prcomp : PCA residuals not zero - r

I have 3 variables on which I ran PCA using prcomp. I tried to reconstruct the variables using the loadings and factors but residuals is not zero. Statistically (I might be wrong here) I was expecting to be able to reconstruct the original data. Am I missing something?
test = read.table(text='0.8728891 0.7403704 0.6655271
0.8697503 0.7447901 0.6629487
0.8569866 0.7321241 0.6493666
0.8824890 0.7405750 0.6505887
0.8912246 0.7334331 0.6508194
0.8930270 0.7381421 0.6448108
0.8721081 0.7173891 0.6355404
0.8649705 0.7326563 0.6493313
0.8976412 0.7249211 0.6437649
0.9233625 0.7406451 0.6454023',sep=' ')
pca = prcomp(test,center=T,scale=F)
pca$x %*% pca$rotation + matrix(1,nrow=nrow(test),ncol=1) %*% pca$center - test
V1 V2 V3
-0.0020186611 0.0071487188 -0.0240478838
-0.0004352159 -0.0005375912 -0.0262594828
0.0008042558 -0.0039840874 -0.0019352850
0.0009905100 -0.0053390749 -0.0067663626
-0.0008375576 0.0041104957 0.0016244986
0.0013586563 -0.0060476694 0.0036526104
0.0004278214 0.0009280342 0.0298641699
0.0005504918 -0.0026885505 -0.0009348334
-0.0011619165 0.0073130849 0.0185829183
0.0003216158 -0.0009033601 0.0062196504

I use the following function for reconstructing data from a prcomp object:
#This function reconstructs a data set using a defined set of principal components.
#arguments "pca" is the pca object from prcomp, "pcs" is a vector of principal components
#to be used for reconstruction (default includes all pcs)
prcomp.recon <- function(pca, pcs=NULL){
if(is.null(pcs)) pcs <- seq(pca$sdev)
recon <- as.matrix(pca$x[,pcs]) %*% t(as.matrix(pca$rotation[,pcs]))
if(pca$scale[1] != FALSE){
recon <- scale(recon , center=FALSE, scale=1/pca$scale)
}
if(pca$center[1] != FALSE){
recon <- scale(recon , center=-pca$center, scale=FALSE)
}
recon
}
I couldn't figure out exactly what was wrong with your code, but using the prcomp.recon function gives the right result:
> prcomp.recon(pca) - test
V1 V2 V3
1 0 0 0
2 0 0 0
3 0 0 0
4 0 0 0
5 0 0 0
6 0 0 0
7 0 0 0
8 0 0 0
9 0 0 0
10 0 0 0

Related

Error: BoxCox error during preprocess imputation R language

I am looking at the answers to the Chapter 6 exercise 3 questions from the Applied Predictive Modeling book Max Kuhn and I am getting an error in the imputation prediction step (despite following their answer exactly). The reproducible code and problem is as follows:
library(AppliedPredictiveModeling)
library(caret)
library(RANN)
data(ChemicalManufacturingProcess)
predictors <- subset(ChemicalManufacturingProcess,select= -Yield)
yield <- subset(ChemicalManufacturingProcess,select="Yield")
# Impute
#Split data into training and test sets
set.seed(517)
trainingRows <- createDataPartition(yield$Yield,
p = 0.7,
list = FALSE)
trainPredictors <- predictors[trainingRows,]
trainYield <- yield[trainingRows,]
testPredictors <- predictors[-trainingRows,]
testYield <- yield[-trainingRows,]
#Pre-process trainPredictors and apply to trainPredictors and testPredictors
pp <- preProcess(trainPredictors,method=c("BoxCox","center","scale","knnImpute"))
ppTrainPredictors <- predict(pp,newdata=trainPredictors)
ppTestPredictors <- predict(pp,newdata=testPredictors) # This results in an error
The error it gives is: Error in RANN::nn2(old[, non_missing_cols, drop = FALSE], new[, non_missing_cols, : NA/NaN/Inf in foreign function call (arg 2)
When I use YeoJohnson transformation instead it seems to work (which i read is able to handle non positive numbers)
However, I dont understand why it isn't working on the test data seeing as its just a different subset of the training data? and it's just used for the imputation step of the question?
I was unable to find any answers to this which seems strange as surely other people who have followed the book would have noticed? or am I being thick?
Thank you
You get that error because the boxcox transformation does not take in zero. If you check the help page for BoxCoxTrans it writes:
If any(y <= 0) or if length(unique(y)) < numUnique, lambda is not
estimated and no transformation is applied.
So if your preProcess() is run on a train set with no zeros in the columns, the boxcox transformation is applied, but it will not work on a test set with zeros.
In the book example above, most likely the seed was set with an older R version, so it works. If you are using a newer version of R, then it doesn't work. So if I check with your example:
cbind(colSums(trainPredictors==0,na.rm=TRUE),colSums(testPredictors==0,na.rm=TRUE))
[,1] [,2]
BiologicalMaterial01 0 0
BiologicalMaterial02 0 0
BiologicalMaterial03 0 0
BiologicalMaterial04 0 0
BiologicalMaterial05 0 0
BiologicalMaterial06 0 0
BiologicalMaterial07 0 0
BiologicalMaterial08 0 0
BiologicalMaterial09 0 0
BiologicalMaterial10 0 0
BiologicalMaterial11 0 0
BiologicalMaterial12 0 0
ManufacturingProcess01 1 2
ManufacturingProcess02 29 6
ManufacturingProcess03 0 0
ManufacturingProcess04 0 0
ManufacturingProcess05 0 0
ManufacturingProcess06 0 0
ManufacturingProcess07 0 0
ManufacturingProcess08 0 0
ManufacturingProcess09 0 0
ManufacturingProcess10 0 0
ManufacturingProcess11 0 0
ManufacturingProcess12 104 38
ManufacturingProcess13 0 0
ManufacturingProcess14 0 0
ManufacturingProcess15 0 0
ManufacturingProcess16 1 0
ManufacturingProcess17 0 0
ManufacturingProcess18 1 0
You can see that ManufacturingProcess16, ManufacturingProcess18 will give you problems.
