My project is about predicting biomarker breast cancer.
I use this function to give me a 2x2 matrix:
Table(gpl96)[1:10,1:4]
I want to take this data that represents the samples of genes in GDS and compare the p-value to know if it is normally distributed or not.
t.test tests whether there is a difference in location between two samples that adhere to normal distributions.
To approximately check the assumed normality, you might inspect whether outputs from qqnorm seem linear, or use ks.test in conjunction with estimating parameters from observations*:
set.seed(1)
x1 <- rnorm(200,40,10) # should follow a normal distribution
ks.test(x1,"pnorm",mean=mean(x1),sd=sd(x1)) # p: 0.647 [qqnorm(x1) looks linear]
x2 <- rexp(200,10) # should *not* follow a normal distribution
ks.test(x2,"pnorm",mean=mean(x2),sd=sd(x2)) # p: 3.576e-05, [qqnorm(x2) seems curved]
I do not know GEO's Table, but I suggest you might want to use its VALUE columns -and not any 2x2 matrices- as inputs for t.test, qqnorm or ks.test; maybe you might provide some additional illustration of your data by posting outputs of head(Table(gpl96)[1:10,1:4]).
(* After https://stat.ethz.ch/pipermail/r-help/2003-October/040692.html, which also appears to demonstrate the more refined Lilliefors test.)
Related
I am using the R package mclust to separate data into clusters. For this, I am using a uni-dimensional that allows for variable variances of the normal distributions underlying the clustering (the "V" model in the package).
The function looks like this: Mclust(dataToCluster, G=possibleClusters, modelNames=c("V")). To define the number of clusters possible, I use an array possibleClusters, e. g. 1:4 to allow for one to four clusters.
As a result of the clustering, after automatic model selection by Mclust using the BIC, I get a result with parameters of a normal distribution. For a model with three clusters, it might look like this:
# output shortened and commented for better readibility
> result$parameters
# proportion of data points per cluster ("lambda")
$pro
[1] 0.3459566 0.3877521 0.2662913
# mean of normal distribution per cluster ("mu")
$mean
1 2 3
110.3197 204.0477 265.0929
# variances per cluster ("sigma sq")
$variance$sigmasq
[1] 342.5032 128.4648 254.9257
However, I do have some knowledge about what these parameters are supposed to look like a priori. For example, I might know that:
sigmasq must be between 100 and 1000 units
the mean value for adjacent clusters must be at least 40 units apart
if there are three clusters, the mean value of the third cluster must be at least 215 units
Here is a graphical example for possible results of the clustering (the x axis corresponds to the units of mean and sigma unsquared):
Taking into account the constraints given above, example plots A1 (according to rules 1 and 2) and B1 (according to rules 2 and 3) can't be correct. Instead, the results should look more like A2 and B2, which were produced using slightly different data. Note that, taking into account these constraints, the “best” number of clusters might change (A1 vs. A2).
I would like to know how to include this kind of a priori information when using the Mclust function. The function does have a parameter prior, which might allow for this but I wasn't able to figure out how this could work. How could I bring the constraints into the function?
I have two data frames cases and controls and I performed two sample t-test as shown below.But I am doing feature extraction from the feature set of (1299 features/columns) so I want to calculate p-values for each feature. Based on the p-value generated for each feature I want to reject or accept the null hypothesis.
Can anyone explain to me how the below output is interpreted and how to calculate the p-values for each feature?
t.test(New_data_zero,New_data_one)
Welch Two Sample t-test
data: New_data_zero_pca and New_data_one_pca
t = -29.086, df = 182840000, p-value < 2.2e-16
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.02499162 -0.02183612
sample estimates:
mean of x mean of y
0.04553462 0.06894849
Look at ?t.test. x and y are supposed to be vectors not matrixes. So the function is automatically converting them to vectors. What you want to do, assuming that columns are features and the two matrixes have the same features, is:
pvals=vector()
for (i in seq(ncol(New_data_zero))){
pvals[i]=t.test(New_data_zero[,i], New_data_one[,i])$p.value
}
Then you can look at pvals (probably in log scale) and after multiple hypothesis testing correction (see ?p.adjust).
Let's also address the enormously bad idea of this approach to finding differences among your features. Even if all of the effects between these 1299 features are literally zero you will find *significant results in 0.05 of all possible 1299 2-way comparisons which makes this strategy effectively meaningless. I would strongly suggest taking a look at an introductory statistics text, especially the section on family-wise type I error rates before proceeding.
