## Gram-schmidt

 Matrix M has three rows and three columns, and the columns form an orthonormal basis. One of the columns is [2/7,3/7,6/7], and another is [6/7, 2/7, -3/7]. Let the third column be [x,y,z]. Since the length of the vector [x,y,z] must be 1, there is a constraint that x2+y2+z2 = 1. However, there are other constraints, and these other constraints can be used to deduce facts about the ratios among x, y, and z. Compute these ratios, and then identify one of them in the list below. 

I viewed the Khan academy course.

But the credit for the Matlab code goes to Vladd. I didn’t follow his explanation but the Khan academy course helped.I used the Matlab online compiler to test Vladd’s code and ported it to R.

I can’t believe the for and if loops in the R code took a full day to debug.


#
###############################################################################

A =  matrix(c(2/7,3/7,6/7,6/7,2/7,-3/7,1,2,3),ncol=3,nrow=3)
r = dim( A)[];
c = dim( A)[];
print(A)

Q = matrix(c(0,0,0,0,0,0,0,0,0),ncol=3,nrow=3)
for (j in 1:3){

u = matrix(A[ ,j  ]);

if( j - 1 != 0 ){
for(i in 1:(j - 1)){
e = Q[,i]
a = as.matrix(A[,j])
p = (t(e) %*% a) / (t(e) %*% e) * e;
u = u - p
}
}
# normalize it to length of 1 and store it
Q[,j] = u / sqrt(u[1,1]^2 + u[2,1]^2 + u[3,1]^2);
print(Q)
}



The result is this. The last column is what I want and that satisfies all constraints.

### Orthonormality $\begin{pmatrix} 0.2857143& 0.8571429& -0.4285714\\ 0.4285714& -0.2857143& 0.8571429\\ 0.8571429& -0.4285714& -0.2857143\\ \end{pmatrix}$

I will check-in the R code into my Git repository.

This is the general market-basket problem. It is an algorithm to find how many items are frequently found across many shoppers’ baskets based on a threshold. The threshold is a minimum number of occurrences of a particular item. Items that are bought a certain number of times(threshold) are considered frequent.

These items can be singletons or pairs of items(doubletons) and tripletons and so on.

Imagine there are 100 baskets, numbered 1,2,...,100, and 100 items, similarly numbered. Item i is in basket j if and only if i divides j evenly. For example, basket 24 is the set of items {1,2,3,4,6,8,12,24}. Describe all the association rules that have 100% confidence. Which of the following rules has 100% confidence?

A brute-force R approach to solve such a problem. This is a small number of items. In fact such data mining algorithms deal with large quantities of data and a fixed amount of memory. One such algorithm is the A-priori algorithm.

Each of the if loop checks for a condition like this.

 {8,10} -> 20

This checks if item 20 is always found in a basket that has items 8 and 10 or not.

library(Hmisc)
for( i in 1:100){
a <- 1
for( j in 1:100){

if( i %% j == 0 ){
a <- append(a,j)
}
}
#print(paste( i, a ))
if( 8 %in% a &&  10 %in% a && 20 %nin% a ){ //{8,10} -> 20
#print (a)
}
if( 3 %in% a &&  1 %in% a && 6 %in% a && 12 %nin% a ){
print (a)
}
if( 8 %in% a &&  12 %in% a &&  96 %nin% a ){
#print (a)
}
if( 3 %in% a &&  5 %in% a &&  1 %nin% a ){
#print (a)
}
}}

## PageRank Suppose we compute PageRank with a β of 0.7, and we introduce the additional constraint that the sum of the PageRanks of the three pages must be 3, to handle the problem that otherwise any multiple of a solution will also be a solution. Compute the PageRanks a, b, and c of the three pages A, B, and C, respectively. $A = \sum_{i\rightarrow{j}} \beta r_i/d_i + (1 - \beta ) e / n$

My R code is this.