Yeo-Johnson transformation can deal with zeros or negative values, so it is not a problem.
If you would like to carry on with the work example, you can try to use another seed:
set.seed(517)
trainingRows <- createDataPartition(yield$Yield,
p = 0.7,
list = FALSE)
trainPredictors <- predictors[trainingRows,]
trainYield <- yield[trainingRows,]
testPredictors <- predictors[-trainingRows,]
testYield <- yield[-trainingRows,]

Ranger Predicted Class Probability of each row in a data frame

With regard to this link Predicted probabilities in R ranger package, I have a question.
Imagine I have a mixed data frame, df (comprising of factor and numeric variables) and I want to do classification using ranger. I am splitting this data frame as test and train sets as Train_Set and Test_Set. BiClass is my prediction factor variable and comprises of 0 and 1 (2 levels)
I want to calculate and attach class probabilities to the data frame using ranger using the following commands:
Biclass.ranger <- ranger(BiClass ~ ., ,data=Train_Set, num.trees = 500, importance="impurity", save.memory = TRUE, probability=TRUE)
probabilities <- as.data.frame(predict(Biclass.ranger, data = Test_Set, num.trees = 200, type='response', verbose = TRUE)$predictions)
The data frame probabilities is a data frame consisting of 2 columns (0 and 1) with number of rows equal to the number of rows in Test_Set.
Does it mean, if I append or attach this data frame, namely, probabilities to the Test_Set as the last two columns, it shows the probability of each row being either 0 or 1? Is my understanding correct?
My second question, when I attempt to calcuate confusion matrix through
pred = predict(Biclass.ranger, data=Test_Set, num.trees = 500, type='response', verbose = TRUE)
table(Test_Set$BiClass, pred$predictions)
I get the following error:
Error in table(Test_Set$BiClass, pred$predictions) :
all arguments must have the same length
What am I doing wrong?
For your first question yes, it shows the probability of each row being 0 or 1. Using the example below:
library(ranger)
idx = sample(nrow(iris),100)
data = iris
data$Species = factor(ifelse(data$Species=="versicolor",1,0))
Train_Set = data[idx,]
Test_Set = data[-idx,]
mdl <- ranger(Species ~ ., ,data=Train_Set,importance="impurity", save.memory = TRUE, probability=TRUE)
probabilities <- as.data.frame(predict(mdl, data = Test_Set,type='response', verbose = TRUE)$predictions)
We can always check whether they agree:
par(mfrow=c(1,2))
boxplot(probabilities[,"0"] ~ Test_Set$Species,ylab="Prob 0",xlab="Actual label")
boxplot(probabilities[,"1"] ~ Test_Set$Species,ylab="Prob 1",xlab="Actual label")
Not the best plot, but sometimes if the labels are flipped you will see something weird. We need to find the column that has the max probability and assign the label, for this we do:
max.col(probabilities) - 1
[1] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 0
[39] 0 0 0 0 0 0 0 0 0 0 0 0
This goes through each row of probabilities returns 1 or 2 depending on which column has maximum probability and we simply subtract 1 from it to get 0,1. For the confusion matrix:
caret::confusionMatrix(table(max.col(probabilities) - 1,Test_Set$Species))
Confusion Matrix and Statistics
0 1
0 31 2
1 0 17
Accuracy : 0.96
95% CI : (0.8629, 0.9951)
No Information Rate : 0.62
P-Value [Acc > NIR] : 2.048e-08
In your case, you can just do:
confusionMatrix(table(max.col(probabilities)-1,Test_Set$BiClass))

modulus values (roots) in VECM model using R?

thanks for reading my question. I am trying to fit a VECM for an economic research, i am using the vars and urca package on R using Rstudio. Considering i have no stationary time series, and both need one difference ,both are I(1), i need to use the VECM approach, but i can not get all the tests i need.
For example:
First i load the libraries
library(vars)
library(urca)
and create my model
data("Canada")
df <- Canada
VARselect(df)
vecm <- urca::ca.jo(df,K = 3)
model <- vec2var(vecm)
The problem is, i can not get the "modules" values to prove stability, i know i can use roots() function to get this values from a "varest" object, for example:
roots(VAR(df,3))
My question is:
how can i get modulus from my vec2var object, roots() doesn't handle this kind of object. I know Gretl can do it (using unit circle to prove stability), so is posible to get this values from a VECM?. How can i do it in R?
Starting with:
data("Canada")
dim(Canada) #84observations x 4 variables
VARselect(Canada) # since in small samples, AIC>BIC; VAR(3) is chosen.
Now, the range of the dataset Canada: 1980.1 - 2000.4 (20 years) is long enough for modeling. This 20-year long period definitely includes lots of crises and interventions. Hence, structural breaks in the data MUST be searched. This is necessary since in structurally-broken series, the existence of SBs changes t values of nonstationarity tests (thereby affects the decision on whether a series is stationary or not).
Since Narayan-Popp 2010 nonstationarity test under multiple structural breaks is statistically very powerful against previous ones (Lee-Strazichic2003, Zivot-Andres1992), and since Joyeux 2007 (in Rao2007) has proven the illogicalness of these previous tests, and NP2013 has proven the superiority of NP2010's statistical power, one MUST use NP2010. Since Gauss code for NP2010 seemed to be ugly to me, I converted it to R code, and with the help of ggplot2, results are presented nicer.