I have a list that looks like this, it is a measure of dispersion for each sample.
1 2 3 4 5
0.11829384 0.24987017 0.08082147 0.13355495 0.12933790
To further analyze this I need it to be a distance structure, the -vegan- package need it as a 'dist' object.
I found some solutions that applies to matrices > dist, but how could I change this current data into a dist object?
I am using the FD package, at the manual I found,
Still, one potential advantage of FDis over Rao’s Q is that in the unweighted case
(i.e. with presence-absence data), it opens possibilities for formal statistical tests for differences in
FD between two or more communities through a distance-based test for homogeneity of multivariate
dispersions (Anderson 2006); see betadisper for more details
I wanted to use vegan betadisper function to test if there are differences among different regions (I provided this using element "region" with column "region" too)
functional <- FD(trait, comun)
mod <- betadisper(functional$FDis, region$region)
using gowdis or fdisp from FD didn't work too.
distancias <- gowdis(rasgo)
mod <- betadisper(distancias, region$region)
dispersion <- fdisp(distancias, presence)
mod <- betadisper(dispersion, region$region)
I tried this but I need a list object. I thought I could pass those results to betadisper.
You cannot do this: FD::fdisp() does not return dissimilarities. It returns a list of three elements: the dispersions FDis for each sampling unit (SU), and the results of the eigen decomposition of input dissimilarities (eig for eigenvalues, vectors for orthonormal eigenvectors). The FDis values are summarized for each original SU, but there is no information on the differences among SUs. The eigen decomposition can be used to reconstruct the original input dissimilarities (your distancias from FD::gowdis()), but you can directly use the input dissimilarities. Function FD::gowdis() returns a regular "dist" structure that you can directly use in vegan::betadisper() if that gives you a meaningful analysis. For this, your grouping variable must be based on the same units as your distancias. In typical application of fdisp, the units are species (taxa), but it seems you want to get analysis for communities/sites/whatever. This will not be possible with these tools.
I want to perform a Shapiro-Wilk Normality Test test. My data is csv format. It looks like this:
heisenberg
HWWIchg
1 -15.60
2 -21.60
3 -19.50
4 -19.10
5 -20.90
6 -20.70
7 -19.30
8 -18.30
9 -15.10
However, when I perform the test, I get:
shapiro.test(heisenberg)
Error in [.data.frame(x, complete.cases(x)) :
undefined columns selected
Why isnt`t R selecting the right column and how do I do that?
What does shapiro.test do?
shapiro.test tests the Null hypothesis that "the samples come from a Normal distribution" against the alternative hypothesis "the samples do not come from a Normal distribution".
How to perform shapiro.test in R?
The R help page for ?shapiro.test gives,
x - a numeric vector of data values. Missing values are allowed,
but the number of non-missing values must be between 3 and 5000.
That is, shapiro.test expects a numeric vector as input, that corresponds to the sample you would like to test and it is the only input required. Since you've a data.frame, you'll have to pass the desired column as input to the function as follows:
> shapiro.test(heisenberg$HWWIchg)
# Shapiro-Wilk normality test
# data: heisenberg$HWWIchg
# W = 0.9001, p-value = 0.2528
Interpreting results from shapiro.test:
First, I strongly suggest you read this excellent answer from Ian Fellows on testing for normality.
As shown above, the shapiro.test tests the NULL hypothesis that the samples came from a Normal distribution. This means that if your p-value <= 0.05, then you would reject the NULL hypothesis that the samples came from a Normal distribution. As Ian Fellows nicely put it, you are testing against the assumption of Normality". In other words (correct me if I am wrong), it would be much better if one tests the NULL hypothesis that the samples do not come from a Normal distribution. Why? Because, rejecting a NULL hypothesis is not the same as accepting the alternative hypothesis.
In case of the null hypothesis of shapiro.test, a p-value <= 0.05 would reject the null hypothesis that the samples come from normal distribution. To put it loosely, there is a rare chance that the samples came from a normal distribution. The side-effect of this hypothesis testing is that this rare chance happens very rarely. To illustrate, take for example:
set.seed(450)
x <- runif(50, min=2, max=4)
shapiro.test(x)
# Shapiro-Wilk normality test
# data: runif(50, min = 2, max = 4)
# W = 0.9601, p-value = 0.08995
So, this (particular) sample runif(50, min=2, max=4) comes from a normal distribution according to this test. What I am trying to say is that, there are many many cases under which the "extreme" requirements (p < 0.05) are not satisfied which leads to acceptance of "NULL hypothesis" most of the times, which might be misleading.