M = matrix(c(0,1/2,1/2,0,0,1,0,0,1),ncol=3)
e = matrix(c(1,1,1),ncol=1)
v1 = matrix(c(1,1,1),ncol=1)
v1 = v1 / 3
for( i in 1:5){

v1 =  ((0.7 * M ) %*% v1 ) + (((1 - 0.7 ) * e ) /3 )
}
v1 = v1 * 3

      [,1]
[1,] 0.300
[2,] 0.405
[3,] 2.295


## Apache Mahout

I followed this tutorial. Mahout seems to be an easy way to test Machine Learning algorithms using the Java API.

But I would use this  this R code instead of the one shown in the tutorial to convert the MovieLens dataset to CSV format.

r<-file("u.data","r")
w<-file("u1.csv","w")

while( length(data <- readLines(r)) > 0 ){
writeLines(gsub("\\s+",",",data),w)
}


## Time series forecast

This code counts how many values from the testing dataset fall within the 95% Confidence Interval range.

library(forecast)
library(lubridate)  # For year() function below
training = dat[year(dat$date) < 2012,] testing = dat[(year(dat$date)) > 2011,]
tstrain = ts(training$visitsTumblr) sum <- 0 fit <- bats(tstrain) fc <- forecast(fit,h=235) mat <- fc$upper
for(i in 1:nrow(mat)){
v <- data.frame(mat[i,])
print(paste(testing$visitsTumblr[i],v[1,] , v[2,])) if(testing$visitsTumblr[i] > v[1,] & testing$visitsTumblr[i] < v[2,]){ sum <- sum + 1 } } print(sum)  The forecast object has this type of data(Lo 95 & Hi 95) which I use. > fc Point Forecast Lo 80 Hi 80 Lo 95 Hi 95 366 207.4397 -124.2019 539.0813 -299.7624 714.6418 367 197.2773 -149.6631 544.2177 -333.3223 727.8769 368 235.5405 -112.0582 583.1392 -296.0658 767.1468 369 235.5405 -112.7152 583.7962 -297.0707 768.1516 370 235.5405 -113.3710 584.4520 -298.0736 769.1546 371 235.5405 -114.0256 585.1065 -299.0747 770.1556 372 235.5405 -114.6789 585.7599 -300.0739 771.1548  ## Lasso fit ### The code I was given set.seed(3523) library(AppliedPredictiveModeling) data(concrete) inTrain = createDataPartition(concrete$CompressiveStrength, p = 3/4)[]
training = concrete[ inTrain,]
testing = concrete[-inTrain,]


### This is the data

<- head(as.matrix(training))
Cement BlastFurnaceSlag FlyAsh Water Superplasticizer CoarseAggregate
47   349.0              0.0      0 192.0              0.0          1047.0
55   139.6            209.4      0 192.0              0.0          1047.0
56   198.6            132.4      0 192.0              0.0           978.4
58   198.6            132.4      0 192.0              0.0           978.4
63   310.0              0.0      0 192.0              0.0           971.0
115  362.6            189.0      0 164.9             11.6           944.7
FineAggregate Age CompressiveStrength
47          806.9   3               15.05
55          806.9   7               14.59
56          825.5   7               14.64
58          825.5   3                9.13
63          850.6   3                9.87
115         755.8   7               22.90