[Processing structural breaks is a MUST for cointegration check as well since Osterwald-Lenum1992 CVs ignore SBs whereas Johansen-Mosconi-Nielsen2000 CVs cares SBs.]
Canada <- as.data.frame(Canada)
head(Canada)
e prod rw U
1 929.6105 405.3665 386.1361 7.53
2 929.8040 404.6398 388.1358 7.70
...................................
# Assign lexiographic row names for dates of observations
row.names(Canada) <- paste(sort(rep(seq(1980, 2000, 1), 4) ), rep(seq(1, 4, 1), 20), sep = ".")
# Insert lexiographic "date" column to the dataframe. This is necessary for creating intervention dummies.
DCanada <- data.frame(date=row.names(Canada),Canada) # dataset with obs dates in a column
head(DCanada)
date e prod rw U
1980.1 1980.1 929.6105 405.3665 386.1361 7.53
1980.2 1980.2 929.8040 404.6398 388.1358 7.70
Perform Narayan-Popp 2010 nonstationarity test to the series:
[H0: "(with 2 structural breaks) series is nonstationary";
H1: "(with 2 structural breaks) series is stationary";
"test stat > critical value" => "hold H0"; "test stat < critical value" => "hold H1"]
library(causfinder)
narayanpopp(DCanada[,2]) # for e
narayanpopp(DCanada[,3]) # for prod
narayanpopp(DCanada[,4]) # for rw
narayanpopp(DCanada[,5]) # for U
Narayan-Popp 2010 nonstationarity test results (with obs #s):
variable t stat lag SB1 SB2 Integration Order
e -4.164 2 37:946.86 43:948.03 I(1)
prod -3.325 1 24:406.77 44:405.43 I(1)
rw -5.087 0 36:436.15 44:446.96 I(0) <trend-stationary>
U -5.737 1 43:8.169 53:11.070 I(0) <stationary pattern> (M2 computationally singular; used M1 model)
(critical values (M2): (1%,5%,10%): -5.576 -4.937 -4.596)
(critical values (M1): (1%,5%,10%): -4.958 -4.316 -3.980
Since in a VAR structure, all variables are treated equally, continue to equal-treatment when determining structural breaks systemwise:
mean(c(37,24,36,43)) # 35; SB1 of system=1988.3
mean(c(43,44,44,53)) # 46; SB2 of system=1990.2
The following is to overcome "In Ops.factor(left, right) : >= not meaningful for factors" error. In some dataset, we need to do the following:
library(readxl)
write.xlsx(Canada, file="data.xlsx", row.names=FALSE) # Take this to the below folder, add "date" column with values 1980.1,....,2000.4
mydata <- read_excel("D://eKitap//RAO 2007 Cointegration for the applied economist 2E//JoyeuxCalisma//Canada//data.xlsx")
# arrange your path accordingly in the above line.
mydata <- as.data.frame(mydata)
library(lubridate); library(zoo)
row.names(mydata) <- as.yearqtr(seq(ymd('1980-01-01'), by = '1 quarter', length.out=(84)))
Dmydata <- mydata # Hold it in a variable
Define intervention dummy matrix with 2 SBs (35:1988.3 and 46: 1990.2) as follows:
library(data.table)
DataTable <- data.table(Dmydata, keep.rownames=FALSE)
Dt <- cbind("bir"=1, # intervention dummies matrix
"D2t" = as.numeric(ifelse( DataTable[,c("date"), with=FALSE] >= "1988.3" & DataTable[,c("date"), with=FALSE] <= "1990.1", 1 , 0)),
"D3t" = as.numeric(ifelse( DataTable[,c("date"), with=FALSE] >= "1990.2" & DataTable[,c("date"), with=FALSE] <= "2000.4", 1 , 0)))
On the fly indicator variables accompanying intervention dummies:
OnTheFlyIndicator <- cbind(
"I2t" = as.numeric(DataTable[, c("date"), with=FALSE] == "1988.3"),
"I3t" = as.numeric(DataTable[, c("date"), with=FALSE] == "1990.2"))
myTimeTrend <- as.matrix(cbind("TimeTrend" = as.numeric(1:nrow(Dt))))
zyDt <- Dt * as.vector(myTimeTrend) # TimeTrendDavranisDegisimleri
colnames(zyDt) <- paste(colnames(myTimeTrend), colnames(Dt), sep="*")
mydata <- mydata[,-1]
Selection of VAR order:
library(vars)
# Lag order selection with the effects of intervention dummies
VARselect(mydata, lag.max=5, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)) # Take VAR(3)
Lagger matrix for Joyeux2007 indexing technique:
lagmatrix <- function(x, maxlag){
x <- as.matrix(x)
if(is.null(colnames(x))== TRUE){ colnames(x) <- "VarCol0" }
DondurulenDizey <- embed(c(rep(NA,maxlag),x),maxlag+1)
dimnames(DondurulenDizey)[[2]] <- c(colnames(x)[1, drop = FALSE], paste(colnames(x)[1,drop=FALSE],".",1:maxlag,"l", sep = ""))
return(DondurulenDizey)
}
Assign VAR lag and no. of subsamples:
VARlag <- 3
Subsamples <- 3 # subsamples = no. of str breaks +1
Dummy matrix for 2 structural breaks:
dummymatrix2SB <- matrix(NA,DataTable[,.N], 10)
dummymatrix2SB <- cbind(myTimeTrend,
lagmatrix(zyDt[,c("TimeTrend*D2t"), drop=FALSE], maxlag=VARlag)[,1+VARlag, drop=FALSE],
lagmatrix(zyDt[,c("TimeTrend*D3t"), drop=FALSE], maxlag=VARlag)[,1+VARlag, drop=FALSE],
lagmatrix(Dt[,c("D2t"), drop=FALSE], maxlag=VARlag)[,1+VARlag, drop=FALSE],
lagmatrix(Dt[,c("D3t"), drop=FALSE], maxlag=VARlag)[,1+VARlag, drop=FALSE],
lagmatrix(OnTheFlyIndicator[,c("I2t"), drop=FALSE], maxlag=VARlag-1),
lagmatrix(OnTheFlyIndicator[,c("I3t"), drop=FALSE], maxlag=VARlag-1))
dummymatrix2SB[is.na(dummymatrix2SB)] <- 0 # replace NAs with 0
dummymatrix2SB # Print dummy matrix for 2 str breaks to make sure all are OK
TimeTrend TimeTrend.D2t.3l TimeTrend.D3t.3l D2t.3l D3t.3l I2t I2t.1l I2t.2l I3t I3t.1l I3t.2l
1 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0
...........................................
