phenotypic data file for the
use with \PA{}:
<<>>=
-write.table(data.frame(id=idnames(df500),rcT=rcT),
- file="rcT.PHE",quote=F,row.names=F)
+write.table(data.frame(id=idnames(df500), rcT=rcT),
+ file="rcT.PHE", quote=FALSE, row.names=FALSE)
@
-next, try command '\texttt{system("head rcT.PHE")}' to check few first
+next, try the command '\texttt{system("head rcT.PHE")}' to check the few first
lines of the file.
At this moment, leave \texttt{R} (or, rather, start new console!),
@@ -129,22 +129,22 @@
yourname at server> palinear --pheno rcT.PHE --info mach1.out.mlinfo /
--dose mach1.mldose.fvi
\end{verbatim}
-Do not forget to check that you start analysis in right directory, i.e.
+Do not forget to check that you start the analysis in right directory, i.e.
all files (\texttt{rcT.PHE}, \texttt{mach1.out.mlinfo},
\texttt{mach1.mldose.fvi}, \texttt{mach1.mldose.fvd})
-are present in the working directory (use command '\texttt{ls}'
+are present in the working directory (use the '\texttt{ls}' command
from the console).
<<echo=false,hide=true>>=
system("cp RData/mach1* .")
system("cp RData/rcT.PHE .")
-palCmd <- "palinear --pheno rcT.PHE --info mach1.out.mlinfo --dose mach1.mldose.fvi"
+palCmd <- "palinear --pheno rcT.PHE --info mach1.out.mlinfo --dose mach1.mldose.fvi -o regression"
system(palCmd)
@
Now, you can return to \texttt{R} and load the analysis results:
<<>>=
-qtsPal <- read.table("regression_add.out.txt",head=T,strings=F)
+qtsPal <- read.table("regression_add.out.txt", head=T, strings=F)
qtsPal[1:5,]
@
As you see, there is not $P$-value produced in the \PA{} output, and
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