PredictiveAdvantage: Plot predictive advantage of LPC vs. T
In lpc: Lassoed Principal Components for Testing Significance of Features
Description
Usage
Arguments
Details
Value
Author(s)
References
Examples
Description
This function plots the predictive advantage of LPC vs. T. The
predictive advantage is the difference between the red and black
curves. If the red curve is higher than the black curve on average, then
LPC should be used instead of T on this data set.
Usage
1
2
PredictiveAdvantage(x,y,type,nreps=20,ngenes=100,
soft.thresh=NULL,censoring.status=NULL)
Arguments
x
The matrix of gene expression values; pxn where n is the
number of observations and p is the number of genes.
y
A vector of
length n, with an outcome for each observation. For two-class
outcome, y's
elements are 1 or 2. For quantitative outcome, y's elements
are
real-valued. For survival
data, y indicates the survival time. For multiclass outcome,
y is coded as 1,2,3,..
type
One of "regression" (for a quantitative outcome), "two
class", "multiclass", or "survival".
nreps
Number of training/test set splits used in computing
predictive advantage.
soft.thresh
Value of soft threshold used in L1 constraint for
LPC. If NULL, then it will be computed adaptively.
ngenes
Number of genes to include in predictive advantage
plot. (E.g., if ngenes=100 (default) then $E(|T_test| |
|L_train|>alpha(L_train))$ and
$E(|T_test| | |T_train|>alpha(T_train))$ will be plotted (see
"Details"), where $alpha(L_train)$ and $alpha(T_train)$ are the 100th
largest (in absolute value) T and LPC scores on the training set.)
censoring.status
For survival outcome only, a vector of length
n which takes on values 0 or 1
depending on whether the observation is complete or censored.
Details
As explained
in the paper, predictive advantage is computed by first splitting the
data into a training set and a test set (each with 50% of the
samples). Then, the following is computed: $E(|T_test| | |L_train|>alpha(L_train)) -
E(|T_test| | |T_train|>alpha(T_train))$, where $T_test$ are the T scores on
the test data, $T_train$ are the T scores on the training data, and
$L_train$ are the LPC scores on the training data. $alpha(L_train)$ and
$alpha(T_train)$ are the $alpha$ quantiles of the LPC and T scores on the training data. A
large value of $E(|T_test| | |L_train|>alpha(L_train)) -
E(|T_test| | |T_train|>alpha(T_train))$ suggests that LPC is superior to
T on this data set.
Value
lpc
A vector of numGenes elements. The ith element is of the form
$E(|T_test| | |L_train|>alpha(L_train))$, where $alpha(L_train)$ is
the ith smallest (in absolute value) training set LPC score. Note that $L_train$
are LPC scores on the training set, $T_test$ are T scores on the
test set.
t
A vector of numGenes elements. The ith element is of the form
$E(|T_test| | |L_train|>alpha(T_train))$, where $alpha(T_train)$ is
the ith smallest (in absolute value) training set T score, and where $T_train$ are
T scores on the training set and $T_test$ are T scores on the test set.
Author(s)
Daniela M. Witten and Robert Tibshirani
References
Witten, D.M. and Tibshirani, R. (2008) Testing significance
of features by lassoed principal components. Annals of Applied
Statistics. http://www-stat.stanford.edu/~dwitten
Examples
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
# Not run due to timing; uncomment to run
#set.seed(2)
#n <- 40 # 40 samples
#p <- 1000 # 1000 genes
#x <- matrix(rnorm(n*p), nrow=p) # make 40x1000 gene expression matrix
#y <- rnorm(n) # quantitative outcome
## make first 50 genes differentially-expressed
#x[1:25,y<(-.5)] <- x[1:25,y<(-.5)]+ 1.5
#x[26:50,y<0] <- x[26:50,y<0] - 1.5
## compute LPC and T scores for each gene
#lpc.obj <- LPC(x,y, type="regression")
## Look at plot of Predictive Advantage
#pred.adv <-
#PredictiveAdvantage(x,y,type="regression",soft.thresh=lpc.obj$soft.thresh)
## Estimate FDRs for LPC and T scores
#fdr.lpc.out <-
#EstimateLPCFDR(x,y,type="regression",soft.thresh=lpc.obj$soft.thresh,nreps=50)
## Estimate FDRs for T scores only. This is quicker than computing FDRs
## for LPC scores, and should be used when only T FDRs are needed. If we
## started with the same random seed, then EstimateTFDR and EstimateLPCFDR
## would give same T FDRs.
#fdr.t.out <- EstimateTFDR(x,y, type="regression")
## print out results of main function
#lpc.obj
## print out info about T FDRs
#fdr.t.out
## print out info about LPC FDRs
#fdr.lpc.out
## Compare FDRs for T and LPC on 6% of genes. In this example, LPC has
## lower FDR.
#PlotFDRs(fdr.lpc.out,frac=.06)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores.