34 0 0 0 0 0 0 0 0 0 0
35 0 0 0 0 1 0 0 0 0 0
36 0 0 0 0 0 1 0 0 0 0
37 0 0 0 0 0 0 1 0 0 0
38 35 0 1 0 0 0 0 0 0 0
39 36 0 1 0 0 0 0 0 0 0
40 37 0 1 0 0 0 0 0 0 0
41 38 0 1 0 0 0 0 0 0 0
42 39 0 1 0 0 0 0 1 0 0
43 40 0 1 0 0 0 0 0 1 0
44 41 0 1 0 0 0 0 0 0 1
45 0 42 0 1 0 0 0 0 0 0
46 0 43 0 1 0 0 0 0 0 0
............................................
83 0 80 0 1 0 0 0 0 0 0
84 0 81 0 1 0 0 0 0 0 0
STABILITY of VAR:
Victor, theoretically you are wrong. Stability is checked from VAR side even in the case of restricted (cointegrated) VAR models. See Joyeux2007 for details. Also, estimations from both sides are same:
"unrestricted VAR = unrestricted VECM" and
"restricted VAR = restricted VECM".
Hence, checking stability of unrestricted VAR is equal to checking stability of unrestricted VECM, and vice versa. They are equal math'ly, they are just different representations.
Also, checking stability of restricted VAR is equal to checking stability of restricted VECM, and vice versa. They are equal math'ly, they are just different representations. But, you do not need this checking for restricted VECM cases since we are surfing in subspace of a feasible VAR. That is to say, if original unr VAR corresponding to restd VeCM is stable, then all are OK.
If your series are cointegrated, you check the stability from VAR side even in that case! If you wonder "whether you should check stability for restricted VECM", the answer is NO. You should not check. Because, in cointegrated case, you are in the subspace of feasible solution. That said, if you insist to check stability of restricted (cointegrated) VECM, you can still do that via urca::ca.jo extentions and vars::vec2var extentions:
print(roots(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)), modulus=TRUE))
# [1] 0.96132524 0.77923543 0.68689517 0.68689517 0.67578368 0.67578368
[7] 0.59065419 0.59065419 0.55983617 0.55983617 0.33700725 0.09363846
print(max(roots(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)), modulus=TRUE)))
#0.9613252
(optional) Check stability via OLS-CUSUM:
plot(stability(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)), type="OLS-CUSUM"))
NON-AUTOCORRELATION of VAR residuals test:
for (j in as.integer(1:5)){
print(paste("VAR's lag no:", j))
print(serial.test(VAR(mydata, p=j, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)), lags.bg=4, type= c("ES")))
# lags.bg: AR order of VAR residuals
}
NORMALITY of VAR residuals test:
print(normality.test(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)), multivariate=TRUE))
library(normtest)
for (i in as.integer(1:4)){ # there are 4 variables
print(skewness.norm.test(resid(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)))[,i]))
print(kurtosis.norm.test(resid(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)))[,i]))
print(jb.norm.test(resid(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator)))[,i]))
}
HOMOSCEDASTICITY of VAR residuals test:
print(arch.test(VAR(mydata, p=3, "both", exogen=cbind(zyDt[drop=FALSE], Dt[drop=FALSE], OnTheFlyIndicator))), lags.multi=6, multivariate.only=TRUE)
Since integration orders of series is different, there is no way that they are cointegrated. That said,
Assume for a while all are I(1) and perform cointegration test with multiple structural breaks with Johansen-Mosconi-Nielsen 2000 CVs:
(extend urca::cajo to causfinder::ykJohEsbInc (i.e., add the functionality to process 1 SB and 2 SBs))
summary(ykJohEsbInc(mydata, type="trace", ecdet="zamanda2yk", K=3, spec="longrun", dumvar=dummymatrix2SB[,c(-1,-2,-3)]))
# summary(ykJohEsbInc(mydata, type="trace", ecdet="zamanda2yk", K=3, spec="transitory", dumvar=dummymatrix2SB[,c(-1,-2,-3)])) gives the exactly same result.
Since there are 2 SBs in the system (1988.3, 1990.2), there are q=2+1=3 subsamples.
1st SB ratio: v1= (35-1)/84= 0.4047619
2nd SB ratio:v2= (46-1)/84= 0.5357143
Hence, JMN2000 CVs for cointegration test with 2 SBs:
(The following is TR-localized. One can find original EN-local code in Giles website)
library(gplots)
# Johansen vd. (2000) nin buldugu, yapisal kirilmalarin varliginda esbutunlesim incelemesinin degistirilmis iz sinamalarinin yanasik p degerleri ve karar degerlerini hesaplama kodu
# Ryan Godwin & David Giles (Dept. of Economics, Univesity of Victoria, Canada), 29.06.2011
# Kullanici asagidaki 4 degeri atamalidir
#======================================
degiskensayisi <- 4 # p
q<- 3 # q: verideki farkli donemlerin sayisi; q=1: 1 donem, hicbir yapisal kirilma yok demek oldugundan v1 ve v2 nin degerleri ihmal edilir
v1<- 0.4047619 # (35-1)/84 # 1.yk anı=34+1=35. Johansen et. al 2000 v1 def'n , v1: SB1 - 1
v2<- 0.5357143 # (46-1)/84 # 2nd SB moment 45+1=46.