Another issue I'd like to quote here from #PaulHiemstra from under comments about the effects on large sample size:
An additional issue with the Shapiro-Wilk's test is that when you feed it more data, the chances of the null hypothesis being rejected becomes larger. So what happens is that for large amounts of data even very small deviations from normality can be detected, leading to rejection of the null hypothesis event though for practical purposes the data is more than normal enough.
Although he also points out that R's data size limit protects this a bit:
Luckily shapiro.test protects the user from the above described effect by limiting the data size to 5000.
If the NULL hypothesis were the opposite, meaning, the samples do not come from a normal distribution, and you get a p-value < 0.05, then you conclude that it is very rare that these samples do not come from a normal distribution (reject the NULL hypothesis). That loosely translates to: It is highly likely that the samples are normally distributed (although some statisticians may not like this way of interpreting). I believe this is what Ian Fellows also tried to explain in his post. Please correct me if I've gotten something wrong!
#PaulHiemstra also comments about practical situations (example regression) when one comes across this problem of testing for normality:
In practice, if an analysis assumes normality, e.g. lm, I would not do this Shapiro-Wilk's test, but do the analysis and look at diagnostic plots of the outcome of the analysis to judge whether any assumptions of the analysis where violated too much. For linear regression using lm this is done by looking at some of the diagnostic plots you get using plot(lm()). Statistics is not a series of steps that cough up a few numbers (hey p < 0.05!) but requires a lot of experience and skill in judging how to analysis your data correctly.
Here, I find the reply from Ian Fellows to Ben Bolker's comment under the same question already linked above equally (if not more) informative:
For linear regression,
Don't worry much about normality. The CLT takes over quickly and if you have all but the smallest sample sizes and an even remotely reasonable looking histogram you are fine.
Worry about unequal variances (heteroskedasticity). I worry about this to the point of (almost) using HCCM tests by default. A scale location plot will give some idea of whether this is broken, but not always. Also, there is no a priori reason to assume equal variances in most cases.
Outliers. A cooks distance of > 1 is reasonable cause for concern.
Those are my thoughts (FWIW).
Hope this clears things up a bit.
You are applying shapiro.test() to a data.frame instead of the column. Try the following:
shapiro.test(heisenberg$HWWIchg)
You failed to specify the exact columns (data) to test for normality.
Use this instead
shapiro.test(heisenberg$HWWIchg)
Set the data as a vector and then place in the function.
I am new to R and cointegration so please have patience with me as I try to explain what it is that I am trying to do. I am trying to find cointegrated variables among 1500-2000 voltage variables in the west power system in Canada/US. THe frequency is hourly (common in power) and cointegrated combinations can be as few as N variables and a maximum of M variables.
I tried to use ca.jo but here are issues that I ran into:
1) ca.jo (Johansen) has a limit to the number of variables it can work with
2) ca.jo appears to force the first variable in the y(t) vector to be the dependent variable (see below).
Eigenvectors, normalised to first column: (These are the cointegration relations)
V1.l2 V2.l2 V3.l2
V1.l2 1.0000000 1.0000000 1.0000000
V2.l2 -0.2597057 -2.3888060 -0.4181294
V3.l2 -0.6443270 -0.6901678 0.5429844
As you can see ca.jo tries to find linear combinations of the 3 variables but by forcing the coefficient on the first variable (in this case V1) to be 1 (i.e. the dependent variable). My understanding was that ca.jo would try to find all combinations such that every variable is selected as a dependent variable. You can see the same treatment in the examples given in the documentation for ca.jo.
3) ca.jo does not appear to find linear combinations of fewer than the number of variables in the y(t) vector. So if there were 5 variables and 3 of them are cointegrated (i.e. V1 ~ V2 + V3) then ca.jo fails to find this combination. Perhaps I am not using ca.jo correctly but my expectation was that a cointegrated combination where V1 ~ V2 + V3 is the same as V1 ~ V2 + V3 + 0 x V4 + 0 x V5. In other words the coefficient of the variable that are NOT cointegrated should be zero and ca.jo should find this type of combination.
I would greatly appreciate some further insight as I am fairly new to R and cointegration and have spent the past 2 months teaching myself.
Thank you.
I have also posted on nabble:
http://r.789695.n4.nabble.com/ca-jo-cointegration-multivariate-case-tc3469210.html
I'm not an expert, but since no one is responding, I'm going to try to take a stab at this one.. EDIT: I noticed that I just answered to a 4 year old question. Hopefully it might still be useful to others in the future.