#PrintGeneList(lpc.obj,numGenes=20)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores and their FDRs for LPC and T.
#PrintGeneList(lpc.obj,numGenes=20,lpcfdr.out=fdr.lpc.out)
## Now, repeating everything that we did before, but using a
## **survival** outcome
#set.seed(2)
#n <- 40 # 40 samples
#p <- 1000 # 1000 genes
#x <- matrix(rnorm(n*p), nrow=p) # make 40x1000 gene expression matrix
#y <- rnorm(n) + 10 # survival times; must be positive
## censoring outcome: 0 or 1
#cens <- rep(1,40) # Assume all observations are complete
## make first 50 genes differentially-expressed
#x[1:25,y<9.5] <- x[1:25,y<9.5]+ 1.5
#x[26:50,y<10] <- x[26:50,y<10] - 1.5
#lpc.obj <- LPC(x,y, type="survival", censoring.status=cens)
## Look at plot of Predictive Advantage
#pred.adv <-
#PredictiveAdvantage(x,y,type="survival",soft.thresh=lpc.obj$soft.thresh,
#censoring.status=cens)
## Estimate FDRs for LPC scores and T scores
#fdr.lpc.out <- EstimateLPCFDR(x,y,type="survival",
#soft.thresh=lpc.obj$soft.thresh,nreps=20,censoring.status=cens)
## Estimate FDRs for T scores only. This is quicker than computing FDRs
## for LPC scores, and should be used when only T FDRs are needed. If we
## started with the same random seed, then EstimateTFDR and EstimateLPCFDR
## would give same T FDRs.
#fdr.t.out <- EstimateTFDR(x,y, type="survival", censoring.status=cens)
## print out results of main function
#lpc.obj
## print out info about T FDRs
#fdr.t.out
## print out info about LPC FDRs
#fdr.lpc.out
## Compare FDRs for T and LPC scores on 10% of genes.
#PlotFDRs(fdr.lpc.out,frac=.1)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores.
#PrintGeneList(lpc.obj,numGenes=20)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores and their FDRs for LPC and T.
#PrintGeneList(lpc.obj,numGenes=20,lpcfdr.out=fdr.lpc.out)
lpc documentation built on May 2, 2019, 2:49 p.m.
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
Description Usage Arguments Details Value Author(s) References Examples
Description
This function plots the predictive advantage of LPC vs. T. The predictive advantage is the difference between the red and black curves. If the red curve is higher than the black curve on average, then LPC should be used instead of T on this data set.
Usage
1 2 | PredictiveAdvantage(x,y,type,nreps=20,ngenes=100,
soft.thresh=NULL,censoring.status=NULL)
|
Arguments
x |
The matrix of gene expression values; pxn where n is the number of observations and p is the number of genes. |
y |
A vector of length n, with an outcome for each observation. For two-class outcome, y's elements are 1 or 2. For quantitative outcome, y's elements are real-valued. For survival data, y indicates the survival time. For multiclass outcome, y is coded as 1,2,3,.. |
type |
One of "regression" (for a quantitative outcome), "two class", "multiclass", or "survival". |
nreps |
Number of training/test set splits used in computing predictive advantage. |
soft.thresh |
Value of soft threshold used in L1 constraint for LPC. If NULL, then it will be computed adaptively. |
ngenes |
Number of genes to include in predictive advantage plot. (E.g., if ngenes=100 (default) then $E(|T_test| | |L_train|>alpha(L_train))$ and $E(|T_test| | |T_train|>alpha(T_train))$ will be plotted (see "Details"), where $alpha(L_train)$ and $alpha(T_train)$ are the 100th largest (in absolute value) T and LPC scores on the training set.) |
censoring.status |
For survival outcome only, a vector of length n which takes on values 0 or 1 depending on whether the observation is complete or censored. |
Details
As explained in the paper, predictive advantage is computed by first splitting the data into a training set and a test set (each with 50% of the samples). Then, the following is computed: $E(|T_test| | |L_train|>alpha(L_train)) - E(|T_test| | |T_train|>alpha(T_train))$, where $T_test$ are the T scores on the test data, $T_train$ are the T scores on the training data, and $L_train$ are the LPC scores on the training data. $alpha(L_train)$ and $alpha(T_train)$ are the $alpha$ quantiles of the LPC and T scores on the training data. A large value of $E(|T_test| | |L_train|>alpha(L_train)) - E(|T_test| | |T_train|>alpha(T_train))$ suggests that LPC is superior to T on this data set.