#======================================
# iz istatistiginin biri veya her ikisi icin p degerlerinin olmasi istendiginde, sonraki 2 satirin biri veya her ikisini degistir
izZ <- 15.09 # Vz(r) istatistiginin degeri
izK <- 114.7 # Vk(r) istatistiginin degeri
#=========================================
enbuyuk_p_r<- degiskensayisi # "p-r > 10" olmasın; bkz: Johansen vd. (2000)
# "a" ve "b" nin değerleri yapısal kırılmaların sayısına (q-1) bağlıdır
# q=1 iken, hiçbir yapısal kırılma olmadığı bu durumda a=b=0 ata
# q=2 iken, 1 yapısal kırılma olduğu bu durumda a=0 (Johansen vd. 2000 4.Tabloda) ve b=min[V1 , (1-V1)] ata
# q=3 iken, 2 yapısal kırılma olduğu bu durumda a=min[V1, (V2-V1), (1-V2)] ve b=min[geriye kalan iki V ifadesi] ata
a = c(0, 0, min(v1, v2-v1, 1-v2))[q]
b = c(0, min(v1, 1-v1), median(c(v1,v2-v1,1-v2)))[q]
# YanDagOrtLog: yanaşık dağılımın ortalamasının logaritması
# YanDagDegLog: yanaşık dağılımın değişmesinin logaritması
# V(Zamanyönsemsi) veya V(Kesme) sınamalarını yansıtmak üzere adlara z veya k ekle.
# Bkz. Johansen vd. (2000) 4. Tablo.
# Önce Vz(r) sınamasının sonra Vk(r) sınamasının karar değerlerini oluştur
pr<- c(1:enbuyuk_p_r)
YanDagOrtLogZ <- 3.06+0.456*pr+1.47*a+0.993*b-0.0269*pr^2-0.0363*a*pr-0.0195*b*pr-4.21*a^2-2.35*b^2+0.000840*pr^3+6.01*a^3-1.33*a^2*b+2.04*b^3-2.05/pr-0.304*a/pr+1.06*b/pr
+9.35*a^2/pr+3.82*a*b/pr+2.12*b^2/pr-22.8*a^3/pr-7.15*a*b^2/pr-4.95*b^3/pr+0.681/pr^2-0.828*b/pr^2-5.43*a^2/pr^2+13.1*a^3/pr^2+1.5*b^3/pr^2
YanDagDegLogZ <- 3.97+0.314*pr+1.79*a+0.256*b-0.00898*pr^2-0.0688*a*pr-4.08*a^2+4.75*a^3-0.587*b^3-2.47/pr+1.62*a/pr+3.13*b/pr-4.52*a^2/pr-1.21*a*b/pr-5.87*b^2/pr+4.89*b^3/pr
+0.874/pr^2-0.865*b/pr^2
OrtalamaZ<- exp(YanDagOrtLogZ)-(3-q)*pr
DegismeZ<- exp(YanDagDegLogZ)-2*(3-q)*pr
# Sinama istatistiginin yanasik dagilimina yaklasmakta kullanilacak Gama dagiliminin sekil ve olcek degiskelerini elde etmek icin yanasik ortalama ve degismeyi kullanarak
# V0 varsayimi altinda istenen quantilelari elde et:
# quantilelar: olasilik dagiliminin araligini veya bir ornekteki gozlemleri, esit olasiliklara sahip birbirlerine bitisik araliklarla bolen kesim noktalari.
tetaZ <- DegismeZ/OrtalamaZ
kZ <- OrtalamaZ^2/DegismeZ
YanDagOrtLogK<- 2.80+0.501*pr+1.43*a+0.399*b-0.0309*pr^2-0.0600*a*pr-5.72*a^2-1.12*a*b-1.70*b^2+0.000974*pr^3+0.168*a^2*pr+6.34*a^3+1.89*a*b^2+1.85*b^3-2.19/pr-0.438*a/pr
+1.79*b/pr+6.03*a^2/pr+3.08*a*b/pr-1.97*b^2/pr-8.08*a^3/pr-5.79*a*b^2/pr+0.717/pr^2-1.29*b/pr^2-1.52*a^2/pr^2+2.87*b^2/pr^2-2.03*b^3/pr^2
YanDagDegLogK<- 3.78+0.346*pr+0.859*a-0.0106*pr^2-0.0339*a*pr-2.35*a^2+3.95*a^3-0.282*b^3-2.73/pr+0.874*a/pr+2.36*b/pr-2.88*a^2/pr-4.44*b^2/pr+4.31*b^3/pr+1.02/pr^2-0.807*b/pr^2
OrtalamaK <- exp(YanDagOrtLogK)-(3-q)*pr
DegismeK <- exp(YanDagDegLogK)-2*(3-q)*pr
# Sinama istatistiginin yanasik dagilimina yaklasmakta kullanilacak Gama dagiliminin sekil ve olcek degiskelerini elde etmek icin yanasik ortalama ve degismeyi kullanarak
# V0 varsayimi altinda istenen quantilelari elde et:
# quantilelar: olasilik dagiliminin araligini veya bir ornekteki gozlemleri, esit olasiliklara sahip birbirlerine bitisik araliklarla bolen kesim noktalari.
