[R] keeping seperate row.names

2008-07-17 Thread Patrick Richardson

I have microarray data with gene names in the first column, gene id in the
second and the expression data in the remaining columns.  When trying use
read.table I get the error, . . . more columns than column names.  Is
there any way to keep both columns of names without having to discard one.
or the other?  The raw data is read from a text file and is in the form
mitogen-activated protein kinase 3  1000_at 946 1928.8  1504.9  722.5   
873.9
836.9   1294.3  631.1   606 1126.6  841.2   833.6   689.6   1256.9  685.8   
755.3   974.8
where, mitogen-activated protein kinase 3 is the first column of this (tab
delimited) text file and 1000_at is the second column.  I want to keep both
columns as labels.  Is there a way to do that?

I've used: 
test - read.table(path, header=T, sep=\t, row.names=1)
and get the error, more columns than column names.

Many Thanks,

Patrick
Van Andel Institute
Grand Rapids, MI
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[R] Extracting row.names

2008-06-23 Thread Patrick Richardson

List,

I'm trying to extract the row names of a table I have read into R.

 data - read.table(mesodata.txt, header=TRUE, row.names=1)

When I try to extract them using,

 names - data$row.names

I get,

 names
NULL

I've tried changing to a matrix, data frame, etc. and still get NULL. 
I've checked ?row.names as well as help on extracting part of an object,
etc. and unless I'm missing something obvious (likely), I can't figure out
how to extract them.

BACKGROUND
I want to extract the row names (which are basically gene id's) as well as
some other columns (which I can do successfully) and cbind them into
another data frame (which I can also do successfully). I just can't get the
row names extracted (Assigned) to an seperate object.

Any suggestions would be appreciated.

Many Thanks,

Patrick



 sessionInfo()
R version 2.7.0 (2008-04-22) 
i386-pc-mingw32 

locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252

attached base packages:
[1] stats graphics  grDevices utils datasets  methods   base 

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[R] ML Estimation Differences with R and SAS

2008-03-10 Thread Patrick Richardson
List,

I'm working on fitting a logistic model for a well known dataset (which is
given below in case anyone wants to try to reproduce).  I used both R and
SAS to fit the model and have some differences in the parameter estimates.
I'm wondering if R calculates the ML estimates differently.  I'm making NO
accusations as to which program is right or wrong.  That is not the focus
of this posting.  As a newer R user I'm trying to understand the algorithm
that R might use to calculate ML estimation.  The largest difference seems
to with the race factors.  R gives a p-value of 0.46995 for race=black and
SAS gives a p-value of 0.0753 for race=black.  Clearly one is borderline
significant and the other is not.  Many thanks to all who might be able to
offer any insight on this.  Both R and SAS code and output are included in
this message (along with the dataset).

Thanks,

Patrick


MY R CODE IS:

Dataset - read.table(path, header=TRUE, sep=, na.strings=NA,
dec=., strip.white=TRUE)
Dataset$race - factor(Dataset$race, levels=c('other','black','white'))
GLM.1 - glm(low  ~ lwt  + ptl  + ht  + race  + smoke ,
family=binomial(logit), data=Dataset)
summary(GLM.1)

MY SAS CODE IS:

PROC LOGISTIC descending DATA=p2;
class race (ref='other');
MODEL LOW = lwt ptl ht race smoke / lackfit parmlabel expb link=logit;
RUN;

MY R OUTPUT IS:

Coefficients:
  Estimate Std. Error z value Pr(|z|)   
(Intercept)0.926190.85549   1.083  0.27897   
lwt   -0.016500.00692  -2.384  0.01712 * 
ptl1.231160.44607   2.760  0.00578 **
ht 1.761970.70748   2.490  0.01276 * 
race[T.black]  0.395520.54739   0.723  0.46995   
race[T.white] -0.862910.43517  -1.983  0.04737 * 
smoke  0.880070.40049   2.197  0.02798 * 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 234.67  on 188  degrees of freedom
Residual deviance: 200.62  on 182  degrees of freedom
AIC: 214.62

Number of Fisher Scoring iterations: 4

MY SAS OUTPUT IS:

  The LOGISTIC Procedure

   Analysis of Maximum Likelihood Estimates

 StandardWald
  ParameterDF  Estimate Error  Chi-Square  Pr  ChiSq  Exp(Est)
Label