Your general understanding is correct. I'm not going to go in great detail about the whole procedure but will try to give some general insight. The first thing that the Johansen procedure does is create a VECM out of the VAR model that best corresponds to the data (This is why you need the lag length for the VAR as input to the procedure as well). The procedure will then investigate the non-lagged component matrix of the VECM by looking at its rank: If the variables are not cointegrated then the rank of the matrix will not be significantly different from 0. A more intuitive way of understanding the johansen VECM equations is to notice the comparibility with the ADF procedure for each distinct row of the model.
Furthermore, The rank of the matrix is equal to the number of its eigenvalues (characteristic roots) that are different from zero. Each eigenvalue is associated with a different cointegrating vector, which
is equal to its corresponding eigenvector. Hence, An eigenvalue significantly different
from zero indicates a significant cointegrating vector. Significance of the vectors can be tested with two distinct statistics: The max statistic or the trace statistic. The trace test tests the null hypothesis of less than or equal to r cointegrating vectors against the alternative of more than r cointegrating vectors. In contrast, The maximum eigenvalue test tests the null hypothesis of r cointegrating vectors against the alternative of r + 1 cointegrating vectors.
Now for an example,
# We fit data to a VAR to obtain the optimal VAR length. Use SC information criterion to find optimal model.
varest <- VAR(yourData,p=1,type="const",lag.max=24, ic="SC")
# obtain lag length of VAR that best fits the data
lagLength <- max(2,varest$p)
# Perform Johansen procedure for cointegration
# Allow intercepts in the cointegrating vector: data without zero mean
# Use trace statistic (null hypothesis: number of cointegrating vectors <= r)
res <- ca.jo(yourData,type="trace",ecdet="const",K=lagLength,spec="longrun")
testStatistics <- res#teststat
criticalValues <- res#criticalValues
# chi^2. If testStatic for r<= 0 is greater than the corresponding criticalValue, then r<=0 is rejected and we have at least one cointegrating vector
# We use 90% confidence level to make our decision
if(testStatistics[length(testStatistics)] >= criticalValues[dim(criticalValues)[1],1])
{
# Return eigenvector that has maximum eigenvalue. Note: we throw away the constant!!
return(res#V[1:ncol(yourData),which.max(res#lambda)])
}
This piece of code checks if there is at least one cointegrating vector (r<=0) and then returns the vector with the highest cointegrating properties or in other words, the vector with the highest eigenvalue (lamda).
Regarding your question: the procedure does not "force" anything. It checks all combinations, that is why you have your 3 different vectors. It is my understanding that the method just scales/normalizes the vector to the first variable.
Regarding your other question: The procedure will calculate the vectors for which the residual has the strongest mean reverting / stationarity properties. If one or more of your variables does not contribute further to these properties then the component for this variable in the vector will indeed be 0. However, if the component value is not 0 then it means that "stronger" cointegration was found by including the extra variable in the model.
Furthermore, you can test test significance of your components. Johansen allows a researcher to test a hypothesis about one or more
coefficients in the cointegrating relationship by viewing the hypothesis as
a restriction on the non-lagged component matrix in the VECM. If there exist r cointegrating vectors, only these linear combinations or linear transformations of them, or combinations of the cointegrating vectors, will be stationary. However, I'm not aware on how to perform these extra checks in R.
Probably, the best way for you to proceed is to first test the combinations that contain a smaller number of variables. You then have the option to not add extra variables to these cointegrating subsets if you don't want to. But as already mentioned, adding other variables can potentially increase the cointegrating properties / stationarity of your residuals. It will depend on your requirements whether or not this is the behaviour you want.
I've been searching for an answer to this and I think I found one so I'm sharing with you hoping it's the right solution.
By using the johansen test you test for the ranks (number of cointegration vectors), and it also returns the eigenvectors, and the alphas and betas do build said vectors.
In theory if you reject r=0 and accept r=1 (value of r=0 > critical value and r=1 < critical value) you would search for the highest eigenvalue and from that build your vector. On this case, if the highest eigenvalue was the first, it would be V1*1+V2*(-0.26)+V3*(-0.64).
This would generate the cointegration residuals for these variables.
Again, I'm not 100%, but preety sure the above is how it works.
Nonetheless, you can always use the cajools function from the urca package to create a VECM automatically. You only need to feed it a cajo object and define the number of ranks (https://cran.r-project.org/web/packages/urca/urca.pdf).
If someone could confirm / correct this, it would be appreciated.