tetaK <- DegismeK/OrtalamaK
kK <- OrtalamaK^2/DegismeK
# (izZ veya izK den biri 0 dan farklı ise) karar değerlerini ve p değerlerini tablolaştır:
windows(6,3.8)
KararDegerleri <- cbind(sapply(c(.90,.95,.99) , function(x) sprintf("%.2f",round(c(qgamma(x, shape=kZ,scale=tetaZ)),2))),
sapply(c(.9,.95,.99) , function(x) sprintf("%.2f",round(c(qgamma(x, shape=kK,scale=tetaK)),2))))
colnames(KararDegerleri) <- rep(c(0.90,0.95,0.99),2)
# rownames(KararDegerleri) <- pr
rownames(KararDegerleri) <- c(sapply((degiskensayisi -1):1, function(i) paste(degiskensayisi - i, " ","(r<=", i, ")",sep="")), paste(degiskensayisi, " ( r=0)", sep=""))
textplot(KararDegerleri, cex=1)
text(.064,.91,"p-r",font=2)
text(.345,1,expression(paste(plain(V)[z],"(r) test")),col=2)
text(.821,1,expression(paste(plain(V)[k],"(r) test")),col=4)
title("Yanasik Karar Degerleri \n (p:duzendeki degisken sayisi; r:esbutunlesim ranki)")
if(izZ!=0){
windows(4,3.8)
pDegerleri <- matrix(sprintf("%.3f",round(1 - pgamma(izZ, shape=kZ, scale = tetaZ),3)))
# rownames(pDegerleri) <- pr
rownames(pDegerleri) <- c(sapply((degiskensayisi -1):1, function(i) paste(degiskensayisi - i, " ","(r<=", i, ")",sep="")), paste(degiskensayisi, " ( r=0)", sep=""))
textplot(pDegerleri,cex=1,show.colnames=F)
text(.69,.96,substitute(paste("Pr(",plain(V)[z],">",nn,")"),list(nn=izZ)),col=2)
text(.45,.96,"p-r",font=2)
title("Yanasik p Degerleri \n (p:duzendeki degisken sayisi; \n r:esbutunlesim ranki)")
}
if(izK!=0){
windows(3,3.8)
pDegerleri <- matrix(sprintf("%.3f",round(1 - pgamma(izK, shape=kK, scale = tetaK),3)))
#rownames(pDegerleri) <- pr
rownames(pDegerleri) <- c(sapply((degiskensayisi -1):1, function(i) paste(degiskensayisi - i, " ","(r<=", i, ")",sep="")), paste(degiskensayisi, " ( r=0)", sep=""))
textplot(pDegerleri,cex=1,show.colnames=F)
text(.78,.96,substitute(paste("Pr(",plain(V)[k],">",nn,")"),list(nn=izK)),col=4)
text(.43,.96,"p-r",font=2)
title("Yanasik p Degerleri \n (p:duzendeki degisken sayisi; \n r:esbutunlesim ranki)")
}
Hence, the according to JMN2000 CVs, there is no cointegration as well. So, your usage of vec2var is meaningless. Because, vec2var is needed in cointegrated cases. Again, assume all series are cointegrated to make you happy (to create need to use vec2var) and continue with the most difficult case (cointegration for series with multiple structural breaks); i.e., we are continueing with "One who pee-pees ambitiously drills the wall" logic.
Extend vars::vec2var to causfinder::vec2var_ykJohEsbInc to handle transformations under "multiple structural breaks" case having relevant intervention dummies. JMN2000 application above showed cointegration rank r is not within [1,4-1]=[1,3] range. Even though that assume JMN2000 CVs resulted r=1 in the above for the sake of argument.
So, to transform restricted VECM to restricted VAR (under multiple=2 structural breaks), apply:
vec2var_ykJohEsbInc(ykJohEsbInc(mydata, type="trace", ecdet="zamanda2yk", K=3, spec="longrun", dumvar=dummymatrix2SB[,c(-1,-2,-3)]),r=1)
These results in:
Deterministic coefficients (detcoeffs):
e prod rw U
kesme 22.6612871 -0.215892151 32.0610121 -9.26649249 #(const)
zyonsemesi 0.2505164 -0.009900004 0.3503561 -0.10494714 #(trend)
zy*D2t_3 0.2238060 -0.008844454 0.3130007 -0.09375756
zy*D3t_3 -0.1234803 0.004879743 -0.1726916 0.05172878
$deterministic
kesme zyonsemesi zy*D2t_3 zy*D3t_3 D2t.3l D3t.3l
e 22.6612871 0.250516390 0.223806048 -0.123480327 -8.8012612 5.3052074
prod -0.2158922 -0.009900004 -0.008844454 0.004879743 -0.1157137 -0.3396206
rw 32.0610121 0.350356063 0.313000702 -0.172691620 -12.5838458 7.2201840
U -9.2664925 -0.104947142 -0.093757559 0.051728781 3.5836119 -2.2921099
I2t I2t.1l I2t.2l I3t I3t.1l I3t.2l
e -0.2584379 0.08470453 0.2102661 -0.51366831 -1.0110891 -2.08728944
prod 0.3013044 0.25103445 -0.8640467 0.08804425 -0.2362783 -0.05606892
rw -0.5838161 0.28400182 1.2073483 -0.67760848 -2.2650094 -0.70586316
U 0.1305258 0.03559119 0.1476985 0.14614290 0.6847273 1.27469940
$A
$A$A1
e.1g prod.1g rw.1g U.1g
e 1.4817704 0.1771082 -0.2274936 0.2332402
prod -0.1605790 1.1846699 0.0406294 -0.9398689
rw -0.8366449 -0.1910611 0.9774874 0.4667430
U -0.4245817 -0.1498295 0.1226085 0.7557885
$A$A2
e.2g prod.2g rw.2g U.2g
e -0.8441175 -0.04277845 0.01128282 -0.01896916
prod -0.3909984 -0.25960184 -0.20426749 0.79420691
rw 1.4181448 -0.03659278 -0.12240211 -0.06579174
U 0.4299422 0.09070905 0.04935195 -0.12691817
$A$A3
e.3g prod.3g rw.3g U.3g
e 0.40149641+0i -0.07067529+0i -0.008175418-0i 0.2286283+0i
prod 0.55003024+0i 0.07241639+0i 0.172505474-0i 0.1281593+0i
rw -0.52674826+0i 0.31667695+0i -0.168897398-0i 0.2184591+0i
U -0.02176108-0i 0.03245409-0i -0.077959841+0i 0.1855889-0i
So, now, check roots:
print(roots(vec2var_ykJohEsbInc(ykJohEsbInc(mydata, type="trace", ecdet="zamanda2yk", K=3, spec="longrun", dumvar=dummymatrix2SB[,c(-1,-2,-3)]),r=1), modulus=TRUE))
This result in "Please provide an object of class 'varest', generated by 'VAR()'." since vars::roots was not extended because: we do NOT need this extention! As I said before, even in the case of restricted VECM, stability is checked from VAR side. You must read Joyeux2007 line by line to see this.