### Lasso fit and plot

predictors <- as.matrix(training)[,-9]
lasso.fit <- lars(predictors,training$CompressiveStrength,type="lasso",trace=TRUE) headings <- names(training[-(9:10)]) plot(lasso.fit, breaks=FALSE) legend("topleft", headings,pch=8, lty=1:length(headings),col=1:length(headings)) According to this graph the last coefficient to be set to zero as the penalty increases is Cement. I think this is correct but I may change this. ## RandomForests I am just posting R code at this time. The explanation is missing but I am making some progress. library(ElemStatLearn) library(randomForest) data(vowel.train) data(vowel.test)  > head(vowel.train) y x.1 x.2 x.3 x.4 x.5 x.6 x.7 x.8 x.9 x.10 1 1 -3.639 0.418 -0.670 1.779 -0.168 1.627 -0.388 0.529 -0.874 -0.814 2 2 -3.327 0.496 -0.694 1.365 -0.265 1.933 -0.363 0.510 -0.621 -0.488 3 3 -2.120 0.894 -1.576 0.147 -0.707 1.559 -0.579 0.676 -0.809 -0.049 4 4 -2.287 1.809 -1.498 1.012 -1.053 1.060 -0.567 0.235 -0.091 -0.795 5 5 -2.598 1.938 -0.846 1.062 -1.633 0.764 0.394 -0.150 0.277 -0.396 6 6 -2.852 1.914 -0.755 0.825 -1.588 0.855 0.217 -0.246 0.238 -0.365 vowel.train$y <- factor(vowel.train$y) set.seed(33833) fit.rf <- randomForest(vowel.train$y ~ .,data=vowel.train)
plot(fit.rf)
varImpPlot(fit.rf) I was asked to find the order of variable importance which this graph shows. ## Principal Component Analysis

library(caret)
library(AppliedPredictiveModeling)
set.seed(3433)
data(AlzheimerDisease)
inTrain = createDataPartition(adData$diagnosis, p = 3/4)[] training = adData[ inTrain,] testing = adData[-inTrain,]  I recently studied Predictive Analytics techniques as part of a course. I was given the code shown above. I generated the following two predictive models to compare their accuracy figures. This might be easy for experts but I found it tricky. So I post the code here for my reference. ### Non-PCA training1 <- training[,grepl("^IL|^diagnosis",names(training))] test1 <- testing[,grepl("^IL|^diagnosis",names(testing))] modelFit <- train(diagnosis ~ .,method="glm",data=training1) confusionMatrix(test1$diagnosis,predict(modelFit, test1))


Confusion Matrix and Statistics

Reference

Prediction Impaired Control

Impaired 2 20

Control 9 51

Accuracy : 0.6463

95% CI : (0.533, 0.7488)

No Information Rate : 0.8659

P-Value [Acc > NIR] : 1.00000

Kappa : -0.0702

Mcnemar’s Test P-Value : 0.06332

Sensitivity : 0.18182

Specificity : 0.71831

Pos Pred Value : 0.09091

Neg Pred Value : 0.85000

Prevalence : 0.13415

Detection Rate : 0.02439

Detection Prevalence : 0.26829

Balanced Accuracy : 0.45006

‘Positive’ Class : Impaired

### PCA

training2 <- training[,grepl("^IL",names(training))]

preProc <- preProcess(training2,method="pca",thresh=0.8)

test2 <- testing[,grepl("^IL",names(testing))]

trainpca <- predict(preProc, training2)

testpca <- predict(preProc, test2)

modelFitpca <- train(training1$diagnosis ~ .,method="glm",data=trainpca) confusionMatrix(test1$diagnosis,predict(modelFitpca, testpca))

Confusion Matrix and Statistics

Reference
Prediction Impaired Control
Impaired 3 19
Control 4 56

Accuracy : 0.7195
95% CI : (0.6094, 0.8132)
No Information Rate : 0.9146
P-Value [Acc > NIR] : 1.000000

Kappa : 0.0889
Mcnemar’s Test P-Value : 0.003509

Sensitivity : 0.42857
Specificity : 0.74667
Pos Pred Value : 0.13636
Neg Pred Value : 0.93333
Prevalence : 0.08537
Detection Rate : 0.03659
Detection Prevalence : 0.26829
Balanced Accuracy : 0.58762

‘Positive’ Class : Impaired

## Open Government Data Platform India I found many datasets in this site but many of them are not useful. Some of them are just junk and others are not useful for predictive analytics.