  Intercept 10.92870.9326  0.9916  0.3193 2.531
Intercept: low=1
  lwt   1   -0.0173   0.00699  6.1425  0.0132 0.983
  ptl   11.19580.4472  7.1493  0.0075 3.306
  ht11.74820.7090  6.0805  0.0137 5.745
  race  black   10.59630.3352  3.1643  0.0753 1.815
race black
  race  white   1   -0.72000.2668  7.2803  0.0070 0.487
race white
  smoke 10.86480.4009  4.6534  0.0310 2.375





0 19 182 black 0 0 1 0 0 2523
0 33 155 other 0 0 0 1 0 2551
0 20 105 white 1 0 0 1 0 2557
0 21 108 white 1 0 1 1 0 2594
0 18 107 white 1 0 1 0 0 2600
0 21 124 other 0 0 0 0 0 2622
0 22 118 white 0 0 0 1 0 2637
0 17 103 other 0 0 0 1 0 2637
0 29 123 white 1 0 0 1 0 2663
0 26 113 white 1 0 0 0 0 2665
0 19 95  other 0 0 0 0 0 2722
0 19 150 other 0 0 0 1 0 2733
0 22 95  other 0 1 0 0 0 2750
0 30 107 other 0 0 1 1 1 2750
0 18 100 white 1 0 0 0 0 2769
0 15 98  black 0 0 0 0 0 2778
0 25 118 white 1 0 0 1 0 2782
0 20 120 other 0 0 1 0 0 2807
0 28 120 white 1 0 0 1 0 2821
0 32 101 other 0 0 0 1 0 2835
0 31 100 white 0 0 1 1 0 2835
0 36 202 white 0 0 0 1 0 2836
0 28 120 other 0 0 0 0 0 2863
0 25 120 other 0 0 1 1 0 2877
0 28 167 white 0 0 0 0 0 2877
0 17 122 white 1 0 0 0 0 2906
0 29 150 white 0 0 0 1 0 2920
0 26 168 black 1 0 0 0 0 2920
0 17 113 black 0 0 0 1 0 2920
0 24 90  white 1 0 0 1 1 2948
0 35 121 black 1 0 0 1 1 2948
0 25 155   

[R] Help with Error Message

2007-11-05 Thread Patrick Richardson
Hoping someone can offer me some assistance.  I'm trying to execute a script
and I keep getting this error message about Error: element 12 is empty.
I'm wondering if my syntax is incorrect within legend.list.  If anyone has
any suggestions to sees something obvious that I am missing, I would greatly
appreciate any help.

Many Thanks,

Patrick

 # These are the symbols and colors to use for each phenotype in the model
and test sets 
 #  model samples:   square symbols
 #  color symbol  phenotype
 legend.list - c(green, 22,# ALL-B 
+  steelblue, 22,# ALL-T 
+  red,   22,# AML   
+ #  test samples:cicle symbols
+ #  color symbol  phenotype 
+  lightgreen,21,# ALL-B 
+  lightblue, 21,# ALL T 
+  orange,21,# AML 
+ )
Error: element 12 is empty;
   the part of the args list of 'c' being evaluated was:
   (22, steelblue, 22, red, 22, lightgreen, 21, lightblue, 21,
orange, 21, )
 
 col - legend.list[seq(1, length(legend.list), 2)]
Error: object legend.list not found
 symbs - as.numeric(legend.list[seq(2, length(legend.list), 2)])
Error: object legend.list not found

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[R] Package Building help

2007-09-24 Thread Patrick Richardson
List,

 

I've used package.skeleton to build my package and am trying to check it
using R CMD check . But can't seem to get anything to work.  When I try to
enter the R CMD command into R I get this message.

 

R CMD check estpkg

Error: syntax error, unexpected SYMBOL, expecting '\n' or ';' in R CMD

 

I'm not sure if I'm using the wrong syntax or what but I can't get R CMD to
check my package from within R.  I've tried to use Rcmd in the /bin folder
of the R root directory and the DOS windows flashes on my screen and
disappears immediately.  If someone could help me get over this, I would
greatly appreciate it.

 

Cheers,

 

Patrick


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