I will supply the ouputs (print-screens) of above functions thouroughly for further clarification.
I will also write extention to vars::root as well just for pedagogical reasons.

EM algorithm for multivariate t mixed models

I'm trying to implement an EM algorithm for family data where I'm assuming my observations have a multivariate t distribution. I have only two siblings per family, so all of the family groups have only two observations. Basically I'm trying to follow the E(C)M steps in this article:
https://pdfs.semanticscholar.org/9445/ef865c4eb1431f9cb2abdb5efc1c361172cc.pdf
However, now I'm not sure if EM is doable for this kind of data, since my correlation matrix Psi should be block diagonal for families.
So here's an R example of how my families are structured
fam_id = sort(rep(1:5, 2))
Z= matrix(0, nrow = length(fam_id), ncol = length(unique(fam_id)))
colnames(Z) = unique(fam_id)
k = 1
i = 1
# Random effects dummy matrix
while (k <= ncol(Z)) {
Z[i:(i+1), k] = c(1, 1)
k = k +1
i = i+2
}
> Z
1 2 3 4 5
[1,] 1 0 0 0 0
[2,] 1 0 0 0 0
[3,] 0 1 0 0 0
[4,] 0 1 0 0 0
...
The EM algorithm chokes after 5th iteration saying that the correlation matrix Psi is not:
Error in solve.default(psi_hat) :
system is computationally singular
If anyone could shed some light to this, I'd be very happy!
Please check this answer in the Statschange website
https://stats.stackexchange.com/questions/76488/error-system-is-computationally-singular-when-running-a-glm
Your are probably ending up with a non invertible matrix in your 5th iteration

PCA multiplot in R

I have a dataset that looks like this:
India China Brasil Russia SAfrica Kenya States Indonesia States Argentina Chile Netherlands HongKong
0.0854026763 0.1389383234 0.1244184371 0.0525460881 0.2945586244 0.0404562539 0.0491597968 0 0 0.0618342901 0.0174891774 0.0634064181 0
0.0519483159 0.0573851759 0.0756806292 0.0207164181 0.0409872092 0.0706355932 0.0664503936 0.0775285039 0.008545575 0.0365674701 0.026595575 0.064280902 0.0338135148
0 0 0 0 0 0 0 0 0 0 0 0 0
0.0943708876 0 0 0.0967733329 0 0.0745076688 0 0 0 0.0427047276 0 0.0583873189 0
0.0149521013 0.0067569437 0.0108914448 0.0229991162 0.0151678343 0.0413174214 0 0.0240999375 0 0.0608951432 0.0076549109 0 0.0291972756
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0.0096710124 0.0095669967 0 0.0678582869 0 0 0.0170707337 0.0096565543 0.0116698364 0.0122773071
0.1002690681 0.0934563916 0.0821680095 0.1349534369 0.1017157777 0.1113249348 0.1713480649 0.0538715423 0.4731833978 0.1956743964 0.6865919069 0.2869189344 0.5364034876
1.5458338337 0.2675380321 0.6229046372 0.5059107039 0.934209603 0.4933799388 0.4259769181 0.3534169521 14.4134845836 4.8817632117 13.4034293299 3.7849346739 12.138551171
0.4625375671 0.320258205 0.4216459567 0.4992764309 0.4115887595 0.4783677078 0.4982410179 0.2790259278 0.3804405781 0.2594924212 0.4542162376 0.3012339384 0.3450847892
0.357614592 0.3932670219 0.3803417257 0.4615355254 0.3807061655 0.4122433346 0.4422282977 0.3053712842 0.297943232 0.2658160167 0.3244018409 0.2523836582 0.3106600754
0.359953567 0.3958391813 0.3828293473 0.4631507073 0.3831961707 0.4138590365 0.4451206879 0.3073685624 0.2046559772 0.2403036541 0.2326305393 0.2269373716 0.2342962436
0.7887404662 0.6545878236 0.7443676393 0.7681244767 0.5938002158 0.5052305973 0.4354571648 0.40511005 0.8372481106 0.5971130339 0.8025313223 0.5708610817 0.8556609579
0.5574207497 1.2175251783 0.8797484259 0.952685465 0.4476585005 1.1919229479 1.03612509 0.5490564488 0.2407034171 0.5675492645 0.4994121344 0.5460544861 0.3779468604
0.5632651223 1.0181714714 1.1253803155 1.228293512 0.6949993291 1.0346288085 0.5955221073 0.5212567091 1.1674901423 1.2442735568 1.207624867 1.3854352274 0.7557131826
0.6914760031 0.7831502333 1.0282730148 0.750270567 0.7072739935 0.8041764647 0.8918512571 0.6998554585 2.3448306081 1.2905783367 2.4295927684 1.3029766224 1.9310763864
0.3459898177 0.7474525109 0.7253451876 0.7182493014 0.3081791886 0.7462088907 0.5950509439 0.4443221541 3.6106852374 2.7647504885 3.3698608994 2.6523062395 1.8016571476
0.4629523517 0.6549211677 0.6158018856 0.7637088814 0.4951554309 0.6277236471 0.6227669055 0.383909839 2.9502307101 1.803480973 2.3083113522 1.668759497 1.7130459012