But I found one that I actually used. The y-axis labels are smudged but that can be fixed.

library(RJSONIO)
library(ggplot2)
library(reshape2)
library(grid)
this.dir <- dirname(parent.frame(2)$ofile) setwd(this.dir) airlines = fromJSON("json") df <- sapply(airlines$data,unlist)
df <- data.frame(t(df))
colnames(df) <- c( (airlines[][])[], (airlines[][])[], (airlines[][])[], (airlines[][])[], (airlines[][])[], (airlines[][])[], (airlines[][])[], (airlines[][])[], (airlines[][])[],(airlines[][])[] )

df.melted <- melt(df, id = "YEAR")
print(class(df.melted$value)) df.melted$value<-as.numeric(df.melted$value) df.melted$value <- format(df.melted$value, scientific = FALSE) print(ggplot(data = df.melted, aes(x = YEAR, y = value, color = variable)) +geom_point() + theme(axis.text.x = element_text(angle = 90,hjust = 0.9)) + theme(axis.text.y = element_text(angle = 360, hjust = 1, size=7.5, vjust=1))+ theme(plot.margin =unit(c(3,1,0.5,1), "cm")) + ylab("") + theme(legend.text=element_text(size=6))) This is the sample data.  head(df) YEAR INTERNATIONAL ACM (IN NOS) DOMESTIC ACM (IN NOS) TOTAL ACM (IN NOS) 1 1995-96 92515 314727 407242 2 1996-97 94884 324462 419346 3 1997-98 98226 317531 415757 4 1998-99 99563 325392 424955 5 1999-00 99701 368015 467716 6 2000-01 103211 386575 489786 INTERNATIONAL PAX (IN NOS) DOMESTIC PAX (IN NOS) TOTAL PAX (IN NOS) 1 11449756 25563998 37013754 2 12223660 24276108 36499768 3 12782769 23848833 36631602 4 12916788 24072631 36989419 5 13293027 25741521 39034548 6 14009052 28017568 42026620 INTERNATIONAL FREIGHT (IN MT) DOMESTIC FREIGHT (IN MT) TOTAL FREIGHT (IN MT) 1 452853 196516 649369 2 479088 202122 681210 3 488175 217405 705580 4 474660 224490 699150 5 531844 265570 797414 6 557772 288373 846145 ## Decision Tree This is a technique used to analyze data for prediction. I came across this when I was studying Machine Learning. Tree models are computationally intensive techniques for recursively partitioning response variables into subsets based on their relationship to one or more (usually many) predictor variables. > head(data) file_id time cell_id d1 d2 fsc_small fsc_perp fsc_big pe chl_small 1 203 12 1 25344 27968 34677 14944 32400 2216 28237 2 203 12 4 12960 22144 37275 20440 32400 1795 36755 3 203 12 6 21424 23008 31725 11253 32384 1901 26640 4 203 12 9 7712 14528 28744 10219 32416 1248 35392 5 203 12 11 30368 21440 28861 6101 32400 12989 23421 6 203 12 15 30032 22704 31221 13488 32400 1883 27323 chl_big pop 1 5072 pico 2 14224 ultra 3 0 pico 4 10704 ultra 5 5920 synecho 6 6560 pico  training <- createDataPartition(data$pop, times=1,p=.5,list=FALSE)
train <- data[training,]
test <- data[,training]

fol <- formula(pop ~ fsc_small + fsc_perp + fsc_big + pe + chl_big + chl_small)
model <- rpart(fol, method="class", data=train)
print(model)

n= 36172

node), split, n, loss, yval, (yprob)
* denotes terminal node

1) root 36172 25742 pico (0.0014 0.18 0.29 0.25 0.28)
2) pe=41300 5175 660 nano (0 0.87 0 0 0.13) *
11) chl_small=5001.5 9856 783 synecho (0.0052 0.054 0.0051 0.92 0.015)
6) chl_small>=38109.5 653 133 nano (0.078 0.8 0 0.055 0.07) *
7) chl_small< 38109.5 9203 166 synecho (0 0.0015 0.0054 0.98 0.011) *