Value
lpc |
A vector of numGenes elements. The ith element is of the form $E(|T_test| | |L_train|>alpha(L_train))$, where $alpha(L_train)$ is the ith smallest (in absolute value) training set LPC score. Note that $L_train$ are LPC scores on the training set, $T_test$ are T scores on the test set. |
t |
A vector of numGenes elements. The ith element is of the form $E(|T_test| | |L_train|>alpha(T_train))$, where $alpha(T_train)$ is the ith smallest (in absolute value) training set T score, and where $T_train$ are T scores on the training set and $T_test$ are T scores on the test set. |
Author(s)
Daniela M. Witten and Robert Tibshirani
References
Witten, D.M. and Tibshirani, R. (2008) Testing significance of features by lassoed principal components. Annals of Applied Statistics. http://www-stat.stanford.edu/~dwitten
Examples
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 | # Not run due to timing; uncomment to run
#set.seed(2)
#n <- 40 # 40 samples
#p <- 1000 # 1000 genes
#x <- matrix(rnorm(n*p), nrow=p) # make 40x1000 gene expression matrix
#y <- rnorm(n) # quantitative outcome
## make first 50 genes differentially-expressed
#x[1:25,y<(-.5)] <- x[1:25,y<(-.5)]+ 1.5
#x[26:50,y<0] <- x[26:50,y<0] - 1.5
## compute LPC and T scores for each gene
#lpc.obj <- LPC(x,y, type="regression")
## Look at plot of Predictive Advantage
#pred.adv <-
#PredictiveAdvantage(x,y,type="regression",soft.thresh=lpc.obj$soft.thresh)
## Estimate FDRs for LPC and T scores
#fdr.lpc.out <-
#EstimateLPCFDR(x,y,type="regression",soft.thresh=lpc.obj$soft.thresh,nreps=50)
## Estimate FDRs for T scores only. This is quicker than computing FDRs
## for LPC scores, and should be used when only T FDRs are needed. If we
## started with the same random seed, then EstimateTFDR and EstimateLPCFDR
## would give same T FDRs.
#fdr.t.out <- EstimateTFDR(x,y, type="regression")
## print out results of main function
#lpc.obj
## print out info about T FDRs
#fdr.t.out
## print out info about LPC FDRs
#fdr.lpc.out
## Compare FDRs for T and LPC on 6% of genes. In this example, LPC has
## lower FDR.
#PlotFDRs(fdr.lpc.out,frac=.06)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores.
#PrintGeneList(lpc.obj,numGenes=20)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores and their FDRs for LPC and T.
#PrintGeneList(lpc.obj,numGenes=20,lpcfdr.out=fdr.lpc.out)
## Now, repeating everything that we did before, but using a
## **survival** outcome
#set.seed(2)
#n <- 40 # 40 samples
#p <- 1000 # 1000 genes
#x <- matrix(rnorm(n*p), nrow=p) # make 40x1000 gene expression matrix
#y <- rnorm(n) + 10 # survival times; must be positive
## censoring outcome: 0 or 1
#cens <- rep(1,40) # Assume all observations are complete
## make first 50 genes differentially-expressed
#x[1:25,y<9.5] <- x[1:25,y<9.5]+ 1.5
#x[26:50,y<10] <- x[26:50,y<10] - 1.5
#lpc.obj <- LPC(x,y, type="survival", censoring.status=cens)
## Look at plot of Predictive Advantage
#pred.adv <-
#PredictiveAdvantage(x,y,type="survival",soft.thresh=lpc.obj$soft.thresh,
#censoring.status=cens)
## Estimate FDRs for LPC scores and T scores
#fdr.lpc.out <- EstimateLPCFDR(x,y,type="survival",
#soft.thresh=lpc.obj$soft.thresh,nreps=20,censoring.status=cens)
## Estimate FDRs for T scores only. This is quicker than computing FDRs
## for LPC scores, and should be used when only T FDRs are needed. If we
## started with the same random seed, then EstimateTFDR and EstimateLPCFDR
## would give same T FDRs.
#fdr.t.out <- EstimateTFDR(x,y, type="survival", censoring.status=cens)
## print out results of main function
#lpc.obj
## print out info about T FDRs
#fdr.t.out
## print out info about LPC FDRs
#fdr.lpc.out
## Compare FDRs for T and LPC scores on 10% of genes.
#PlotFDRs(fdr.lpc.out,frac=.1)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores.
#PrintGeneList(lpc.obj,numGenes=20)
## Print out names of 20 genes with highest LPC scores, along with their
## LPC and T scores and their FDRs for LPC and T.
#PrintGeneList(lpc.obj,numGenes=20,lpcfdr.out=fdr.lpc.out)
|
R Package Documentation
Browse R Packages
We want your feedback!
Note that we can't provide technical support on individual packages. You should contact the package authors for that.
Embedding an R snippet on your website
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.