0.301548861 0.5961888126 0.4027007075 0.5540290853 0.4078662541 0.5108773106 0.4610682726 0.3712800134 0.3813402422 0.7391417247 1.0935364978 0.691857974 0.4416304953
2.5038287529 3.2005148394 2.9181517373 3.557918333 1.8868234768 2.9369926312 0.4117894127 0.3074815035 3.9187777037 7.3161555954 6.9586996112 5.7096144353 2.7007439732
2.5079707359 3.2058093222 2.9229791182 3.563804054 1.8899447728 2.9418511798 0.4124706194 0.269491388 3.9252603798 7.3282584169 6.9702111077 5.7190596205 2.7052117051
2.6643724791 1.2405320493 2.0584120188 2.2354369334 1.7199730388 2.039829709 1.7428132997 0.9977029725 8.9650886611 4.6035139163 8.1430131464 5.2450639988 6.963309864
0.5270581435 0.8222128903 0.7713479951 0.8785815313 0.624993821 0.7410405193 0.5350834321 0.4797121891 1.3753525725 1.2219267886 1.397221881 1.2433155977 0.8647136903
0.2536079475 0.5195514789 0.0492623195 0.416102668 0.2572670724 0.4805482899 0.4866090738 0.4905212099 0.2002506403 0.5508609827 0.3808572148 0.6276294938 0.3191452919
0.3499009885 0.5837491529 0.4914807442 0.5851537888 0.3638549977 0.537655052 0.5757185943 0.4730102035 0.9098072064 0.6197285737 0.7781825654 0.6424684366 0.6424429128
0.6093076876 0.9456457011 0.8518013605 1.1360347777 0.511960743 0.9038104168 0.5048413575 0.2777622235 0.2915840525 0.6628516415 0.4600364351 0.7996524113 0.3765721177
0.9119207879 1.2363073271 1.3285269752 1.4027039939 0.9250782309 2.1599381031 1.312307839 0 0 0.8253250513 0 0 0.8903632354
It is stored in a data.txt file.
I want to have a PCA multiplot that looks like this:
What I am doing:
d <- read.table("data.txt", header=TRUE, as.is=TRUE)
model <- prcomp(d, scale=TRUE)
After this I am lost.
How can I cluster the dataset according to the PCA projections and obtain the pictures similar to those above?
You are actually asking two different questions:
How to cluster the data after PCA projections.
How to obtain the above plots.
However before getting to those I would like to add that if your samples are in columns, then you are not doing PCA correctly. You should do it on transposed dataset instead like so:
model <- prcomp(t(d), scale=TRUE)
But for that to work you would have to remove all the constant rows in your data.
Now I assume that you did your PCA step how you wanted.
prcomp returns the rotated matrix when you specify retX=TRUE (it's true by default). So you will want to use model$x.
Your next step is clustering the data based on principal components. This can be done in various ways. One is hierarchical clustering. If you want 5 groups in the end here is one way:
fit <- hclust(dist(model$x[,1:3]), method="complete") # 1:3 -> based on 3 components
groups <- cutree(fit, k=5) # k=5 -> 5 groups
This step will get you groups that will be later used for coloring.
The final step is plotting. Here I wrote a simple function to do all in one shot:
library(rgl)
plotPCA <- function(x, nGroup) {
n <- ncol(x)
if(!(n %in% c(2,3))) { # check if 2d or 3d
stop("x must have either 2 or 3 columns")
}
fit <- hclust(dist(x), method="complete") # cluster
groups <- cutree(fit, k=nGroup)
if(n == 3) { # 3d plot
plot3d(x, col=groups, type="s", size=1, axes=F)
axes3d(edges=c("x--", "y--", "z"), lwd=3, axes.len=2, labels=FALSE)
grid3d("x")
grid3d("y")
grid3d("z")
} else { # 2d plot
maxes <- apply(abs(x), 2, max)
rangeX <- c(-maxes[1], maxes[1])
rangeY <- c(-maxes[2], maxes[2])
plot(x, col=groups, pch=19, xlab=colnames(x)[1], ylab=colnames(x)[2], xlim=rangeX, ylim=rangeY)
lines(c(0,0), rangeX*2)
lines(rangeY*2, c(0,0))
}
}
This function is simple: it takes two arguments: 1) a matrix of scores, with principal components in columns and your samples in rows. You can basically use model$x[,c(1,2,4)] if you want (for example) 1st, 2nd and 4th components. 2) number of groups for clustering.
Then it cluster the data based on passed principal components and plots (either 2D or 3D depending on the number of columns passed)
Here are few examples:
plotPCA(model$x[,1:2], 5)
And 3D example (based on 3 first principal components):
plotPCA(model$x[,1:3], 5)
This last plot will be interactive so you can rotate it to or zoom in/out.
Hope this helps.

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